🧠:Okay, so there's this circuit with 2000 contacts, and every pair of contacts is initially connected by a separate wire. Vasya and Petya take turns cutting wires. Vasya goes first, cutting one wire each turn, and Petya can cut either one or three wires on his turn. The person who cuts the last wire connected to any contact loses. We need to figure out who has a winning strategy with optimal play.First, let me try to understand the problem. Each contact is connected to 1999 other contacts initially, right? Because there are 2000 contacts, and each has a wire to every other contact. So each contact starts with 1999 wires. The game continues until someone is forced to cut the last wire of any contact, which would make them lose. So, the losing condition is when a player, after their move, there exists at least one contact that has all its wires cut except the one they just cut. Wait, actually, the exact wording is "the prankster who cuts the last wire connected to any contact loses." Hmm. So, if a player cuts a wire that is the last remaining wire for either of the two contacts it connects, then that player loses. Wait, maybe more precisely: when a player cuts a wire, if after cutting that wire, either of the two contacts involved in that wire has no remaining wires left, then that player loses. Because cutting the last wire connected to a contact would make them lose. So the losing condition is when a player's move results in a contact losing its last wire. So, during a player's turn, if they cut a wire, and either end of that wire had only that one wire left, then the player who cut it loses.Therefore, the goal is to avoid being the one who cuts the last wire of any contact. So, players take turns cutting wires, and you lose if your move causes a contact to lose its last wire. So, the key is to force your opponent into a position where any wire they cut will result in them cutting the last wire of a contact.Given that initially, each contact has 1999 wires. The total number of wires is C(2000, 2) = 1999*1000 = 1999000 wires. But the exact number might not matter as much as the parity or some other invariant.Wait, but maybe it's better to model this as a graph. Each contact is a vertex, and each wire is an edge. The game is played on a complete graph with 2000 vertices. Players alternately remove edges, with Vasya removing one edge per turn and Petya removing one or three edges per turn. The loser is the one who removes the last edge adjacent to any vertex.Alternatively, think of each vertex having a degree. The game ends when any vertex's degree becomes zero, and the player who caused that loses.But since all vertices start with the same degree, 1999, which is odd. Wait, 1999 is an odd number. Each move affects the degrees of two vertices. When you cut a wire between contact A and contact B, both A and B lose one degree. So each move reduces the degree of two vertices by one. Therefore, each turn, Vasya removes one edge, so reducing two degrees by one each. Then Petya removes either one or three edges. If he removes one edge, that's reducing two degrees by one each. If he removes three edges, that's reducing six degrees by one each.Wait, perhaps another way: each edge removal affects two vertices. So each player's move can be characterized by how many edges they remove, which affects how many vertices' degrees are decremented. Vasya removes 1 edge (affects 2 vertices), Petya removes 1 or 3 edges (affects 2 or 6 vertices).But the losing condition is when a player's move causes a vertex's degree to drop to zero. So, when a player removes an edge, if either endpoint of that edge had degree 1 before the removal, then removing that edge would bring the degree to 0, causing the player to lose. Similarly, for each edge removed in a move, you have to check if any of the endpoints was at degree 1 before cutting.But since players are cutting edges one by one, but in the case of Petya cutting three edges, does he cut them one after another, potentially causing intermediate states where a vertex might be reduced to zero in between? Wait, the problem says "cut either one or three wires per turn." So when Petya cuts three wires, he does so all at once, right? So the game is turn-based, and during a player's turn, they remove their allotted number of wires (1 for Vasya, 1 or 3 for Petya), and if during their turn, any of the wires they cut result in a contact losing its last wire, then they lose immediately.So, for example, if it's Petya's turn and he decides to cut three wires, he selects three wires and cuts them. If any of those three wires had an endpoint with degree 1 before cutting, then cutting that wire would reduce that endpoint's degree to 0, so Petya would lose. So even if he cuts three wires, if any one of them causes a contact to lose its last wire, he loses.Therefore, players must be careful not to cut any wire that is the last wire for either of its two contacts. So the game is about avoiding cutting such "last wires" while forcing the opponent into a position where they have no choice but to do so.Given that, the initial degrees are all 1999, which is high, so the game will go on for a long time. But with optimal play, perhaps we can find a pattern or a parity argument.First, note that all contacts start with odd degrees (1999). Each time a wire is cut, the degree of two contacts is reduced by 1, turning them from odd to even (since 1999 is odd, subtracting 1 makes it 1998, which is even). However, if a contact's degree is even, cutting a wire connected to it turns it to odd. Wait, actually, each time you cut a wire, you subtract 1 from two degrees. So if a degree is even, it becomes odd; if it's odd, it becomes even.Therefore, each move flips the parity of two vertices. Since the game starts with all vertices having odd degrees (1999). After the first move (Vasya cuts one wire), two vertices become even (1998). Then Petya's turn: if he cuts one wire, two more vertices become even (or flips two from even to odd, depending on which he cuts). Wait, but he can choose which wires to cut. So players can choose which edges to cut, so they can target specific vertices.However, since all vertices are symmetric at the start, perhaps the key is to maintain symmetry or some invariant.Alternatively, maybe we can model this as a Nim game or some impartial game, but with the specific move structure.Alternatively, perhaps consider that each contact can be considered as a pile with 1999 stones, and each edge corresponds to two stones in two piles. But cutting an edge removes one stone from each of two piles. However, the losing condition is when you take the last stone from any pile. But this is different because in standard Nim, taking the last stone wins, but here it's losing.Alternatively, it's similar to a game where you have multiple heaps, and players take turns removing stones, but if you take the last stone from any heap, you lose. That's called a variant of Nim, sometimes called the "poisoned piece" game or similar.In such games, the strategy often revolves around avoiding taking the last stone. If all heaps have size 1, the player to move loses because they have to take the last stone. If heaps have larger sizes, players can reduce heaps to 1, forcing the opponent to take them.But in our problem, instead of heaps, we have edges connected to contacts, and cutting an edge reduces the degree of two contacts. So it's a different structure.Alternatively, since each move affects two contacts, perhaps the game is equivalent to a bipartite game where each move affects two components. However, I'm not sure.Alternatively, think of it as each contact is a node with a certain number of lives (degree), and each edge connects two nodes. Cutting an edge reduces both connected nodes' lives by 1. The loser is the one who reduces any node's lives to 0.This seems similar to the game of Kayles or similar impartial games, but again, the interaction between nodes complicates things.Alternatively, perhaps we can model the game as a misère game, where the last move loses. However, standard misère Nim has a different analysis.Given the complexity, maybe we can simplify the problem by considering smaller numbers and look for a pattern.Let's try smaller cases.Case 1: n = 2 contacts. Then there is 1 wire. Vasya starts, he must cut the only wire. Cutting this wire will disconnect both contacts, so both contacts lose their only wire. Therefore, Vasya loses immediately. So with n=2, Vasya loses.Case 2: n=3 contacts. Complete graph has 3 wires. Each contact has degree 2. Vasya starts by cutting one wire. Suppose he cuts wire AB. Then contacts A and B now have degree 1 each, contact C still has degree 2. Then Petya's turn. He can cut 1 or 3 wires. If he cuts 1 wire: he could cut AC or BC. If he cuts AC, then contact A's degree goes to 0, so Petya loses. Similarly, if he cuts BC, contact B's degree goes to 0, Petya loses. If he cuts the remaining wire AB (but it's already cut). Wait, no, after Vasya cuts AB, the remaining wires are AC and BC. So Petya has to cut one of them. Either way, cutting AC or BC would reduce contact A or B to degree 1, wait no: contact A had degree 1 after AB was cut, then if Petya cuts AC, contact A's degree becomes 0, so Petya loses. Therefore, Petya cannot cut a single wire; he would lose. Alternatively, can Petya cut three wires? But there are only two wires left. He can't cut three. So Petya has to cut one wire and loses. Therefore, in n=3, Vasya can force a win. Wait, but let's confirm:Vasya cuts AB. Now degrees: A=1, B=1, C=2. Petya's turn. He must cut one wire. If he cuts AC, A's degree becomes 0: Petya loses. Similarly, if he cuts BC, B's degree becomes 0: Petya loses. So Vasya wins. Alternatively, could Vasya have a different first move? No, in n=3, all wires are symmetric. So regardless, Vasya can force Petya to lose. Therefore, n=3, Vasya wins.Wait, but what if n=4? Let's see.n=4 contacts, complete graph has 6 wires. Each contact has degree 3. Vasya starts by cutting one wire, say AB. Then degrees: A=2, B=2, C=3, D=3. Petya's turn. He can cut 1 or 3 wires.If he cuts 1 wire, say AC. Then degrees: A=1, C=2, others remain. Now Vasya's turn. He needs to cut a wire. If he cuts AD, then A's degree becomes 0, so Vasya loses. Similarly, if he cuts BC, B's degree becomes 1, etc. But this seems complicated.Alternatively, maybe there's a pattern based on parity. Since Vasya starts, and the total number of wires is C(n,2). But with n=2000, which is even. Wait, but the degrees start at n-1, which for n=2000 is 1999, odd.Each move by Vasya removes 1 wire, affecting two vertices. Each move by Petya removes 1 or 3 wires, affecting two or six vertices. The total number of wires is huge, but the key is the parity of the degrees.Since all degrees start odd. Each move flips the parity of two (or six) vertices. So after each player's move, the number of odd-degree vertices changes by an even number. Therefore, the parity of the number of odd-degree vertices remains the same. Initially, all 2000 vertices have odd degrees, so number of odd degrees is 2000, which is even. After each move, the number of odd-degree vertices changes by ±0, ±2, or ±6 (since each edge cut affects two vertices, flipping their parity). So the total number of odd-degree vertices remains even throughout the game.Therefore, when the game ends, there must be at least two vertices with degree 0, or one vertex with degree 0 and the rest with even degrees, but since degrees can't be negative, but wait, if one vertex has degree 0, then the number of odd-degree vertices would be odd (since the rest would be even, maybe). Wait, no: if one vertex has degree 0 (even), and the others have degrees that are even or odd. But since total number of odd-degree vertices must remain even, if one vertex is even (0), then the remaining 1999 vertices must have an even number of odd degrees. But 1999 is odd. So this is impossible. Therefore, the game cannot end with exactly one vertex at degree 0. Therefore, the game must end when two vertices reach degree 0 simultaneously. Because if two vertices reach degree 0, that would mean their last wires were cut in the same move. But since each wire connects two vertices, cutting a wire that is the last wire for both of its endpoints would cause both to reach degree 0 at the same time. Therefore, the losing player is the one who cuts a wire which is the last wire for either or both of its endpoints. However, if a player cuts a wire that is the last for both endpoints, then both are reduced to 0, and the player loses.Therefore, the game ends when a player cuts a wire(s) such that at least one endpoint is reduced to 0. But due to the parity argument, it's impossible to have exactly one vertex with degree 0. Therefore, the game must end when two vertices are reduced to 0 simultaneously. Therefore, the losing move is when a player cuts the last wire of two contacts at the same time.But how does that happen? For example, if there's a wire between A and B, and both A and B have only that one wire left. Cutting that wire would reduce both to 0, so the player who cuts it loses.Therefore, the key is to avoid being forced to cut such a wire. So the players will try to avoid leaving pairs of contacts connected by a single wire, because cutting that wire would lose the game.But how does the game progress? Let's think about the endgame. When most wires have been cut, and many contacts have low degrees. The players must avoid cutting the last wire of any contact.Given that all degrees start odd, and each move changes the parity of two vertices, the degrees will oscillate between odd and even. For a contact to reach degree 1, it must have been reduced from degree 3, 5, etc., each time by having wires cut. Once a contact has degree 1, the next time a wire connected to it is cut, it will go to 0, and the player who cuts that wire loses.But due to the parity argument, we can't have a single contact at degree 1. Because the number of odd-degree vertices must remain even. So if one contact is at degree 1 (odd), there must be at least one other contact at odd degree. Therefore, the endgame scenario would involve at least two contacts with degree 1, connected by a wire. Cutting that wire would reduce both to 0, causing the player to lose.Therefore, the last move of the game would be cutting the wire between two contacts that each have degree 1, thereby losing. So the player who is forced to cut that last wire between two degree 1 contacts loses.Therefore, the game reduces to avoiding being the one to cut that last wire. So the players will try to force the opponent into a position where they have to make that move.Given that, perhaps the key is whether the number of such critical wires (connecting degree 1 contacts) is odd or even, and who is forced to make the last move.But given the initial number of contacts, 2000, which is even, and all starting with odd degrees, we might need to consider parity arguments.Another angle: consider that each move by Vasya removes 1 wire, and Petya removes 1 or 3 wires. The total number of wires is C(2000, 2) = 1999000. Let's calculate the total number of moves.But since the losing condition can occur before all wires are cut, it's not directly about the total number of wires. However, maybe we can think in terms of normal play vs misère play, but it's not straightforward.Alternatively, since Petya can remove 1 or 3 wires, he has some flexibility. Vasya removes 1 each turn, Petya can adjust to make the total number of wires removed per full round (Vasya + Petya) either 2 or 4. But how does this affect the game?Alternatively, think of the game in terms of moves modulo some number. For example, if the total number of moves required to remove all wires is such that depending on who makes the last move, but again, the game ends prematurely when a player cuts a last wire.Alternatively, since the game ends when a player cuts a last wire, which could happen well before all wires are removed, we need to consider the strategies to avoid that.Another idea: since all contacts start with odd degrees, and each move changes two degrees from odd to even or vice versa. To reach a state where two contacts have degree 1 (both odd), you need to have an even number of odd degrees. Since we start with 2000 (even) odd degrees, every time you reduce two odds to evens, you keep the count even. If you turn two evens to odds, you also keep the count even. Similarly for six degrees. Therefore, the number of odd degrees remains even throughout the game.Therefore, the game must end when two contacts are reduced to 1 (odd), and they are connected by a wire. Cutting that wire would reduce both to 0, causing the player to lose. So the last move is cutting that wire.Therefore, the total number of moves until the end is related to reducing all contacts from 1999 to 1. Each wire cut reduces two contacts' degrees by 1. So to reduce a contact from 1999 to 1, you need to cut 1998 wires connected to it. Since each wire cut affects two contacts, the total number of wire cuts needed is (2000 * 1998)/2 = 1000 * 1998 = 1,998,000. Then, there are 2000/2 = 1000 wires left, each connecting pairs of contacts with degree 1. Cutting any of these 1000 wires would end the game.But this assumes that all contacts are reduced uniformly, which might not be the case with optimal play. However, if both players play optimally, they would try to mirror each other's moves or balance the degrees to avoid creating a contact with degree 1 prematurely.Alternatively, if we consider that the game ends when the last two contacts with degree 1 are connected by a wire, the number of moves required to reach that state is fixed, and depending on who makes the last move, that player loses.But how does the turn order work? The players alternately cut wires, with Vasya going first. If the total number of moves required to reach the endgame is odd or even, that determines who makes the last move.But the total number of wires is C(2000, 2) = 1999000. However, the game ends when the last wire of any contact is cut, which would happen before all wires are gone. However, if we consider the hypothetical scenario where all wires are cut except the 1000 wires connecting pairs of degree 1 contacts, then those 1000 wires are the ones that would cause a loss if cut. The player who is forced to cut the first of these 1000 wires loses, because cutting it would reduce two contacts to 0.But wait, actually, each of those 1000 wires connects two contacts with degree 1. Therefore, cutting any of them would lose the game. Therefore, the first player to cut any of these 1000 wires loses.Therefore, the key is to determine who is forced to make the first move among these 1000 wires.To get to that state, the players must have reduced all contacts from degree 1999 to degree 1. Each reduction of degree requires cutting wires connected to them. Since each wire cut reduces two contacts' degrees by 1, the total number of cuts needed to reduce all contacts from 1999 to 1 is (1999 - 1) * 2000 / 2 = 1998 * 1000 = 1,998,000 cuts. So before the endgame, there are 1,998,000 cuts made, and the remaining 1000 wires are the ones connecting degree 1 contacts.Now, the total number of cuts made to reach the endgame is 1,998,000. Each full round (Vasya and Petya) removes 1 + 1 = 2 or 1 + 3 = 4 wires. However, Petya can choose to remove 1 or 3 wires each turn. So the total number of moves can vary.Wait, but this approach may not be correct because the players might not cooperate to reduce all contacts to degree 1. Instead, they might try to create a situation where the opponent is forced to cut a last wire earlier. Therefore, the game could end before all contacts are reduced to degree 1.However, under optimal play, both players would avoid creating contacts with degree 1 unless forced. Therefore, they would symmetrically reduce the degrees to maintain as many contacts as possible, avoiding isolating any contact.But given the high number of contacts, 2000, which is even, perhaps they can pair up the contacts and mirror moves. For example, Vasya cuts a wire between A and B; Petya can mirror by cutting a wire between C and D, etc. But this requires that the graph remains in some symmetric state, which might be possible.Alternatively, since the degrees are all odd initially, and each move changes the parity of two contacts, the players could be trying to control the parity.Another approach: consider that the game is equivalent to a Nim game where each contact is a heap, and each wire is a stone shared between two heaps. Cutting a wire removes a stone from two heaps. The losing condition is taking the last stone from any heap.But this is similar to a bipartite Nim game or the game of 2-dimensional Nim. However, such games are complex and don't have straightforward solutions.Alternatively, think of each contact as a node, and the wires as edges. The game is an edge removal game where players take turns removing edges, and losing if they remove the last edge incident to any node.This type of game is known as a "node kayles" game or similar, but I'm not sure of the standard name.In any case, the key is to determine the Grundy number or nimber for this game, but with the given starting position and move options.However, calculating Grundy numbers for such a large graph is impractical. Therefore, we need a different approach.Let's consider parity again. Vasya starts, and removes 1 wire per turn. Petya removes 1 or 3 wires. The total number of wires is C(2000,2)=1999000. The parity of the total number of wires is even (1999000 is even). Vasya removes 1 (odd), Petya can remove 1 or 3 (both odd). So each turn, the total number of wires removed is odd + odd = even, or odd + odd = even. Wait, no: Vasya's turn: he removes 1 (odd). Then Petya removes 1 or 3 (both odd). So each full round (Vasya + Petya) removes 2 or 4 wires, both even numbers. Therefore, the parity of the total number of wires remaining is preserved modulo 2 after each full round. Since the initial number is even, it remains even after each full round. However, the game ends when a player removes a wire that is the last one for any contact, which could happen at any time, not necessarily when all wires are gone.But this parity might not directly determine the winner, unless we can relate the losing move to the parity.Alternatively, think about the number of remaining wires modulo something. Since Petya can choose to remove 1 or 3 wires, he can control the modulo 4. For example, if Vasya removes 1 wire, Petya can remove 1 or 3 to make the total reduction per round 2 or 4. Therefore, Petya can control the pace.But how does this affect the endgame? If the total number of wires is even, and each round removes 2 or 4 wires, then the number of full rounds is floor(total / 2) or similar. But this might not help.Alternatively, consider that in order to force the opponent to make the last move, the player needs to control the number of remaining "safe" moves. If Petya can always mirror Vasya's moves or adjust his move (cutting 1 or 3) to set up a situation where Vasya is forced to make the losing move.Another angle: since Vasya starts, and both can remove odd numbers of wires per turn, but Petya has a choice between removing 1 or 3. This gives Petya more flexibility. Maybe Petya can always mirror Vasya's moves in such a way to maintain an advantage.For example, suppose Vasya removes 1 wire, and Petya removes 1 wire, keeping the total removed per round at 2. Or if Vasya removes 1, Petya removes 3 to make the total per round 4. By adjusting this, Petya can control the game's progression.But I'm not sure how this directly leads to a winning strategy.Wait, let's think about the endgame again. When there are 1000 wires left, each connecting two contacts of degree 1. The next player to move must cut one of these wires and lose. Therefore, the player who is forced to move when there are 1000 wires left will lose, because any move they make will cut one of those wires.Therefore, the goal is to force the opponent to be the one to face the 1000 wires. So, if a player can make the number of wires reduce to 1000 on the opponent's turn, they win.To get from the initial 1999000 wires down to 1000, you need to remove 1998000 wires. The players take turns removing wires: Vasya removes 1, Petya removes 1 or 3, etc.Let’s consider how many moves are required to remove 1998000 wires. However, the number of moves depends on how many wires each player removes. But since Petya can choose to remove 1 or 3 wires per turn, he can control the total number of wires removed per full round (Vasya + Petya) as either 2 or 4.If Petya always removes 1 wire, then each full round removes 2 wires. To remove 1998000 wires, it would take 1998000 / 2 = 999000 full rounds. Since Vasya starts, the number of full rounds is 999000, and then Vasya would have the next turn, but the wires would already be at 1000. Wait, no. Let's clarify.Total wires to remove: 1999000 - 1000 = 1998000.If each full round (Vasya + Petya) removes 2 wires (1+1), then the number of full rounds needed is 1998000 / 2 = 999000 full rounds. After that, the number of wires left is 1000, and it would be Vasya's turn to move. But he needs to cut 1 wire, which would be one of the 1000, causing him to lose. Therefore, if Petya always removes 1 wire, then Vasya is forced to make the losing move.Alternatively, if Petya sometimes removes 3 wires, making the total per round 4 wires. Let's see: if they alternate between removing 1 and 3 wires, but since Vasya can only remove 1, the total per round would vary.But Petya can choose each turn how many to remove. So Petya can control the total number of wires removed per round. If he wants to reach a certain number of wires, he can adjust his moves.The key is that Petya can choose to remove 1 or 3 wires to force the remaining wires to a multiple that puts Vasya in a losing position.This resembles a classic subtraction game where players can remove certain numbers, and the key is to reach a target number.In this case, the target is 1000 wires. Petya wants to force the game so that when there are 1000 wires left, it's Vasya's turn.To do this, Petya can adjust his moves so that after each of his turns, the number of wires remaining is congruent to 0 modulo 4, for example. Let's see:Suppose the total number of wires to remove is 1998000. If Petya can force the game such that after each of his turns, the number of wires remaining is a multiple of 4, then Vasya is always faced with a multiple of 4, and whatever Vasya does (remove 1), Petya can remove 3 to get back to a multiple of 4.For example, if there are 4k wires left, Vasya removes 1, leaving 4k -1. Then Petya removes 3, leaving 4k -4 = 4(k-1). Repeating this, they keep reducing by 4 each round until they reach 4 wires. Then Vasya removes 1, leaving 3, Petya removes 3, leaving 0. But this is if the target is 0, but in our case, the target is 1000.Alternatively, if we consider that the game needs to reduce 1998000 wires to 1000, which is a difference of 1997000. If Petya can control the reduction rate to force Vasya into the losing position.But this is complicated because the number is huge, and the target is not zero. However, if we think of the 1000 wires as the losing position, then Petya wants to make sure that Vasya is faced with the 1000 wires.To do this, Petya can use a strategy where he ensures that after his move, the number of wires remaining is congruent to 1000 modulo the maximum he can force. For example, if the remaining wires are N, Petya wants to choose a move that makes N - (Vasya's move + Petya's move) = N - (1 + x) = target, where x is 1 or 3.But I'm not sure.Alternatively, consider that the total number of wires to remove is 1998000. If we divide this by the maximum per round (4 wires: 1+3), then the number of full rounds is 1998000 /4 = 499,500. But this is not exact since 1998000 is divisible by 4 (1998000 ÷4=499,500). If Petya always removes 3 wires after Vasya's 1, then they can remove 4 wires per round. After 499,500 rounds, all 1998000 wires would be removed, and the remaining 1000 wires would be faced by Vasya, forcing him to lose.But wait, in this scenario, after 499,500 rounds of removing 4 wires each, totaling 1,998,000 wires removed, the remaining wires would be 1999000 - 1998000 = 1000. Since each round is Vasya then Petya, after 499,500 rounds, it would be Petya's turn to move. Wait, no. Each round is Vasya then Petya. So after 499,500 rounds, it's Petya who made the last move. Therefore, the next move would be Vasya's, but there are 1000 wires left. Vasya has to cut one, losing.Yes, exactly. If Petya can force the removal of 4 wires per round (1 by Vasya, 3 by Petya), then after 499,500 rounds, 499,500 *4=1,998,000 wires are removed, leaving 1000 wires. It would be Vasya's turn, he must cut one, losing.Therefore, if Petya can always respond to Vasya's 1 wire by cutting 3 wires, maintaining the 4 wires per round, he can force Vasya into the losing position.But can Petya always do this? Is there any reason he cannot cut 3 wires each time?Petya can choose to cut 1 or 3 wires on his turn. However, cutting 3 wires requires that there are at least 3 wires remaining. But since the total number of wires is initially 1,999,000, and they are removing 4 per round, there will always be enough wires until the end.But wait, when approaching the end, when there are only 1000 wires left, Vasya's move would cut to 999, then Petya can cut 3 to make 996, but this doesn't apply because the losing condition is already triggered at 1000.Wait, no. The losing condition isn't about the total number of wires, but about cutting the last wire of any contact. However, our previous analysis suggested that the game would end when there are 1000 wires left, each connecting two degree 1 contacts. Therefore, if Petya can force the game to reach 1000 wires on Vasya's turn, Vasya loses.Therefore, if Petya can always respond by cutting 3 wires when Vasya cuts 1, maintaining a removal rate of 4 wires per round, then they would reach 1000 wires on Vasya's turn.Therefore, Petya's winning strategy is to always cut 3 wires after Vasya's 1, thereby removing 4 wires per round. This way, after 499,500 rounds, they remove 1,998,000 wires, leaving 1000 for Vasya to cut and lose.However, we need to confirm that Petya can always cut 3 wires without prematurely causing a contact to lose its last wire. In other words, when Petya cuts 3 wires, he must be able to do so without any of those 3 wires being the last wire for their respective contacts.But during the game, until the very end (the 1000 wires), all contacts have degree at least 2. Because to reach degree 1, a contact must have had its degree reduced from 1999 by an even number (since each cut reduces degree by 1, and each contact starts at odd degree 1999). So after an even number of cuts to a contact, its degree is odd, and after an odd number, even. But since we need to reach degree 1 (odd), that requires 1998 cuts (even number) to that contact. Therefore, all contacts will have degree 1 only after an even number of cuts, which corresponds to the 1000 wires remaining.Therefore, until the last 1000 wires, all contacts have degree ≥2, so cutting any 3 wires is safe, as none of them will be the last wire for any contact. Therefore, Petya can safely cut 3 wires each time, provided that he doesn't target wires connected to contacts with degree 2. Wait, but if a contact has degree 2, cutting a wire connected to it would reduce it to 1, which is safe as long as you don't cut its last wire. But if all contacts have degree ≥2 until the very end, then cutting any wire is safe.Wait, but if a contact has degree 2, cutting one of its wires reduces it to 1. But since the game only loses when you cut the last wire (degree 1 to 0), cutting a wire that reduces a contact to degree 1 is allowed. The next time someone cuts a wire connected to that contact, if it's reduced to 0, they lose.Therefore, as long as the players avoid cutting the last wire of a contact, they can continue. So during the main phase of the game, players can cut any wires, but must avoid cutting wires that are the last ones for any contact.But if all contacts are kept at degree ≥2 until the final 1000 wires, then Petya can safely cut 3 wires each time. However, if by cutting 3 wires, he might inadvertently reduce a contact's degree to 1, but if he spreads out his cuts to different contacts, he can avoid this.Alternatively, if Petya mirrors Vasya's moves in a way that maintains symmetry. For example, if Vasya cuts a wire between contacts A and B, Petya can cut three wires in a symmetric fashion, such as cutting wires between C and D, E and F, G and H, maintaining some balance.But this requires a high degree of symmetry and coordination, which might be possible given the even number of contacts.Therefore, if Petya can maintain the symmetry and ensure that all contacts have the same degree, then when Vasya cuts one wire, Petya can cut three wires in a symmetric way, keeping all degrees equal. This way, they all get reduced uniformly.For example, suppose all contacts have degree d. Vasya cuts a wire, reducing two contacts to d-1. Petya then cuts three wires, each connecting pairs of contacts, reducing six contacts by 1. But to maintain symmetry, he would need to ensure that the degrees remain as uniform as possible.However, with 2000 contacts, which is divisible by 2 and 4, perhaps Petya can partition the contacts into groups and balance the degree reductions.Alternatively, since the total number of contacts is even, Petya can pair them up and for each wire Vasya cuts between a pair, Petya cuts three wires between other pairs, maintaining some kind of parity.This is getting quite abstract. Let's return to the earlier idea where Petya can force the total number of wires to decrease by 4 each round (1 by Vasya, 3 by Petya), leading to 1000 wires left on Vasya's turn. If this is possible without Petya ever being forced to cut a last wire, then Petya wins.Since the total number of wires to remove is 1998000, which is divisible by 4 (1998000 ÷4 = 499500), this is feasible. Each round removes 4 wires, and after 499500 rounds, exactly 1998000 wires are removed, leaving 1000. Since Vasya starts, the rounds go Vasya-Petya-Vasya-Petya-...-Vasya-Petya, total of 499500 Petya moves and 499500 Vasya moves, but wait, each round is one Vasya and one Petya move. So after 499500 rounds, both have made 499500 moves, and the total wires removed are 499500 *4=1998000. Therefore, the remaining wires are 1000, and it's Vasya's turn to move. He must cut one wire, which is one of the 1000 remaining, each connecting two degree 1 contacts. Cutting any of these would reduce two contacts to degree 0, so Vasya loses.Therefore, if Petya can always respond to Vasya's 1 wire by cutting 3 wires, and this is possible without causing a loss during the process, then Petya can force a win.But is there any reason Petya can't cut 3 wires each time? He needs to ensure that none of those 3 wires are the last wire for any contact. As long as all contacts have degree ≥2 before his move, he can safely cut any 3 wires. Since we start with all degrees at 1999, and each time Vasya cuts 1 wire (reducing two degrees to 1998), and Petya cuts 3 wires (reducing six degrees by 1 each to 1997), then the degrees are still high enough to avoid any contact getting to degree 1.Wait, but after the first round: Vasya cuts 1 wire, two contacts go to 1998. Petya cuts 3 wires, six contacts go to 1997. Next round: Vasya cuts another wire, two more contacts go to 1997 (or maybe some already reduced ones). Wait, but the degrees are being reduced non-uniformly.This might be a problem. If Petya cuts three wires connected to different contacts, he could be reducing some contacts multiple times, leading to lower degrees earlier.For example, suppose Petya cuts three wires all connected to the same contact. That contact's degree would go from 1998 (after Vasya's first move) to 1998 -3 = 1995. But other contacts would also be reduced. However, this could lead to some contacts being reduced more than others.But to minimize the risk of creating a contact with degree 1, Petya should spread out his cuts across different contacts.But since there are 2000 contacts, and Petya is cutting three wires (affecting six contacts) each time, as long as he distributes the cuts evenly, he can ensure that no contact is reduced too quickly.But this requires careful strategy. However, given the large number of contacts and the high initial degrees, it's feasible that Petya can spread out the cuts to avoid reducing any contact to degree 1 prematurely.Therefore, under optimal play, Petya can mirror Vasya's moves in a way that maintains a balanced reduction across all contacts, ensuring that the degrees are reduced uniformly, and that the game proceeds until the final 1000 wires are left, forcing Vasya to make the losing move.Therefore, the winner is Petya with optimal play.

✅:With 2000 contacts initially forming a complete graph, Vasya starts by cutting 1 wire per turn, while Petya can cut 1 or 3 wires. The loser is the player who cuts the last wire connected to any contact. Key Analysis:1. Initial State: Each contact has 1999 wires (odd). Total wires: ( binom{2000}{2} = 1999000 ).2. Parity Argument: Each move reduces the degree of two contacts by 1, flipping their parity. The number of odd-degree contacts remains even throughout the game, preventing a single contact from being isolated until the end.3. Endgame Scenario: The game ends when two contacts with degree 1 are connected by a wire. Cutting this wire reduces both to 0, causing the player to lose.4. Optimal Play Strategy: Petya can enforce a removal rate of 4 wires per round (Vasya cuts 1, Petya cuts 3). This ensures that after ( frac{1998000}{4} = 499500 ) rounds, exactly 1000 wires remain. These wires connect pairs of degree 1 contacts.5. Final Move: Vasya is forced to cut one of these 1000 wires on his turn, resulting in two contacts losing their last wire and his defeat.Conclusion: Petya can strategically cut 3 wires per turn, maintaining control over the game's progression and forcing Vasya into the losing position. Winner: Petya. boxed{Petya}

🔑:1. Setup and Initial Condition: - We have 2000 contacts, initially each pair of contacts is connected by a separate wire. - Vasya and Petya take turns cutting wires. Vasya starts and he cuts one wire per turn. Petya can cut either one or three wires per turn. - The game ends when one player cuts the last wire from any contact, and that player loses the game.2. Divide and Label Groups: - Mentally divide the contacts into four equal groups labeled ( A, B, C, ) and ( D ). - Each group ( A, B, C, ) and ( D ) contains 500 contacts, numbered from 1 to 500.3. Petya's Strategy: - For any move Vasya makes, Petya will respond in such a way that for each number ( k ) from 1 to 500, the contacts ( A_k, B_k, C_k, ) and ( D_k ) will have an equal number of wires removed up to Petya’s last turn.4. Maintenance of Condition: - Before the game starts, this condition is trivially satisfied.5. Different Cases for Vasya's and Petya's Moves: - Case 1: If Vasya cuts a wire between two contacts within the same group (e.g., ( A_i A_j )): - Petya will respond by cutting wires ( B_i B_j, C_i C_j, ) and ( D_i D_j ). - Case 2: If Vasya cuts a wire between contacts from different groups with different indices (e.g., ( A_i B_j )): - Petya will respond by cutting wires ( A_j B_i, C_i D_j, ) and ( C_j D_i ). - Case 3: If Vasya cuts a wire between contacts from different groups with the same index (e.g., ( A_k B_k )): - Petya will respond by cutting wire ( C_k D_k ).6. Ensuring Petya’s Strategy Completeness: - Note that in each of the described cases, Petya’s strategy ensures that equal numbers of wires remain connected to each ( k ) indexed contact in groups ( A, B, C, ) and ( D ). - The wires that Petya plans to cut are not already severed before his turn.7. Equal Number of Wires Condition: - After Petya’s move, for each number ( k ), contacts ( A_k, B_k, C_k, ) and ( D_k ) will still have equal numbers of wires removed. - The situation where the last wire from any contact is cut will occur first after a move by Vasya.8. Game Conclusion: - Because the total number of wires is finite, eventually, the case of the last wire being cut will happen. - Since this condition must first occur after Vasya’s move, Vasya will lose the game. Conclusion: [boxed{text{Petya wins}}]

🧠:Okay, let me try to tackle this problem step by step. The question is asking for the number of positive integers x (where x is not equal to 9) that satisfy the inequality:log base (x/9) of (x²/3) < 6 + log base 3 of (9/x)Hmm, logarithms can sometimes be tricky, especially with different bases. Let me first recall some logarithm properties to see how I can simplify this inequality.First, remember that log_b(a) < c is equivalent to a < b^c when b > 1, but if 0 < b < 1, the inequality direction reverses because the logarithm function is decreasing. So the base of the logarithm here is (x/9), which can be greater than 1 or between 0 and 1 depending on the value of x. That’s important—so I need to consider two cases: when the base (x/9) is greater than 1 and when it's between 0 and 1.Also, note that x is a positive integer not equal to 9. So x is in N, x ≠ 9.Let me write down the inequality again:log_{x/9}(x²/3) < 6 + log_3(9/x)Maybe I can express both logarithms with the same base to make it easier. Since the right side has log base 3, perhaps converting the left side to log base 3 as well. Let me try that.Using the change of base formula: log_b(a) = log_c(a) / log_c(b). So log_{x/9}(x²/3) can be written as log_3(x²/3) / log_3(x/9).So let me rewrite the inequality:[log_3(x²/3) / log_3(x/9)] < 6 + log_3(9/x)Simplify the numerator and denominator:First, log_3(x²/3) = log_3(x²) - log_3(3) = 2 log_3(x) - 1.Similarly, log_3(x/9) = log_3(x) - log_3(9) = log_3(x) - 2.So the left side becomes (2 log_3(x) - 1) / (log_3(x) - 2).On the right side, 6 + log_3(9/x) can be simplified as 6 + [log_3(9) - log_3(x)] = 6 + [2 - log_3(x)] = 8 - log_3(x).So substituting back into the inequality:(2 log_3(x) - 1) / (log_3(x) - 2) < 8 - log_3(x)Let me denote y = log_3(x) to simplify the expressions. Let’s set y = log_3(x). Since x is a positive integer, y is a real number, and x = 3^y. But since x is an integer, y must be such that 3^y is an integer. So y must be an integer or a logarithm that results in an integer when exponentiated. Wait, actually, x is a positive integer, so x can be written as 3^k where k is a non-negative integer, unless x has prime factors other than 3. Wait, no, x could be any positive integer, not necessarily a power of 3. For example, x could be 2, 4, 5, etc. Hmm, maybe that approach complicates things. Let me stick with y = log_3(x), which is just a real number, and proceed algebraically.So substituting y into the inequality:Left side: (2y - 1)/(y - 2)Right side: 8 - ySo the inequality becomes:(2y - 1)/(y - 2) < 8 - yNow, let's solve this inequality for y. But we have to remember that the base of the original logarithm is (x/9) = (3^y)/9 = 3^{y - 2}. For the logarithm to be defined, the base must be positive and not equal to 1. Since x is a positive integer not equal to 9, then (x/9) is positive (as x is positive) and not equal to 1 (since x ≠ 9). So the base is valid as long as x ≠ 9, which is given.Additionally, the argument of the logarithm must be positive. The argument of log_{x/9} is (x²)/3, which is positive since x is a positive integer. So that's okay.So, going back to the inequality in terms of y:(2y - 1)/(y - 2) < 8 - yLet me rearrange this inequality. First, bring all terms to the left side:(2y - 1)/(y - 2) - (8 - y) < 0Compute the left side:[(2y - 1) - (8 - y)(y - 2)] / (y - 2) < 0Let me expand the numerator:First, expand (8 - y)(y - 2):= 8(y - 2) - y(y - 2)= 8y - 16 - y² + 2y= (8y + 2y) - y² -16= 10y - y² -16So the numerator becomes:(2y - 1) - [10y - y² -16] = 2y -1 -10y + y² +16 = y² -8y +15Therefore, the inequality simplifies to:(y² -8y +15)/(y -2) < 0So we have the rational inequality (y² -8y +15)/(y -2) < 0. Let's factor the numerator:y² -8y +15 factors into (y -3)(y -5). Let me check:(y -3)(y -5) = y² -8y +15. Yes, correct.So, the inequality is:(y -3)(y -5)/(y -2) < 0Now, to solve this inequality, we need to find the intervals where the expression is negative. Let's find the critical points where the expression is zero or undefined.The numerator is zero when y = 3 or y = 5.The denominator is zero when y = 2.So critical points at y = 2, 3, 5. These divide the real line into four intervals:1. y < 22. 2 < y < 33. 3 < y < 54. y > 5Now, we can test each interval.But before that, note that the original logarithm base is (x/9) = 3^{y - 2}. Since the base of a logarithm must be positive and not equal to 1, we already have that covered since x ≠ 9 (so y ≠ 2, because x = 9 would mean y = log_3(9) = 2). Also, x is a positive integer, so y is defined.But we have to remember that x is a positive integer, so y = log_3(x). However, x doesn't have to be a power of 3. Wait, but x is an integer, so y can be any real number such that 3^y is an integer. But 3^y is an integer only when y is an integer because 3^k is an integer for integer k, and if y is not an integer, 3^y would not be an integer. Wait, is that correct?Wait, for example, if x = 2, then y = log_3(2), which is irrational. But in the original problem, x is a positive integer, but there is no restriction on y. However, in our substitution, we set y = log_3(x), which is defined for any positive x. But in our case, x is an integer, so y can take values such that 3^y is integer. However, 3^y is an integer only when y is an integer because 3^y is injective. For example, if y is not an integer, 3^y would not be an integer. So if x is an integer, then y must be log_3(x), which is either an integer (if x is a power of 3) or an irrational number (if x is not a power of 3). However, in our substitution above, we treated y as a real variable, but perhaps x being an integer imposes that y must be such that x = 3^y is an integer. So y must be log_3(n), where n is a positive integer. Hmm, this complicates things because y can be both integers and non-integers. However, maybe since we need x to be integer, after solving for y, we need to check which y correspond to integer x. Alternatively, perhaps it's easier to proceed with the inequality in terms of y, solve it, and then map back to x via x = 3^y, considering x must be an integer. However, since x must be an integer, and x = 3^y, then y must be such that 3^y is integer. Thus, y must be integers. Because 3^y is an integer only when y is a non-negative integer. For example, y = 0: x = 1; y =1: x=3; y=2: x=9; but x ≠9; y=3: x=27; etc. Wait, but x can be other integers not powers of 3. For example, x=2: then y=log_3(2), which is not an integer, but x=2 is allowed. Wait, the problem states x is a positive integer, but doesn't restrict x to be a power of 3. So x can be any positive integer except 9. So y can be log_3(x) for any positive integer x ≠9, which can be any positive real number except y=2 (since x=9 corresponds to y=2). So perhaps my initial substitution is okay, but when solving for y, we have to remember that x must be a positive integer, so after finding the range for y, we have to find the integers x=3^y that lie within that range. But 3^y is an integer only when y is an integer, as I thought earlier. Wait, is that true? Let's check:If y is an integer, then 3^y is an integer. If y is not an integer, then 3^y is not an integer. For example, 3^{1} =3, 3^{2}=9, 3^{0}=1, 3^{-1}=1/3, which is not integer. So if x is a positive integer, then y must be a non-negative integer. Because if y is a non-integer, then x=3^y would not be an integer. Therefore, even though x can be any positive integer, the corresponding y must be non-negative integers. Because x must be an integer. Therefore, y must be integers such that x=3^y is an integer.Wait, but hold on: x can be any positive integer, not necessarily a power of 3. For example, x=2, which is not a power of 3, so log_3(2) is irrational. But in that case, how does that fit into our substitution?Wait, perhaps there's a confusion here. Let me clarify:When we set y = log_3(x), then x = 3^y. But if x is an integer, then 3^y must be an integer, which only occurs when y is a non-negative integer. Therefore, even if x is not a power of 3, like x=2, then y would be log_3(2), but 3^{log_3(2)} = 2, which is an integer. Wait, that's correct! So actually, x can be any positive integer, and y = log_3(x) can be any real number such that 3^y is integer. But 3^y is integer not only when y is integer. For example, 3^{log_3(2)} =2, which is an integer. However, log_3(2) is not an integer. So my earlier assertion was wrong. So in fact, y can be any real number such that x=3^y is an integer. Therefore, x can be any positive integer, and y is just log_3(x). So y is real, not necessarily integer. Therefore, in solving the inequality, y can be any real number except y=2 (since x≠9), and then we have to find x=3^y where x is a positive integer. Therefore, after solving for y, we need to find all positive integers x (x≠9) such that y=log_3(x) satisfies the inequality.This complicates things because x must be an integer, but y is real. Therefore, the steps would be:1. Solve the inequality for y: (y -3)(y -5)/(y -2) < 0.2. Find the intervals of y where this inequality holds.3. Then, for each interval, find the values of x=3^y that are positive integers (x ≠9).But since x must be an integer, and x=3^y, then y must be such that 3^y is integer. However, except for powers of 3, other integers like 2,4,5 etc., when expressed as 3^y, would require y to be log_3(2), log_3(4), etc., which are real numbers but not integers. Therefore, x can be any positive integer, but y is log_3(x), which is real. Therefore, the approach should be:First, solve the inequality for y, find the range of y where (y -3)(y -5)/(y -2) < 0, then for those y, x=3^y must be a positive integer (x ≠9). Therefore, we need to find all positive integers x (other than 9) such that log_3(x) is in the solution set of the inequality.Alternatively, since x is a positive integer, maybe we can avoid dealing with y and instead manipulate the original inequality in terms of x directly. Let me try that approach as well, to see if it's simpler.But first, let's proceed with the y substitution, since we already started.We had the inequality (y -3)(y -5)/(y -2) < 0. Let's determine the intervals where this is true.Critical points at y=2, y=3, y=5.We can make a sign chart:- For y < 2: Let's pick y=1. Then numerator: (1-3)(1-5)=(-2)(-4)=8; denominator: 1-2=-1. So 8/-1 = -8 <0. So the expression is negative here.- For 2 < y <3: Let's pick y=2.5. Numerator: (2.5-3)(2.5-5)=(-0.5)(-2.5)=1.25; denominator: 2.5-2=0.5. So 1.25/0.5=2.5 >0. So positive here.- For 3 < y <5: Let's pick y=4. Numerator: (4-3)(4-5)=(1)(-1)=-1; denominator: 4-2=2. So -1/2=-0.5 <0. So negative here.- For y >5: Let's pick y=6. Numerator: (6-3)(6-5)=3*1=3; denominator:6-2=4. So 3/4 >0. Positive here.Therefore, the expression is negative in intervals y <2 and 3 < y <5.But we have to remember that y = log_3(x), and x is a positive integer, x ≠9.Therefore, the solution set for y is y <2 or 3 < y <5.But x must be a positive integer, so let's translate these y intervals into x values.First interval: y <2. Since y = log_3(x), this means log_3(x) <2 => x <3^2=9.But x is a positive integer, so x can be 1,2,3,4,5,6,7,8. However, x ≠9, which is already excluded here.Second interval: 3 < y <5. So log_3(x) >3 and log_3(x) <5 => x >3^3=27 and x <3^5=243. Therefore, x can be 28,29,...,242. But x must be an integer, so integers from 28 up to 242 inclusive.Therefore, combining both intervals, the possible x are:From first interval: x=1,2,3,4,5,6,7,8.From second interval: x=28,29,...,242.Now, we need to check if all these x satisfy the original inequality. However, we must remember that the base of the logarithm in the original inequality is (x/9). The base must be positive and not equal to 1, which is already satisfied since x is positive and x≠9. However, depending on whether (x/9) is greater than 1 or between 0 and 1, the direction of the inequality when we converted it might have changed.Wait a second! In the initial step, when we converted log_{x/9}(x²/3) < 6 + log_3(9/x), we changed the base to 3 and manipulated the inequality. However, when dealing with logarithmic inequalities, the direction of the inequality can reverse depending on the base. So when we rewrote the left side as [log_3(x²/3)] / [log_3(x/9)], and then proceeded to solve the inequality, we assumed that log_3(x/9) is positive or negative? Wait, no—actually, the original logarithm's base is (x/9), which, when we converted using change of base formula, became log_3(x/9) in the denominator. So the sign of log_3(x/9) affects the direction of the inequality when we multiply both sides by it. However, in our process above, we didn't consider that because we moved all terms to one side and formed a rational inequality. However, when solving rational inequalities, we have to consider the sign of the denominator, which is log_3(x/9) - 2 = log_3(x) - 2 -2? Wait, no. Wait, the original denominator in the left side after substitution was log_3(x/9) = log_3(x) - log_3(9) = log_3(x) -2. Therefore, the denominator in the left side expression is (log_3(x) -2). Therefore, when we set up the inequality:(2 log_3(x) -1)/(log_3(x) -2) < 8 - log_3(x)Then, moving everything to the left:[(2y -1)/(y -2) - (8 - y)] <0Which became (y² -8y +15)/(y -2) <0But during this process, we didn't multiply both sides by (y -2) because we combined the terms over a common denominator. Therefore, the critical points are considered, and the sign chart is valid. However, in this process, we didn't have to flip the inequality sign because we didn't multiply both sides by a variable expression; instead, we formed a rational expression and found where it is negative.Therefore, the solution set for y is indeed y <2 and 3 < y <5, as we found.But now, we need to check for x=1,2,...,8 and x=28,...,242.But wait, let me verify this with some test values. For example, take x=3 (which is in the first interval). Let's check if it satisfies the original inequality.Original inequality:log_{(3/9)}(3²/3) < 6 + log_3(9/3)Simplify:log_{1/3}(9/3) < 6 + log_3(3)log_{1/3}(3) < 6 +1log_{1/3}(3) = -1, since (1/3)^-1 =3. So left side is -1 <7, which is true.Another test: x=8 (in first interval).log_{(8/9)}(64/3) <6 + log_3(9/8)Hmm, let's compute both sides.Left side: log_{8/9}(64/3). The base is 8/9 which is less than 1, so the log is decreasing. Let's compute it using change of base:log(64/3)/log(8/9). Since base <1, the log will be negative. Let's compute approximate values:log_{8/9}(64/3) = ln(64/3)/ln(8/9) ≈ (ln(21.333)/ln(0.888...)) ≈ (3.06)/(-0.1178) ≈ -26.0Right side: 6 + log_3(9/8) =6 + [log_3(9) - log_3(8)] =6 + [2 - log_3(8)]. log_3(8)=log_3(2^3)=3 log_3(2)≈3*0.6309≈1.8927. So 6 +2 -1.8927≈6.1073. Therefore, left side ≈-26 <6.1073, which is true.Another test: x=27, which is not in the solution set (since 27 is 3^3, y=3, which is at the boundary). Wait, but our intervals are y <2 and 3 < y <5. So x=27 corresponds to y=3, which is excluded. Wait, but according to the inequality solution, 3 < y <5, so x=27 is y=3, which is not included. Let's check x=28, which is y=log_3(28). Let's compute:Left side: log_{28/9}(28²/3). Let's compute:28/9 is the base, approximately 3.111.28²=784, so argument is 784/3 ≈261.333.log_{3.111}(261.333). Let's compute this using natural logs:ln(261.333)/ln(3.111) ≈5.566/1.135 ≈4.905.Right side:6 + log_3(9/28). Compute log_3(9/28)= log_3(9) - log_3(28)=2 - log_3(28). log_3(28)=ln(28)/ln(3)≈3.332/1.0986≈3.033. So 2 -3.033≈-1.033. Therefore, right side=6 -1.033≈4.967.So left side≈4.905 <4.967, which is true.Now, check x=243, which is y=5. Since the upper interval is y <5, x=243 is excluded. Let's check x=242:log_{(242/9)}(242²/3) <6 + log_3(9/242)Compute left side:242/9≈26.888, argument≈242²/3≈58564/3≈19521.333log_{26.888}(19521.333). Since 26.888^3≈26.888*26.888=722.95*26.888≈19455, which is close to 19521.333. So log base 26.888 of 19521≈3. So left side≈3.Right side:6 + log_3(9/242)=6 + [log_3(9) - log_3(242)]=6 + [2 - log_3(242)]. log_3(242)=ln(242)/ln(3)≈5.489/1.0986≈5.0. So 6 +2 -5=3. Therefore, left side≈3 <3? No, equality. But the inequality is strict, so x=242 would be at the boundary where left side approaches 3 from below, which would be less than 3? Wait, but since y approaches 5 from below, x approaches 243 from below. Let me compute more accurately.x=242, log_3(242)=log_3(3^5 -1)=5 - log_3(1 + 1/242). Approximately, log_3(242)≈5 - (1/(242 ln3))≈5 - 0.002. So log_3(242)≈4.998. Therefore, y≈4.998.Thus, the original left side expression (y -3)(y -5)/(y -2)≈(4.998 -3)(4.998 -5)/(4.998 -2)= (1.998)(-0.002)/(2.998)≈-0.00133, which is negative. Therefore, the inequality holds for x=242.Similarly, x=28: log_3(28)≈3.033, so y≈3.033. Then (3.033 -3)(3.033 -5)/(3.033 -2)= (0.033)(-1.967)/(1.033)≈-0.063, which is negative. So the inequality holds.Therefore, the solution set for x is x in {1,2,3,4,5,6,7,8} and x in {28,29,...,242}.Now, we need to count these integers.First interval: x <9, x≠9. So x=1,2,3,4,5,6,7,8. That's 8 numbers.Second interval: x >27 and x <243. So x=28,29,...,242. The number of integers from 28 to 242 inclusive is 242 -28 +1= 215.Therefore, total number of solutions is 8 +215=223.But wait, hold on. Let me verify if all x in these intervals satisfy the original inequality. Especially because when we converted the logarithmic inequality to the rational inequality, we might have introduced some extraneous solutions or missed some constraints.But earlier, we considered the critical points and tested some values, which worked. However, let's check x=3, which is a power of 3.x=3: log_{(3/9)}(9/3)=log_{1/3}(3)= -1 <6 + log_3(3)=6 +1=7. True.x=1: log_{1/9}(1/3)=log_{1/9}(1/3). Let's compute this:(1/9)^k =1/3 =>9^{-k}=3^{-1} =>3^{-2k}=3^{-1} =>-2k=-1 =>k=1/2. So log_{1/9}(1/3)=1/2. Then 1/2 <6 + log_3(9/1)=6 +2=8. 1/2 <8, true.x=4: log_{4/9}(16/3). Let's compute both sides.Left side: log_{4/9}(16/3). Let's use change of base:ln(16/3)/ln(4/9)= (ln(16) - ln(3))/(ln(4) - ln(9))≈ (2.7726 -1.0986)/(1.3863 -2.1972)=1.674/(-0.8109)≈-2.065.Right side:6 + log_3(9/4)=6 + [log_3(9) - log_3(4)]=6 +[2 - log_3(4)]≈6 +2 -1.2619≈6.7381.-2.065 <6.7381, true.Another test in the second interval: x=81 (which is 3^4=81, so y=4). Then check original inequality:log_{81/9}(81² /3)=log_{9}(6561/3)=log_{9}(2187).2187=9^3.5? Wait, 9^3=729, 9^4=6561. So 2187=729*3=9^3*3=3^6*3=3^7=2187. Wait, 3^7=2187. So log_9(2187)=log_{3^2}(3^7)=7/2=3.5.Right side:6 + log_3(9/81)=6 + log_3(1/9)=6 + (-2)=4.So 3.5 <4, which is true.Another test: x=243 is excluded, but x=242, which is just below 3^5=243. As we calculated earlier, the left side is approximately 3 and the right side is approximately 3, but slightly less. Wait, when x=242, left side≈3, right side≈3. But let's compute more precisely.x=242:Left side: log_{242/9}(242² /3) = log_{26.888...}(242² /3). Let's compute 242²=58564, so 58564/3≈19521.333.Now, 26.888^3≈26.888*26.888=722.95; 722.95*26.888≈19455. So 26.888^3≈19455, which is slightly less than 19521.333. Therefore, log_{26.888}(19521.333)≈3 + log_{26.888}(19521.333/26.888^3). Compute 19521.333/19455≈1.0034. So log_{26.888}(1.0034)=ln(1.0034)/ln(26.888)≈0.0034/3.292≈0.001. So total≈3.001.Right side:6 + log_3(9/242)=6 + [log_3(9) - log_3(242)]=6 + [2 - log_3(242)]. log_3(242)=ln(242)/ln(3)≈5.489/1.0986≈5.0. Therefore, 6 +2 -5=3. So right side≈3. Therefore, left side≈3.001 <3? No, that would be false. But wait, this contradicts our previous calculation.Wait, perhaps there was an error in approximation. Let me compute more accurately.log_3(242)=ln(242)/ln(3)≈5.489/1.0986≈5.0. But actually, 3^5=243, so log_3(243)=5. Therefore, log_3(242)=5 - log_3(243/242)=5 - log_3(1 +1/242)≈5 - (1/(242 ln3))≈5 - (1/(242*1.0986))≈5 -0.0037≈4.9963.Therefore, log_3(242)≈4.9963, so 6 +2 -4.9963≈3.0037.Left side: log_{242/9}(242² /3)=log_{26.888}(19521.333). As calculated earlier, 26.888^3≈19455, so the value is slightly more than 3. But if we compute:Let’s use natural logs:ln(19521.333)/ln(26.888)= ln(19521.333)/ln(26.888).Compute ln(19521.333)≈9.879.ln(26.888)=ln(27 -0.112)=approx ln(27) - (0.112)/27≈3.296 -0.004≈3.292.Therefore, 9.879/3.292≈3.000. So the left side≈3.000, and the right side≈3.0037. Therefore, 3.000 <3.0037, which is true. So x=242 is valid.Similarly, at x=243, which is not included, the left side would be log_{243/9}(243² /3)=log_{27}(19683/3)=log_{27}(6561). Since 27^3=19683, so 6561=27^(2.333...). Wait, 27^2=729, 27^3=19683. 729*9=6561. So 6561=9^3=27^2. Wait, 27^2=729. Wait, no. Wait, 27^2=27*27=729. 729*9=6561. So 6561=9^3*9=9^4=6561. So log_{27}(6561)=log_{27}(9^4)=log_{27}(9^4)=4*log_{27}(9)=4*(2/3)=8/3≈2.6667.Right side:6 + log_3(9/243)=6 + log_3(1/27)=6 + (-3)=3. So left side≈2.6667 <3, which is true. But x=243 is excluded because x≠9? Wait, x=243 is excluded only if x=9, but 243≠9. Wait, the original problem says x≠9. So x=243 is allowed. Wait, but in our solution earlier, we said the upper interval is y <5, so x <3^5=243. Therefore, x=243 is not included. But x=243 is allowed since x≠9. However, according to the inequality solution, y must be less than5, so x=243 corresponds to y=5, which is not in the interval. Therefore, x=243 is excluded.But according to the original inequality, if we plug x=243:Left side: log_{243/9}(243² /3)=log_{27}(243² /3)=log_{27}(59049 /3)=log_{27}(19683)=3, since 27^3=19683.Right side:6 + log_3(9/243)=6 + log_3(1/27)=6 + (-3)=3.So the inequality becomes 3 <3, which is false. Therefore, x=243 does not satisfy the inequality. Hence, our interval y <5 (x <243) is correct.Therefore, the total number of solutions is 8 +215=223.But let's confirm the count from 28 to242 inclusive.242 -28 +1=215. Yes, that's correct. For example, from 1 to n inclusive, it's n -1 +1=n. So from 28 to242:242-28=214, plus1=215.Therefore, total solutions:8+215=223.But wait, the problem says positive integers x where x≠9. So x=9 is excluded, but in our first interval, x=1 to8, which doesn't include9, so that's fine. In the second interval, x=28 to242, which also doesn't include9. So all counted x are valid.But let me check if there are any other constraints that might exclude some x values.For example, in the original inequality, the base of the logarithm is x/9. Since x is a positive integer, x/9 must be positive and ≠1. Which is already satisfied since x≠9 and x is positive. So all x considered are valid.Therefore, the number of positive integers x (x≠9) satisfying the inequality is 223.Wait, but this seems like a large number. Let me check again.First interval: x=1,2,3,4,5,6,7,8. 8 numbers.Second interval: x=28 to242 inclusive. How many numbers is that?From28 to242: 242-28=214, plus1=215 numbers. So total 8+215=223.But the answer seems quite big. Maybe I made a mistake in translating the y intervals to x.Wait, y <2 corresponds to x <3^2=9, so x=1-8.3 < y <5 corresponds to x between3^3=27 and3^5=243, i.e., x=28-242. Because 3^3=27 is excluded (since y>3), so next integer is28. 3^5=243 is excluded (since y<5), so up to242.Yes, that's correct.Therefore, the total number is indeed8+215=223.But the problem is from a math competition or similar, and the answer is likely to be smaller. Maybe I messed up in the sign chart.Wait, let's re-examine the rational inequality solution.We had (y -3)(y -5)/(y -2) <0.Critical points at y=2,3,5.Sign chart:- For y <2: (y -3)=negative, (y -5)=negative, (y -2)=negative. So (-)(-)/(-)= (-)/(-)=positive? Wait, hold on. Wait, the numerator is (y-3)(y-5), denominator (y -2).If y <2: y -3 <0, y -5 <0, so numerator is positive. Denominator y -2 <0. Therefore, positive/negative=negative. So the expression is negative in y <2.For 2 <y <3: numerator (y -3)(y -5)=negative*(negative)=positive, denominator positive. So expression positive.For3 <y <5: numerator (positive)(negative)=negative, denominator positive. So expression negative.Fory >5: numerator positive*positive=positive, denominator positive. So expression positive.Therefore, the expression is negative in y <2 and 3 <y <5. Therefore, correct.Therefore, mapping to x: x=1-8 and x=28-242.Hence, 223.But maybe I have to check if in the original inequality, there are any hidden constraints.Wait, for example, take x=3 (which is in the first interval). The base of the logarithm is3/9=1/3. The argument is3²/3=3. So log_{1/3}(3)= -1. Right side:6 +log_3(9/3)=6 +log_3(3)=6 +1=7. So -1 <7, which is true.Another test: x=27 (which is at y=3, excluded). log_{27/9}(27²/3)=log_3(729/3)=log_3(243)=5. Right side:6 +log_3(9/27)=6 +log_3(1/3)=6 -1=5. So 5 <5? No, equality. Hence, x=27 is excluded correctly.x=28: as checked earlier, satisfies the inequality.x=243: doesn't satisfy, as checked.So the count seems correct. Therefore, the number of solutions is223.But since the problem is likely expecting a smaller answer, perhaps I made a mistake in the translation from y to x.Wait, but y=log_3(x). For y <2: x <9. For3 <y <5: x between27 and243.But x must be a positive integer. So x=1-8 andx=28-242. Thus, counts are8 and215.But let's verify if all these x satisfy the original inequality. Given that we tested several cases and they worked, and the rational inequality solution seems correct, it's likely correct.However, perhaps in the original problem statement, there is a typo or something, but assuming the problem is as stated, the answer should be223.Wait, but let me think again: x is a positive integer, x≠9. The solutions are x=1-8 andx=28-242. Total numbers:8 + (242-28+1)=8+215=223. Yes.But maybe I made an error in counting. Let's compute 242-28+1.242-28=214. 214+1=215. Correct.Therefore, the total number is223.But since this is a competition-style problem, maybe there's a trick I missed. Let me check if x=9 is excluded, but since we already excluded it, it's fine.Alternatively, maybe when converting the original inequality, I missed some condition.Original inequality:log_{x/9}(x²/3) <6 +log_3(9/x)We converted the left side to base3:log_3(x²/3)/log_3(x/9)= (2 log_3x -1)/(log_3x -2)Right side:6 +log_3(9/x)=6 +2 -log_3x=8 -log_3xThus, the inequality becomes:(2y -1)/(y -2) <8 -y where y=log_3x.Then moving terms:(2y -1)/(y -2) + y -8 <0Which becomes:[2y -1 + (y -8)(y -2)]/(y -2) <0Wait, this is the same as before. Wait, no—wait, when I bring everything to the left:(2y -1)/(y -2) - (8 -y) <0Which is:(2y -1 - (8 -y)(y -2))/(y -2) <0Wait, previously I expanded (8 -y)(y -2)= -y² +10y -16, hence 2y -1 -(-y² +10y -16)=2y -1 +y² -10y +16=y² -8y +15.Therefore, correct.Thus, the steps are correct. Therefore, the answer should be223. However, this seems very large, which makes me wonder if there is a miscalculation.Alternatively, maybe I misapplied the logarithm properties. Let me check the conversion again.Original left side: log_{x/9}(x²/3)Using change of base formula: log_a(b)=log_c(b)/log_c(a). So here, log_{x/9}(x²/3)=log_3(x²/3)/log_3(x/9)log_3(x²/3)=log_3x² -log_33=2 log_3x -1.log_3(x/9)=log_3x -log_39=log_3x -2.Correct.Right side:6 +log_3(9/x)=6 +log_39 -log_3x=6 +2 -log_3x=8 -log_3x.Correct.Therefore, inequality is (2y -1)/(y -2) <8 -y, which leads to (y² -8y +15)/(y -2) <0, factored as (y -3)(y -5)/(y -2) <0. Correct.Therefore, the solution is y <2 or 3 <y <5, which translates to x=1-8 andx=28-242. Correct.Therefore, the answer should be223. Since I can't find any mistake in the process, I think this is the correct answer.Final Answerboxed{223}

✅:To solve the inequality (log_{frac{x}{9}}left(frac{x^2}{3}right) < 6 + log_3left(frac{9}{x}right)) for positive integers (x neq 9), we start by converting the logarithms to the same base using the change of base formula and simplifying:1. Convert (log_{frac{x}{9}}left(frac{x^2}{3}right)) to base 3: [ frac{log_3left(frac{x^2}{3}right)}{log_3left(frac{x}{9}right)} = frac{2 log_3 x - 1}{log_3 x - 2} ]2. Simplify the right-hand side: [ 6 + log_3left(frac{9}{x}right) = 6 + 2 - log_3 x = 8 - log_3 x ]3. Let (y = log_3 x). The inequality becomes: [ frac{2y - 1}{y - 2} < 8 - y ]4. Rearrange and simplify the inequality: [ frac{(2y - 1) - (8 - y)(y - 2)}{y - 2} < 0 ] Expanding and simplifying the numerator: [ frac{y^2 - 8y + 15}{y - 2} < 0 ] Factoring the numerator: [ frac{(y - 3)(y - 5)}{y - 2} < 0 ]5. Determine the intervals where the inequality holds by analyzing the critical points (y = 2), (y = 3), and (y = 5). The solution intervals for (y) are (y < 2) and (3 < y < 5).6. Convert these intervals back to (x) values: - For (y < 2), (x < 3^2 = 9), so (x) can be (1, 2, 3, 4, 5, 6, 7, 8). - For (3 < y < 5), (3^3 < x < 3^5), so (x) can be (28, 29, ldots, 242).7. Count the valid integers: - From (1) to (8): 8 integers. - From (28) to (242): (242 - 28 + 1 = 215) integers.Summing these, the total number of positive integers (x neq 9) that satisfy the inequality is (8 + 215 = 223).[boxed{223}]

🔑:Given the inequality:[log _{frac{x}{9}}left(frac{x^{2}}{3}right) < 6 + log _{3}left(frac{9}{x}right)]we need to transform the logarithmic expressions to standardize the bases and solve for ( x ).1. Transform the left-hand side:[log _{frac{x}{9}}left(frac{x^{2}}{3}right)]Using the change of base formula ( log_{a}(b) = frac{log_{c}(b)}{log_{c}(a)} ):[log _{frac{x}{9}}left(frac{x^{2}}{3}right) = frac{log_{3}left(frac{x^{2}}{3}right)}{log_{3}left(frac{x}{9}right)}]2. Simplify the logarithmic expressions:[log_{3}left(frac{x^{2}}{3}right) = log_{3}(x^{2}) - log_{3}(3) = 2 log_{3}(x) - 1][log_{3}left(frac{x}{9}right) = log_{3}(x) - log_{3}(9) = log_{3}(x) - 2]Thus, the left-hand side becomes:[frac{log_{3}(x^{2}) - log_{3}(3)}{log_{3}(x) - log_{3}(9)} = frac{2 log_{3}(x) - 1}{log_{3}(x) - 2}]3. Transform the right-hand side:[6 + log_{3}left(frac{9}{x}right)][log_{3}left(frac{9}{x}right) = log_{3}(9) - log_{3}(x) = 2 - log_{3}(x)]So the right-hand side is:[6 + 2 - log_{3}(x) = 8 - log_{3}(x)]4. Combine the transformations into inequality:[frac{2 log_{3}(x) - 1}{log_{3}(x) - 2} < 8 - log_{3}(x)]Let ( u = log_{3}(x) ). The inequality now becomes:[frac{2u - 1}{u - 2} < 8 - u]5. Solve the transformed inequality:First, bring everything to one side to form a common inequality:[frac{2u - 1}{u - 2} + u - 8 < 0][frac{2u - 1 + (u - 8)(u - 2)}{u - 2} < 0]Simplify the numerator:[2u - 1 + u^2 - 10u + 16 = u^2 - 8u + 15]Thus:[ frac{u^2 - 8u + 15}{u - 2} < 0]6. Analyze the quadratic inequality:First, find the roots of ( u^2 - 8u + 15 = 0 ) by factoring:[(u - 3)(u - 5) = 0]So the roots are ( u = 3 ) and ( u = 5 ). Analyze the sign of the quadratic expression around these critical points and the root of the denominator (u = 2).By testing intervals ( (-infty, 2) ), ( (2, 3) ), ( (3, 5) ), and ( (5, infty) ), we determine ( frac{u^2 - 8u + 15}{u - 2} < 0 ) in the intervals:[u < 2 quad text{or} quad 3 < u < 5]7. Convert intervals back to ( x ) values:[log_{3}(x) < 2 quad Rightarrow quad x < 3^2 quad Rightarrow quad x < 9][3 < log_{3}(x) < 5 quad Rightarrow quad 3^3 < x < 3^5 quad Rightarrow quad 27 < x < 243]8. Count the number of valid positive integers ( x ):[text{For } x < 9: quad text{Integers are } 1, 2, ldots, 8 quad (8 text{ values})][text{For } 27 < x < 243: quad text{Integers are } 28, 29, ldots, 242 quad (215 text{ values})]Subtract the disallowed value ( x = 9 ):[8 + 215 = 223] Conclusion:boxed{223}

🧠:Okay, so I need to find all natural numbers ( k ) such that ( 3^{k} + 5^{k} ) is a power of a natural number with an exponent of at least 2. Hmm, let's start by understanding the problem properly. First, let me rephrase the question: find all positive integers ( k ) where the sum ( 3^k + 5^k ) equals ( m^n ) for some natural numbers ( m ) and ( n ), where ( n geq 2 ). So, the sum has to be a perfect square, cube, fourth power, etc. I should start by checking small values of ( k ) to see if any obvious solutions pop up. Let's compute ( 3^k + 5^k ) for ( k = 1, 2, 3, ldots ) and see if any of these sums are perfect powers.For ( k = 1 ):( 3^1 + 5^1 = 3 + 5 = 8 ). Is 8 a power with exponent ≥2? Yes, 8 is ( 2^3 ). So, ( k = 1 ) is a solution. Wait, but the problem says "natural numbers ( k )", and natural numbers usually start at 1. So that's valid.Wait, but hold on. Let me confirm if ( m ) has to be a natural number as well. The problem states "a power of a natural number", so ( m ) must be natural, and exponent ( n geq 2 ). Since 8 is ( 2^3 ), that's valid, so ( k = 1 ) is indeed a solution. But the question says "exponent ≥2", so 3 is ≥2, so it's okay. So, ( k=1 ) is a solution. Hmm, but maybe they want exponents ≥2, but if the exponent itself is allowed to be 3, then 8 is okay. But let's check the next ones.For ( k = 2 ):( 3^2 + 5^2 = 9 + 25 = 34 ). 34 is not a perfect square or cube. 5^2 is 25, 6^2 is 36, so 34 is in between. 3^3 is 27, 4^3 is 64, so 34 is not a cube. So, no.For ( k = 3 ):( 3^3 + 5^3 = 27 + 125 = 152 ). 152 is between 12^2=144 and 13^2=169. Not a square. Cubes: 5^3=125, 6^3=216, so no. So, 152 isn't a perfect power.For ( k = 4 ):( 3^4 + 5^4 = 81 + 625 = 706 ). 706 is between 26^2=676 and 27^2=729. Not a square. Cubes: 8^3=512, 9^3=729, so no. Similarly, higher exponents would be even larger, so no.For ( k = 5 ):( 3^5 + 5^5 = 243 + 3125 = 3368 ). 3368 is way between squares: 58^2=3364 and 59^2=3481. So, 3368 is 58^2 +4, not a square. Cubes: 15^3=3375, which is just 7 more than 3368, so not a cube. So, no.For ( k = 6 ):( 3^6 + 5^6 = 729 + 15625 = 16354 ). Let's see, squares: sqrt(16354) ≈ 127.8, so 127^2=16129, 128^2=16384. 16354 is 30 less than 16384. Not a square. Cubes: 25^3=15625, 26^3=17576. 16354 is in between, so not a cube. Similarly, higher exponents would be even larger, so no.For ( k = 7 ):( 3^7 + 5^7 = 2187 + 78125 = 80312 ). Squares: 283^2=80089, 284^2=80656. 80312 is between them, not a square. Cubes: 43^3=79507, 44^3=85184. So, no. For ( k = 8 ):( 3^8 + 5^8 = 6561 + 390625 = 397186 ). Square: Let's compute sqrt(397186). 630^2=396900, 631^2=398161. So 397186 is between them. Cubes: 73^3=389017, 74^3=405224. Not a cube. So far, only ( k=1 ) works. Let's check higher k maybe up to 10, but this might take too long. Alternatively, perhaps there's a mathematical approach here.Wait, the problem is to find all natural numbers ( k ). So maybe ( k=1 ) is the only solution. But let's see if that's the case.But first, let me check ( k=0 ). Wait, natural numbers usually start at 1, but sometimes 0 is included. However, if ( k=0 ), ( 3^0 + 5^0 = 1 + 1 = 2 ), which is not a perfect power with exponent ≥2. So even if 0 were allowed, it wouldn't work. So, sticking with ( k geq 1 ), only k=1 works so far.But maybe there's a larger k where it works. Let me think of another approach.Suppose that ( 3^k + 5^k = m^n ), where ( n geq 2 ). Let's analyze this equation modulo some numbers to see possible constraints.First, modulo 3. Let's compute ( 5^k mod 3 ). Since 5 ≡ 2 mod 3, so 5^k ≡ 2^k mod 3. Then, 3^k ≡ 0 mod 3. So, the sum is ( 0 + 2^k mod 3 ). Therefore, ( m^n ≡ 2^k mod 3 ). But m^n mod 3 must be 0, 1, or 2. Let's see possible residues for m^n mod 3. If m ≡ 0 mod 3, then m^n ≡ 0 mod 3. If m ≡ 1 mod 3, then m^n ≡ 1 mod 3. If m ≡ 2 mod 3, then m^n ≡ 2^n mod 3. Since 2^1=2, 2^2=4≡1, 2^3=8≡2, etc. So 2^n mod 3 cycles with period 2. So if n is even, 2^n ≡1 mod3, if n is odd, 2^n≡2 mod3.Therefore, m^n mod3 can be 0,1, or 2 depending on m and n. But in our case, the sum is 2^k mod3. Therefore, 2^k mod3 must equal m^n mod3. Let's see:If m ≡0 mod3, then m^n ≡0 mod3. So we have 2^k ≡0 mod3, which is impossible because 2 and 3 are coprime. So m cannot be divisible by 3.If m ≡1 mod3, then m^n ≡1 mod3. Therefore, 2^k ≡1 mod3. Since 2^1=2≡2, 2^2=4≡1, 2^3=8≡2, 2^4=16≡1, etc. So 2^k ≡1 mod3 iff k is even.If m ≡2 mod3, then m^n ≡2^n mod3. Therefore, 2^k ≡2^n mod3. So if n is even, 2^n≡1 mod3, which would require 2^k≡1 mod3, so k even. If n is odd, 2^n≡2 mod3, which would require 2^k≡2 mod3, so k odd.So, depending on m mod3, we have different conditions. Let's summarize:If m ≡0 mod3: impossible.If m ≡1 mod3: k must be even.If m ≡2 mod3: If n is even, k must be even; if n is odd, k must be odd.Hmm, so depending on m and n, different parity conditions on k. However, this might not be enough. Let's check modulo 4.Compute ( 3^k + 5^k mod4 ). 3 ≡ -1 mod4, so 3^k ≡ (-1)^k mod4. 5 ≡1 mod4, so 5^k ≡1^k ≡1 mod4. Thus, sum ≡ (-1)^k +1 mod4.So, if k is even: (-1)^k =1, so sum ≡1+1=2 mod4.If k is odd: (-1)^k =-1≡3 mod4, so sum≡3 +1=0 mod4.Now, m^n must be congruent to either 0 or 2 mod4. Let's see possible residues for m^n mod4.If n ≥2:- If m is even, then m is 0 or 2 mod4. If m is even, m^n ≡0 mod4 if n≥2. If m is 2 mod4, then m^2=4≡0 mod4, so m^n≡0 mod4 for n≥2.- If m is odd, then m ≡1 or 3 mod4. Then, 1^n ≡1 mod4, and 3^n ≡(3^2)^{n/2}≡1^{n/2}≡1 mod4 if n even, or 3*1^{(n-1)/2}≡3 mod4 if n odd. However, n≥2, so if n is even, m^n ≡1 mod4; if n is odd, m^n ≡3 mod4.So, combining:If m is even: m^n ≡0 mod4.If m is odd and n even: m^n≡1 mod4.If m is odd and n odd: m^n≡3 mod4.But our sum is either 2 mod4 (if k even) or 0 mod4 (if k odd).Therefore:If k is even: sum ≡2 mod4. So m^n ≡2 mod4. But m^n can only be 0,1, or3 mod4. 2 mod4 is impossible. Therefore, if k is even, sum≡2 mod4, which cannot be a perfect power (since m^n can't be 2 mod4). Therefore, k cannot be even.If k is odd: sum≡0 mod4. Therefore, m^n must be ≡0 mod4, which implies that m is even. So m is even, and n≥2.So from modulo 3 and 4 analysis:From modulo4: k must be odd.From modulo3: since k is odd, let's revisit that. Earlier, we saw that if k is odd, then modulo3, the sum is 2^k +0 ≡2^k mod3. And m^n ≡ either 1 mod3 (if m≡1 mod3 and any n) or 2^n mod3 (if m≡2 mod3). But since m is even (from modulo4), m must be ≡0, 2 mod3. But since m cannot be ≡0 mod3 (from earlier modulo3 analysis), so m must be ≡2 mod3, and so m^n ≡2^n mod3. Therefore, 2^k ≡2^n mod3.Which implies that 2^{k} ≡2^{n} mod3. Since 2 and 3 are coprime, we can divide both sides by 2^{min(k,n)}, leading to 2^{|k -n|} ≡1 mod3. But 2^1=2≡2 mod3, 2^2=4≡1 mod3, so 2^m ≡1 mod3 iff m is even. Therefore, |k -n| must be even. That is, k -n is even. Therefore, k and n have the same parity.But since k is odd (from modulo4), this implies that n is also odd. So n is an odd integer ≥2, which means n≥3.So, in summary:- k must be odd.- m is even and ≡2 mod3.- n is odd and ≥3.- |k -n| is even, so k ≡n mod2. Since k is odd, n must be odd.Therefore, possible solutions have k odd and n odd ≥3.But how does this help? Let's try to see if we can find more constraints.Another approach: Let's suppose that ( 3^k + 5^k = m^n ), where n≥2, m is even, k is odd.Let’s consider the case when n=3. Maybe the sum is a cube. Let's test for small k odd.We already checked k=1: 3+5=8=2^3. So that's a solution.k=3: 27+125=152. Is 152 a cube? 5^3=125, 6^3=216, so no.k=5: 243+3125=3368. 15^3=3375, which is close but not 3368.k=7: 2187+78125=80312. 43^3=79507, 44^3=85184. So no.k=9: 3^9=19683, 5^9=1953125. Sum=19683+1953125=1972808. Let's check cubes: cube root of 1972808 is approximately 125.4 since 125^3=1953125, 126^3=2000376. So no.k=11: 3^11=177147, 5^11=48828125. Sum=177147+48828125=48905272. Cube root of this is approx 366.8, since 366^3=48905256, which is very close. Wait, 366^3=366*366=133956; 133956*366= let's compute:First, 133956*300=40,186,800133,956*60=8,037,360133,956*6=803,736Adding up: 40,186,800 +8,037,360=48,224,160 +803,736=49,027,896. But 366^3=49,027,896. But the sum at k=11 is 48,905,272, which is less. So, not a cube.So, n=3 seems not to work beyond k=1.What about n=5? Maybe the sum is a fifth power. Let's check k=1: 8 is not a fifth power. k=3: 152 is not a fifth power. k=5: 3368. The fifth powers are 2^5=32, 3^5=243, 4^5=1024, 5^5=3125, 6^5=7776. 3368 is between 5^5 and 6^5, not a fifth power. Similarly, k=7: 80312. 6^5=7776, 7^5=16807, 8^5=32768, 9^5=59049, 10^5=100000. 80312 is between 9^5 and 10^5, not a fifth power.This doesn't look promising either. Maybe the only solution is k=1.But let's think algebraically. Suppose that 3^k +5^k =m^n. Let's factor this if possible. For example, when k is 1, it's 3+5=8=2^3. For k odd, 3^k +5^k can be factored as (3 +5)(3^{k-1} -3^{k-2}*5 + ... +5^{k-1}). Since k is odd, the sum is divisible by 3+5=8, which we saw for k odd, the sum is divisible by 4, and actually, for k odd, since 3 +5 divides 3^k +5^k, which is 8. Therefore, 8 divides the sum. But we can say more: for k odd, 3^k +5^k = (3 +5)(3^{k-1} -3^{k-2}*5 + ... +5^{k-1}) =8 * [some integer]. Therefore, the sum is 8 times another number. So m^n=8*N, where N is the other factor. Since m is even (from modulo4 analysis), let's write m=2*a, where a is a natural number. Then m^n=(2a)^n=2^n *a^n. Therefore, 2^n *a^n=8*N. So, 2^{n} *a^n=8*N ⇒ 2^{n-3}*a^n=N. So, N must be divisible by 2^{n-3} if n≥3.But since N=(3^{k} +5^{k})/8, which is an integer only when k is odd. Let's check for k=1: N= (3+5)/8=1. So, 2^{n-3}*a^n=1. Since a and n are natural numbers, the only possibility is a=1 and 2^{n-3}=1 ⇒ n-3=0 ⇒n=3. Which matches the case k=1: 8=2^3.For k=3: N=(27 +125)/8=152/8=19. So 19=2^{n-3}*a^n. Since 19 is prime, possible factors are 1 and 19. Since 2^{n-3}*a^n=19, and a and n are natural numbers. Let's check possible n:If n=3: 2^{0}*a^3=1*a^3=19 ⇒ a^3=19 ⇒ not integer.If n=4: 2^{1}*a^4=2*a^4=19 ⇒a^4=19/2 ⇒ not integer.n=5: 2^{2}*a^5=4*a^5=19 ⇒a^5=19/4 ⇒ no.Similarly, no solutions. Thus, for k=3, there is no n≥3 such that 152=8*19 is a perfect power.For k=5: sum=3368, N=3368/8=421. 421 is a prime number (I think so). Check if 421 is prime: 421 divided by primes up to sqrt(421)≈20.5. 421 ÷2=210.5, ÷3=140.333, ÷5=84.2, ÷7=60.14, ÷11=38.27, ÷13=32.38, ÷17=24.76, ÷19=22.15. None divide, so 421 is prime. Therefore, 421=2^{n-3}*a^n. Similarly, no solution since 421 is prime.For k=7: sum=80312, N=80312/8=10039. Check if 10039 is prime. sqrt≈100.19. Divide by primes up to 100. 10039 ÷2=5019.5, ÷3≈3346.33, ÷5=2007.8, ÷7≈1434.14, ÷11≈912.63, ÷13≈772.23, ÷17≈590.52, ÷19≈528.37, ÷23≈436.48, ÷29≈346.17, ÷31≈323.8, etc. Not divisible, so likely prime. So 10039=2^{n-3}*a^n. Again, prime, so only possible if a^n=10039 and 2^{n-3}=1 ⇒ n=3. But a^3=10039, which is not a cube. So no.Therefore, for these k=3,5,7, etc., the factor N=(3^k +5^k)/8 is prime or a product of primes in such a way that 2^{n-3}*a^n=N is not possible unless N is a perfect cube (for n=3), but even then, N would need to be a cube. For k=1, N=1, which is 1^3. For higher k, N is prime or composite but not a cube.Therefore, perhaps the only solution is k=1. But let's check k=9: sum=1972808. Divide by8: 1972808/8=246601. Check if 246601 is prime. sqrt≈496.59. Test divisibility by primes up to 500. 246601 ÷2=123300.5, ÷3=82200.333, ÷5=49320.2, ÷7≈35228.714, ÷11≈22418.27, etc. Maybe it's prime. If so, same issue: 246601=2^{n-3}*a^n. Not possible unless n=3 and a≈62.7 (not integer). So no.Thus, it seems that for k>1, the factor N=(3^k +5^k)/8 is either prime or a composite number that isn't a perfect power. Therefore, the only solution is k=1.But wait, let's think deeper. Suppose that for some k>1, 3^k +5^k is a perfect power. What if n is even? Wait, earlier analysis said that k must be odd. Because if k is even, the sum is 2 mod4, which can't be a perfect power. So k must be odd, and n must be odd ≥3. However, even if n is odd, the sum might not necessarily be a perfect power. But maybe there's a case where N=(3^k +5^k)/8 is a perfect power. For example, if N is a square, then 8*N=2^3*N, which would need to be a perfect power. But 8*N=2^3*N. If N is a square, then 2^3*square. To be a perfect power, the exponents in prime factorization must all be multiples of n. For example, if 8*N=2^{3}*N, and if N has factors of 2, then the exponent of 2 would be 3 plus the exponent in N. If N is a square, but let's say N is 2^{a}*other primes. Then 8*N=2^{3 +a}*other primes. For this to be a perfect power, all exponents must be multiples of n. So n divides 3 +a and the exponents in N. But this is getting too abstract.Alternatively, let's consider that 3^k +5^k =m^n, with m even. So m=2a, then m^n=2^n a^n. Then 3^k +5^k=2^n a^n. Since 3 and5 are both odd, their sum is even, so m must be even, as we saw. Let's write the equation as 5^k = 2^n a^n -3^k. Then 5^k ≡ -3^k mod2^n. Let's explore this for certain n. For example, when n=3, modulus 8. 5≡5 mod8, 3≡3 mod8. So 5^k +3^k ≡0 mod8. Let's check:For k=1: 5+3=8≡0 mod8.For k=3: 5^3=125≡5 mod8, 3^3=27≡3 mod8. 5+3=8≡0 mod8.For k=5: 5^5=3125≡5 mod8, 3^5=243≡3 mod8. 5+3=8≡0 mod8.So in general, for odd k, 5^k≡5 mod8 and 3^k≡3 mod8, so their sum≡8≡0 mod8. Therefore, 8 divides the sum, which we already knew.But for higher powers, let's consider modulo 16.Compute 5^k +3^k mod16.For k odd:5^1=5 mod16, 3^1=3 mod16. Sum=8 mod16.5^3=125≡13 mod16, 3^3=27≡11 mod16. Sum=13+11=24≡8 mod16.5^5=5^4*5=625*5≡1*5=5 mod16, 3^5=3^4*3=81*3≡1*3=3 mod16. Sum=5+3=8 mod16.Similarly, 5^7=5^6*5=(5^4*5^2)=625*25≡1*9=9 mod16, 3^7=3^4*3^3=81*27≡1*11=11 mod16. Sum=9+11=20≡4 mod16. Wait, that's different. Wait, no:Wait, 5^1=5, 5^2=25≡9, 5^3=5*9=45≡13, 5^4=5*13=65≡1, 5^5=5*1=5, 5^6=5*5=25≡9, 5^7=5*9=45≡13, 5^8=5*13=65≡1 mod16. So the cycle for 5^k mod16 is 4.Similarly, 3^k mod16:3^1=3, 3^2=9, 3^3=27≡11, 3^4=33≡1, 3^5=3, etc. So cycle of 4.So for odd k:When k=1: 5 mod16 and 3 mod16, sum=8.k=3: 5^3=13, 3^3=11, sum=24≡8 mod16.k=5:5^5=5, 3^5=3, sum=8.k=7:5^7=13, 3^7=11, sum=24≡8 mod16.Wait, that contradicts my previous calculation. Wait, 5^7 mod16:Since 5^4≡1, so 5^7=5^(4+3)=5^4*5^3≡1*13=13 mod16.Similarly, 3^7=3^(4+3)=3^4*3^3≡1*11=11 mod16. 13+11=24≡8 mod16. So actually, for any odd k, 5^k +3^k ≡8 mod16. Because:Since 5^k mod16 cycles every 4, and for k odd: k=1,3,5,7,...If k=1 mod4: 5^k≡5 mod16.k=3 mod4:5^k≡13 mod16.k=5 mod4:5^k≡5 mod16.k=7 mod4:5^k≡13 mod16.Similarly, 3^k mod16 for odd k:k=1 mod4:3.k=3 mod4:11.k=5 mod4:3.k=7 mod4:11.So when k is 1 mod4: sum=5+3=8 mod16.k=3 mod4:13+11=24≡8 mod16.Similarly, k=5 mod4:5+3=8, etc.Therefore, for any odd k, 5^k +3^k ≡8 mod16. Therefore, m^n=5^k +3^k ≡8 mod16. But m=2a, so m^n=2^n a^n. So 2^n a^n ≡8 mod16.Divide both sides by8: 2^{n-3} a^n ≡1 mod2.But 2^{n-3} a^n is an integer. If n≥3, then 2^{n-3} is an integer. If n=3: 2^{0}a^3 ≡1 mod2 ⇒a^3≡1 mod2 ⇒a is odd.If n>3, say n=4: 2^{1}a^4≡1 mod2 ⇒2a^4≡1 mod2 ⇒0≡1 mod2, which is impossible. Similarly, n=5:2^{2}a^5≡1 mod2 ⇒4a^5≡1 mod2 ⇒0≡1 mod2, impossible. Therefore, for n≥4, this congruence is impossible. Therefore, only possible when n=3.Therefore, the only possible exponent n is 3. Therefore, the equation reduces to 3^k +5^k= m^3.So, now we have to solve 3^k +5^k= m^3. And from previous analysis, k must be odd. Also, from modulo3: m≡2 mod3, because 3^k +5^k≡0 +2^k mod3. Since k is odd, 2^k≡2 mod3, so m^3≡2 mod3. But cubes mod3: 0^3=0,1^3=1,2^3=8≡2 mod3. Therefore, m≡2 mod3. So m=3b+2 for some integer b.Now, let's consider this equation 3^k +5^k =m^3. Let's see if there are any solutions beyond k=1.For k=1: 3+5=8=2^3, which is a solution.For k=3:27+125=152. Is 152 a cube? No, as 5^3=125,6^3=216.For k=5:243+3125=3368. 14^3=2744,15^3=3375. Close, but no.k=7:2187+78125=80312. 43^3=79507,44^3=85184. No.k=9:19683+1953125=1972808. 125^3=1953125,126^3=2000376. No.k=11:177147+48828125=48905272. 366^3≈48905256, which is very close. Let me check 366^3=366*366=133956; 133956*366:Calculate 133956*300=40,186,800133,956*60=8,037,360133,956*6=803,736Total:40,186,800+8,037,360=48,224,160+803,736=49,027,896. Which is 366^3=49,027,896. But our sum is48,905,272, which is less. So not a cube.Similarly, higher k would produce sums even further between consecutive cubes. It seems no solutions.But to be thorough, let's try to see if there's a mathematical reason why no solutions exist beyond k=1.Assume that k ≥3 and 3^k +5^k =m^3.We can use inequalities to bound m. For large k, 5^k dominates 3^k, so m^3≈5^k. Therefore, m≈5^{k/3}. Let's write m=5^{k/3} +δ, where δ is small compared to5^{k/3}.But this might not be helpful. Alternatively, let's consider taking logarithms. If 3^k +5^k =m^3, then taking natural logs:ln(3^k +5^k)=3 ln m.But for large k, 5^k >>3^k, so ln(5^k)=k ln5 ≈3 ln m ⇒ m≈5^{k/3}. Thus, m is approximately5^{k/3}.Let’s write m=5^{k/3} +t, where t is an integer. Then:m^3=5^k +3*5^{2k/3}t +3*5^{k/3}t^2 +t^3.But 3^k +5^k =m^3 ⇒3^k=3*5^{2k/3}t +3*5^{k/3}t^2 +t^3.This suggests that t must be very small, as the left side is 3^k and the right side involves terms with 5^{2k/3} which is much larger than 3^k for large k. For example, 5^{2k/3} grows faster than3^k because 5^{2/3}≈2.924>3^{1}≈2.098? Wait, 5^{2/3}=e^{(2/3 ln5)}≈e^{(2/3*1.609)}≈e^{1.073}≈2.924, while3=3. So 5^{2k/3}=(5^{2/3})^k≈2.924^k, which is larger than3^k for large k? Wait, no: 2.924 <3, so 5^{2/3}<3. Therefore, 5^{2k/3}= (5^{2/3})^k <3^k. Therefore, 3*5^{2k/3}t <3*3^k t. So the right side is dominated by 3*5^{2k/3}t, but since 5^{2k/3} <3^k, the right side is less than3^{k+1} t. If t is positive, but left side is3^k. So we have 3^k ≈3^{k+1} t ⇒t≈1/3. But t must be an integer, so t=0 or1. But t=0 gives m=5^{k/3}, which is not integer unless k is multiple of3. Let’s check t=1:Then:3^k ≈3*5^{2k/3}*1 +3*5^{k/3}*1 +1.But 5^{2k/3} >5^{k/3} >1, so dominant term is3*5^{2k/3}. Therefore:3^k ≈3*5^{2k/3} ⇒5^{2k/3}≈3^{k-1} ⇒(5^{2/3})^k ≈3^{k-1} ⇒(5^{2/3}/3)^k ≈1/3.But 5^{2/3}/3≈2.924/3≈0.975 <1. Therefore, (0.975)^k≈1/3. Taking log both sides:k ln(0.975)≈-ln3 ⇒k≈-ln3 /ln(0.975)≈-1.0986/(-0.0253)≈43.4. So around k≈43, this approximation suggests possible equality. But this is very rough. However, for large k, the left side (3^k) is exponentially smaller than5^{2k/3}, which is (5^{2/3})^k≈2.924^k. Since3^k=3^k, and2.924^k <3^k, but wait, no: 2.924 <3, so5^{2k/3} <3^k. Therefore, 3*5^{2k/3} <3^{k+1}. But 3^{k+1} is larger than3^k. So the equation 3^k ≈3*5^{2k/3} implies that3^{k} is approximately equal to3*5^{2k/3}, which is possible only if5^{2k/3}≈3^{k-1}. But since5^{2/3}≈2.924 <3, 5^{2k/3}= (5^{2/3})^k≈2.924^k, and3^{k-1}=3^{k}/3. Therefore, 2.924^k ≈3^{k}/3 ⇒(2.924/3)^k≈1/3 ⇒(0.9747)^k≈1/3. As before, this would require k≈43. But for k=43, let's check:Left side:3^43.Right side:3*5^{2*43/3}=3*5^{28.666...}. Wait, but 5^{28.666}=5^{28}*5^{2/3}≈5^{28}*2.924. Similarly, 3^43=3^{43}. Comparing 5^{28.666} vs3^{43-1}=3^{42}. Take log base10:log10(5^{28.666})=28.666*log10(5)≈28.666*0.69897≈20.03.log10(3^{42})=42*log10(3)≈42*0.4771≈20.0382.Wow, they are almost equal. Therefore, 5^{28.666}≈3^{42} ⇒3*5^{28.666}≈3^{43}. Therefore, for k=43, the approximation suggests that3^43 +5^43≈3^{43} +5^{43}≈5^{43}≈5^{43} which is much larger than3^{43}, but our earlier approximation says that3^43≈3*5^{28.666}≈3^{43}, which would mean that3^43≈3^{43}, which is true. But 5^{43} is much larger than3^{43}, so 3^43 +5^{43}≈5^{43}, which would be roughly (5^{43/3})^3≈(5^{14.333})^3=5^{43}. So m≈5^{14.333}, which is not an integer. Therefore, m^3=5^{43} +3^{43} is not a perfect cube because the cube root is not an integer.This suggests that even though for k=43, the approximation makes it seem close, in reality, the sum is dominated by5^k and thus not a perfect cube.Therefore, it's unlikely there are solutions for k>1.Another angle: Use bounds. Suppose that k≥3, and 3^k +5^k =m^3.We can write 5^k <3^k +5^k <5^k +5^k=2*5^k. So 5^k <m^3 <2*5^k. Taking cube roots:5^{k/3} <m < (2)^{1/3}*5^{k/3}.Since (2)^{1/3}≈1.26, so m is between5^{k/3} and1.26*5^{k/3}. Therefore, m must be roughly equal to5^{k/3}. Let's define m=5^{k/3} +t, where t is a small integer. Then:(m)^3=5^k +3*5^{(2k)/3}t +3*5^{k/3}t^2 +t^3.Thus:3^k=3*5^{(2k)/3}t +3*5^{k/3}t^2 +t^3.Divide both sides by5^{(2k)/3}:3^k /5^{(2k)/3} =3t +3*5^{-k/3}t^2 +5^{-2k/3}t^3.The left side is (3/5^{2/3})^k. Since 5^{2/3}≈2.924, and3/5^{2/3}≈3/2.924≈1.026>1. Therefore, as k increases, the left side grows exponentially, while the right side is linear in t plus terms that decay exponentially. Therefore, for large k, this equation is impossible because the left side grows while the right side is bounded if t is small. Therefore, there can only be solutions for small k.Thus, the only possible solution is k=1.To confirm, let's check k=1:3+5=8=2^3. For k=3:152, not a cube. k=5:3368, not a cube. k=7:80312, not a cube. k=9:1972808, not a cube. So no other solutions.Therefore, the only natural number k is 1.

✅:The natural numbers ( k ) for which ( 3^k + 5^k ) is a power of a natural number with an exponent ( geq 2 ) are determined as follows:1. Initial Testing for Small ( k ): - For ( k = 1 ): ( 3^1 + 5^1 = 8 = 2^3 ), which is a valid solution. - For ( k = 2, 3, 4, ldots ): Computations show none of these sums (34, 152, 706, etc.) are perfect powers.2. Modular Analysis: - Modulo 3: ( 3^k + 5^k equiv 2^k mod 3 ). This implies constraints on ( m ) and ( n ) depending on ( k )'s parity. - Modulo 4: ( 3^k + 5^k equiv 0 mod 4 ) if ( k ) is odd, and ( 2 mod 4 ) if even. Since ( m^n equiv 2 mod 4 ) is impossible, ( k ) must be odd.3. Factorization and Structural Insights: - For odd ( k ), ( 3^k + 5^k = 8N ), where ( N ) is an integer. This implies ( m^n = 8N ), requiring ( N ) to compensate for factors of 2. Testing shows ( N ) often becomes prime or non-power composites for ( k > 1 ).4. Bounds and Asymptotic Behavior: - For large ( k ), ( 5^k ) dominates ( 3^k ), making ( 3^k + 5^k ) approximate ( 5^k ). However, ( 5^k ) itself isn't a perfect cube unless ( k ) aligns with specific exponents, which it doesn't beyond ( k = 1 ).5. Conclusion: - Only ( k = 1 ) satisfies all conditions, yielding ( 8 = 2^3 ). All higher ( k ) values fail modular checks, factorization requirements, and asymptotic bounds.Final Answer: The only natural number ( k ) is (boxed{1}).

🔑:We are tasked with finding all natural numbers ( k ) such that ( 3^k + 5^k ) is a power of a natural number with exponent (geq 2). 1. Assume the Desired Form: Assume that ( 3^k + 5^k = n^t ) where ( t geq 2 ).2. Considering ( k ) Even: If ( k ) is even, say ( k = 2m ), then [ 3^{2m} + 5^{2m} equiv 1 + 1 = 2 pmod{4}. ] The left-hand side is even but not divisible by 4, which contradicts ( t geq 2 ). Hence, ( k ) cannot be even.3. Considering ( k ) Odd: Let ( k ) be odd. Then we decompose the expression: [ 3^k + 5^k = (3 + 5) left(3^{k-1} - 3^{k-2} cdot 5 + ldots - 3 cdot 5^{k-2} + 5^{k-1}right). ] Notice that the second factor in the product is odd because it consists of an odd number of terms, all of which are odd.4. Divisibility Considerations: As ( k ) is odd, ( 3^k + 5^k ) is divisible by 8 but not by 16, which implies that ( t = 3 ). Therefore, [ 3^k + 5^k = n^3. ]5. Check Small Values of ( k ): Obviously, ( k = 1 ) is a solution because [ 3^1 + 5^1 = 3 + 5 = 8 = 2^3. ]6. Argument for Absence of Larger Solutions: Now, suppose ( k geq 2 ). Consider the equation modulo 9. Since ( k ) is odd, [ 5^k pmod{9} ] cycles every 6. We need to evaluate the possible residues: [ 3^k + 5^k pmod{9}. ] We have: [ 3^k equiv 0 pmod{9} quad text{since} quad 3^2 equiv 0 pmod{9}, ] and for ( 5^k ): [ begin{cases} 5^1 equiv 5 pmod{9}, 5^3 equiv 5 cdot 5 cdot 5 = 125 equiv 8 pmod{9}, 5^5 equiv 5 cdot 8 equiv 40 equiv 4 pmod{9}. end{cases} ] Thus, ( 5^k equiv 5, 8, text{or } 2 pmod{9} ). However, ( n^3 pmod{9} ) can only be ( 0, 1, text{or } 8 ). Therefore, ( k equiv 3 pmod{6} ).7. Refinement: Consider Modulo 7: Since ( k ) is odd, and considering modulo 7: [ begin{cases} 3^1 + 5^1 equiv 8 equiv 1 pmod{7}, 3^3 + 5^3 equiv 2 pmod{7}. end{cases} ] We need to match: [ 3^k + 5^k equiv 0, 1, text{or } 6 pmod{7}. ] This implies ( k equiv 1, 5 pmod{6} ). Since it contradicts ( k equiv 3 pmod{6} ), there is no solution other than ( k = 1 ).8. Conclusion: Therefore, the only natural number ( k ) such that ( 3^k + 5^k ) is a power of a natural number with an exponent (geq 2) is boxed{1}.

🧠:Okay, let's try to figure out how to solve this problem. So, there are three cubic pools: large, medium, and small. Their inner edge lengths are 6 meters, 3 meters, and 2 meters respectively. Then, there are two piles of gravel. When the first pile is submerged in the medium pool, the water level rises by 6 centimeters. The second pile, when submerged in the small pool, causes the water level to rise by 4 centimeters. The question is, if both piles are submerged together into the large pool, by how many centimeters will the water level rise there?Hmm, let me think. I remember that when you submerge an object in water, the volume of the displaced water is equal to the volume of the object. So, the rise in water level is related to the volume of the gravel piles. So, maybe I need to calculate the volume of each gravel pile based on the water rise in their respective pools, then add those volumes together and find out how much the water would rise in the large pool.First, let's get the units straight. The edge lengths are in meters, but the water level rise is in centimeters. Probably need to convert everything to the same unit. Let's convert centimeters to meters because the edge lengths are given in meters. So, 6 centimeters is 0.06 meters, and 4 centimeters is 0.04 meters. Wait, but when dealing with volume, maybe it's better to convert everything to centimeters? Let me see. The edge lengths are 6 meters (which is 600 cm), 3 meters (300 cm), and 2 meters (200 cm). The water level rise is 6 cm and 4 cm. Maybe working in centimeters would be better here to avoid decimal confusion.Okay, let's convert the pool dimensions to centimeters. So:- Large pool: 6 meters = 600 cm, so each edge is 600 cm. Volume is 600 x 600 x 600, but maybe we don't need the total volume, just the cross-sectional area or something?Wait, no. The volume displaced by the gravel is equal to the volume of the gravel. When you submerge the gravel, the water level rises by a certain height. The volume displaced is equal to the base area of the pool multiplied by the rise in water level. So, for each gravel pile, the volume is (base area of the pool) * (water level rise). So, for the medium pool, which has edges of 3 meters (300 cm), the base area is 300 cm * 300 cm. Then, the volume of the first gravel pile is 300 * 300 * 6 cm³. Similarly, the volume of the second gravel pile is 200 * 200 * 4 cm³, since the small pool has edges of 2 meters (200 cm).Then, when both gravel piles are submerged in the large pool, their combined volume will cause the water level to rise. The base area of the large pool is 600 cm * 600 cm. So, the total volume of gravel is (300*300*6 + 200*200*4). To find the rise in water level (let's call it h), we set that equal to the base area of the large pool multiplied by h: 600*600*h = 300*300*6 + 200*200*4. Then solve for h.Let me compute that step by step.First, calculate the volume of the first gravel pile (submerged in medium pool):Medium pool base area: 300 cm * 300 cm = 90,000 cm²Water rise: 6 cmVolume of gravel pile 1: 90,000 * 6 = 540,000 cm³Then, volume of the second gravel pile (submerged in small pool):Small pool base area: 200 cm * 200 cm = 40,000 cm²Water rise: 4 cmVolume of gravel pile 2: 40,000 * 4 = 160,000 cm³Total volume of both gravel piles: 540,000 + 160,000 = 700,000 cm³Now, submerged in the large pool:Large pool base area: 600 cm * 600 cm = 360,000 cm²The rise in water level h is given by:Total volume = base area * h700,000 = 360,000 * hSo, h = 700,000 / 360,000Let me compute that. 700 divided by 360 is approximately 1.944... cm. So, approximately 1.944 centimeters. But let me do the exact division.700,000 divided by 360,000. Both numerator and denominator can be divided by 1000 first, so 700 / 360. Then, divide numerator and denominator by 20: 700 ÷ 20 = 35, 360 ÷20=18. So, 35/18 ≈1.944... cm. So, 35 divided by 18 is 1.944..., which is 1 and 17/18 cm. To convert 17/18 cm to a decimal, 17 divided by 18 is approximately 0.944. So, 1.944 cm.But the question might expect an exact answer in fraction form or maybe rounded to a certain decimal. But since the problem gives the rises as 6 cm and 4 cm, which are whole numbers, perhaps the answer is a fraction. Let me check.35/18 cm can be simplified as a mixed number: 1 and 17/18 cm. But maybe we can express it as a decimal to two decimal places, 1.94 cm, but the problem doesn't specify. Alternatively, maybe there's a calculation error. Let me verify all steps again.First, the volumes:Gravel pile 1: 300*300*6. 300*300 is 90,000, times 6 is 540,000 cm³. Correct.Gravel pile 2: 200*200*4. 200*200 is 40,000, times 4 is 160,000 cm³. Correct.Total volume: 540,000 + 160,000 = 700,000 cm³. Correct.Large pool base area: 600*600 = 360,000 cm². Correct.Then, h = 700,000 / 360,000 = 7/3.6 ≈1.944 cm. Wait, 700,000 divided by 360,000 is equal to 700/360, which simplifies to 70/36, then 35/18. 35 divided by 18 is indeed approximately 1.944 cm. So, 35/18 cm is the exact value.But maybe we can simplify 35/18. 35 divided by 18 is 1.944... which is 1 and 17/18. But the problem asks for the answer in centimeters, so likely to be expressed as a decimal, maybe rounded to the nearest tenth or as a fraction. However, in math problems like this, unless specified, fractions are often preferred. Let me see if 35/18 reduces. 35 and 18 have no common factors besides 1, so it's 35/18. Alternatively, 1.944... cm.But let me check if I did everything correctly. Maybe the edge lengths were given in meters, and I converted them to centimeters, but is that the right approach?Wait, the gravel is submerged in the pool, so the displacement depends on the cross-sectional area. The cross-sectional area of the pool is (length × width), which for a cube is the same as the base area. So, if the pool is a cube with inner edge length L meters, then the base area is L² square meters. But when you put gravel in, the water level rises by h meters, so the volume displaced is L² × h. So, the volume of gravel is L² × h.But in the problem, the water levels rise by 6 cm and 4 cm, which is 0.06 meters and 0.04 meters. Wait, perhaps I should compute everything in meters to be consistent. Let me try that approach and see if I get the same answer.First, convert the water level rises to meters:6 cm = 0.06 m4 cm = 0.04 mNow, calculate the volume of gravel pile 1 submerged in the medium pool (edge length 3 m):Volume1 = 3m * 3m * 0.06m = 0.54 m³Volume of gravel pile 2 submerged in the small pool (edge length 2 m):Volume2 = 2m * 2m * 0.04m = 0.16 m³Total volume = 0.54 + 0.16 = 0.70 m³Now, submerged in the large pool (edge length 6 m). The rise in water level h (in meters) is given by:Volume = (6m * 6m) * hSo, 0.70 = 36 * hh = 0.70 / 36 ≈0.019444... metersConvert that to centimeters: 0.019444... meters * 100 cm/m = 1.9444... cm, which is the same as before. So, same answer. So, 35/18 cm is 1.944... cm. So, the answer is 35/18 cm, which is approximately 1.94 cm. But since the problem uses centimeters with whole numbers (6 cm and 4 cm), maybe the answer should be expressed as a fraction. 35/18 is an improper fraction, but as a mixed number, it's 1 17/18 cm. However, in the context of the problem, it might be acceptable to present it as a decimal rounded to two decimal places, but the problem doesn't specify. Alternatively, maybe they expect an exact fraction.Alternatively, maybe I made a miscalculation. Wait, let's check 35/18. 18*1=18, 35-18=17, so yes, 1 17/18. So, 1.944... cm. Hmm.Alternatively, maybe the problem expects the answer in a different way. Let me check once again.Another way: compute each gravel's volume, sum them, then divide by the base area of the large pool. Which is exactly what I did. So, seems correct.So, the answer is 35/18 centimeters, which is approximately 1.94 cm. But maybe we can write it as 35/18 or simplified. Wait, 35 and 18 share a common factor? 35 is 5*7, 18 is 2*3*3. No common factors, so 35/18 is the simplest form.Alternatively, the problem might expect the answer in centimeters as a whole number, but since 35/18 is roughly 1.94, which is about 2 cm. But that's an approximation. However, given that the original rises were 6 and 4 cm, which are exact, the answer should be exact as 35/18 cm. So, I think the correct answer is 35/18 cm, which can be written as boxed{dfrac{35}{18}} or approximately 1.94 cm. But in the context of a math competition or textbook problem, fractions are preferred unless specified otherwise.Wait, but let me check once more:Large pool's base area: 6m x 6m = 36 m²Total gravel volume: 0.54 + 0.16 = 0.70 m³So, h = 0.70 / 36 = 0.019444... m = 1.944... cm. Yes, correct.Alternatively, maybe there's a trick here. The problem says "submerged in the medium and small pools", so two separate gravel piles. So, each gravel pile is in their respective pool, causing the respective water level rises. Then, when both gravel piles are moved to the large pool, the total volume is the sum, leading to the total rise as calculated.Alternatively, maybe there's a unit conversion error. Wait, let's check the units again.Original edge lengths in meters. Water level rise in centimeters. When we calculated in meters, converted 6 cm to 0.06 m, 4 cm to 0.04 m. Then calculated the volumes in cubic meters. Then converted the final answer back to centimeters. All steps seem correct.Alternatively, perhaps a different approach: calculate the volume in cubic centimeters directly.First, convert pool edge lengths to centimeters:Medium pool: 3 m = 300 cm. So, base area = 300 * 300 = 90,000 cm². Water rise 6 cm, so volume = 90,000 * 6 = 540,000 cm³.Small pool: 2 m = 200 cm. Base area = 200 * 200 = 40,000 cm². Water rise 4 cm, volume = 40,000 * 4 = 160,000 cm³.Total volume: 540,000 + 160,000 = 700,000 cm³.Large pool: 6 m = 600 cm. Base area = 600 * 600 = 360,000 cm².Water rise h = total volume / base area = 700,000 / 360,000 = 7/3.6 = 1.944... cm.Same result. So, yeah, 35/18 cm.But 7 divided by 3.6 is the same as 70 divided by 36, which simplifies to 35/18. So, same fraction.Therefore, the answer should be 35/18 cm, which is approximately 1.94 cm. Since the question asks for the answer in centimeters, I think expressing it as a fraction is acceptable. So, the box should be boxed{dfrac{35}{18}}.Final AnswerThe water level in the large pool will rise by boxed{dfrac{35}{18}} centimeters.

✅:To determine the rise in water level when both piles of gravel are submerged in the large pool, we need to calculate the volume of each gravel pile based on the water level rise in their respective pools and then find the combined effect in the large pool.1. Convert pool edge lengths to centimeters: - Medium pool: 3 meters = 300 cm - Small pool: 2 meters = 200 cm - Large pool: 6 meters = 600 cm2. Calculate the volume of each gravel pile: - Volume of the first gravel pile (submerged in the medium pool): - Base area of the medium pool: (300 , text{cm} times 300 , text{cm} = 90000 , text{cm}^2) - Water level rise: 6 cm - Volume: (90000 , text{cm}^2 times 6 , text{cm} = 540000 , text{cm}^3) - Volume of the second gravel pile (submerged in the small pool): - Base area of the small pool: (200 , text{cm} times 200 , text{cm} = 40000 , text{cm}^2) - Water level rise: 4 cm - Volume: (40000 , text{cm}^2 times 4 , text{cm} = 160000 , text{cm}^3)3. Total volume of both gravel piles: - Total volume: (540000 , text{cm}^3 + 160000 , text{cm}^3 = 700000 , text{cm}^3)4. Calculate the water level rise in the large pool: - Base area of the large pool: (600 , text{cm} times 600 , text{cm} = 360000 , text{cm}^2) - Water level rise (h): [ h = frac{700000 , text{cm}^3}{360000 , text{cm}^2} = frac{700}{360} = frac{35}{18} , text{cm} ]Thus, the water level in the large pool will rise by boxed{dfrac{35}{18}} centimeters.

🔑:1. Determine the volume of the cubes and the amount of water displacement: - For the middle cube: [ text{Side length} = 3 text{ meters}, quad text{Height increase} = 6 text{ millimeters} ] Convert the height increase to meters: [ 6 text{ millimeters} = 0.006 text{ meters} ] Volume of water displaced in the middle cube: [ V_{text{middle}} = text{side}^2 times text{height increase} = 3^2 times 0.006 = 9 times 0.006 = 0.054 text{ cubic meters} ] - For the small cube: [ text{Side length} = 2 text{ meters}, quad text{Height increase} = 4 text{ millimeters} ] Convert the height increase to meters: [ 4 text{ millimeters} = 0.004 text{ meters} ] Volume of water displaced in the small cube: [ V_{text{small}} = text{side}^2 times text{height increase} = 2^2 times 0.004 = 4 times 0.004 = 0.016 text{ cubic meters} ]2. Calculate the total volume of the rocks: [ V_{text{total rocks}} = V_{text{middle}} + V_{text{small}} = 0.054 + 0.016 = 0.07 text{ cubic meters} ]3. Determine the height rise in the large cube: - For the large cube: [ text{Side length} = 6 text{ meters} ] Let the height rise in the large cube be ( h ) meters. Using the volume displacement formula: [ text{Volume of water displaced} = text{side}^2 times h = 6^2 times h = 36h ] Set this equal to the total volume of the rocks: [ 36h = 0.07 ] Solve for ( h ): [ h = frac{0.07}{36} = frac{7}{360} text{ meters} ] Convert to millimeters: [ h text{ in millimeters} = left( frac{7}{360} right) times 1000 = frac{7000}{360} = frac{350}{18} = frac{35}{1.8} = frac{35}{18} text{ millimeters} ]Conclusion:[boxed{frac{35}{18}}]

🧠:Okay, so I need to find the limit of this sequence as n approaches infinity: [(3n² -5n)/(3n² -5n +7)]^(n+1). Hmm, limits involving exponents can be tricky. Let me think step by step.First, let me look at the base of the exponent, which is (3n² -5n)/(3n² -5n +7). As n becomes very large, both the numerator and the denominator are dominated by the 3n² terms. So, intuitively, the base should approach 1 because the leading terms in the numerator and denominator are the same. But since there's a +7 in the denominator, maybe the base is slightly less than 1? Let me check that.Let me compute the limit of the base first as n approaches infinity. Let's divide numerator and denominator by n² to make it easier:Numerator: (3n² -5n)/n² = 3 - 5/nDenominator: (3n² -5n +7)/n² = 3 -5/n +7/n²So the base becomes [3 -5/n]/[3 -5/n +7/n²]. As n approaches infinity, the terms with 1/n and 1/n² go to zero. So the base approaches 3/3 = 1. So the base is approaching 1, and the exponent is n+1, which is going to infinity. So this is a 1^∞ indeterminate form. For limits of the form [1 + f(n)]^g(n), where f(n) approaches 0 and g(n) approaches infinity, we can use the fact that the limit is e^{lim_{n→∞} f(n)*g(n)} } provided that lim f(n)*g(n) exists. So maybe I can rewrite the expression in this form.Let me try to express the base as 1 + something small. Let's see:The base is (3n² -5n)/(3n² -5n +7). Let's write this as 1 - 7/(3n² -5n +7). Because:(3n² -5n) = (3n² -5n +7) -7, so the base becomes [ (3n² -5n +7) -7 ] / (3n² -5n +7) = 1 - 7/(3n² -5n +7). So the base is 1 - 7/(3n² -5n +7). Then, the entire expression is [1 - 7/(3n² -5n +7)]^{n+1}.Now, according to the limit formula, if we have [1 + a(n)]^{b(n)} and a(n) → 0 and b(n) → ∞, then the limit is e^{lim a(n)*b(n)}. But in this case, a(n) is -7/(3n² -5n +7), which for large n behaves like -7/(3n²). So a(n) ≈ -7/(3n²). The exponent b(n) is n+1 ≈ n. So multiplying them gives (-7/(3n²)) * n = -7/(3n). Then the limit would be e^{-7/(3n) * n} = e^{-7/3}? Wait, that can't be right. Wait, no, because the limit of a(n)*b(n) would be lim (-7/(3n²)) * (n) ) = lim (-7/(3n)) ) = 0. Then the overall limit would be e^0 = 1. But that's conflicting with the initial intuition.Wait, maybe my approach here is wrong. Let's check again.Alternatively, perhaps I should use the logarithm to simplify. Taking the natural logarithm of the sequence:Let L = lim_{n→∞} [(3n² -5n)/(3n² -5n +7)]^{n+1}Then ln L = lim_{n→∞} (n+1) * ln[(3n² -5n)/(3n² -5n +7)]Let me compute this limit. Let's denote the expression inside the log as (3n² -5n)/(3n² -5n +7). As n approaches infinity, as before, this ratio approaches 1. So ln(1) is 0, and we have (n+1)*0, which is an indeterminate form 0*∞. To resolve this, I can rewrite it as a fraction and apply L'Hospital's Rule.Let me write ln L = lim_{n→∞} [ ln((3n² -5n)/(3n² -5n +7)) ] / [1/(n+1) ]So this is 0/0 form? Wait, as n approaches infinity, the numerator ln(1 - 7/(3n² -5n +7)) ≈ -7/(3n² -5n +7) by the approximation ln(1 + x) ≈ x when x is small. The denominator is 1/(n+1) ≈ 1/n. So the expression becomes [ -7/(3n²) ] / (1/n ) = -7/(3n) * n = -7/3. Wait, but let me check that again carefully.First, approximate ln((3n² -5n)/(3n² -5n +7)) = ln(1 - 7/(3n² -5n +7)). Let me denote x = -7/(3n² -5n +7). Then ln(1 + x) ≈ x - x²/2 + x³/3 - ... For very small x, this is approximately x. So ln(1 -7/(3n² -5n +7)) ≈ -7/(3n² -5n +7). So then the numerator is approximately -7/(3n² -5n +7), and the denominator is 1/(n+1). Therefore, the entire expression becomes:lim_{n→∞} [ -7/(3n² -5n +7) ] / [1/(n+1) ] = lim_{n→∞} -7(n+1)/(3n² -5n +7)Divide numerator and denominator by n²:Numerator: -7(n+1)/n² = -7(1/n + 1/n²)Denominator: 3 -5/n +7/n²So the limit becomes (-7*(0 +0))/(3 -0 +0) = 0/3 = 0. Therefore, ln L = 0, so L = e^0 = 1. Wait, that contradicts my previous thought where I thought maybe it's e^{-7/3} but that was wrong.Wait, maybe I made a mistake in the approximation. Let me compute it again.Alternatively, let me compute the limit ln L = lim_{n→∞} (n+1) * ln[(3n² -5n)/(3n² -5n +7)]Express the fraction as 1 - 7/(3n² -5n +7). So ln(1 -7/(3n² -5n +7)) ≈ -7/(3n² -5n +7) - (7/(3n² -5n +7))² /2 - ... So the leading term is -7/(3n² -5n +7). Therefore, ln L ≈ lim_{n→∞} (n+1)*(-7)/(3n² -5n +7)Compute numerator: -7(n+1)Denominator: 3n² -5n +7Divide numerator and denominator by n²:Numerator: -7(n+1)/n² = -7/n -7/n²Denominator: 3 -5/n +7/n²So as n approaches infinity, numerator approaches 0 and denominator approaches 3. Therefore, the entire expression approaches 0 / 3 = 0. Thus, ln L = 0 implies L = e^0 = 1. So the limit is 1?But wait, let me verify this with another approach to be sure.Another approach: Let's recall that if we have lim_{n→∞} [1 - a/n²]^n, where a is a constant, then as n approaches infinity, this would approach 1, because a/n² *n = a/n →0, so the exponent's product tends to zero, so e^{0}=1.Alternatively, if the base is 1 - a/n, then [1 -a/n]^n tends to e^{-a}. But in our case, the base is 1 - 7/(3n² -5n +7). For large n, denominator is ~3n², so the base is 1 - 7/(3n²). So [1 -7/(3n²)]^{n}. Then, using the approximation (1 + c/n^k)^n ~ e^{c/n^{k-1}}}. Here k=2, so exponent is e^{-7/(3n^{2-1})} = e^{-7/(3n)} which tends to e^0 =1 as n approaches infinity. Therefore, this confirms that the limit is indeed 1.But let's check with specific large n. Let's take n=1000. Then the base is (3*1000² -5*1000)/(3*1000² -5*1000 +7) ≈ (3,000,000 -5,000)/(3,000,000 -5,000 +7) = 2,995,000 / 2,995,007 ≈ 0.99999766. Raising this to the power of 1001. Let's compute ln(0.99999766) ≈ -0.00000234. Then multiply by 1001: -0.00000234 *1001 ≈ -0.00234. Then exponentiating gives e^{-0.00234} ≈ 0.99766, which is close to 1 but less than 1. But as n increases, say n=1,000,000, then the exponent would be 1,000,001, and the base would be even closer to 1. So ln(base) ≈ -7/(3n²), multiplied by n gives -7/(3n), which tends to 0. Therefore, the limit is indeed 1.Wait, but when I computed for n=1000, the approximate value was around 0.997, which is not 1. But as n increases, that approaches 1. For example, if n=10^6, then the exponent is 10^6 +1, and the base is approximately 1 - 7/(3*10^{12}), so ln(base) ≈ -7/(3*10^{12}), multiplied by 10^6 +1 gives ≈ -7/(3*10^6), so e^{-7/(3*10^6)} ≈ 1 - 7/(3*10^6) + ... which is very close to 1. So as n approaches infinity, the value gets closer to 1. Therefore, the limit is indeed 1.But wait, another thought: sometimes even though each term approaches 1, raising it to the nth power might not go to 1. For example, (1 + a/n)^n approaches e^a. But in this case, we have (1 - 7/(3n²))^n. Let's see, the exponent is n, and the term inside is 1 - a/n². Then, using the limit as n→∞, (1 - a/n²)^n ≈ e^{-a/n} → 1. So yes, in that case, it approaches 1. So the key is that the term inside the base approaches 1 faster than the exponent can make it tend to something else.Therefore, the conclusion is that the limit is 1.But let me check with another method. Maybe expanding the fraction:(3n² -5n)/(3n² -5n +7) = 1 - 7/(3n² -5n +7). Let's write the expression as [1 - 7/(3n² -5n +7)]^{n+1}.Let me denote k = 3n² -5n +7. Then, as n approaches infinity, k ~ 3n². So 7/k ~ 7/(3n²). Then, [1 - 7/k]^{n+1}. Taking logarithm: (n+1) ln(1 -7/k) ≈ (n+1)*(-7/k - (7/k)^2/2 - ... ) ≈ -7(n+1)/k. Since k ~3n², this is ~ -7n/(3n²) = -7/(3n) →0 as n→∞. Therefore, the logarithm tends to 0, so the original expression tends to e^0 =1. Alternatively, using the expansion ln(1 - x) ≈ -x -x²/2 - ..., but for very small x, higher terms are negligible. So ln(1 -7/k) ≈ -7/k. Then (n+1)*ln(1 -7/k) ≈ -7(n+1)/k ≈ -7n/(3n²) = -7/(3n) →0. So exponent is 0, so limit is e^0=1.Therefore, after multiple approaches and checks, the limit is indeed 1. Wait, but let me check with an alternative substitution. Let me set m = n, and as m approaches infinity. Let me see if there's another perspective.Alternatively, I can write the expression as [1 + (-7)/(3n² -5n +7)]^{n+1}. Let's denote a = -7/(3n² -5n +7) and b = n+1. So the expression is (1 + a)^b. If a*b approaches a finite limit L as n→∞, then (1 +a)^b → e^L. Here, a ~ -7/(3n²), b ~n. So a*b ~ -7/(3n²) *n = -7/(3n) →0. Therefore, the limit is e^0=1. Yes, that's consistent with previous results. So regardless of the method, it seems the limit is 1. Therefore, the answer should be 1.But just to be thorough, let me try to use the natural logarithm and L’Hospital’s Rule properly. Let's consider the expression:ln L = lim_{n→∞} (n+1) * ln[(3n² -5n)/(3n² -5n +7)]Let me write this as lim_{n→∞} [ln(3n² -5n) - ln(3n² -5n +7)] * (n+1)Let me denote f(n) = ln(3n² -5n) - ln(3n² -5n +7). So f(n) = ln[(3n² -5n)/(3n² -5n +7)] = ln(1 -7/(3n² -5n +7)) as before.We can expand f(n) using Taylor series. Let’s let h(n) = -7/(3n² -5n +7). Then f(n) = ln(1 + h(n)) ≈ h(n) - h(n)^2/2 + h(n)^3/3 - … For very small h(n), this is approximately h(n). Therefore, f(n) ≈ -7/(3n² -5n +7). Then multiplying by (n+1):f(n)*(n+1) ≈ -7(n+1)/(3n² -5n +7) ≈ -7n/(3n²) = -7/(3n) →0 as n→infty.Thus, ln L =0, so L=1. Therefore, the limit is 1.Alternatively, if I want to apply L’Hospital’s Rule more formally, I can consider the expression as:lim_{n→∞} [ln(3n² -5n) - ln(3n² -5n +7)] / [1/(n+1)]Let’s set t = 1/n, so as n→∞, t→0+. Rewriting the expression in terms of t:Numerator: ln(3(1/t)^2 -5(1/t)) - ln(3(1/t)^2 -5(1/t) +7)= ln(3/t² -5/t) - ln(3/t² -5/t +7)= ln[(3 -5t)/t²] - ln[(3 -5t +7t²)/t²]= ln(3 -5t) - ln(t²) - [ln(3 -5t +7t²) - ln(t²)]= ln(3 -5t) - ln(3 -5t +7t²)Denominator: 1/(1/t +1) = t/(1 + t)So the limit becomes:lim_{t→0+} [ln(3 -5t) - ln(3 -5t +7t²)] / (t/(1 + t))Multiply numerator and denominator by (1 + t):lim_{t→0+} [ln(3 -5t) - ln(3 -5t +7t²)] * (1 + t) / tAs t→0, (1 + t) ≈1, so focus on [ln(3 -5t) - ln(3 -5t +7t²)] / tLet me compute this limit. Let’s denote the numerator as ln(A) - ln(B) where A=3-5t and B=3-5t +7t². So ln(A/B) = ln(1 -7t²/B). As t→0, B≈3, so ln(1 -7t²/3) ≈ -7t²/3. Therefore, the numerator ≈ -7t²/3. Divide by t: (-7t²/3)/t = -7t/3. So the expression becomes (-7t/3)*1 = -7t/3 →0 as t→0. Therefore, the limit is 0. Hence, ln L=0, so L=1.Therefore, after multiple methods and checks, all leading to the conclusion that the limit is 1. So I think the final answer is 1.But just to ensure there's no miscalculation, let me verify with another substitution.Suppose I let n approach infinity and write the expression as [1 -7/(3n² -5n +7)]^{n+1}. Let's approximate 3n² -5n +7 ≈3n² for large n. Then, the expression becomes approximately [1 -7/(3n²)]^{n}. The limit of this as n→∞ is e^{lim n*(-7/(3n²))} = e^{lim -7/(3n)} = e^0=1.Yes, so this again confirms it. Therefore, I'm confident the limit is 1.Final Answerboxed{1}

✅:To find the limit of the sequence:[lim _{n rightarrow infty}left(frac{3 n^{2}-5 n}{3 n^{2}-5 n+7}right)^{n+1}]we start by analyzing the base of the exponent. For large ( n ), both the numerator and the denominator are dominated by the ( 3n^2 ) terms. Dividing the numerator and denominator by ( n^2 ), we get:[frac{3 - frac{5}{n}}{3 - frac{5}{n} + frac{7}{n^2}}]As ( n to infty ), the terms (frac{5}{n}) and (frac{7}{n^2}) approach 0, so the base approaches 1. This results in a ( 1^infty ) indeterminate form. We rewrite the base as:[1 - frac{7}{3n^2 - 5n + 7}]Using the approximation ( ln(1 + x) approx x ) for small ( x ), we take the natural logarithm of the sequence:[ln L = lim_{n to infty} (n + 1) lnleft(1 - frac{7}{3n^2 - 5n + 7}right)]Approximating the logarithm:[lnleft(1 - frac{7}{3n^2 - 5n + 7}right) approx -frac{7}{3n^2 - 5n + 7}]Thus, the expression becomes:[ln L approx lim_{n to infty} -(n + 1) cdot frac{7}{3n^2 - 5n + 7}]For large ( n ), the denominator ( 3n^2 - 5n + 7 approx 3n^2 ), simplifying the limit to:[ln L approx lim_{n to infty} -frac{7(n + 1)}{3n^2} = lim_{n to infty} -frac{7}{3n} = 0]Therefore, ( ln L = 0 ), implying ( L = e^0 = 1 ).After verifying through multiple methods including logarithmic transformation, Taylor series expansion, and substitution, we conclude the limit is:[boxed{1}]

🔑:1. Consider the given limit: [ lim_{{n to infty}} left( frac{{3n^2 - 5n}}{{3n^2 - 5n + 7}} right)^{n+1} ]2. Simplify the expression inside the limit by dividing the numerator and the denominator by (n^2): [ lim_{{n to infty}} left( frac{{3 - frac{5}{n}}}{{3 - frac{5}{n} + frac{7}{n^2}}} right)^{n+1} ]3. As (n to infty), (frac{5}{n} to 0) and (frac{7}{n^2} to 0). Thus, the fraction inside the limit approaches: [ frac{3}{3} = 1 ]4. To analyze the behavior more closely, rewrite the expression using the technique for evaluating limits of the form ( left( 1 + frac{a}{b} right)^b ): [ lim_{{n to infty}} left( frac{{3n^2 - 5n}}{{3n^2 - 5n + 7}} right)^{n+1} = lim_{{n to infty}} left( 1 + frac{7}{3n^2 - 5n} right)^{-(n+1)} ]5. Recognize this as a limit involving (e). We use the logarithm and exponential limit properties (the second remarkable limit): [ left( 1 + frac{1}{x} right)^x to e text{ as } x to infty ]6. Set the expression in a form that matches the remarkable limit. We rewrite the expression with (frac{3n^2 - 5n + 7}{7}) as follows: [ lim_{{n to infty}} left( 1 + frac{1}{frac{3n^2 - 5n}{7}} right)^{-left(n+1right)} ]7. Let (y = frac{3n^2 - 5n}{7}), which means (y to infty) as (n to infty). Therefore, we have: [ left( 1 + frac{1}{y} right)^{-y cdot frac{7}{3n^2 - 5n}(n+1)} ]8. Simplify the exponent: [ -y cdot frac{7}{3n^2 - 5n}(n+1) = - (n+1) cdot frac{7}{3n^2 - 5n} ]9. Now evaluate the exponential limit: [ e^{lim_{{n to infty}} frac{7(-n - 1)}{3n^2 - 5n}} = e^{lim_{{n to infty}} frac{7(-n - 1)}{3n^2 - 5n}} ]10. Consider the limit as ( frac{7(-n - 1)}{3n^2 - 5n} ): [ lim_{{n to infty}} frac{7(-n - 1)}{3n^2 - 5n} = lim_{{n to infty}} frac{7(-1 - frac{1}{n})}{3 - frac{5}{n}} = lim_{{n to infty}} frac{-7(1 + frac{1}{n})}{3} = frac{-7(1 + 0)}{3} = 0 ]11. Simplify the exponential term to: [ e^0 = 1 ]In conclusion:[boxed{1}]

🧠:Alright, let's try to tackle this problem. The goal is to prove that the sum of all the edge lengths of a polyhedron is greater than 3d, where d is the distance between the two most distant vertices. Hmm, okay. Let me start by understanding the problem better.First, I need to recall some basics about polyhedrons. A polyhedron is a three-dimensional solid with flat polygonal faces, straight edges, and sharp corners or vertices. The edges are the line segments where two faces meet. The distance between the two most distant vertices is called the diameter of the polyhedron, right? So d is the diameter here.Now, the problem states that the sum of all edge lengths is greater than 3d. That means if I add up the lengths of every edge in the polyhedron, the total should be more than three times the maximum distance between any two vertices. Interesting. I need to find a way to relate the sum of edges to this maximum distance.Let me think about some simple polyhedrons to get an intuition. Take a cube, for example. In a cube, all edges are of equal length, let's say length a. The diameter of the cube is the space diagonal, which is a√3. The cube has 12 edges, so the total edge length is 12a. Let's compute 3d for the cube: 3 * a√3 ≈ 5.196a. Comparing that to 12a, we see that 12a is indeed greater than 5.196a. So in the case of a cube, the inequality holds. That's a good sign.Another example: a regular tetrahedron. All edges are equal here as well. The diameter would be the edge length itself, since each pair of vertices is connected by an edge. So if edge length is a, the diameter d is a. The tetrahedron has 6 edges, so total edge length is 6a. Then 3d is 3a, and 6a > 3a, which also holds. Okay, so for these regular polyhedrons, the inequality is true.But the problem is for any polyhedron, not just regular ones. So I need a general proof. Let's think about the properties of polyhedrons and how edges relate to the diameter.The key idea might be to relate the edges to paths connecting the two farthest vertices. Since d is the distance between the farthest vertices, any path connecting them along the edges must have a length at least d. But how does this help with the sum of all edges?Wait, perhaps we can use some kind of averaging argument or consider multiple paths. Alternatively, maybe consider projecting edges onto the line connecting the two farthest vertices.Let me try to formalize this. Let’s denote the two vertices that are farthest apart as A and B, with distance AB = d. For any polyhedron, there must be a path along the edges connecting A to B. The shortest path along the edges from A to B must have a length that is at least d, but actually, the shortest path along edges can't be shorter than the straight-line distance. However, in 3D space, the straight line might not be an edge, so the path along edges could be longer. But this might not directly help.Alternatively, maybe we can use the concept of graph theory. A polyhedron can be considered as a 3-connected planar graph (by Steinitz's theorem). The diameter in graph terms is the maximum number of edges in the shortest path between any two vertices. But here, the diameter d is the geometric distance, not the number of edges. Hmm.Another approach: think about the projections of edges onto the line AB. If we project each edge onto the line connecting A and B, the sum of these projections should relate to the distance d. Let me elaborate.Suppose we have a line segment AB of length d. For any edge in the polyhedron, we can project it onto the line AB. The projection of an edge's length onto AB would be the length of the edge multiplied by the cosine of the angle between the edge and AB. If we sum all these projections, what do we get?But how does that relate to the total edge length? If we can find a relationship between the sum of projections and the distance d, maybe we can bound the total edge length.Wait, here's a thought. Let's consider the three coordinate axes. If we project all edges onto each of the x, y, and z axes, then the total sum of projections on each axis must be at least d, since the projections of the line segment AB onto each axis would sum up to d (by the triangle inequality). But I'm not sure if that's directly applicable here.Wait, actually, if we take the projections of all edges onto the line AB, the total sum of these projections must be at least the length of AB, which is d. Because the path from A to B along the edges must have projections that add up to at least d. But since there might be multiple paths or overlapping projections, this could be tricky.But maybe for each edge, we can consider its contribution in three different directions. If we use three mutually orthogonal directions, perhaps aligned with the coordinate system, and apply some inequality.Alternatively, think about the fact that in 3D space, moving from A to B requires moving in three dimensions. Each edge can contribute to movement along the three axes. But I need to make this precise.Let me recall that in any connected graph, the number of edges is at least one less than the number of vertices, but that's probably not directly helpful here.Wait, here's an idea from the book "Proofs from THE BOOK". There's a proof that in any connected graph, the sum of the distances between adjacent vertices is at least the diameter. But that's in graph terms, not geometric. Wait, maybe not directly applicable.Alternatively, maybe use the Cauchy-Schwarz inequality. If I can relate the sum of edge lengths to d via some inequality.But let's consider all pairs of vertices. The diameter d is the maximum distance between any two vertices. For each edge, it connects two vertices, and the length of the edge is the distance between those two vertices. So each edge contributes to the distance between its endpoints. However, the sum over all edges doesn't directly relate to the maximum distance d.Wait, maybe think about it in terms of the traveling salesman problem. The sum of the edges would be related to the path that visits all vertices, but again, not sure.Alternatively, maybe use the concept of the convex hull. Since any polyhedron can be convex, and the diameter would be the same as the diameter of its convex hull. Maybe consider convex polyhedrons without loss of generality.Assuming the polyhedron is convex, which is a common assumption unless specified otherwise. In a convex polyhedron, the line segment connecting any two vertices lies inside the polyhedron. So the diameter d is just the maximum distance between two vertices.Now, let's try to use vector geometry. Let’s place the two farthest vertices A and B on the x-axis, with coordinates A = (0, 0, 0) and B = (d, 0, 0). Now, consider all the edges of the polyhedron. Each edge can be represented as a vector in 3D space. The sum of all edge lengths is the sum of the magnitudes of these vectors.But how does this relate to the x-axis? If we project each edge vector onto the x-axis, the sum of these projections could be related to d. Let me formalize this.For each edge vector e, its projection onto the x-axis is e ⋅ i, where i is the unit vector in the x-direction. The sum over all edges of these projections would be the sum of (e ⋅ i) over all edges. However, this sum can be related to the movement along the x-axis required to traverse the polyhedron.But in particular, consider the two vertices A and B. Any path from A to B along the edges must have a total projection on the x-axis of at least d. However, since there are multiple paths and edges can be traversed in different directions, this might not directly give us the sum over all edges.Wait, but maybe if we consider that each edge can be part of multiple paths. But this seems vague.Alternatively, think about the fact that for each edge not on the line AB, its projection onto AB is less than its length, but edges along AB have full projection. However, the problem is that we don't know how many edges are aligned with AB.Alternatively, consider that the sum of the projections of all edges onto AB must be at least 3d. Then, since each edge's projection is at most its length, the sum of the lengths must be at least the sum of the projections. Wait, but that would require that sum of projections >= 3d, hence sum of edge lengths >= 3d. But why would the sum of projections be at least 3d?Wait, let's think in three dimensions. If we take three mutually orthogonal directions, say the x, y, and z axes. For each direction, the sum of the projections of all edges onto that direction must be at least d, since the diameter in each direction is at least a component of d. Wait, but the diameter is the maximum distance between two points, which is a straight line, not necessarily aligned with any axis.Hmm, maybe not. Let's suppose that we align AB along the x-axis. Then, the projection of AB onto the x-axis is d, and the projections onto y and z axes are zero. But how does that help with other edges?Alternatively, consider that for any three mutually orthogonal directions, the sum of the projections of all edges onto each direction must be at least d. Then, summing over the three directions, the total sum would be at least 3d, and since each edge contributes to three directions, maybe we can relate this to the total edge length.Wait, here's a method I recall from some inequality proofs. For any vector in 3D space, the sum of the absolute values of its projections onto three mutually orthogonal directions is at least its magnitude. Specifically, for a vector v, |v_x| + |v_y| + |v_z| ≥ |v|. This is due to the Cauchy-Schwarz inequality: (|v_x| + |v_y| + |v_z|)^2 ≤ 3(v_x^2 + v_y^2 + v_z^2), so |v_x| + |v_y| + |v_z| ≤ √3 |v|. Wait, that's the opposite direction. Hmm.Wait, actually, the inequality |v_x| + |v_y| + |v_z| ≥ |v| is not generally true. For example, if v = (1,1,1), then |v_x| + |v_y| + |v_z| = 3, and |v| = √3 ≈ 1.732, so 3 ≥ √3, which is true. But if v is aligned with one axis, say (1,0,0), then |v_x| + |v_y| + |v_z| = 1 = |v|. So in that case, equality holds. So in general, |v_x| + |v_y| + |v_z| ≥ |v|, because the maximum component is at least |v| divided by √3, but summed over three components, it's definitely more. Wait, actually, if v is in an arbitrary direction, then |v_x| + |v_y| + |v_z| ≥ |v|. Let me verify.Suppose v = (a, b, c). Then |a| + |b| + |c| ≥ √(a² + b² + c²). Is this true? Let's square both sides. Left side squared: (|a| + |b| + |c|)² = a² + b² + c² + 2|ab| + 2|ac| + 2|bc|. Right side: a² + b² + c². So the left side squared is greater than or equal to the right side squared. Therefore, |a| + |b| + |c| ≥ √(a² + b² + c²). Yes, this inequality holds. Therefore, for any vector, the sum of the absolute values of its components is at least its magnitude.Therefore, if we take three orthogonal directions, say the x, y, and z axes, then for each edge, |v_x| + |v_y| + |v_z| ≥ |v|. Therefore, the sum over all edges of (|v_x| + |v_y| + |v_z|) is greater than or equal to the sum of |v| over all edges, which is the total edge length. Wait, but we have:Sum_{edges} (|v_x| + |v_y| + |v_z|) ≥ Sum_{edges} |v|But the left-hand side is equal to Sum_{edges} |v_x| + Sum_{edges} |v_y| + Sum_{edges} |v_z|. So, Sum_{edges} |v_x| + Sum_{edges} |v_y| + Sum_{edges} |v_z| ≥ Sum_{edges} |v|.But how does this help us? We need to relate Sum_{edges} |v| to 3d. If we can show that each of Sum_{edges} |v_x|, Sum_{edges} |v_y|, Sum_{edges} |v_z| is at least d, then their sum would be at least 3d, which would give the result. Is that possible?Let me think. Suppose we choose a coordinate system such that the line connecting A and B is along the x-axis. So A is at (0, 0, 0) and B is at (d, 0, 0). Then, the projection of the polyhedron onto the x-axis must cover the interval [0, d]. Now, consider the sum of the x-components of all edges. Each edge's x-component is the difference in the x-coordinates of its endpoints. If we sum over all edges, the total sum of x-components would be equal to twice the sum over all vertices of their x-coordinates times the degree of the vertex, but this seems complicated.Wait, but in any connected graph, the sum of the edge directions must satisfy some conservation. Wait, maybe not. Alternatively, consider that when you traverse the polyhedron from A to B, the total x-component you accumulate must be at least d. But since the edges can go back and forth, the total sum of x-components over all edges could be larger. However, how does this relate to the total sum of |v_x|?Alternatively, think of the polyhedron as a graph. The difference in x-coordinates between A and B is d. The sum of the x-components of the edges along a path from A to B is at least d. But since the edges can have positive or negative x-components depending on direction, the absolute sum is tricky.Wait, here's another approach inspired by electrical networks. If we think of each edge as a conductor and apply a potential difference between A and B, the current flowing would relate to the edges. But I'm not sure how this helps.Wait, going back to projections. If we project the entire polyhedron onto the x-axis, the projection is the interval [0, d]. Each edge, when projected onto the x-axis, contributes a segment whose length is the absolute difference in x-coordinates of its endpoints. The sum of these projections over all edges would be the total "coverage" on the x-axis. However, since edges can overlap in projection, this sum could be larger than d. But how much larger?Similarly, if we project onto the y-axis and z-axis, the projections might also cover some intervals. However, the maximum projection onto the y and z axes could be less than d, but I need to relate this.Alternatively, consider that for each coordinate axis, the sum of the absolute differences (i.e., the projections) along that axis for all edges must be at least d. If we can show this for three orthogonal axes, then summing them up gives 3d, and since each edge's contribution to the total sum is at most its length (by the triangle inequality), the total edge length must be at least 3d.Wait, let's formalize this. Suppose we have three orthogonal directions, say x, y, z. For each direction, consider the sum of the absolute differences (projections) of all edges along that direction. If for each direction, this sum is at least d, then the total sum over all three directions is at least 3d. Since each edge's length is at least the maximum of its projections onto any single axis, but we have that the sum over all three projections for each edge is at least the edge length (from the earlier inequality |v_x| + |v_y| + |v_z| ≥ |v|). Therefore, the total sum over all edges of (|v_x| + |v_y| + |v_z|) is equal to the sum over x, y, z of the total projections, which is ≥ 3d. But since each edge's total projections are ≥ edge length, the sum over all edge lengths is ≤ total projections sum. Wait, no, the total projections sum is ≥ sum of edge lengths. Therefore, if total projections sum ≥ 3d, then sum of edge lengths ≤ total projections sum ≥ 3d? Wait, no, this is reversed. Wait, if for each edge, |v_x| + |v_y| + |v_z| ≥ |v|, then Sum_{edges} (|v_x| + |v_y| + |v_z|) ≥ Sum_{edges} |v|. Therefore, Sum_{x,y,z} Sum_{edges} |v_dir| ≥ Sum_{edges} |v|. But we need Sum_{edges} |v| ≥ 3d. So if we can show that for each direction, Sum_{edges} |v_dir| ≥ d, then Sum_{x,y,z} Sum_{edges} |v_dir| ≥ 3d, hence Sum_{edges} |v| ≤ Sum_{x,y,z} Sum_{edges} |v_dir| ≥ 3d. Wait, but that would mean Sum_{edges} |v| ≤ something ≥3d. That doesn't directly give us the needed inequality. Hmm, maybe this approach isn't working.Let me backtrack. Let's consider projecting all edges onto the line AB. The sum of these projections should be related to the number of times edges cross certain planes or something like that.Alternatively, use the fact that in a polyhedron, any two vertices are connected by a path of edges, and maybe the sum of edges relates to the distances between vertices. But I'm not sure.Wait, here's a different idea inspired by the fact that in any graph, the sum of degrees is twice the number of edges. But not sure.Wait, consider that in 3D space, the polyhedron is a 3-connected graph. Maybe use some combinatorial argument.Alternatively, think of the polyhedron as composed of vertices connected in space, and the edges as connections. The furthest pair is A and B, separated by d. Then, every other vertex must be connected to at least three edges (since it's a polyhedron, each vertex has at least three edges). But not sure.Wait, maybe apply Jung's theorem, which relates the diameter of a set to the radius of the smallest enclosing ball. But I don't see a direct connection.Alternatively, think about the fact that the polyhedron must have a certain number of edges. For a convex polyhedron, Euler's formula states V - E + F = 2. But unless we know more about the number of vertices or faces, this might not help.Wait, let's try using vectors again. Let’s place vertex A at the origin and vertex B at (d, 0, 0). Now, consider all the edges incident to A. Each of these edges connects A to some other vertex, say C_i. The vectors from A to C_i have certain lengths. Similarly, edges incident to B connect B to other vertices D_j.But how does this help? The edges from A and B contribute to moving towards or away from B.Alternatively, consider that to get from A to B, you need to traverse edges that have components in the positive x-direction. Each edge can either increase, decrease, or not change the x-coordinate. But the total sum of x-components over a path from A to B must be d. However, since edges can go back and forth, the total x-components over all edges in the polyhedron must account for all possible movements.Wait, maybe use the concept of net flow. If we consider the polyhedron as a network where each edge allows movement in both directions, the net movement from A to B is d. But the total movement, considering all edges in all directions, must be greater than 3d. Not sure.Wait, another idea. For each edge, consider its contribution to each of the three coordinates. If we can show that for each coordinate axis, the total "movement" along that axis (sum of absolute differences) is at least d, then adding them up gives 3d. Since each edge contributes to three coordinates, the total edge length would be at least 3d.But why would the total movement along each axis be at least d? Let's think about the x-axis first. If we have two points A and B separated by d along the x-axis, then any path from A to B must have a total x-component of at least d. But the sum of x-components over all edges in the polyhedron includes not just the path from A to B but all other edges as well. However, even if other edges cancel out some x-components, their absolute contributions still add up. So maybe the total sum of absolute x-components is at least d, similarly for y and z.Wait, here's a possible approach inspired by the 1-dimensional case. Suppose we have points on a line, and the total distance between the leftmost and rightmost points is d. Then, the sum of all pairwise distances between adjacent points (i.e., edges) must be at least d. This is obvious because the edges connect the points from left to right, and the total sum would cover the distance d. But in higher dimensions, it's more complex because edges can go in various directions.But perhaps, for each coordinate axis, we can consider the "span" of the polyhedron along that axis, which is the difference between the maximum and minimum coordinates. For the x-axis, this span is d (since A and B are at 0 and d). For the y and z axes, the spans might be smaller or larger, but we can perform a rotation to align AB with any axis. Wait, but if we rotate the polyhedron, the edge lengths remain the same, but the coordinate system changes. If we can choose three orthogonal directions such that the span in each direction is at least d, then the sum over each direction's total edge contributions would be at least d, leading to a total of 3d.But how can we guarantee that there exist three orthogonal directions with span at least d? Actually, in 3D space, for any convex set, the maximum width in any direction is at least the diameter divided by √3. But I need to recall some theorem here. Wait, maybe the theorem that in any convex body, there exists three mutually orthogonal directions where the width in each direction is at least d/√3. Then, the sum over these three widths would be d/√3 * 3 = √3 d, which is less than 3d. Hmm, not helpful.Alternatively, perhaps by the Cauchy projection theorem, which states that the average width of a convex body over all directions is proportional to its surface area or something. Not sure.Wait, let's think differently. Suppose we have a convex polyhedron with diameter d. Take any three mutually orthogonal vectors u, v, w. The width of the polyhedron in each of these directions is the distance between the two supporting hyperplanes perpendicular to that vector. The sum of the widths in these three directions might relate to the diameter. However, I don't recall a direct relationship.Alternatively, consider that in the coordinate system where A is at the origin and B is at (d, 0, 0), the projections of all vertices onto the x-axis lie between 0 and d. Now, the edges of the polyhedron connect these vertices. For each edge, the absolute difference in x-coordinates is part of its projection. If we sum all these absolute differences over all edges, can we show that this sum is at least d?Similarly for y and z axes. If we can show that for each axis, the sum of absolute differences (projections) is at least d, then summing over three axes gives 3d. Then, since each edge's length is at least the maximum of its projections onto the three axes, but the sum over all projections is 3d, then the total edge length must be at least 3d. Wait, not exactly. Because the sum of projections is 3d, but each edge's length is at least the maximum projection, so the total edge length would be at least the maximum projection sum, but not necessarily the total.Wait, no. Let me think again. If each edge contributes to three projections (x, y, z), and the sum of all x-projections is S_x, y-projections S_y, z-projections S_z, then S_x + S_y + S_z ≥ sum of edge lengths, because for each edge, |v_x| + |v_y| + |v_z| ≥ |v|. So S_x + S_y + S_z ≥ total edge length. Therefore, if we can show that S_x ≥ d, S_y ≥ d, S_z ≥ d, then S_x + S_y + S_z ≥ 3d, hence total edge length ≥ 3d. Therefore, the key is to show that for each coordinate axis, the sum of the absolute projections of all edges onto that axis is at least d.So, how to show S_x ≥ d, S_y ≥ d, S_z ≥ d? Let's focus on the x-axis. Since A is at 0 and B is at d on the x-axis. The polyhedron is connected, so there is a path from A to B along edges. Each edge in this path has a projection onto the x-axis, either positive or negative. However, the total sum of projections along this path must be d. But since we take absolute values for S_x, each edge's contribution is its absolute x-projection. Therefore, even if the path goes back and forth, the sum of absolute x-projections along the path is at least the net movement, which is d. But the total S_x is the sum over all edges, not just the ones in the path. So S_x is at least the sum of absolute x-projections along the path, which is at least d. Therefore, S_x ≥ d.Similarly, for the y and z axes, even though the span might be less than d, we can rotate the coordinate system such that the three orthogonal directions each have a span of d. Wait, but how?Wait, actually, no. The diameter is the maximum distance between any two points, so in the original coordinate system, the span along the x-axis is d, but along y and z, it could be less. However, we can choose any three orthogonal directions. Perhaps we can choose the three principal axes of the polyhedron, but I don't think that necessarily ensures each has a span of d.Wait, but hold on. Suppose we align AB along the x-axis. Then, the span along x is d. The spans along y and z could be anything. But even if they are small, say s and t, how does that affect the sum of projections?But earlier, we considered that along the x-axis, the sum of absolute x-projections S_x is at least d, because of the path from A to B. Similarly, even if the spans in y and z are smaller, maybe the sums S_y and S_z are still each at least d? That doesn't seem right. For example, if the polyhedron is very flat in the y and z directions, then the projections onto those axes might be small.Wait, this suggests a flaw in the previous reasoning. If the polyhedron is almost a line segment along the x-axis, then the sum of y and z projections would be very small. However, such a "polyhedron" would just be a line segment with two vertices, which isn't a polyhedron. A polyhedron must have at least four faces (tetrahedron). So, perhaps in any genuine polyhedron, there must be sufficient edges in different directions to make S_y and S_z each at least d. But how?Wait, perhaps even if the polyhedron is flat, like a prism with a very small height, but still, it has to have some edges in the y and z directions. For example, a very flat hexagonal prism would have two hexagonal bases and rectangles connecting them. The edges along the height (z-axis) would be very short, but there are many of them. So the sum S_z would be the number of vertical edges times their small height. If the height is h, and there are, say, 6 vertical edges, then S_z = 6h. If h is very small, 6h could still be larger than d? Wait, no, because in this case, the diameter d is the distance between two opposite vertices on the hexagonal base, which is 2a if the hexagon has side length a. The height h is independent. So if h approaches zero, S_z = 6h approaches zero, but d remains 2a. So in this case, S_z can be made arbitrarily small, while d is fixed. Therefore, the earlier reasoning that S_x, S_y, S_z are each at least d is incorrect.Therefore, this approach is flawed. Need to think differently.Hmm. Back to square one. Let's think of another strategy.Perhaps consider that in any polyhedron, each vertex has degree at least 3. For a polyhedron, each vertex is where at least three edges meet. Therefore, the number of edges is at least (3V)/2, by the handshake lemma. But without knowing V, this might not help.Alternatively, think of the polyhedron's edges forming a 3-connected graph. Maybe use Menger's theorem, which states that there are three disjoint paths between any two vertices. But how does that relate to edge lengths?Wait, if there are three disjoint paths from A to B, each path has a length of at least d (since the straight-line distance is d). If we can show that the total length of the three paths is at least 3d, then since these paths are made up of edges, the total edge length would be at least 3d. But wait, each path's length is at least d, so three paths would give a total of at least 3d. However, edges can be shared between paths. So the total edge length might be less than the sum of the path lengths. Therefore, this approach also doesn't directly work.Alternatively, consider that each edge can be part of multiple paths, but I need a way to account for overlaps.Wait, here's a different angle. Let's use the triangle inequality. For any three vertices, the triangle inequality holds. But with multiple vertices, perhaps use induction on the number of vertices.Alternatively, think of the polyhedron as embedded in 3D space with diameter d. Then, for each vertex, its distance from A plus its distance from B is at least d (by triangle inequality). But not sure.Wait, let's try to use the concept of geometric graphs. In geometric graph theory, there's a result called the Erdos's distinct distances problem, but not sure.Alternatively, recall that in a complete graph with n vertices, the sum of all edge lengths is related to the minimum spanning tree, but again, not directly helpful.Wait, another idea. The average length of an edge in the polyhedron. If we can show that the average edge length is greater than 3d divided by the number of edges, but unless we can bound the number of edges, this might not help.Alternatively, consider that the polyhedron has to have enough edges to ensure that the sum exceeds 3d. For example, in the cube, which has 12 edges, each of length a, and diameter a√3. Then 12a > 3a√3, which simplifies to 4 > √3, which is true. For a tetrahedron, 6 edges, diameter a, 6a > 3a, which is true. For an octahedron, 12 edges, diameter 2a (distance between two opposite vertices), sum is 12a, 12a > 3*2a=6a, which holds. So for regular polyhedrons, the inequality holds. For a polyhedron that's a prism with a square base, say, edges include 8 base edges of length a, 4 vertical edges of length h, and the diameter is sqrt(a² + h²) (distance between two opposite vertices on different bases). The total edge length is 8a + 4h. To check if 8a + 4h > 3 sqrt(a² + h²). Let’s set h very small. Then, 8a > 3a, which is true. If h is large, say h=ka, then 8a +4ka >3a sqrt(1 +k²). For k=1: 12a >3a√2≈4.24a, true. For k approaching infinity: 4ka >3a k, which is 4k >3k, true. So in all cases, the inequality holds. So maybe it's always true. But how to prove it generally.Perhaps consider that the polyhedron must have three edge-disjoint paths from A to B, each of length at least d, so total edge length is at least 3d. But why would there be three edge-disjoint paths?By Menger's theorem, in a 3-connected graph (which a convex polyhedron is), there are three vertex-disjoint paths between any two vertices. However, these paths may share edges, so not necessarily edge-disjoint. Therefore, this approach might not work.Alternatively, since the graph is 3-connected, the number of edge-disjoint paths between A and B is at least three. But wait, Menger's theorem says the number of edge-disjoint paths is equal to the minimum number of edges that must be removed to disconnect A from B. In a 3-connected graph, the edge connectivity is at least 3, so there are at least three edge-disjoint paths. Therefore, there are three edge-disjoint paths from A to B. Each of these paths has a length of at least d (since the straight-line distance is d). Therefore, the total length of these three paths is at least 3d. Since the paths are edge-disjoint, their edges are distinct, so the sum of all edge lengths is at least the sum of the lengths of these three paths, which is at least 3d. Hence, the total edge length is greater than or equal to 3d.But wait, the problem states "greater than 3d", not greater than or equal. So perhaps in reality, the paths can't be exactly length d, because the straight line is not an edge (unless the polyhedron has an edge between A and B, but even then, the other paths would be longer). So maybe the total sum is strictly greater than 3d.But let me verify this. Suppose there is a polyhedron where the sum of edge lengths is exactly 3d. Is this possible? If the three edge-disjoint paths each have length exactly d, then all edges in those paths must be along the straight line between A and B. But in 3D, three edge-disjoint paths from A to B that are all along the straight line would require multiple edges overlapping, which is not allowed in a polyhedron. Hence, in reality, each path must have length greater than d, so the total sum would be greater than 3d.Therefore, using Menger's theorem, we can argue that there are three edge-disjoint paths from A to B, each longer than d, hence their total length is greater than 3d, and since these paths are edge-disjoint, the sum of all edge lengths is at least the sum of these paths, hence greater than 3d.This seems like a solid approach. Let me check if this holds.First, by Steinitz's theorem, the graph of a convex polyhedron is 3-connected. Therefore, between any two vertices A and B, there are three edge-disjoint paths. Each of these paths must traverse a sequence of edges connecting A to B.Now, the straight-line distance between A and B is d. Each path, being a sequence of edges, must have a total Euclidean length greater than or equal to d, by the triangle inequality. However, equality holds only if the path is the straight line itself. But in a polyhedron, unless AB is an edge, the path must go through other vertices, making its length strictly greater than d. Even if AB is an edge, the other two edge-disjoint paths cannot be the same edge, so they must go through other vertices, hence their lengths are greater than d.Therefore, each of the three edge-disjoint paths has length > d. Thus, the total length of these three paths is > 3d. Since these paths are edge-disjoint, their edges are distinct, so the sum of all edge lengths of the polyhedron is at least the sum of these three paths, which is > 3d. Therefore, the result follows.Yes, this seems to work. Let me verify with an example. Take the cube. The three edge-disjoint paths from A to B (two opposite vertices) could be:1. The direct edge (if it exists). Wait, in a cube, the edge connecting A and B doesn't exist; the edges are along the axes. The space diagonal is not an edge. So the three edge-disjoint paths would each consist of three edges. For example, moving along three different faces. Each path would have length 3a, and there are three such paths. But in reality, the cube's graph is 3-connected, so there are three edge-disjoint paths. However, each path has three edges of length a, so total length 3a. Three such paths would have total length 9a. The diameter d is a√3 ≈1.732a. 3d ≈5.196a, and 9a >5.196a. But the total edge length of the cube is 12a, which is greater than 9a. So the inequality holds.Another example: a tetrahedron. Any two vertices are connected by an edge. So the three edge-disjoint paths from A to B are: the direct edge, and two more paths each consisting of two edges. The direct edge has length d, and the other two paths each have length 2d (since each edge is length d). Thus, total length for three paths is d + 2d + 2d =5d, which is greater than 3d. The total edge length is 6d, which is also greater than 3d.Therefore, this approach using Menger's theorem and edge-disjoint paths seems valid. Therefore, the proof would be:By Steinitz's theorem, the graph of a convex polyhedron is 3-connected. Therefore, between any two vertices, there exist three edge-disjoint paths. Each path's length is at least the distance between A and B, which is d. However, since the polyhedron is embedded in 3D space, any path other than the straight line (if it exists as an edge) must have a length strictly greater than d. Even if one path is the straight line (an edge), the other two paths must have lengths greater than d. Hence, the total length of these three edge-disjoint paths is greater than 3d. Since these paths are edge-disjoint, their edges are distinct, so the sum of all edge lengths of the polyhedron is greater than 3d.Therefore, the sum of the edge lengths of a polyhedron is greater than 3d.

✅:To prove that the sum of the edge lengths of a polyhedron is greater than (3d), where (d) is the distance between the most distant vertices, we utilize graph-theoretic properties of polyhedrons and geometric inequalities:1. Graph Connectivity: By Steinitz's theorem, the graph of a convex polyhedron is 3-connected. This ensures there exist three edge-disjoint paths between any two vertices (A) and (B).2. Path Lengths: Each of these paths must traverse edges whose total Euclidean length is at least (d), the straight-line distance between (A) and (B). Since these paths are not colinear (unless (AB) is an edge), their lengths strictly exceed (d) due to the triangle inequality.3. Summing Paths: The three edge-disjoint paths collectively contribute a total length greater than (3d). As these paths share no common edges, their lengths are distinct and additive in the total edge sum.Conclusion: The sum of all edge lengths of the polyhedron exceeds (3d), as it includes at least three edge-disjoint paths each longer than (d). Thus, the inequality holds for any polyhedron.[boxed{text{The sum of the edge lengths of a polyhedron exceeds } 3d}]

🔑: Problem:Prove that the sum of the lengths of the edges of a polyhedron is greater than 3d, where d is the distance between the most distant vertices of the polyhedron.1. Identify the Most Distant Vertices: Let A and B be the two vertices of the polyhedron that are at the maximum distance d from each other.2. Constructing Perpendicular Sections: Through points A and B, construct planes that are perpendicular to the line segment AB.3. Restriction by Planes: The entire polyhedron is enclosed between these two perpendicular planes.4. Vertices and Cross-Sections: Construct planes perpendicular to AB through each vertex of the polyhedron.5. Analyzing Neighboring Sections: Consider two consecutive planes from the construction. Between these two planes, there would be, at the very least, three segments of the edges of the polyhedron.6. Projection Analysis: The projection of each of these segments onto the line AB will not be less than the length of the segment itself. More crucially, among the projections, there will definitely be segments that are either parallel or not parallel to AB.7. Summing Up Projections: Thus, by examining throughout the polyhedron, the total length sum of the edges must be greater than 3 cdot AB = 3d. This ensures that collectively, these edge segments will cover the distance between A and B at least three times.8. Conclusion: Therefore, the sum of the lengths of the edges of the polyhedron is greater than 3d.boxed{ }