🧠:Okay, so there's this game between two people, A and B. A starts by choosing a positive integer, and then each player takes turns following a rule. If the last number was odd, they add 7 to it. If it was even, they divide it by 2. The goal is to be the one who repeats the first number, which is the number A chose initially. We need to find all the numbers A can choose to guarantee a win.First, let me make sure I understand the rules correctly. The game starts with a number n chosen by A. Then B has to apply the rule: if n is odd, B says n + 7; if n is even, B says n / 2. Then it's A's turn again, using the same rule based on the number B just said. They keep alternating, and whoever repeats the original number n wins. So the game continues until someone says n again, and that person is the winner.So, for example, if A starts with an odd number, then B will add 7, making it even. Then A's turn: since it's even, divide by 2. Then B's turn again, depending on whether that result is odd or even. If the chain ever loops back to the original number, the person who says it wins.Our task is to find all starting numbers n such that A can force a win, regardless of B's moves. Wait, but in this game, B doesn't have choices, right? Each move is determined by the previous number. So actually, once A picks a number, the entire sequence is determined. Because each player must follow the rule based on the previous number. So there's no strategy involved beyond choosing the starting number. Wait, but the problem says "the winner is the player that repeats the first number said." So the game continues until someone repeats the starting number, and that player wins. So, depending on the starting number, the sequence of numbers could cycle, and whoever lands on the starting number again wins.But since the players alternate turns, whether the starting number is repeated on an even or odd step (relative to the starting point) determines if A or B wins. So the key is to find numbers n such that when you follow the sequence of operations (add 7 if odd, divide by 2 if even), the first repetition of n occurs on A's turn. Since A starts, the turns are: A (1st move), B (2nd), A (3rd), B (4th), etc. So if the starting number n reappears on an odd-numbered step (1st, 3rd, 5th, ...), then A wins; if it's on an even step, B wins.Therefore, the problem reduces to finding all numbers n for which, when you follow the sequence starting from n, the first time n reappears is at an odd position in the sequence.So we can model this as a function f(n) which gives the next number in the sequence. Then, starting from n, we generate the sequence n, f(n), f(f(n)), etc., until we see n again. The number of steps it takes to return to n is the cycle length. If the cycle length is odd, then the starting player A will be the one to repeat n (since the cycle would go through 1, 3, 5, ... steps). If it's even, then player B would be the one to repeat it.Wait, but actually, the cycle length is the number of steps to return to the starting number. For example, if starting at n, after k steps, we get back to n. Then, since players alternate, if k is even, then the (k+1)th step would be player A again. Wait, maybe I need to think more carefully.Wait, the first number is said by A: that's step 1. Then step 2 is B, step 3 is A, etc. So if the cycle length is k, meaning that after k moves, we return to the starting number. Then, the player who says the k+1 th number would be the one to repeat the starting number. Wait, no. If the cycle is k steps, then the (k+1)th number is the same as the first number. But since the starting player is A, step 1 is A, step 2 is B, ..., step k+1 is A if k is even, B if k is odd. Wait, let me check:Number of steps to return: k.If starting at step 1: A says n.Then step 2: B says f(n).Step 3: A says f(f(n))....If after k steps, we return to n at step k+1. So the player at step k+1 is A if k is even (since step numbers: 1 (A), 2 (B), ..., k+1: if k is even, k+1 is odd, so A). If k is odd, then step k+1 is even, so B.Wait, no. If k is the number of steps taken after the initial number. Wait, maybe confusion here.Alternatively, consider the sequence as starting with n (step 0), then step 1 is f(n) (said by B), step 2 is f(f(n)) (said by A), step 3 is f^3(n) (said by B), etc. Wait, maybe the confusion is in how we count the steps.Wait, the problem states: "A starts choosing a positive integer number and then, each player in its turn, say a number due to the following rule."So the first number is chosen by A, let's call that turn 0? Or is the first move A choosing the number, and then B goes next. So the sequence is:- Turn 1: A picks n.- Turn 2: B applies the rule to n.- Turn 3: A applies the rule to the result from B.- Turn 4: B applies the rule again.So the turns are alternating, starting with A choosing the number, then B, then A, etc. So the initial number is said by A on turn 1. Then B says the next number on turn 2, A on turn 3, etc. So if the starting number is repeated on turn k, then if k is odd, A wins; if even, B wins.Therefore, the question is: for each starting number n, trace the sequence of numbers generated by the rules (add 7 if odd, divide by 2 if even), and find the smallest k where the number returns to n. Then, if k is even (since starting from turn 1, the next occurrence would be turn 1 + k, which is 1 + k. So if k is odd, 1 + k is even; if k is even, 1 + k is odd). Wait, that seems conflicting. Let me formalize:Let’s define the sequence S_1, S_2, S_3, ..., where S_1 = n (chosen by A on turn 1). Then S_2 is computed by B according to the rule, S_3 by A, S_4 by B, etc. We need to find the smallest t >= 2 such that S_t = n. The player who says S_t wins. Since turns alternate, if t is odd, it's A's turn; if t is even, it's B's turn.Wait, actually, S_1 is n (turn 1, A). Then S_2 is B's move, S_3 is A's move, etc. So the first time n reappears is at some turn t. If t is odd (A's turn), A wins; if even (B's turn), B wins.Therefore, for each n, we need to determine the minimal t such that starting from n, after t-1 applications of the function f, we get back to n. Then, if t is odd, A can win by choosing n; if t is even, B would win, so A should not choose such n.Thus, the problem reduces to finding all positive integers n where the minimal period t of the sequence starting at n is odd.Therefore, we need to find all n where the minimal t such that f^t(n) = n and t is minimal, then t is odd.Hence, the solution requires analyzing the function f(n) defined as:f(n) = n + 7, if n is odd;f(n) = n / 2, if n is even.We need to find all n for which the minimal period of n under f is odd.So, to approach this, perhaps we can model the function f as a directed graph, where each node is a positive integer, and there's an edge from n to f(n). Then, the problem reduces to finding all nodes n where the cycle containing n has an odd length. However, since the function f might not necessarily be cyclic for all n; some numbers might lead into a cycle, but not necessarily return to themselves. However, in this problem, since we are looking for numbers that eventually return to n, those numbers must be part of a cycle.Wait, but the function f is deterministic, so starting from any n, the sequence will eventually enter a cycle. However, the question is about numbers n where this cycle includes n itself, i.e., n is part of a cycle. Otherwise, if n leads into a cycle that doesn't include n, then n will never be repeated. Therefore, for the problem, we only need to consider numbers n that are in a cycle (i.e., the sequence starting at n eventually returns to n), and then among those, determine the length of the cycle. If the cycle length is odd, then A can choose n to win.But first, we need to characterize all numbers n that are part of a cycle under the function f. Then, for each such cycle, if the length is odd, all numbers in that cycle are winning starting numbers for A.Alternatively, since the function f is defined differently for odd and even numbers, perhaps we can analyze the possible cycles.Let’s consider small numbers first and see if we can find any cycles.Starting with n = 1:1 is odd, so next number is 1 + 7 = 8.8 is even, next is 4.4 is even, next is 2.2 is even, next is 1.So the sequence is 1 → 8 → 4 → 2 → 1. That's a cycle of length 4. So starting at 1, the cycle length is 4. Therefore, the minimal t such that f^t(1) = 1 is 4. Then, the first repetition occurs at t = 4 steps. Since starting at turn 1 (A), the next occurrences would be at turn 1 + 4 = 5, which is A's turn again. Wait, but 4 steps after turn 1 would be turn 5. So t = 4, which is the number of steps from 1 to 1 again. Therefore, since 4 is even, the player would be B if we count turns. Wait, confusion arises here.Wait, perhaps we need to clarify:If starting at n, the sequence takes k steps to return to n, then the number of turns between the first occurrence and the repeat is k. Since players alternate, the parity of k determines who gets to say the repeated number. Since A starts, after k steps (i.e., k turns), the (k+1)th turn would be the one repeating n. Wait, maybe not. Let's take an example.Take n = 1:Sequence:Turn 1: A says 1.Turn 2: B says 8.Turn 3: A says 4.Turn 4: B says 2.Turn 5: A says 1.So here, after 4 turns (B's turn 4), the next turn is 5, which is A's turn, and they say 1 again. So the cycle length from turn 1 to turn 5 is 4 steps (turns 2-5), but the number of steps between the first 1 and the next 1 is 4 turns (B, A, B, A). So the cycle length in terms of function applications is 4, leading to the repeat on the 5th turn, which is A's turn. Therefore, in this case, the cycle length is 4, but the number of turns between the original and the repeat is 4, which is even, meaning the repeat happens on an odd turn (5), which is A's turn. Wait, but 4 steps after turn 1 is turn 5. So if the cycle length (number of function applications) is 4, then the repeat occurs on turn 1 + 4 = 5, which is A's turn. Therefore, for a cycle length k, the repeat occurs on turn 1 + k. So, if k is even, 1 + k is odd, meaning A wins. If k is odd, 1 + k is even, meaning B wins. Wait, that seems counterintuitive.Wait, in the example, cycle length is 4 (steps), leading to turn 5. So if k is the cycle length (number of function applications), then the repeat occurs on turn 1 + k. So if k is even, 1 + k is odd, so A wins. If k is odd, 1 + k is even, so B wins. Therefore, A wants cycles where k is even? But in the example, n = 1 has a cycle length of 4, which is even, leading to A winning. Similarly, if a cycle length is odd, then 1 + k is even, B wins. Therefore, the numbers n that A can choose to win are those that are in cycles of even length. Because then 1 + k is odd, so A gets to repeat the number.Wait, but in the problem statement, it's said that the winner is the player who repeats the first number. So if starting number is n, when the sequence comes back to n, whoever is the current player wins. Therefore, if the cycle has length k (number of steps to return), then the player is A if k is even (since starting from A, after k steps: step 1: A, step 2: B, ..., step k+1: if k is even, k+1 is odd, so A). Therefore, if k is even, the player who repeats is A; if k is odd, it's B. Therefore, A can win by choosing numbers in cycles of even length.Thus, the problem reduces to finding all numbers n that are part of a cycle with even length under the function f(n).Therefore, our task is to find all positive integers n that lie in a cycle of even length for the function f(n) = n + 7 if odd, n / 2 if even.So, first, we need to identify all cycles in this function, determine their lengths, and then collect all numbers in cycles of even length.Let’s attempt to find cycles.Starting with n = 1:1 → 8 → 4 → 2 → 1 (cycle length 4, even)n = 3:3 is odd, so next is 10.10 is even, next is 5.5 is odd, next is 12.12 is even, next is 6.6 is even, next is 3.So the cycle here is 3 → 10 → 5 → 12 → 6 → 3. That's a cycle of length 5, which is odd. So n = 3 is in a cycle of length 5. Therefore, A cannot choose 3, as B would win.n = 5:From the above, 5 → 12 → 6 → 3 → 10 → 5. So cycle length 5 again.n = 2:2 → 1 → 8 → 4 → 2. Cycle length 4.n = 4:4 → 2 → 1 → 8 → 4. Cycle length 4.n = 6:6 → 3 → 10 → 5 → 12 → 6. Cycle length 5.n = 7:7 is odd, next is 14.14 is even, next is 7.So cycle: 7 → 14 → 7. Cycle length 2, which is even. Therefore, starting at 7:Turn 1: A says 7.Turn 2: B says 14.Turn 3: A says 7. So A wins.Thus, n = 7 is a winning number.n = 9:9 is odd, next is 16.16 is even, next is 8.8 is even, next is 4.4 is even, next is 2.2 is even, next is 1.1 is odd, next is 8.So 9 → 16 → 8 → 4 → 2 → 1 → 8 → ... This seems to enter the cycle 1 → 8 → 4 → 2 → 1. So 9 is not in a cycle itself; it leads into another cycle. Therefore, 9 is not part of a cycle, so starting at 9 would never return to 9, meaning A cannot win by choosing 9.n = 10:From earlier, 10 is part of the cycle 3 → 10 → 5 → 12 → 6 → 3, cycle length 5.n = 12:Similarly, part of the 5-length cycle.n = 14:14 is part of the cycle 7 → 14 → 7, length 2.n = 15:15 is odd, next is 22.22 is even, next is 11.11 is odd, next is 18.18 is even, next is 9.9 is odd, next is 16.16 is even, next is 8.8 is even, next is 4.4 is even, next is 2.2 is even, next is 1.1 is odd, next is 8.So 15 leads into the cycle starting at 1. So 15 is not in a cycle.n = 17:17 is odd, next is 24.24 is even, next is 12.12 is part of the 5-length cycle. So 17 → 24 → 12 → 6 → 3 → 10 → 5 → 12... So again, leads into the existing cycle.n = 19:19 → 26 → 13 → 20 → 10 → 5 → 12 → 6 → 3 → 10... So enters the 5-length cycle.n = 21:21 → 28 → 14 → 7 → 14... Wait, 21 is odd, add 7: 28. 28 is even, divide by 2: 14. 14 is even, divide by 2: 7. 7 is odd, add 7: 14. So the cycle here is 7 → 14 → 7. But 21 leads into this cycle. So 21 is not part of a cycle itself.Wait, so let's check if 21 is part of a cycle. Starting at 21:21 → 28 → 14 → 7 → 14 → 7... So once it reaches 7, it cycles between 7 and 14. So 21 is not in a cycle, it's a pre-cycle number. Therefore, starting at 21 would never return to 21, so A cannot choose 21.n = 23:23 is odd, next is 30.30 is even, next is 15.15 is odd, next is 22.22 is even, next is 11.11 is odd, next is 18.18 is even, next is 9.9 is odd, next is 16.16 is even, next is 8.8 is even, next is 4.4 is even, next is 2.2 is even, next is 1.Then 1 leads into the 4-length cycle. So 23 leads into the 4-length cycle.n = 25:25 is odd, next is 32.32 is even, next is 16.16 is even, next is 8.8 leads into the 4-length cycle.So 25 → 32 → 16 → 8 → 4 → 2 → 1 → 8... So again, entering the 4-length cycle.n = 7 and n = 14 form a cycle of length 2. Similarly, the cycle 1,8,4,2 has length 4. Then the cycle 3,10,5,12,6 has length 5. Are there any other cycles?Check n = 7: cycle length 2.n = 14: same cycle.Check n = 7, 14, 7, 14... yes.What about n = 28:28 is even, next is 14. Then 7, then 14. So n = 28 is part of the cycle 7-14.But 28 itself is not in a cycle, since starting at 28: 28 →14 →7 →14... So the cycle is 7-14, so 28 is pre-cycle.Similarly, n = 16: 16 →8 →4 →2 →1 →8... So 16 is part of the 4-length cycle.Wait, earlier n = 16 was considered, but from n =1, the cycle is 1 →8 →4 →2 →1. Wait, but 16 →8 →4 →2 →1 →8... So 16 leads into the cycle. Therefore, 16 is not part of a cycle itself.Wait, but if starting at 8: 8 →4 →2 →1 →8, which is a cycle. So 8 is part of the cycle. Similarly, 4, 2, 1 are part of the cycle.Wait, n = 8: starting at 8, we get 8 →4 →2 →1 →8. So cycle length 4.Similarly, n = 4: 4 →2 →1 →8 →4. Cycle length 4.n = 2: cycle length 4.n = 1: cycle length 4.n = 3: cycle length 5.n = 10: cycle length 5.n = 5: cycle length 5.n = 12: cycle length 5.n = 6: cycle length 5.n = 7: cycle length 2.n = 14: cycle length 2.So so far, we have three cycles:1. The 4-cycle: 1 ↔8 ↔4 ↔2.2. The 5-cycle: 3 ↔10 ↔5 ↔12 ↔6.3. The 2-cycle:7 ↔14.So these are the cycles. All other numbers either lead into one of these cycles or potentially into another cycle we haven't found yet. Let's check some higher numbers.n = 11:11 is odd, next is 18.18 is even, next is 9.9 is odd, next is 16.16 is even, next is 8.Then 8 is part of the 4-cycle. So 11 leads into the 4-cycle.n = 13:13 is odd, add 7: 20.20 is even, divide by 2:10.10 is part of the 5-cycle. So 13 leads into the 5-cycle.n = 17:17 →24→12, which is part of the 5-cycle.n = 19:19→26→13→20→10, into the 5-cycle.n = 23:23→30→15→22→11→18→9→16→8, into the 4-cycle.n = 25:25→32→16→8, into the 4-cycle.n = 27:27 is odd, add 7:34.34 is even, divide by 2:17.17→24→12, into the 5-cycle.n = 29:29→36→18→9→16→8, into the 4-cycle.n = 31:31→38→19→26→13→20→10, into the 5-cycle.So it seems that all numbers eventually lead into one of the three cycles: the 4-cycle, 5-cycle, or 2-cycle. Therefore, the only numbers that are part of cycles are those in the three cycles mentioned: {1,8,4,2}, {3,10,5,12,6}, and {7,14}.Therefore, these are the numbers that eventually repeat. Now, among these cycles, the cycle lengths are 4, 5, and 2. So cycles of even length are the 4-cycle and the 2-cycle. The 5-cycle is odd.Therefore, the numbers in the 4-cycle (1,2,4,8) and the 2-cycle (7,14) are the numbers where A can choose them and win, since the cycle length is even, leading to the repetition on A's turn.But wait, for the 2-cycle: starting at 7, cycle length 2.So starting with A choosing 7 (turn 1). Then B says 14 (turn 2). Then A says 7 again (turn 3). So the repetition occurs on turn 3, which is A's turn. Therefore, A wins. Similarly, starting at 14: turn 1 (A) says 14, turn 2 (B) says 7, turn 3 (A) says 14 again. So A wins. So both 7 and 14 are winning numbers.Similarly, numbers in the 4-cycle:Take n=1: turn 1 (A) says 1, turn 2 (B) says 8, turn 3 (A) says 4, turn 4 (B) says 2, turn 5 (A) says 1. So repetition on turn 5, which is A's turn. Therefore, A wins. Similarly, starting at 8: turn 1 (A) says 8, turn 2 (B) says 4, turn 3 (A) says 2, turn 4 (B) says 1, turn 5 (A) says 8. So repetition on turn 5, A's turn. Similarly for 4 and 2.Therefore, numbers in the 4-cycle and 2-cycle are winning for A, while those in the 5-cycle are losing, as the repetition occurs on B's turn.Therefore, the answer is all numbers in {1,2,4,7,8,14}.But let's verify with starting number 7:A chooses 7 (turn 1). Then B must say 7 +7=14 (turn 2). Then A must say 14 /2=7 (turn 3). So A repeats 7 on turn 3 and wins. Correct.Starting number 14:A chooses 14 (turn 1). B says 14 /2=7 (turn 2). A says 7 +7=14 (turn3). So A wins.Starting number 1:A chooses 1. B says 8. A says 4. B says 2. A says 1. So A wins.Similarly, starting at 8:A chooses 8. B says 4. A says 2. B says 1. A says 8. A wins.Starting at 4:A chooses 4. B says 2. A says 1. B says 8. A says 4. A wins.Starting at 2:A chooses 2. B says 1. A says 8. B says 4. A says 2. A wins.So all these numbers result in A winning. Now, numbers in the 5-cycle, like 3:A chooses 3. B says 10. A says 5. B says 12. A says 6. B says 3. So repetition occurs on B's turn (turn 6), so B wins. Therefore, A cannot choose 3.Similarly, 10 would lead to B winning.Thus, the numbers A can choose to win are exactly those in the 4-cycle {1,2,4,8} and the 2-cycle {7,14}.Therefore, the final answer is all numbers in {1,2,4,7,8,14}.Final AnswerThe numbers A can choose to win are boxed{1}, boxed{2}, boxed{4}, boxed{7}, boxed{8}, and boxed{14}.

✅:To determine the numbers that player ( A ) can choose to guarantee a win, we need to analyze the sequences generated by the given rules and identify cycles. The key is to find numbers ( n ) such that the first repetition of ( n ) occurs on an odd-numbered turn (A's turn).1. Cycle Analysis: - 4-cycle: The sequence starting at 1 is ( 1 rightarrow 8 rightarrow 4 rightarrow 2 rightarrow 1 ). This cycle has a length of 4 (even), so numbers 1, 8, 4, and 2 are winning numbers. - 5-cycle: The sequence starting at 3 is ( 3 rightarrow 10 rightarrow 5 rightarrow 12 rightarrow 6 rightarrow 3 ). This cycle has a length of 5 (odd), so numbers 3, 10, 5, 12, and 6 are not winning numbers. - 2-cycle: The sequence starting at 7 is ( 7 rightarrow 14 rightarrow 7 ). This cycle has a length of 2 (even), so numbers 7 and 14 are winning numbers.2. Verification: - For numbers in the 4-cycle (1, 2, 4, 8), the repetition occurs on the 5th turn (A's turn). - For numbers in the 2-cycle (7, 14), the repetition occurs on the 3rd turn (A's turn). - Numbers in the 5-cycle (3, 10, 5, 12, 6) repeat on the 6th turn (B's turn), so they are not winning numbers.3. Conclusion: - The winning numbers are those in the 4-cycle and 2-cycle: 1, 2, 4, 7, 8, and 14.Thus, the numbers ( A ) can choose to win are (boxed{1}), (boxed{2}), (boxed{4}), (boxed{7}), (boxed{8}), and (boxed{14}).

🔑:To solve this problem, we need to analyze the sequence of numbers generated by the rules of the game and determine which starting numbers will eventually lead back to the starting number itself. Let's break down the steps:1. Define the rules and initial conditions: - If the last number said was odd, the player adds 7 to this number. - If the last number said was even, the player divides it by 2. - The winner is the player who repeats the first number said.2. Analyze the behavior of the sequence: - Let's denote the starting number by ( n ). - We need to determine the sequence generated by the rules and check if it eventually returns to ( n ).3. Check small values of ( n ): - For ( n = 1 ): - Sequence: ( 1 rightarrow 8 rightarrow 4 rightarrow 2 rightarrow 1 ) - The sequence returns to 1, so ( n = 1 ) works. - For ( n = 2 ): - Sequence: ( 2 rightarrow 1 rightarrow 8 rightarrow 4 rightarrow 2 ) - The sequence returns to 2, so ( n = 2 ) works. - For ( n = 4 ): - Sequence: ( 4 rightarrow 2 rightarrow 1 rightarrow 8 rightarrow 4 ) - The sequence returns to 4, so ( n = 4 ) works. - For ( n = 7 ): - Sequence: ( 7 rightarrow 14 rightarrow 7 ) - The sequence returns to 7, so ( n = 7 ) works. - For ( n = 8 ): - Sequence: ( 8 rightarrow 4 rightarrow 2 rightarrow 1 rightarrow 8 ) - The sequence returns to 8, so ( n = 8 ) works. - For ( n = 14 ): - Sequence: ( 14 rightarrow 7 rightarrow 14 ) - The sequence returns to 14, so ( n = 14 ) works.4. Generalize for larger values of ( n ): - For ( n > 21 ), we need to show that the sequence will not return to ( n ). - Consider ( n ) is odd and greater than 21: - The next number will be ( n + 7 ), which is even and greater than 28. - Dividing by 2 will give a number greater than 14. - Adding 7 again will give a number greater than 21. - This process will not return to the original ( n ). - Consider ( n ) is even and greater than 21: - Dividing by 2 will give a number greater than 10.5. - Adding 7 will give a number greater than 17.5. - Dividing by 2 again will give a number greater than 8.75. - Adding 7 will give a number greater than 15.75. - This process will not return to the original ( n ).Therefore, the only starting numbers that ( A ) can choose in order to win are ( 1, 2, 4, 7, 8, ) and ( 14 ).The final answer is ( boxed{1, 2, 4, 7, 8, 14} ).

🧠:Okay, let's tackle this problem step by step. The question states that there exist positive integers N and M such that the remainders when N is divided by 6, 36, 216, and M form an increasing nonzero geometric sequence in that order. We need to find the smallest possible value of M.First, let's make sure I understand the problem correctly. We have four divisors: 6, 36, 216, and M. When we divide N by each of these, the remainders form a geometric sequence. The sequence must be increasing and each term must be nonzero. Our goal is to find the smallest such M.Let's recall that a geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, r. Since the sequence is increasing and nonzero, the common ratio r must be a positive integer greater than 1. Because if r were 1, the sequence would be constant, which is not increasing, and if r were a fraction, the sequence would decrease, which contradicts the "increasing" part.So, the remainders when N is divided by 6, 36, 216, and M are a geometric sequence with ratio r > 1. Let's denote the four remainders as a, ar, ar², ar³, where a is the first term and r is the common ratio.But wait, there's a catch here. When you take a number modulo different divisors, the remainder must be less than the divisor. So, for example, the remainder when N is divided by 6 must be less than 6, the remainder modulo 36 must be less than 36, and so on. Therefore, the remainders must satisfy:a < 6,ar < 36,ar² < 216,ar³ < M.Additionally, since the remainders are part of a geometric sequence, each term must be an integer. Therefore, a, ar, ar², ar³ must all be integers, which implies that a must be a positive integer (since the remainders are nonzero), and r must be a rational number such that each multiplication by r results in an integer. However, since r is the ratio between two integers (ar / a = r), r must be a rational number. But given that we need r to multiply a to get integers each time, and a is an integer, r must be an integer. Because if r were a fraction, say p/q, then ar would be a*(p/q), which would need to be an integer, implying that q divides a. But since a is less than 6, and the denominators q can only be factors of a, but this might complicate things. However, given that the problem states that N and M are positive integers, and the remainders are also integers, it's more straightforward to assume that r is a positive integer. Let's verify this assumption.Suppose r is a fraction. Let r = p/q where p and q are coprime positive integers. Then the terms would be a, a*(p/q), a*(p²/q²), a*(p³/q³). For these to all be integers, q must divide a, but since a < 6, q must be a divisor of a. However, since p and q are coprime, this would mean that q divides a, but then a must be at least q. If q is 2, for example, then a must be a multiple of 2. However, this complicates things because the ratio could be fractional but the terms must still be integers. However, given that the problem is about integer remainders, and modulus operations, it's more natural to think of the common ratio as an integer. Therefore, I'll proceed under the assumption that r is an integer greater than 1.Therefore, our four remainders are a, ar, ar², ar³, with a < 6, ar < 36, ar² < 216, and ar³ < M. Also, since the remainders are in the order modulo 6, 36, 216, M, each subsequent remainder is modulo a larger number, so the remainder can be larger than the previous modulus. Wait, but when you take modulo 36, the remainder is less than 36, but modulus 6 is less than 6. However, the problem states that the remainders form an increasing sequence. So, a < ar < ar² < ar³. Therefore, since each term is larger than the previous, and the first term a is less than 6, the next term ar must be greater than a, so r > 1, which we already have.Now, let's note that when N is divided by 6, the remainder is a. Then, when N is divided by 36, the remainder is ar. But in reality, the remainder when N is divided by 36 must satisfy certain congruence conditions. Specifically, since 36 is a multiple of 6, the remainder modulo 36 must be congruent to a modulo 6. Similarly, the remainder modulo 216 must be congruent to the remainder modulo 36 when divided by 36, which in turn must be congruent to a modulo 6. Similarly, the remainder modulo M must be congruent to the remainder modulo 216 when divided by 216, and so on.So perhaps there's a relationship here between the remainders and the divisors. Let's formalize this.Let’s denote:N ≡ a mod 6N ≡ ar mod 36N ≡ ar² mod 216N ≡ ar³ mod MWe need to find integers N, M, a, r (with a, r positive integers, r ≥ 2, a < 6, ar < 36, ar² < 216) such that these congruences hold, and M is minimized.First, let's note that N ≡ a mod 6 implies that N = 6k + a for some integer k ≥ 0.Similarly, N ≡ ar mod 36. Since 36 is a multiple of 6, the congruence N ≡ ar mod 36 must be compatible with N ≡ a mod 6. Therefore, ar ≡ a mod 6. Since a is non-zero modulo 6 (because a < 6 and the remainder is non-zero), we can divide both sides by a (mod 6). Thus, r ≡ 1 mod (6/gcd(a,6)). Wait, let's think carefully.Let’s write ar ≡ a mod 6. Then, a(r - 1) ≡ 0 mod 6. Since a and 6 may have a common factor, let's let d = gcd(a,6). Then, we have (a/d)(r - 1) ≡ 0 mod (6/d). Because we can divide both sides by d. Therefore, since a/d and 6/d are coprime, we must have r - 1 ≡ 0 mod (6/d). Therefore, r ≡ 1 mod (6/d). So, the value of r is 1 + t*(6/d) for some integer t ≥ 1 (since r > 1). But since r must be an integer greater than 1, t must be at least 1.Similarly, since a < 6 and d = gcd(a,6), possible values for a are 1, 2, 3, 4, 5. Let's consider each possible a:Case 1: a = 1Then, gcd(1,6) = 1. Therefore, r ≡ 1 mod 6/1 = 6. Therefore, r ≡ 1 mod 6. Hence, r = 1 + 6t. But since r must be at least 2, the smallest possible r is 7. Let's check if this works.But then, ar = 1*7 = 7. But ar must be less than 36, which is true. Then ar² = 1*7² = 49, which needs to be less than 216, which is true. ar³ = 1*7³ = 343. Then M must be greater than 343. But we need the smallest possible M. However, M must be such that N ≡ 343 mod M. But this would require M > 343. But maybe there is a smaller M that satisfies the congruence. Wait, but how does M relate to the previous terms?Alternatively, perhaps we need to find M such that the remainder ar³ is less than M, and M must be such that N ≡ ar³ mod M. But M has to be larger than ar³, otherwise the remainder would be ar³ mod M, which is less than M, but the problem states that the remainders form the sequence a, ar, ar², ar³. So, if M is larger than ar³, then the remainder is ar³. If M is less than or equal to ar³, then the remainder would be ar³ mod M, which is different. Therefore, to have the remainder be exactly ar³, M must be greater than ar³. Therefore, M must be at least ar³ + 1. However, the problem states that the remainders form an increasing geometric sequence, so ar³ must be less than M. Therefore, M must be greater than ar³.But in the case where a=1 and r=7, ar³=343, so M must be at least 344. However, maybe there's a smaller M that allows the remainder to be 343 even if M is less than 343? Wait, no. Because the remainder when divided by M must be less than M. So if M is less than 343, the remainder would be 343 mod M, which is less than M, but we need the remainder to be exactly 343. Therefore, M must be greater than 343, so the minimal M in this case would be 344. But 344 is quite large, and maybe there are smaller M for other a and r.Let's check other possible values of a.Case 2: a=2gcd(2,6)=2. Therefore, from above, r ≡1 mod (6/2)=3. So, r ≡1 mod 3. Therefore, possible values of r are 1 + 3t, t ≥1. The smallest r is 4 (since r must be greater than 1). Let's check r=4.Then, ar=2*4=8 <36, ar²=2*16=32 <216, ar³=2*64=128. So M must be greater than 128. So minimal M here is 129. But let's see if this works.But we need to check if there exists an N such that:N ≡2 mod 6,N ≡8 mod 36,N ≡32 mod 216,N ≡128 mod M.We need to check if such an N exists. Let's first check the congruences modulo 6, 36, 216.Starting with N ≡2 mod 6. Then N=6k+2.Then N ≡8 mod 36. So 6k +2 ≡8 mod 36. Therefore, 6k ≡6 mod 36 ⇒6(k -1)≡0 mod 36 ⇒k -1≡0 mod 6 ⇒k=6m +1. Therefore, N=6*(6m +1)+2=36m +6 +2=36m +8. So N≡8 mod 36, which matches the second condition.Now, N must also ≡32 mod 216. So 36m +8 ≡32 mod 216 ⇒36m ≡24 mod 216 ⇒Divide both sides by 12: 3m ≡2 mod 18 ⇒3m ≡2 mod 18. Let's solve for m.Multiply both sides by the inverse of 3 modulo 18. Since gcd(3,18)=3, but 3 doesn't divide 2, so there is no solution. Therefore, there is no integer m that satisfies 3m ≡2 mod 18. Therefore, this case is impossible. So r=4 for a=2 doesn't work.Next, try the next possible r for a=2, which is r=1+3*2=7.So r=7. Then ar=14, ar²=98, ar³=686. But ar²=98 must be less than 216, which is true. ar³=686, so M must be at least 687. But before that, let's check congruences.N ≡2 mod 6,N ≡14 mod 36,N ≡98 mod 216,N ≡686 mod M.Again, start with N=6k +2.N ≡14 mod 36: 6k +2 ≡14 mod 36 ⇒6k ≡12 mod 36 ⇒k ≡2 mod 6 ⇒k=6m +2. Thus, N=6*(6m +2)+2=36m +12 +2=36m +14. So N≡14 mod 36.Now, N ≡98 mod 216. So 36m +14 ≡98 mod 216 ⇒36m ≡84 mod 216 ⇒Divide both sides by 12: 3m ≡7 mod 18. Again, 3m ≡7 mod 18. Since gcd(3,18)=3, and 3 doesn't divide 7, no solution. Therefore, this is impossible.Similarly, r=10 (next in the sequence 1 +3t: 4,7,10,...). ar=20, ar²=200, ar³=2000. But ar²=200 <216. Check congruence:N=6k +2 ≡20 mod 36 ⇒6k +2 ≡20 mod36 ⇒6k≡18 mod36 ⇒k≡3 mod6. So N=6*(6m +3)+2=36m +18 +2=36m +20. Then N≡20 mod36. Now, N≡200 mod216. So 36m +20 ≡200 mod216 ⇒36m≡180 mod216 ⇒Divide by 12: 3m≡15 mod18 ⇒m≡5 mod6. So m=6n +5. Therefore, N=36*(6n +5)+20=216n +180 +20=216n +200. Therefore, N=216n +200. Then, N mod M =686 when M=687? Wait, ar³=2*7³=686. Wait, no, if r=10, then ar³=2*10³=2000. Wait, confusion here. Let's correct that.Wait, if a=2 and r=10, then ar=20, ar²=200, ar³=2000. So M must be greater than 2000, so minimal M is 2001. But even before that, when we tried r=10, the congruence N≡200 mod216 is possible. So N=216n +200. Then, the remainder when N is divided by M must be 2000. Therefore, 216n +200 ≡2000 mod M. So 216n ≡1800 mod M. But since M must be greater than 2000, and 216n +200 must be congruent to 2000 mod M. However, unless M divides 216n +200 -2000=216n -1800, which is complicated. Alternatively, maybe there's no solution here. But even if there is, M would need to be a divisor of 216n -1800, but since n can be any integer, we can choose n such that 216n ≡1800 mod M. However, this seems too vague, and given that M must be greater than 2000, the minimal M here would be 2001. But this is larger than the previous 344, so not helpful.Therefore, for a=2, r=4 and r=7 lead to no solution because of congruence inconsistencies. Maybe there's a different approach here.Alternatively, maybe the problem is not only about the congruences but also about the actual remainders. Let's think again. The remainder when N is divided by 6 is a. The remainder when divided by 36 is ar. But since 36 is a multiple of 6, the remainder ar modulo 36 must satisfy ar ≡ a mod 6. Which we already considered. Similarly, the remainder when divided by 216 (which is 36*6) must satisfy ar² ≡ ar mod 36. Because 216 is a multiple of 36, so the remainder modulo 216 when divided by 36 gives the remainder modulo 36, which must be ar. Therefore, ar² ≡ ar mod 36. Similarly, ar² ≡ ar mod 36 ⇒ ar² - ar ≡0 mod36 ⇒ar(r -1) ≡0 mod36. Since we already have ar ≡ a mod6, which led to r ≡1 mod (6/gcd(a,6)).But perhaps we need to use the Chinese Remainder Theorem or solve these congruences step by step.Let me try to formalize the system:1. N ≡ a mod 62. N ≡ ar mod363. N ≡ ar² mod2164. N ≡ ar³ mod MWe need to find N, M, a, r such that these hold, with a <6, ar <36, ar² <216, ar³ <M, and the sequence a, ar, ar², ar³ is increasing.Let’s start with the first three congruences and see what constraints they impose on a and r.From 1 and 2: N ≡ a mod6 and N ≡ ar mod36. Since 36 is a multiple of 6, the second congruence implies the first. Specifically, N ≡ ar mod36 ⇒ N ≡ ar mod6. But N ≡ a mod6. Therefore, ar ≡ a mod6 ⇒a(r -1) ≡0 mod6. As before.Similarly, from 2 and 3: N ≡ ar mod36 and N ≡ ar² mod216. Since 216 is a multiple of 36, the third congruence implies the second. Specifically, N ≡ ar² mod216 ⇒ N ≡ ar² mod36. But N ≡ ar mod36. Therefore, ar² ≡ ar mod36 ⇒ar(r -1) ≡0 mod36.Similarly, from 1 and 3: N ≡ a mod6 and N ≡ ar² mod216. Then N ≡ ar² mod6. But N ≡a mod6 ⇒ ar² ≡a mod6 ⇒a(r² -1) ≡0 mod6.So we have three congruence conditions:1. a(r -1) ≡0 mod62. ar(r -1) ≡0 mod363. a(r² -1) ≡0 mod6Let’s analyze each of these.Starting with the first: a(r -1) ≡0 mod6. This means that 6 divides a(r -1). Since a <6, possible a=1,2,3,4,5. Let's analyze for each a:For a=1:1*(r -1) ≡0 mod6 ⇒r ≡1 mod6For a=2:2*(r -1) ≡0 mod6 ⇒(r -1) ≡0 mod3 ⇒r ≡1 mod3For a=3:3*(r -1) ≡0 mod6 ⇒(r -1) ≡0 mod2 ⇒r ≡1 mod2For a=4:4*(r -1) ≡0 mod6 ⇒2*(r -1) ≡0 mod3 ⇒(r -1) ≡0 mod3 ⇒r ≡1 mod3For a=5:5*(r -1) ≡0 mod6 ⇒5 and 6 are coprime ⇒(r -1) ≡0 mod6 ⇒r ≡1 mod6So this gives us the conditions on r for each a.Now, moving to the third congruence: a(r² -1) ≡0 mod6. Let's see:a(r² -1) =a(r -1)(r +1). We already know that 6 divides a(r -1) from the first condition. Therefore, 6 divides a(r -1), and thus a(r -1)(r +1) is divisible by 6*(r +1). Therefore, 6 divides a(r² -1). Therefore, the third condition is automatically satisfied if the first condition is. So maybe the third condition doesn't add new information.Now the second congruence: ar(r -1) ≡0 mod36.Given that a(r -1) ≡0 mod6, let's denote a(r -1)=6k for some integer k. Then, ar(r -1)=6k *r. Therefore, 6k *r ≡0 mod36 ⇒6kr ≡0 mod36 ⇒kr ≡0 mod6. So kr must be divisible by6.Given that k = a(r -1)/6, and a and r are integers with a <6.Let's consider each a case by case.Case 1: a=1, r ≡1 mod6So r=6t +1, t≥1 (since r>1). The minimal r is 7.Then, a=1, r=7:Check second congruence: ar(r -1)=1*7*6=42. 42 ≡6 mod36. So 42 ≡6 mod36 ≠0 mod36. Therefore, this does not satisfy the second condition. Therefore, contradiction.Wait, but according to our earlier substitution:k =a(r -1)/6=1*6/6=1. Then kr=1*7=7. So kr=7, which must be ≡0 mod6. But 7≡1 mod6≠0. Therefore, the second condition fails. Hence, a=1, r=7 does not work.But why? Let's check:ar(r -1)=1*7*6=42. 42 mod36=6≠0. Therefore, the second condition is not satisfied.Therefore, even though the first condition is met (r≡1 mod6), the second condition fails. Therefore, we need to find r such that kr ≡0 mod6. For a=1, k= (1*(r-1))/6. Therefore, kr= [ (r-1)/6 ] * r. Since r≡1 mod6, let r=6t +1. Then, kr= [ (6t)/6 ]*(6t +1)= t*(6t +1). Therefore, kr=6t² +t. This must be ≡0 mod6. So 6t² +t ≡t ≡0 mod6. Therefore, t≡0 mod6. Therefore, t=6s. Then r=6*(6s)+1=36s +1. Therefore, the minimal r in this case is r=37. But then ar=1*37=37, which is not less than36, violating the ar <36 condition. Therefore, impossible. Thus, for a=1, there is no solution.Case 2: a=2, r≡1 mod3So possible r=4,7,10,... Let's take r=4 first.Then, ar(r -1)=2*4*3=24. 24 mod36=24≠0. So 24≡24 mod36≠0. Therefore, second condition fails.Compute kr= [a(r-1)/6]*r= [2*3/6]*4= [6/6]*4=1*4=4≡4 mod6≠0. So kr≡4 mod6≠0. Therefore, fails.Next r=7:ar(r -1)=2*7*6=84. 84 mod36=12≠0. kr= [2*6/6]*7=2*7=14≡2 mod6≠0. Fails.Next r=10:ar(r -1)=2*10*9=180. 180 mod36=180-5*36=0. So 180≡0 mod36. Therefore, this works.Check kr= [2*(10-1)/6]*10= [18/6]*10=3*10=30≡0 mod6. Yes, 30≡0 mod6. Therefore, satisfies.So for a=2, r=10:ar=20 <36,ar²=200 <216,ar³=2000 <M.So M must be greater than2000. However, let's check the congruences:N ≡2 mod6,N≡20 mod36,N≡200 mod216,N≡2000 modM.Let's find such N.From N≡2 mod6 and N≡20 mod36:As before, N=36m +20. Then N≡20 mod36.Then N≡200 mod216:36m +20 ≡200 mod216 ⇒36m ≡180 mod216 ⇒Divide by 12: 3m ≡15 mod18 ⇒m≡5 mod6. Therefore, m=6n +5.Thus, N=36*(6n +5)+20=216n +180 +20=216n +200.Therefore, N=216n +200. Now, this N must ≡2000 modM. So:216n +200 ≡2000 modM ⇒216n ≡1800 modM.But since M must be greater than2000 (as the remainder is2000), the minimal M is2001. But we need to check if there exists n such that216n ≡1800 mod2001. However, this is a bit involved. However, since M must be such that2000 is the remainder when N is divided byM, we must have M>2000 and M divides (N -2000). Given that N=216n +200, then M divides (216n +200 -2000)=216n -1800. Therefore, M divides216n -1800. To have M>2000, we can choose n such that216n -1800 ≥2001. Therefore,216n ≥3801 ⇒n≥3801/216≈17.58 ⇒n≥18. Then, M=216*18 -1800=3888 -1800=2088. Wait, but M can be any divisor of216n -1800 greater than2000. The minimal such M is the minimal divisor of216n -1800 greater than2000. However, this seems too vague. Maybe M is2001, but 2001 divides216n -1800. Let's check if2001 divides216n -1800.Find n such that216n ≡1800 mod2001. We can write this as216n ≡1800 mod2001.First, simplify the equation. Let's compute gcd(216,2001).2001 ÷216=9.27... Let's compute 2001=216*9 + 57 (since 216*9=1944, 2001-1944=57)Then gcd(216,57). 57=216-3*57=216-171=45. Then gcd(57,45)=3, since57=45*1 +12; gcd(45,12)=3; gcd(12,3)=3. Therefore, gcd(216,2001)=3.Now, the equation216n ≡1800 mod2001 has solutions iff3 divides1800. Since1800/3=600, yes. Therefore, solutions exist.The general solution is n ≡(1800/3)*( (216/3)^(-1) mod (2001/3)) mod (2001/3)Wait, to solve216n ≡1800 mod2001:Divide both sides by3:72n ≡600 mod667.Now, find the inverse of72 mod667.Compute gcd(72,667):667=72*9 +1972=19*3 +1519=15*1 +415=4*3 +34=3*1 +13=1*3 +0So gcd=1. Therefore, inverse exists.Now, find inverse of72 mod667.Using extended Euclidean algorithm:We need to find x such that72x ≡1 mod667.From the above steps:1=4 -3*1But 3=15 -4*3So 1=4 - (15 -4*3)*1=4*4 -15*1But 4=19 -15*1So 1= (19 -15)*4 -15=19*4 -15*5But15=72 -19*3So 1=19*4 - (72 -19*3)*5=19*4 -72*5 +19*15=19*19 -72*5But19=667 -72*9So 1= (667 -72*9)*19 -72*5=667*19 -72*171 -72*5=667*19 -72*176Therefore, -176*72 ≡1 mod667 ⇒x≡-176 mod667≡667-176=491. So inverse of72 mod667 is491.Therefore, solution is n ≡600*491 mod667.Compute600*491 mod667:First compute600 mod667=600491 mod667=491Now compute600*491:Breakdown:600*491=600*(500 -9)=600*500 -600*9=300,000 -5,400=294,600Now compute294,600 mod667:Divide294,600 by667:667*441=667*(400 +40 +1)=667*400=266,800; 667*40=26,680; 667*1=667. Total=266,800+26,680=293,480 +667=294,147.Subtract from294,600:294,600 -294,147=453.Therefore, n≡453 mod667.Therefore, the minimal positive n is453. Therefore, N=216*453 +200= let's compute that:216*453=216*(400 +50 +3)=216*400=86,400; 216*50=10,800; 216*3=648. Total=86,400+10,800=97,200 +648=97,848. Then N=97,848 +200=98,048.Then M must divide98,048 -2000=96,048. So M is a divisor of96,048 greater than2000. The smallest such divisor is the smallest prime factor of96,048 greater than2000. Let's factorize96,048.96,048 ÷2=48,02448,024 ÷2=24,01224,012 ÷2=12,00612,006 ÷2=6,0036,003 ÷3=2,0012,001 ÷3=667667 is a prime number (since it's not divisible by2,3,5,7,11,13,17,19,23,29, etc. up to sqrt(667)≈25.8).Therefore,96,048=2⁴ *3² *667. Therefore, the divisors greater than2000 are:First, check 667*3=2001, which is a divisor. Then667*2=1334, which is less than2000. Then3²*667=6003, which is larger. So the smallest divisor greater than2000 is2001. Therefore, the minimal M in this case is2001. However, let's check if M=2001 works.N=98,048. Then N mod2001=98,048 mod2001.Compute98,048 ÷2001:2001*49=98,049. Therefore,98,048=2001*49 -1. Therefore,98,048 mod2001= -1 mod2001=2000. Which matches ar³=2000. Therefore, M=2001 works.So for a=2, r=10, M=2001 is possible. However, we need to check if there is a smaller M possible with other a and r.Case3: a=3, r≡1 mod2 (since a=3, gcd(3,6)=3, so r≡1 mod2)So r must be odd. Possible r=3,5,7,...Check r=3:ar=9 <36, ar²=27 <216, ar³=81. So M>81, minimal M=82.Check congruence conditions:N≡3 mod6,N≡9 mod36,N≡27 mod216,N≡81 modM.First, N=6k +3.N≡9 mod36: 6k +3 ≡9 mod36 ⇒6k ≡6 mod36 ⇒k≡1 mod6 ⇒k=6m +1. Therefore, N=6*(6m +1)+3=36m +6 +3=36m +9.Now, N≡27 mod216: 36m +9 ≡27 mod216 ⇒36m ≡18 mod216 ⇒Divide by 18:2m ≡1 mod12 ⇒2m ≡1 mod12 ⇒m≡ (1* inverse of2 mod12). But inverse of2 mod12 does not exist since gcd(2,12)=2. Therefore, no solution. Therefore, r=3 is invalid.Next r=5:ar=15 <36, ar²=75 <216, ar³=375. So M>375, minimal M=376.Check congruences:N≡3 mod6,N≡15 mod36,N≡75 mod216,N≡375 modM.N=6k +3.N≡15 mod36 ⇒6k +3 ≡15 mod36 ⇒6k ≡12 mod36 ⇒k≡2 mod6 ⇒k=6m +2. Thus, N=6*(6m +2)+3=36m +12 +3=36m +15.N≡75 mod216: 36m +15 ≡75 mod216 ⇒36m ≡60 mod216 ⇒Divide by12:3m ≡5 mod18. Solve3m ≡5 mod18. gcd(3,18)=3, which does not divide5. No solution. Therefore, impossible.Next r=7:ar=21 <36, ar²=147 <216, ar³=1029. So M>1029.Check congruences:N=6k +3 ≡21 mod36 ⇒6k +3 ≡21 mod36 ⇒6k≡18 mod36 ⇒k≡3 mod6 ⇒k=6m +3 ⇒N=36m +18 +3=36m +21.N≡147 mod216 ⇒36m +21 ≡147 mod216 ⇒36m ≡126 mod216 ⇒Divide by18:2m ≡7 mod12 ⇒No solution, since2m≡7 mod12 is impossible. Therefore, no solution.Similarly, r=9:ar=27 <36, ar²=243. But ar²=243 must be <216? No, 243>216. Therefore, invalid. So a=3, r=9 is invalid.Therefore, a=3 does not yield a valid solution.Case4: a=4, r≡1 mod3 (since gcd(4,6)=2, so r≡1 mod3)Possible r=4,7,10,...Check r=4:ar=16 <36, ar²=64 <216, ar³=256. So M>256, minimal M=257.Check congruences:N≡4 mod6,N≡16 mod36,N≡64 mod216,N≡256 modM.Start with N=6k +4.N≡16 mod36:6k +4 ≡16 mod36 ⇒6k≡12 mod36 ⇒k≡2 mod6 ⇒k=6m +2 ⇒N=36m +12 +4=36m +16.N≡64 mod216:36m +16 ≡64 mod216 ⇒36m ≡48 mod216 ⇒Divide by12:3m ≡4 mod18. Solve3m ≡4 mod18. gcd(3,18)=3, which does not divide4. No solution.Next r=7:ar=28 <36, ar²=196 <216, ar³=1372. So M>1372, minimal M=1373.Check congruences:N=6k +4 ≡28 mod36 ⇒6k +4≡28 mod36 ⇒6k≡24 mod36 ⇒k≡4 mod6 ⇒k=6m +4 ⇒N=36m +24 +4=36m +28.N≡196 mod216:36m +28 ≡196 mod216 ⇒36m ≡168 mod216 ⇒Divide by12:3m ≡14 mod18. 3m≡14 mod18. gcd(3,18)=3, which does not divide14. No solution.Next r=10:ar=40. But ar=40 must be <36? No, 40>36. Invalid.Therefore, a=4, r=10 is invalid. So no solution for a=4.Case5: a=5, r≡1 mod6 (since gcd(5,6)=1, so r≡1 mod6)Possible r=7,13,19,...Check r=7:ar=35 <36, ar²=245 <216? No, 245>216. Invalid.Next r=13:ar=65 >36. Invalid.Wait, minimal r=7, but ar²=5*7²=245>216. Therefore, no solution for a=5.Thus, from all cases, the only viable solution is when a=2, r=10, leading to M=2001. However, we need to check if there are other possible a and r with smaller M.Wait, let's revisit the case when a=2, r=10 gives M=2001. Are there other combinations where ar³ is smaller, leading to smaller M? Let's check other a and r.Wait, for a=2, r=4 didn't work due to congruence issues, but let's check other values.Wait, when a=3, r=3 gives ar³=81, which would require M=82. However, the congruence N≡27 mod216 had no solution. Similarly, a=3, r=5 gives M=376, but no solution.Another approach: perhaps the geometric sequence is not necessarily with integer ratio. Wait, but the problem states that the remainders form a geometric sequence. Since remainders are integers, the ratio must be rational. However, if the ratio is rational, say p/q in reduced form, then the terms would be a, a*(p/q), a*(p/q)², a*(p/q)³, which must all be integers. This requires that q³ divides a. But since a <6, the only possible q is 1, because if q≥2, then q³ ≥8 >a. Therefore, q must be1, meaning that the ratio must be an integer. Therefore, the initial assumption that r is an integer is valid. Hence, our previous analysis holds.Thus, the only possible solution is M=2001. However, the problem says "the smallest possible value of M". But maybe there's another case we missed.Wait, let's revisit a=2, r=10. The sequence is 2,20,200,2000. These are the remainders modulo6,36,216,M. So:N mod6=2,N mod36=20,N mod216=200,N modM=2000.We found that N=98,048 and M=2001 works.But wait, is there a smaller M? For example, if M divides98,048 -2000=96,048 and M>2000, then the smallest M is the smallest divisor of96,048 greater than2000. As previously factorized,96,048=2⁴*3²*667. The divisors greater than2000 are:3²*667=9*667=6003,2*3²*667=2*6003=12006,2²*3²*667=4*6003=24012,2³*3²*667=8*6003=48024,2⁴*3²*667=16*6003=96048,667*2=1334 (too small),667*3=2001,667*2²=2668,667*2*3=4002,etc.The smallest divisor greater than2000 is2001. Therefore, M=2001 is indeed the minimal possible.But wait, let's check if there's another a and r combination that gives a smaller M.Let's check a=4, r=5.Wait, a=4, gcd(4,6)=2. Therefore, r≡1 mod3. So r=4,7,10,...For a=4, r=4:ar=16, ar²=64, ar³=256. Then M>256. But earlier we saw that for a=4, r=4, the congruence N≡64 mod216 has no solution.Alternatively, a=5, r=2. But a=5, r must be≡1 mod6. So r=7,13,... But r=7 leads to ar=35 <36, ar²=245>216 invalid.Wait, no other cases seem to work.Therefore, the minimal M is2001. However, let's double-check if there's an alternative with a different a and r that we might have missed.Suppose a=1, r= something where the congruences can be satisfied.Earlier, for a=1, r=7 leads to no solution, but what if r=13?ar=13 <36, ar²=169 <216, ar³=2197. So M>2197.Check congruences:N≡1 mod6,N≡13 mod36,N≡169 mod216,N≡2197 modM.Let’s try:N=6k +1.N≡13 mod36 ⇒6k +1≡13 mod36 ⇒6k≡12 mod36 ⇒k≡2 mod6 ⇒k=6m +2 ⇒N=36m +12 +1=36m +13.N≡169 mod216 ⇒36m +13≡169 mod216 ⇒36m≡156 mod216 ⇒Divide by12:3m≡13 mod18 ⇒No solution. Therefore, invalid.Similarly, a=1, r=19 gives ar=19 <36, ar²=361>216 invalid.No solution for a=1.Thus, the only viable solution is M=2001. However, this seems quite large. Maybe there's a mistake in the earlier analysis.Wait, going back to the problem statement: "N's remainders modulo the four integers 6, 36, 216, and M form an increasing nonzero geometric sequence in that order."We interpreted the sequence as a, ar, ar², ar³. But maybe the sequence is N mod6, N mod36, N mod216, N modM. However, the remainder when divided by36 is not necessarily related to the remainder when divided by6, except that the remainder modulo36 must congruent to the remainder modulo6 modulo6.But perhaps the geometric sequence is formed by the four remainders in the order given. So let's denote:Let’s call the remainders as follows:r₁ = N mod6,r₂ = N mod36,r₃ = N mod216,r₄ = N modM.These four remainders form a geometric sequence: r₁, r₂, r₃, r₄ with r₂/r₁ = r₃/r₂ = r₄/r₃ = r (common ratio).But the key point is that each subsequent remainder is modulo a larger number, so the remainder can be larger than the previous modulus. For example, r₂ = N mod36 is a number between0 and35, but r₁ = N mod6 is between0 and5. However, the problem states that the sequence is increasing and nonzero. Therefore, each term must be strictly larger than the previous.Given that, the sequence must be r₁ < r₂ < r₃ < r₄, and each is a positive integer, hence nonzero.Moreover, r₁ = N mod6 must be at least1, and similarly for the others.But perhaps there's a different approach. Let's suppose that the remainders are a, ar, ar², ar³, forming a geometric sequence. Then:Since a = N mod6, we have a <6.ar = N mod36 ⇒ ar <36.ar² = N mod216 ⇒ar² <216.ar³ = N modM ⇒ar³ <M.Moreover, since N ≡a mod6, and N ≡ar mod36, we have that ar ≡a mod6 ⇒a(r -1) ≡0 mod6.Similarly, N ≡ar mod36 and N ≡ar² mod216 ⇒ar² ≡ar mod36 ⇒ar(r -1) ≡0 mod36.And N ≡ar² mod216 and N ≡ar³ modM ⇒ar³ ≡ar² mod216 ⇒ar²(r -1) ≡0 mod216.But since we are to find M such that ar³ <M, and M must be the smallest possible, we need to ensure that the sequence a, ar, ar², ar³ is valid with the given congruence conditions.Let’s try to find a and r such that a <6, ar <36, ar² <216, and the congruences hold.Let’s consider possible a and r:Start with a=2, r=3.Then the sequence is 2,6,18,54. Check if these can be remainders:- 2 mod6=2 OK.- 6 mod36=6, but the remainder must be6, but N mod36=6. Then N=36k +6. Then N mod6=6 mod6=0, but we need N mod6=2. Contradiction. So invalid.Next, a=2, r=5.Sequence:2,10,50,250. Check congruences:N≡2 mod6,N≡10 mod36,N≡50 mod216,N≡250 modM.From N=6k +2 ≡10 mod36 ⇒6k +2≡10 ⇒6k≡8 ⇒k≡8/6 mod36. Not an integer. So invalid.Next, a=3, r=2.Sequence:3,6,12,24. But N mod6=3, N mod36=6. But 6 mod6=0≠3. Contradiction.a=1, r=5.Sequence:1,5,25,125. Check congruences:N≡1 mod6,N≡5 mod36,N≡25 mod216,N≡125 modM.From N=6k +1 ≡5 mod36 ⇒6k ≡4 mod36 ⇒k ≡4/6 mod6 ⇒k≡2/3 mod6. No solution.a=4, r=3.Sequence:4,12,36,108. But 36 is the modulus for the second term, so the remainder must be less than36. So 36 mod36=0≠36. Invalid.a=3, r=4.Sequence:3,12,48,192. Check congruences:N≡3 mod6,N≡12 mod36,N≡48 mod216,N≡192 modM.From N=6k +3 ≡12 mod36 ⇒6k ≡9 mod36 ⇒k ≡9/6 mod6 ⇒k≡3/2 mod6. No solution.a=4, r=2.Sequence:4,8,16,32. Check congruences:N≡4 mod6,N≡8 mod36,N≡16 mod216,N≡32 modM.From N=6k +4 ≡8 mod36 ⇒6k≡4 mod36 ⇒k≡4/6 mod6 ⇒k≡2/3 mod6. No solution.a=2, r=6.Sequence:2,12,72,432. But ar=12 <36, ar²=72 <216, ar³=432. So M>432.Check congruences:N≡2 mod6,N≡12 mod36,N≡72 mod216,N≡432 modM.N=6k +2 ≡12 mod36 ⇒6k≡10 mod36 ⇒k≡10/6 mod6 ⇒k≡5/3 mod6. No solution.a=2, r=7:Sequence:2,14,98,686. ar=14 <36, ar²=98 <216, ar³=686. Then M>686.Check congruences:N=6k +2 ≡14 mod36 ⇒6k +2=14 mod36 ⇒6k=12 mod36 ⇒k=2 mod6 ⇒k=6m +2 ⇒N=36m +14.N≡98 mod216 ⇒36m +14=98 mod216 ⇒36m=84 mod216 ⇒m=84/36 mod6 ⇒m=7/3 mod6. No solution.a=2, r=9:Sequence:2,18,162,1458. ar=18 <36, ar²=162 <216, ar³=1458. So M>1458.Check congruences:N=6k +2≡18 mod36 ⇒6k=16 mod36 ⇒k=16/6 mod6 ⇒k=8/3 mod6. No solution.a=2, r=5:Sequence:2,10,50,250. Already checked, no solution.a=2, r=10:Sequence:2,20,200,2000. As before, this works with M=2001.Therefore, after checking various possibilities, the only feasible solution is M=2001.But wait, the problem says "the smallest possible value of M". The initial answer we found is M=2001. However, the problem might have a smaller M. Let's think again.Wait, perhaps there's a different sequence where the common ratio is not an integer. Although earlier we concluded that the ratio must be integer because the terms must be integers and a is small. But let's check.Suppose the ratio is a fraction, say r=p/q where p and q are coprime integers. Then the terms would be a, a(p/q), a(p²/q²), a(p³/q³). For all these to be integers, q must divide a. Since a <6, possible q=1,2,3,4,5. But since p and q are coprime, q must divide a. So if q=2, a must be a multiple of2. Similarly for other q.Let’s try a=2, q=2, r=3/2.Then the sequence is:a=2,ar=2*(3/2)=3,ar²=3*(3/2)=9/2=4.5 Not integer. Invalid.a=4, q=2, r=3/2:a=4,ar=4*(3/2)=6,ar²=6*(3/2)=9,ar³=9*(3/2)=13.5 Not integer.a=3, q=3, r=4/3:a=3,ar=3*(4/3)=4,ar²=4*(4/3)=16/3≈5.33 Not integer.a=6, but a<6. No.a=9, no.So this seems impossible. Therefore, ratio must be integer.Hence, the minimal M is indeed2001.But let's check if the problem might have a smaller M. Wait, the problem states that the remainders form an increasing nonzero geometric sequence in the order modulo6,36,216,M. The key point is that the sequence is increasing, so each term must be larger than the previous. For a=2, r=10, the sequence is2,20,200,2000, which is increasing. For other a and r combinations, we either can't satisfy the congruences or get a larger M.Thus, the answer is M=2001. But wait, the problem says "positive integers N, M", so as long as such N and M exist, and M=2001 is the smallest such M.However, let's verify once more. The sequence we found requires M=2001. But maybe there's a different geometric sequence that we missed. For example:Suppose the common ratio is r=3. Let's see if there's a possible a.a=1:Sequence:1,3,9,27. Check congruences:N≡1 mod6,N≡3 mod36,N≡9 mod216,N≡27 modM.From N=6k +1≡3 mod36 ⇒6k≡2 mod36 ⇒k≡2/6 mod6 ⇒k≡1/3 mod6. No solution.a=2:Sequence:2,6,18,54. But N≡2 mod6, N≡6 mod36 ⇒N≡6 mod36 ⇒N≡0 mod6. Contradiction.a=3:Sequence:3,9,27,81. N≡3 mod6,N≡9 mod36,N≡27 mod216,N≡81 modM.From N=6k +3≡9 mod36 ⇒6k≡6 mod36 ⇒k≡1 mod6 ⇒k=6m +1 ⇒N=36m +9.N≡27 mod216 ⇒36m +9≡27 mod216 ⇒36m≡18 mod216 ⇒m≡18/36 mod6 ⇒m≡1/2 mod6. No solution.a=4:Sequence:4,12,36,108. But N≡4 mod6, N≡12 mod36 ⇒N≡0 mod12 ⇒N≡0 mod6. Contradiction.a=5:Sequence:5,15,45,135. N≡5 mod6,N≡15 mod36,N≡45 mod216,N≡135 modM.From N=6k +5≡15 mod36 ⇒6k≡10 mod36 ⇒k≡10/6 mod6 ⇒k≡5/3 mod6. No solution.Thus, no solution for ratio r=3.Similarly, try r=4:a=1:Sequence:1,4,16,64.Check congruences:N≡1 mod6,N≡4 mod36,N≡16 mod216,N≡64 modM.N=6k +1≡4 mod36 ⇒6k≡3 mod36 ⇒k≡3/6 mod6 ⇒k≡1/2 mod6. No solution.a=2:Sequence:2,8,32,128.Check congruences:N≡2 mod6,N≡8 mod36,N≡32 mod216,N≡128 modM.N=6k +2≡8 mod36 ⇒6k≡6 mod36 ⇒k≡1 mod6 ⇒k=6m +1 ⇒N=36m +8.N≡32 mod216 ⇒36m +8≡32 mod216 ⇒36m≡24 mod216 ⇒m≡24/36 mod6 ⇒m≡2/3 mod6. No solution.a=3:Sequence:3,12,48,192.N≡3 mod6,N≡12 mod36,N≡48 mod216,N≡192 modM.N=6k +3≡12 mod36 ⇒6k≡9 mod36 ⇒k≡9/6 mod6 ⇒k≡3/2 mod6. No solution.Similarly, other a's will have issues.Thus, no solution for r=4.After checking various possibilities, it seems the only valid solution is with a=2, r=10, leading to M=2001. Therefore, the smallest possible value of M is2001.

✅:The smallest possible value of ( M ) is (boxed{2001}).Step-by-Step Explanation:1. Understanding the Problem: - We need positive integers ( N ) and ( M ) such that the remainders of ( N ) modulo ( 6 ), ( 36 ), ( 216 ), and ( M ) form an increasing nonzero geometric sequence. - The sequence must be ( a, ar, ar^2, ar^3 ) where ( a ) is the first term and ( r ) is the common ratio (an integer greater than 1).2. Key Constraints: - Each remainder must be less than its respective modulus. - ( a < 6 ), ( ar < 36 ), ( ar^2 < 216 ), and ( ar^3 < M ). - Congruence conditions must be satisfied for each modulus.3. Analyzing Possible Values of ( a ) and ( r ): - For ( a = 2 ): - ( gcd(2, 6) = 2 ), so ( r equiv 1 mod 3 ). - The smallest valid ( r ) is 10 (since lower values like 4, 7 led to congruence issues). - Sequence: ( 2, 20, 200, 2000 ). - Check congruences: - ( N equiv 2 mod 6 ) - ( N equiv 20 mod 36 ) - ( N equiv 200 mod 216 ) - ( N equiv 2000 mod M ) - Solve for ( N ) using the Chinese Remainder Theorem, leading to ( N = 216n + 200 ). - ( M ) must be greater than 2000 and a divisor of ( 216n - 1800 ). The smallest such ( M ) is 2001.4. Verification: - ( M = 2001 ) satisfies all conditions with ( N = 98048 ), ensuring the remainders form the geometric sequence ( 2, 20, 200, 2000 ).Thus, after thorough analysis and verification, the smallest ( M ) is (boxed{2001}).

🔑:1. Define the remainders: Let ( r_6 ), ( r_{36} ), ( r_{216} ), and ( r_M ) be the remainders of ( N ) when divided by ( 6 ), ( 36 ), ( 216 ), and ( M ) respectively. These remainders form an increasing nonzero geometric sequence.2. Geometric sequence properties: Since ( r_6 ), ( r_{36} ), ( r_{216} ), and ( r_M ) form a geometric sequence, there exists a common ratio ( r ) such that: [ r_{36} = r cdot r_6, quad r_{216} = r cdot r_{36} = r^2 cdot r_6, quad r_M = r cdot r_{216} = r^3 cdot r_6 ]3. Constraints on remainders: - ( 1 leq r_6 < 6 ) - ( r_6 < r_{36} < 36 ) - ( r_{36} < r_{216} < 216 ) - ( r_{216} < r_M )4. Case analysis for ( r_6 ): - When ( r_6 = 1 ): [ r_{36} = r, quad r_{216} = r^2, quad r_M = r^3 ] - ( r_{36} ) must be a multiple of 6 plus 1 (i.e., ( 7, 13, 19, 25, 31 )). - Check each value: - ( r = 7 ): ( r_{216} = 49 ), ( r_M = 343 ) (not valid as ( 49 notequiv 7 pmod{36} )) - ( r = 13 ): ( r_{216} = 169 ), ( r_M = 2197 ) (not valid as ( 169 notequiv 13 pmod{36} )) - Other values exceed 36. - When ( r_6 = 2 ): [ r_{36} = 2r, quad r_{216} = 2r^2, quad r_M = 2r^3 ] - ( r_{36} ) must be a multiple of 6 plus 2 (i.e., ( 8, 14, 20, 26, 32 )). - Check each value: - ( r = 4 ): ( r_{36} = 8 ), ( r_{216} = 32 ), ( r_M = 128 ) (not valid as ( 32 notequiv 8 pmod{36} )) - ( r = 7 ): ( r_{36} = 14 ), ( r_{216} = 98 ), ( r_M = 686 ) (not valid as ( 98 notequiv 14 pmod{36} )) - ( r = 10 ): ( r_{36} = 20 ), ( r_{216} = 200 ), ( r_M = 2000 ) (valid as ( 200 equiv 20 pmod{36} )) - When ( r_6 = 3 ): [ r_{36} = 3r, quad r_{216} = 3r^2, quad r_M = 3r^3 ] - ( r_{36} ) must be a multiple of 6 plus 3 (i.e., ( 9, 15, 21, 27, 33 )). - Check each value: - ( r = 3 ): ( r_{36} = 9 ), ( r_{216} = 27 ), ( r_M = 81 ) (not valid as ( 27 notequiv 9 pmod{36} )) - ( r = 5 ): ( r_{36} = 15 ), ( r_{216} = 75 ), ( r_M = 375 ) (not valid as ( 75 notequiv 15 pmod{36} )) - ( r = 7 ): ( r_{36} = 21 ), ( r_{216} = 147 ), ( r_M = 1029 ) (not valid as ( 147 notequiv 21 pmod{36} )) - When ( r_6 = 4 ): [ r_{36} = 4r, quad r_{216} = 4r^2, quad r_M = 4r^3 ] - ( r_{36} ) must be a multiple of 6 plus 4 (i.e., ( 10, 16, 22, 28, 34 )). - Check each value: - ( r = 4 ): ( r_{36} = 16 ), ( r_{216} = 64 ), ( r_M = 256 ) (not valid as ( 64 notequiv 16 pmod{36} )) - ( r = 7 ): ( r_{36} = 28 ), ( r_{216} = 196 ), ( r_M = 1372 ) (not valid as ( 196 notequiv 28 pmod{36} )) - When ( r_6 = 5 ): [ r_{36} = 5r, quad r_{216} = 5r^2, quad r_M = 5r^3 ] - ( r_{36} ) must be a multiple of 6 plus 5 (i.e., ( 11, 17, 23, 29, 35 )). - Check each value: - ( r = 7 ): ( r_{36} = 35 ), ( r_{216} = 245 ), ( r_M = 1715 ) (not valid as ( 245 notequiv 35 pmod{36} ))5. Conclusion: The only valid sequence is ( (r_6, r_{36}, r_{216}) = (2, 20, 200) ), which implies ( r_M = 2000 ). Therefore, the smallest possible value of ( M ) is ( 2001 ).The final answer is ( boxed{2001} )

🧠:Okay, so I need to find all polynomials f(x) with integer coefficients such that for a fixed integer k, f(n) divides (n!)^k for every positive integer n. Hmm, let me start by breaking down the problem.First, the polynomial has integer coefficients. That's important because it imposes certain restrictions on the structure of the polynomial. For example, if f(n) must divide (n!)^k for all n, then the values of f(n) have to be integers, which they would be if the coefficients are integers. So that's already satisfied.Now, the key condition is that f(n) divides (n!)^k for every positive integer n. That means for each n, (n!)^k / f(n) must be an integer. So f(n) has to be a divisor of (n!)^k for all n. I need to find all such polynomials f(x).Let me think about what kind of polynomials could satisfy this. For example, if f(x) is a product of polynomials of the form x, x-1, x-2, etc., but raised to some powers, maybe that could work? Because n! is the product of numbers from 1 to n, so perhaps if f(n) is a product of terms that divide into that product, raised to the kth power?Wait, but f(x) is a polynomial, so it has to work for all n. That is, no matter what n we plug in, f(n) must divide (n!)^k. So maybe f(x) has to be a product of terms like x^a, (x-1)^b, ..., but adjusted such that when evaluated at any n, the exponents don't exceed k times the exponents in n! ?But polynomials are fixed; their structure doesn't change with n. So for example, if we take f(x) = x^m, then f(n) = n^m. For n^m to divide (n!)^k for all n, we need that the exponent of each prime in n^m is less than or equal to the exponent in (n!)^k.But primes in n! are roughly n/log n by prime number theorem, but maybe that's too vague. Let's think about specific primes. For a prime p ≤ n, the exponent of p in n! is given by the sum floor(n/p) + floor(n/p^2) + floor(n/p^3) + ... So in (n!)^k, that exponent is k times that sum.On the other hand, the exponent of p in n^m is m if p divides n (i.e., if p ≤ n), but actually, if p divides n, then n = p * t for some integer t, so n^m = p^m * t^m. So the exponent of p in n^m is m. So for n^m to divide (n!)^k, we must have m ≤ k * (floor(n/p) + floor(n/p^2) + ...).But this has to hold for all primes p and all n. Wait, but n is variable. For example, take p=2. For a fixed m, when n becomes large, floor(n/2) + floor(n/4) + ... is approximately n - 1. So k times that is about k(n - 1). But m is fixed. So as n increases, k(n - 1) will eventually exceed m. Therefore, for sufficiently large n, m ≤ k(n - 1) is true. So maybe n^m divides (n!)^k for sufficiently large n? Wait, but the problem states that f(n) must divide (n!)^k for every positive integer n, including n=1,2,3,...Wait, let's check n=1. Then f(1) must divide (1!)^k = 1. Therefore, f(1) must be ±1. Similarly, for n=2, f(2) divides (2!)^k = 2^k. For n=3, f(3) divides (6)^k, etc. So the polynomial must satisfy f(n) divides (n!)^k for each n ≥ 1.But if we take f(x) = x^m, then f(1)=1^m=1, which divides 1. f(2)=2^m must divide 2^k, so m ≤ k. f(3)=3^m must divide 6^k. Since 3^m divides 6^k = 2^k * 3^k, so m ≤ k. Similarly, f(4)=4^m divides 24^k. 4^m = 2^{2m}, and 24^k = 2^{3k} * 3^k, so 2m ≤ 3k. If m ≤ (3/2)k. But since m has to be an integer, m ≤ floor(3k/2). But for higher n, say n=5, f(5)=5^m divides 120^k. Since 5^m divides 120^k = (2^3 * 3 * 5)^k, so 5^m divides 5^k, so m ≤ k.Wait, so if f(x)=x^m, then m has to be ≤ k, but also for n=4, 2m ≤ 3k. So for k=1, m must be 0? Because if k=1, then for n=4, 2m ≤ 3*1=3 ⇒ m ≤ 1.5, but m must be integer. So m ≤1. But then for n=2, m ≤1. For n=3, m ≤1. For n=5, m ≤1. So if k=1, m can be at most 1. But let's check if x divides n! for all n ≥1. For n=1, x=1 divides 1. For n=2, 2 divides 2. For n=3, 3 divides 6. For n=4, 4 divides 24. But 4 divides 24, yes. Wait, 4 divides 24. Wait, but if k=1, then (n!)^1 = n!. So x divides n! for all n ≥1? For n=1, x evaluated at 1 is 1, which divides 1. For n=2, 2 divides 2. For n=3, 3 divides 6. For n=4, 4 divides 24. For n=5, 5 divides 120, etc. Wait, but 4 divides 24, which is 4!. 5 divides 120, which is 5!. So yes, x divides n! for all n ≥1. Therefore, f(x)=x is a valid polynomial when k=1. Similarly, if k=1, then f(x)=x^m with m ≤1. But m=1 works. But if m=2, then for n=4, f(4)=16, and 16 divides 24? 24/16=1.5, which is not an integer. So f(x)=x^2 wouldn't work for k=1. Therefore, m must be ≤1. So f(x)=x or constants ±1. Wait, constants ±1 would also work because 1 divides everything, and -1 also divides everything since we're talking about divisibility in integers, where divisors can be positive or negative. So for constants, any constant c with c | (n!)^k for all n. Since (n!)^k is at least 1, so c must be ±1.So for the case of f(x) being a monomial x^m, m must be ≤k and also satisfy other conditions for higher n. Wait, but actually, when n increases, the prime factors of n! include more primes. For example, take f(x) = x(x-1). Then f(n) = n(n-1). For n ≥2, n(n-1) divides n! because n! = n(n-1)(n-2)...1. So n(n-1) is a factor. For n=2, f(2)=2*1=2 divides 2. For n=3, 3*2=6 divides 6. For n=4, 4*3=12 divides 24. Yes, works. For n=1, f(1)=1*0=0. But 0 doesn't divide 1. So f(1) must divide 1, but 0 cannot divide anything. So f(1) must not be zero. Therefore, such a polynomial cannot have a factor of x-1, because when x=1, it would be zero. Wait, but in the previous example, f(x)=x(x-1) would have f(1)=0, which doesn't divide 1!^k=1. So f(x) cannot have any factors that result in f(n)=0 for some n. Because zero does not divide any number. Therefore, f(n) must be non-zero for all positive integers n. So in particular, f(n) must never be zero for any n ≥1. Therefore, the polynomial f(x) cannot have any integer roots. So f(x) must have no positive integer roots. Therefore, the polynomial must be non-vanishing at all positive integers n. That's a crucial point.So f(n) ≠ 0 for all n ≥1, and f(n) divides (n!)^k. So f(n) must be ± a divisor of (n!)^k. But since f(n) is a polynomial, it has to consistently pick divisors in such a way that the polynomial structure is maintained. This seems restrictive.Another thought: perhaps f(x) must be a product of binomial coefficients? Because binomial coefficients are polynomials in n and they divide n! For example, the binomial coefficient C(n, m) = n!/(m!(n-m)!) is an integer, so m!(n-m)! divides n!. But binomial coefficients are polynomials only if m is fixed. For example, C(n, 2) = n(n-1)/2, which is a polynomial. But then C(n, 2) divides n! for all n ≥2. However, when evaluated at n=1, C(1,2)=0, which again would not divide 1!^k. So similar issue as before. So perhaps not binomial coefficients.Alternatively, maybe f(x) can be expressed as a product of terms like (x)_m, the falling factorial, which is x(x-1)...(x-m+1). This is a polynomial of degree m. Then (x)_m divides n! when evaluated at x=n, since (n)_m = n!/(n-m)! which divides n!. So if we take f(x) as a product of falling factorials, each raised to some power, such that the product divides (n!)^k.But again, we need to make sure that when we evaluate f(n), it divides (n!)^k for all n. Let's think. Suppose f(x) is a product of falling factorials (x)_m1^{a1} (x)_m2^{a2} ... (x)_mr^{ar} }, where m1, m2, ..., mr are non-negative integers, and a1, a2, ..., ar are exponents such that the total exponents in the prime factors do not exceed k times those in n!.But how does this work? Let's take a simple case. Suppose f(x) = (x)_m = x(x-1)...(x - m +1). Then f(n) = n(n-1)...(n - m +1) = n! / (n - m)! So for this to divide (n!)^k, we need that n!/(n - m)! divides (n!)^k. Which is equivalent to (n - m)! divides (n!)^{k -1}. For all n ≥ m. Wait, but (n - m)! divides (n!)^{k -1} for all n ≥ m. Let's see for k=1. Then (n - m)! divides (n!)^0 =1. So (n - m)! must divide 1, so (n - m)! =1, which implies n - m ≤1. But n can be arbitrary greater than or equal to m. So unless m = n -1, but n is variable. Therefore, this can't hold. Therefore, even for f(x) = (x)_m, unless m=0, which is trivial, it doesn't work for k=1. So maybe my previous thought is incorrect.Alternatively, if k is large enough. Suppose k ≥2. Then (n - m)! divides (n!)^{k -1}. For fixed m, as n grows, (n!)^{k -1} has a lot of factors, but (n - m)! is much smaller. For example, (n - m)! divides (n!)^{k -1} if for each prime p, the exponent of p in (n - m)! is ≤ (k -1) times the exponent in n!.But the exponent of p in n! is roughly n/p, and in (n - m)! it's roughly (n - m)/p. So (n - m)/p ≤ (k -1)(n/p). So (n - m) ≤ (k -1)n. Which would be true for large n if k -1 ≥1, i.e., k ≥2. But even so, for small n, like n = m, (n - m)! =0! =1, which divides anything. For n = m +1, (n - m)! =1! =1. For n = m +2, 2! divides (n!)^{k -1}, which it does. So maybe for k ≥2, f(x) = (x)_m would work? Wait, but when n is large, we need exponents of primes in (n - m)! to be ≤ (k -1) exponents in n!.But take a prime p dividing (n - m)! but not dividing n!. Wait, but if p ≤ n - m, then p ≤ n, so p divides n! So exponents in (n - m)! are exponents in n! minus something. Wait, the exponent of p in (n - m)! is equal to exponent in n! minus the exponent in the last m numbers: n - m +1 to n. So if p is ≤ n - m, then exponent in (n - m)! is the same as in n! minus the exponents in n - m +1 to n. So it's actually less. Therefore, exponent in (n - m)! ≤ exponent in n!. Therefore, exponent in (n - m)! ≤ (k -1) exponent in n! If k -1 ≥1, i.e., k ≥2. Wait, exponent in (n - m)! ≤ exponent in n! ≤ (k -1) exponent in n! if k -1 ≥1. Therefore, if k ≥2, then exponent in (n - m)! ≤ (k -1) exponent in n!.Wait, but exponent in (n - m)! ≤ exponent in n! So if k -1 ≥1, meaning k ≥2, then exponent in (n - m)! ≤ exponent in n! ≤ (k -1) exponent in n! only if exponent in n! ≥0, which it is. Wait, no. If k -1 ≥1, then (k -1) exponent in n! ≥ exponent in n! So exponent in (n - m)! ≤ exponent in n! ≤ (k -1) exponent in n! Therefore, if k ≥2, then (n - m)! divides (n!)^{k -1}, so f(n) = n!/(n - m)! divides (n!)^k. Therefore, if k ≥2, then f(x) = (x)_m divides (n!)^k for all n ≥m. But for n < m, f(n) =0, which is not allowed. So to avoid f(n)=0 for any n ≥1, we need to have m=0, which gives f(x)=1, but that's trivial. Therefore, unless we take products of such falling factorials with variable m, but adjusted so that f(n) is never zero.Alternatively, perhaps the polynomial can be a product of terms like x, x-1, ..., x - t, each raised to some power, but such that t is less than the minimal n. But since n starts at 1, t would have to be negative, which complicates things. Wait, but x - t for t negative would be x + |t|, so factors like x + c for c positive. Then, when evaluated at n ≥1, x + c becomes n + c, which is a positive integer. Then, to have (n + c) divides (n!)^k for all n ≥1. But n + c divides (n!)^k? For example, take c=1, so x +1. Then f(n) = n +1. Does n +1 divide (n!)^k for all n? For n=1, 2 divides 1^k=1? No. So that doesn't work. Similarly, c=2, n +2 divides (n!)^k. For n=1, 3 divides 1, no. So that approach might not work.Alternatively, maybe the polynomial is a product of factorial-like terms but shifted. Wait, but I can't think of a standard polynomial that would have such properties.Alternatively, let's consider the structure of (n!)^k. For each prime p, the exponent of p in (n!)^k is k*(floor(n/p) + floor(n/p²) + ...). So if f(n) is to divide (n!)^k, then for each prime p, the exponent of p in f(n) must be at most k*(floor(n/p) + floor(n/p²) + ...) for all n.But f(n) is a polynomial, so the exponents of primes in f(n) can be related to the multiplicity of p in f(n). However, f(n) is a polynomial, so the prime factors of f(n) depend on the prime factors of n, n-1, etc., depending on the polynomial's structure.Wait, maybe the polynomial must be of the form ± (x^{a_1})(x - 1)^{a_2}...(x - m)^{a_{m+1}}} where the exponents a_i are such that for each term (x - i)^{a_{i+1}}, when evaluated at n, the exponent of each prime in (n - i) is accounted for in (n!)^k.But this seems vague. Let me try small k and see if I can find a pattern.Case k=1: Find all polynomials f(x) with integer coefficients such that f(n) divides n! for all n ≥1.We already saw that f(x)=x works, as well as constants ±1. What about f(x)=x(x -1)? Then f(n)=n(n -1). For n ≥2, n(n -1) divides n! because n! = n(n -1)(n -2)...1. But for n=1, f(1)=1*0=0, which doesn't divide 1!. So f(x) cannot have a factor of (x -1). Similarly, if we take f(x)=x^m, then for n=1, f(1)=1, which is okay. For n=2, 2^m divides 2. So m ≤1. For n=3, 3^m divides 6. So m ≤1. For n=4, 4^m divides 24. 4^1=4 divides 24, 4^2=16 doesn't. So m=1. So f(x)=x is allowed. Similarly, constants ±1. What about f(x)=x times a constant? For example, f(x)=2x. Then f(1)=2, which must divide 1!, but 2 doesn't divide 1. So constants must be ±1. So for k=1, possible polynomials are ±1 and ±x, but wait, let's check.Wait, if f(x) = -x, then f(n) = -n. Since divisibility allows negative numbers, -n divides n! which is true because n! is divisible by n. So yes, f(x)=-x works as well. Similarly, constants ±1. Are there any other polynomials?Suppose f(x) = x(x - 2). Then f(n) = n(n - 2). For n=1: 1*(-1) = -1 divides 1! =1: yes, since -1 divides 1. For n=2: 2*0=0, which doesn't divide 2! =2. So invalid. Similarly, f(x)=x - 2: f(1)= -1 divides 1, okay; f(2)=0 divides 2, no. So any polynomial with a root at positive integer n ≥1 is invalid.So to avoid zeros, f(x) must not have any roots at positive integers. So for example, f(x) = x^2 +1. Then f(n) =n^2 +1. For n=1: 2 divides 1: No. So nope. Similarly, f(x)=x^2 +x +1. For n=1:3 divides 1: No. So constants and linear terms?Wait, let's consider f(x) = 1. Then it always divides. f(x)=-1, same. f(x)=x: divides n! for all n. f(x)=-x: same. What about f(x)=x^m where m ≤1. So for m=0, constant 1 or -1. For m=1, x or -x. Are there any other polynomials?Suppose f(x) = x times a constant. If the constant is ±1, then it's covered. If the constant is something else, like 2, then f(1)=2, which doesn't divide 1. So constants must be ±1. What about quadratic polynomials? For example, f(x)=x(x -1) +1. Then f(n) =n(n -1) +1 =n² -n +1. For n=1:1 -1 +1=1 divides 1. For n=2:4 -2 +1=3 divides 2!=2: 3 doesn't divide 2. So no. Another quadratic: f(x)=x² +1. As before, n=1:2 doesn't divide 1. So no. How about f(x) =x^m + c, with c=±1? For m=2: x² +1. Fails at n=1. For m=2: x² -1. Then f(n)=n² -1. For n=1:0 divides 1: No. So not useful.So maybe the only possibilities are constants ±1 and linear terms ±x. Let's check if there are more.Suppose f(x) = 2. Then f(n)=2 for all n. For n=2, 2 divides 2: yes. For n=1, 2 divides 1: no. So constants other than ±1 are invalid.What about f(x) = x(x -1) +1 evaluated at n=3: 3*2 +1=7 divides 6: No. So invalid.Another approach: Since f(n) divides n! for all n, and n! grows super-exponentially, while a polynomial f(n) grows polynomially. So perhaps only polynomials of degree 0 or 1 can satisfy this. Because higher-degree polynomials would grow too fast? Wait, but x^m grows polynomially, and n! grows faster, so x^m divides n! for all n if m=1? Wait, no. For example, x^2: n^2 divides n! for all n ≥4. Because for n ≥4, n! has factors n, n-1, ..., 1. n^2 =n *n. Since n ≥4, n! includes 4*3*2*1, and for n ≥4, n! contains two copies of n only if n is a prime power? Wait, not necessarily. Wait, the exponent of a prime p in n! is floor(n/p) + floor(n/p²) + ... So for example, take p=2 and n=4: exponent of 2 in 4! is 3. So 2^3 divides 4!. So 4^2=16 divides 4! if 16 divides 24, which it doesn't. So 4^2 does not divide 4!. Therefore, x^2 doesn't work. But for n=5: 5^2=25 divides 120? No. So x^2 doesn't work. Therefore, indeed, only linear terms or constants.So for k=1, the only polynomials are ±1 and ±x. Wait, but what about products? Like f(x)=x(x -1). But that has a root at x=1, which makes f(1)=0, which is invalid. So products that introduce roots are bad. How about f(x) =x *1? Which is x. But that's covered.Alternatively, maybe a product of x shifted by constants that aren't positive integers? For example, f(x)=x + c, where c is not a negative integer. But then f(n) =n +c. To have n +c divides n! for all n. For example, c=0: f(x)=x, which works. For c= -1: f(x)=x -1, which at n=1 is 0, invalid. For c=1: f(x)=x +1. Then f(n)=n +1. For n=1, 2 divides 1: No. So invalid. For c=2: f(n)=n +2. For n=1, 3 divides 1: No. So that doesn't work. Therefore, only c=0, which is f(x)=x, and constants.Wait, but even f(x)=x + c for c non-integer? Wait, the polynomial must have integer coefficients. So c must be integer. Therefore, shifts by integers. So only possibility is c=0.Thus, for k=1, the only solutions are constants ±1 and linear polynomials ±x.Now, generalizing to arbitrary k. For a general k, we need f(n) divides (n!)^k for all n. So the exponents of primes in f(n) must be at most k times their exponents in n!.So perhaps the polynomial can be a product of terms of the form (x)_m, each raised to a power such that the total exponents don't exceed k. Wait, but how?Alternatively, perhaps f(x) must be of the form ± (x)^{a_1} (x -1)^{a_2} ... (x - t)^{a_{t+1}}} where for each prime p, the sum of exponents a_i corresponding to multiples of p in the terms (x - i) is at most k * (floor(n/p) + floor(n/p²) + ...). Wait, this seems complicated.Alternatively, note that (n!)^k is a multiplicative function (though not completely multiplicative), and f(n) must divide it for all n. So perhaps f(n) must be a multiplicative function as well, and given that f is a polynomial, it must be a product of terms of the form x, x-1, etc., each raised to some power.Wait, another angle. Suppose f(x) is such a polynomial. Then for each prime p, the number of times p divides f(n) must be ≤ k*(number of times p divides n!). Since f(n) is a polynomial, the exponent of p in f(n) can be written as a function of n. For example, if f(x) = x^m, then the exponent of p in f(n) is m * exponent of p in n. For this to be ≤ k * exponent of p in n! for all n.But exponent of p in n! is roughly n/p + n/p² + ..., which is greater than or equal to (n -1)/(p -1). So we need m * exponent of p in n ≤ k*(n/p + n/p² + ...). But exponent of p in n is either 0 or 1, unless n is a multiple of p², etc. Wait, no. For a fixed prime p, exponent of p in n is the largest integer e such that p^e divides n. So for example, if n is p, exponent is 1; if n is p², exponent is 2, etc.But since the polynomial f(x) has to work for all n, including n being any multiple of p^e. So if f(x) =x^m, then for n=p^e, the exponent of p in f(n)= (p^e)^m is m*e. The exponent of p in (n!)^k is k*(floor(p^e /p) + floor(p^e /p²) + ... + floor(p^e /p^e)) =k*(p^{e -1} + p^{e -2} + ... +1)=k*( (p^e -1)/(p -1) )).So we need m*e ≤ k*( (p^e -1)/(p -1) ) for all primes p and integers e ≥1.Let’s analyze this inequality. For fixed p and e, how does m relate to k?For example, take e=1. Then m*1 ≤k*( (p -1)/(p -1) )=k. So m ≤k.For e=2, m*2 ≤k*(p^2 -1)/(p -1)=k*(p +1). So 2m ≤k(p +1). For this to hold for all primes p.Since p can be any prime, the smallest prime is 2. Then 2m ≤k(2 +1)=3k ⇒ m ≤ (3/2)k. But since m must be an integer, m ≤ floor(3k/2). But for the next prime p=3, e=2: 2m ≤k*(3 +1)=4k ⇒ m ≤2k. Which is less restrictive if k≥1. Wait, but for p=2, e=2 gives m ≤ (3/2)k, and for p=3, e=2 gives m ≤2k. So the strictest condition is p=2, e=2. Similarly, for higher e and p=2:For e=3, m*3 ≤k*(2^3 -1)/(2 -1)=k*7 ⇒ m ≤ (7/3)k ≈2.333k.But we need this to hold for all e and all p. The minimal p is 2, and as e increases, the right-hand side grows exponentially. For example, p=2, e: m*e ≤k*(2^e -1). For large e, the RHS is roughly k*2^e, so m*e ≤k*2^e ⇒ m ≤k*2^e /e. As e increases, 2^e /e increases, so the bound on m becomes looser. Therefore, the most restrictive cases are for small e and p=2.For e=1: m ≤k.For e=2: m ≤ (3/2)k.For e=3: m ≤ (7/3)k ≈2.33k.But m must satisfy all these simultaneously. However, m is a fixed integer. So for example, if k=1:- e=1: m ≤1- e=2: m ≤3/2 ⇒m ≤1- e=3: m ≤7/3 ⇒m ≤2, but since m must also satisfy e=1 and e=2, m=1.If k=2:- e=1: m ≤2- e=2: m ≤3- e=3: m ≤14/3≈4.66 ⇒m ≤4.But m has to satisfy all, so m ≤2 (from e=1). Wait, but e=2 allows m=3, but e=1 requires m ≤2. So m is limited by the most restrictive condition, which is e=1: m ≤k.Wait, but if k=2, then for e=2, p=2: 2m ≤3*2=6 ⇒m ≤3. But since m ≤k=2, the actual limit is m=2. But if we take m=2, check for p=2, e=3: 3*2 ≤2*(2^3 -1)/(2 -1)=2*7=14 ⇒6 ≤14: true. So m=2 works for k=2.Wait, so the most restrictive condition is m ≤k. Because for e=1, we have m ≤k, and for higher e and p=2, the allowed m increases, but since m must satisfy all primes p and exponents e, the only universal bound is m ≤k. Because for any prime p and e=1, m ≤k. Therefore, m must be ≤k.Similarly, for other primes p >2, e=1 would require m ≤k, and higher e would allow larger m, but since p=2 with e=1 is the most restrictive, m must be ≤k.Therefore, if f(x) =x^m, then m ≤k. Then f(n)=n^m divides (n!)^k for all n. Wait, but earlier with k=1 and m=1, f(n)=n divides n! for all n. For k=2, m=2: f(n)=n². Does n² divide (n!)^2 for all n? For n=1:1 divides 1. For n=2:4 divides 4. For n=3:9 divides 36. Yes. For n=4:16 divides 576. Yes. For n=5:25 divides 14400. Yes. But wait, does n^m divide (n!)^k for m ≤k?Wait, for a prime p, the exponent of p in n^m is m*e where e is the exponent of p in n. The exponent of p in (n!)^k is k*(sum_{i=1}^∞ floor(n/p^i)). So we need m*e ≤k*(sum floor(n/p^i)).But when n is a prime p itself, then exponent of p in n is 1, so m*1 ≤k*(1 +0 +0 +...)=k. Hence, m ≤k. For n=p^2, exponent of p in n is 2, so m*2 ≤k*(floor(p^2/p) + floor(p^2/p²) + ...) =k*(p +1 +0 +...). For p=2, this is k*(2 +1)=3k. So m*2 ≤3k ⇒m ≤1.5k. But since m must be ≤k from the previous condition, this is automatically satisfied.Similarly, for n=p^e, exponent of p in n is e. The exponent in (n!)^k is k*(floor(p^e/p) + floor(p^e/p²)+...+floor(p^e/p^e))=k*(p^{e-1} +p^{e-2}+...+1). So the inequality is m*e ≤k*(p^{e} -1)/(p -1). Since p ≥2, the RHS is at least k*(2^{e} -1). So m*e ≤k*(2^{e} -1). For e=1, m ≤k. For e=2, m ≤k*(3)/2. For e=3, m ≤k*(7)/3, etc. Since m must be an integer ≤k, the only restriction that applies universally is m ≤k. Therefore, as long as m ≤k, n^m divides (n!)^k for all n. Because even though for higher e, the required m could be higher, since m is fixed ≤k, and the RHS increases exponentially with e, the inequality will hold for large e.Wait, let's verify this for m=k. Take n=p, prime. Then exponent of p in n^k is k*1. Exponent of p in (n!)^k is k*1. So equality holds. For n=p^2, exponent of p in n^k is k*2. Exponent in (n!)^k is k*(p +1). For p=2, this is k*3. So k*2 ≤k*3 ⇒2 ≤3, which is true. For p=3, exponent in (n!)^k is k*(3 +1)=4k, and k*2 ≤4k ⇒2 ≤4, true. Similarly, for higher e, the exponent in n^k is k*e, and in (n!)^k is k*(p^{e-1} +p^{e-2} +...+1). Since p ≥2, p^{e-1} +...+1 = (p^e -1)/(p -1) ≥ (2^e -1). For e ≥1, 2^e -1 ≥e. Therefore, k*e ≤k*(2^e -1) ⇒e ≤2^e -1, which is true for e ≥1. Therefore, m=k satisfies the inequality for all primes p and exponents e. Therefore, f(x)=x^k is a valid polynomial.Similarly, if m <k, then it's even better. Therefore, any monomial x^m with m ≤k would work. But also, products of such monomials with other terms.Wait, but if we take f(x)=x^m * g(x), where g(x) is another polynomial with integer coefficients, then f(n) divides (n!)^k only if both x^m divides (n!)^k and g(n) divides (n!)^k. So if we can multiply together several polynomials, each of which divides (n!)^k, then their product would also divide (n!)^k.But we have to ensure that the product doesn't introduce any roots (i.e., f(n) ≠0 for any n≥1). So for example, if we take f(x)=x^a * (x +1)^b, then for n≥1, f(n) =n^a (n +1)^b. We need this to divide (n!)^k. But how?For example, take a=1, b=1, k=2. Then f(n)=n(n+1). For n=1:1*2=2 divides (1!)^2=1? No. So invalid. Therefore, even if each term individually could divide (n!)^k for n≥1, their product may not.Alternatively, suppose f(x)=x^a where a ≤k. Then f(n) divides (n!)^k. Similarly, if we take f(x)=(x)_m =x(x-1)...(x -m +1). Then, as before, for k ≥m, perhaps this works? Wait, no, because when evaluated at n <m, f(n)=0, which is invalid. So unless m=0, which is trivial.Wait, but maybe if we take f(x) as a product of terms like x^a_1, (x -1)^a_2, ..., (x -t)^a_{t+1} where t is some integer and each a_i ≤k. But even then, for example, f(x)=x^a*(x -1)^b. For n=1, f(1)=1^a *0^b. If b>0, then f(1)=0, invalid. So all exponents on terms like (x -i) must be zero, i.e., such terms cannot be present. Therefore, the polynomial cannot have any factors that vanish at positive integers n≥1. Therefore, the polynomial must be of the form ±x^m, where m ≤k, or constants ±1. Because any other factor would introduce a root at some integer n≥1, leading to f(n)=0, which is invalid.Wait, let's test this hypothesis. Suppose f(x)=x^m where m ≤k. Then f(n)=n^m divides (n!)^k. As we saw earlier, this holds. Similarly, constants ±1 trivially divide anything. Are there any other polynomials that do not vanish at any n≥1 and have integer coefficients?For example, f(x)=x^m +1. Then f(n)=n^m +1. For n=1:1 +1=2 divides (1!)^k=1? No. So invalid.Another example: f(x)=x^2 +x +1. For n=1:3 divides 1? No.Alternatively, f(x)=x^2 -x +1. For n=1:1 -1 +1=1 divides 1. For n=2:4 -2 +1=3 divides 2^k. So 3 divides 2^k only if k=0, but k is fixed. So invalid.Alternatively, f(x)=x(x +1). Then for n=1:1*2=2 divides 1? No.Alternatively, f(x)=x^m where m ≤k and multiplied by a constant ±1. So f(x)=±x^m. But constants other than ±1 would fail, as we saw. So constants must be ±1. Therefore, the only possibilities are f(x)=±x^m where 0 ≤m ≤k (with m being an integer), or f(x)=±1.Wait, but wait. What about products of different monomials? For example, f(x)=x^a*(x +1)^b. Then for n=1:1^a *2^b divides 1. So 2^b must divide 1, which implies b=0. So only possible if b=0. Similarly, any factor that introduces a term like (x +c) would, when evaluated at n=1, become (1 +c), which must divide 1. Therefore, 1 +c must be ±1, so c=0 or c=-2. But c=0 gives x, which we already have. c=-2 gives x -2. Then f(x)=x^a*(x -2)^b. For n=2:2^a*0^b=0, which is invalid. So again, any such factor introduces a root at some n, unless its exponent is zero.Therefore, the only polynomials that do not vanish at any positive integer n and have integer coefficients are the constants ±1 and ±x^m for m ≥0, but with the condition that ±x^m must divide (n!)^k for all n≥1. As we saw earlier, x^m divides (n!)^k if and only if m ≤k. Therefore, the polynomials are precisely the constants ±1 and ±x^m for integers m with 0 ≤m ≤k. But wait, x^0=1, which is covered by the constants. So essentially, the polynomials are all integer constants ±1 and ±x^m for 1 ≤m ≤k.But wait, can we have products of different monomials? For example, f(x)=x^a * (x +1)^b. As above, this would require that when evaluated at any n≥1, the product divides (n!)^k. But as shown earlier, introducing any factor beyond x^m and constants leads to divisions that fail for small n. Therefore, the only solutions are ±x^m where 0 ≤m ≤k, with m integer.Wait, let me confirm with k=2. The possible polynomials are ±1, ±x, ±x^2. Check if these work.For f(x)=x^2: f(n)=n^2. Does n^2 divide (n!)^2 for all n? Yes, because (n!)^2 is divisible by n^2. For example, n=4: 16 divides 576=24^2. Yes. For n=5:25 divides 14400=120^2. Yes. For prime n=p: p^2 divides (p!)^2. Since p! includes p and numbers less than p, so p appears once in p!, so (p!)^2 has p^2, which divides it. Yes. So indeed, x^2 works for k=2.What about a product like x*(x -1). For k=2. Then f(n)=n(n -1). For n=1:0 divides 1. No. So invalid. So products introducing roots are invalid.Another example: f(x)=x^2 +x +1. For n=1:3 divides 1. No.Therefore, the conclusion seems to be that the only polynomials f(x) with integer coefficients such that f(n) divides (n!)^k for all positive integers n are the constants ±1 and the monomials ±x^m where m is an integer with 0 ≤m ≤k. But since x^0=1, which is included in the constants, the distinct polynomials are ±x^m for 0 ≤m ≤k. However, constants ±1 correspond to m=0.But wait, constants ±1 are already included when m=0. So the answer is all polynomials of the form ±x^m where m is an integer between 0 and k inclusive.But wait, the problem says "polynomials with integer coefficients". So indeed, these are all such polynomials. But to confirm, let's see:- For m=0: ±1, which work.- For 1 ≤m ≤k: ±x^m. These work because, as shown, x^m divides (n!)^k for all n when m ≤k.- Any other polynomial would either have a root at some n (making f(n)=0, which is invalid), or include terms that cause f(n) to not divide (n!)^k for some n.For example, suppose f(x)=x^{k} +x^{k-1}. Then for n=1:1 +1=2 divides 1. No. So invalid.Another example: f(x)=x^{k} +1. For n=1:1 +1=2 divides 1. No. Invalid.Therefore, the only valid polynomials are indeed of the form ±x^m with 0 ≤m ≤k.Hence, the solution is all constant polynomials ±1 and all monomials ±x^m with integer exponents m satisfying 0 ≤m ≤k.But the problem states "polynomials with integer coefficients". So exponents m must be non-negative integers, since otherwise it's not a polynomial. So m is integer between 0 and k.Thus, the final answer is that all such polynomials are the ones of the form ±x^m where m is an integer with 0 ≤m ≤k, including the constants ±1 (when m=0).Final AnswerThe polynomials are precisely those of the form boxed{pm x^m} where ( m ) is an integer with ( 0 leq m leq k ).

✅:To determine all polynomials ( f(x) ) with integer coefficients such that ( f(n) ) divides ( (n!)^k ) for every positive integer ( n ), we need to consider the structure of ( (n!)^k ) and the divisibility requirements for ( f(n) ).1. Key Observations: - The polynomial ( f(x) ) must have integer coefficients. - ( f(n) ) must divide ( (n!)^k ) for all positive integers ( n ). - ( f(n) ) cannot be zero for any positive integer ( n ), hence ( f(x) ) cannot have any positive integer roots.2. Prime Factor Analysis: - For each prime ( p ), the exponent of ( p ) in ( (n!)^k ) is ( k left( leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + cdots right) ). - The polynomial ( f(x) ) must ensure that the exponents of all primes in ( f(n) ) do not exceed those in ( (n!)^k ).3. Monomial Consideration: - Consider ( f(x) = pm x^m ). For ( f(n) = n^m ) to divide ( (n!)^k ), the exponent of each prime in ( n^m ) must be at most ( k ) times the exponent in ( n! ). - For a prime ( p ), the exponent in ( n^m ) is ( m cdot e ) where ( e ) is the exponent of ( p ) in ( n ). This must be at most ( k cdot left( leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + cdots right) ).4. Verification: - For ( f(x) = x^m ), the most restrictive condition occurs when ( n ) is a prime ( p ). Here, ( m leq k ). - For higher powers and primes, the condition ( m leq k ) is sufficient because the right-hand side of the inequality grows exponentially with ( e ), ensuring ( m cdot e ) does not exceed ( k cdot text{(exponent in } (n!)^k) ).5. General Form: - Polynomials must be non-vanishing at all positive integers. This restricts ( f(x) ) to monomials and constants. - Constants must be ( pm 1 ) since they divide any integer. - Monomials ( pm x^m ) with ( 0 leq m leq k ) satisfy the divisibility condition for all ( n ).Thus, the polynomials ( f(x) ) with integer coefficients such that ( f(n) ) divides ( (n!)^k ) for every positive integer ( n ) are precisely those of the form ( pm x^m ) where ( m ) is an integer with ( 0 leq m leq k ).[boxed{pm x^m}]where ( 0 leq m leq k ).

🔑:1. Initial Claim and Verification: We claim that the only solutions are ( f(x) = pm x^a ) for ( a = 0, 1, ldots, k ). To verify that these work, note that ( n mid n! ) for all positive integers ( n ). Therefore, ( n^a mid (n!)^k ) for ( a leq k ).2. Constant Polynomial Case: If ( f ) is a constant polynomial, it must be ( pm 1 ). This is because any constant ( c ) must divide ( (n!)^k ) for all ( n ). The only integers that divide all factorials are ( pm 1 ).3. Non-constant Polynomial Case: Assume ( f ) is not constant. We claim that only prime factors of ( n ) may divide ( f(n) ). 4. Modulo Argument: Note that for integer polynomials, ( a equiv b pmod{c} implies f(a) equiv f(b) pmod{c} ). Let ( d ) be equivalent to ( n ) modulo some prime ( p ) such that ( 1 leq d < p ). Thus, ( f(n) equiv f(d) pmod{p} ). Since ( f(d) mid (d!)^k ), it is not divisible by ( p ) (as ( d < p )), so ( f(n) ) is not divisible by any prime ( p ) that does not divide ( n ).5. Prime Evaluation: Now, consider ( f(p) ) for a prime ( p ). We have ( f(p) = pm p^l ) for some ( 0 leq l leq k ). Since ( f ) is non-constant, ( l geq 1 ) for infinitely many primes ( p ). 6. Constant Term Analysis: Since ( p ) divides all the non-constant terms of ( f(p) ), the constant term of ( f ) must be divisible by infinitely many primes. Therefore, the constant term of ( f ) is zero.7. Inductive Argument: Since the constant term is zero, we can factor out an ( x ) from ( f(x) ). By induction, we conclude that ( f(x) = pm x^a ) for some ( a ).8. Conclusion: Thus, the only polynomials ( f(x) ) with integer coefficients such that ( f(n) ) divides ( (n!)^k ) for every positive integer ( n ) are ( f(x) = pm x^a ) for ( a = 0, 1, ldots, k ).The final answer is ( boxed{ f(x) = pm x^a } ) for ( a = 0, 1, ldots, k ).

🧠:Okay, let's see. So the problem is about 12 dentists in a clinic, and there are 29 fifth-year students who go to the clinic. Each dentist has to have at least 2 students. We need to find the maximum number of students that can attend a single dentist. Hmm, right. So, essentially, we want to distribute 29 students among 12 dentists with each dentist getting at least 2 students, and we need to figure out the largest number of students that any one dentist can have under these conditions.Alright, let's break this down. If there are 12 dentists and each must have at least 2 students, then the minimum number of students required just to satisfy that condition is 12 times 2, which is 24. But we have 29 students in total. So, 29 minus 24 is 5. That means after assigning 2 students to each dentist, there are 5 students left over. These extra students can be distributed to the dentists, and we want to know how many can go to a single dentist to maximize that number.So, if we have 5 extra students, and we want to give as many as possible to one dentist, we would just add those 5 to one of the dentists who already has 2. So, 2 + 5 = 7. Therefore, the maximum number of students that can attend a single dentist is 7. Wait, but let me think again. Is that right?Alternatively, maybe I need to distribute the remaining 5 students one by one to different dentists. But no, because the problem is asking for the greatest number of students that can attend a single dentist. So, to maximize the number at one dentist, we should allocate all the extra students to that one dentist. So, starting with 2 per dentist, then adding all 5 to one, making that dentist have 7. Then the rest have 2 each. Let me check: 1 dentist has 7, and the remaining 11 dentists have 2 each. Total students would be 7 + 11*2 = 7 + 22 = 29, which matches the total. So that works.But wait, let me make sure there isn't a different way to distribute them where a dentist could have more. Suppose instead, is there a way to have a dentist with more than 7? Let's see. If we tried to give 8 students to one dentist, then the remaining 11 dentists would have to have at least 2 each. So, total students would be 8 + 11*2 = 8 + 22 = 30. But we only have 29 students. 30 is more than 29, so that's impossible. Therefore, 8 is too much. So 7 must be the maximum.Alternatively, maybe another distribution. Suppose we give one dentist 7, another dentist 3, and the rest 2. Then total would be 7 + 3 + 10*2 = 7 + 3 + 20 = 30, which again is too much. So that's not possible. So even if we tried to spread the extra students, but we only have 5 extra. Wait, the initial allocation is 2 each, 24 total. Then adding 5 more, so each additional student is an extra. So, those 5 can be given to any dentists. To maximize one dentist, we give all 5 to one dentist. Thus, 2 + 5 = 7. So 7 is the answer.Wait, but perhaps another approach. Let's use the pigeonhole principle. The idea is to minimize the maximum number per dentist, but here we want to maximize it given the constraints. So, in order to find the maximum possible number at one dentist, we need to minimize the number of students allocated to the other dentists as much as possible. Since each dentist must have at least 2, the minimal total for the other 11 dentists is 11*2 = 22. Therefore, the maximum number that can be allocated to the remaining dentist is 29 - 22 = 7. So that's the same answer as before.So, both approaches give 7. Therefore, the answer should be 7.Wait, but just to double-check, maybe there's a trick here. Let me think again. Suppose we have 12 dentists, each with at least 2 students. So the minimum total is 24. The actual total is 29, which is 5 more than 24. These 5 have to be distributed, and each additional student can go to any dentist. To maximize the number at a single dentist, we assign all 5 to one dentist. Hence, 2 + 5 = 7. So that's correct.Another way: If each dentist starts with 2 students, we have 24 students placed. Then, the remaining 5 can be distributed as "extras". The maximum number of extras any one dentist can have is 5, so the maximum total for a dentist is 2 + 5 = 7.Alternatively, if we tried to give 6 extras to a dentist, that would require 6 extra students, but we only have 5. So 6 is impossible. Hence, 7 is the maximum.Therefore, the answer is 7. So the greatest number of students that can attend a single dentist is 7. Seems straightforward.But wait, let me check once more. Suppose I tried to assign 7 to one dentist, and then the rest 29 -7 = 22 students to the remaining 11 dentists. Each of those 11 dentists would need to have at least 2. 22 divided by 11 is exactly 2. So each of the other dentists would have exactly 2. That works. So total is 7 + 11*2 = 7 + 22 = 29. Perfect. So yes, that works. Therefore, it's possible. If you tried to assign 8, then 29 -8 =21, which needs to be divided among 11 dentists. 21 divided by 11 is about 1.9, but since each dentist needs at least 2, that's not possible. Because 11*2 =22, which would require 22 students, but we only have 21. So impossible. Hence, 8 is too many.Therefore, 7 is indeed the maximum.Final AnswerThe greatest number of students that can attend a single dentist is boxed{7}.

✅:To determine the greatest number of students that can attend a single dentist, we start by noting that there are 12 dentists and each must serve at least 2 students. The minimum number of students required to satisfy this condition is (12 times 2 = 24). Given there are 29 students in total, we subtract the minimum required students from the total number of students to find the remaining students that can be distributed: (29 - 24 = 5). To maximize the number of students at a single dentist, we allocate all 5 remaining students to one dentist. This dentist would then have (2 + 5 = 7) students. We verify this by checking the total number of students: one dentist has 7 students, and the remaining 11 dentists each have 2 students. This gives (7 + 11 times 2 = 7 + 22 = 29), which matches the total number of students. Attempting to allocate more than 5 extra students to a single dentist would exceed the total number of students available. Therefore, the maximum number of students that can attend a single dentist is (boxed{7}).

🔑:1. We start by noting that there are 12 dentists and 29 students. Each dentist serves at least 2 students. Therefore, the minimum number of students served by all dentists is: [ 12 times 2 = 24 text{ students} ]2. Since there are 29 students in total, the number of students left after each dentist serves at least 2 students is: [ 29 - 24 = 5 text{ students} ]3. To maximize the number of students that a single dentist can serve, we need to assign these 5 remaining students to one of the dentists who already serves 2 students. This means that one dentist will serve: [ 2 + 5 = 7 text{ students} ]4. Therefore, the greatest number of students that can attend a single dentist is 7.The final answer is (boxed{7}).

🧠:Okay, so I need to find the greatest possible area of triangle PIE where the circle ω is inscribed in the unit square PLUM, and points I and E are on ω such that U, I, and E are collinear. Hmm, let's start by visualizing the problem.First, the unit square PLUM. Let me clarify the order of the vertices. Since it's a square, the vertices are probably labeled in order, like P, L, U, M. Let me assume that PLUM is a square with sides of length 1. So, coordinates might help here. Let's place the square on a coordinate system. Let me assign coordinates to each vertex to make it easier.If PLUM is a unit square, let's set point P at (0,0), L at (1,0), U at (1,1), and M at (0,1). That way, the square spans from (0,0) to (1,1). The inscribed circle ω would then be tangent to all four sides of the square. The center of the inscribed circle in a square is at the center of the square, which is (0.5, 0.5), and the radius is half the side length, so 0.5. Therefore, the equation of the circle ω is (x - 0.5)^2 + (y - 0.5)^2 = 0.25.Points I and E are on this circle, and they are colinear with U (1,1). So, the line passing through U, I, and E must intersect the circle at points I and E. Since U is a vertex of the square, which is outside the circle (since the circle has radius 0.5 and is centered at (0.5,0.5), so the distance from U to the center is sqrt((0.5)^2 + (0.5)^2) = sqrt(0.5) ≈ 0.707, which is greater than the radius 0.5), so the line through U and any two points on the circle must pass through U and intersect the circle at two points I and E.We need to find the maximum area of triangle PIE, where P is (0,0), I and E are points on the circle, and U, I, E are colinear.Let me recall that the area of a triangle given three points can be found using the determinant formula. If P is (0,0), and I is (x1, y1), E is (x2, y2), then the area is 0.5 * |x1*y2 - x2*y1|. Alternatively, using vectors or coordinates.But since I and E lie on the circle and are colinear with U, maybe parametrize the line through U and find points I and E where it intersects the circle. Then express the coordinates of I and E in terms of a parameter, and compute the area of triangle PIE as a function of that parameter, then maximize it.Let's start by parametrizing the line through U (1,1). Let's consider lines passing through U (1,1) with some slope m. However, parametrizing by slope might be tricky because vertical lines would have undefined slope. Alternatively, use a parametric line equation.Let me use a parameter t to represent points along the line through U. Let’s set the direction vector of the line. Let’s suppose the line has direction vector (a, b), so any point on the line can be written as (1 + a*t, 1 + b*t). Then, points I and E correspond to two values of t where this line intersects the circle ω.But since the line passes through U (1,1), and intersects the circle centered at (0.5,0.5) with radius 0.5, we can substitute the parametric equations into the circle's equation and solve for t.Alternatively, maybe use a different parameter. Let's consider the angle θ that the line makes with the horizontal axis. Then, the slope m = tan(θ), and the line equation is y - 1 = m(x - 1). Then, substituting into the circle's equation.But maybe that's more complicated. Let me try the parametric approach.Let me denote the parametric line through U (1,1) as:x = 1 + t*cos(θ)y = 1 + t*sin(θ)Where θ is the angle the line makes with the positive x-axis, and t is the parameter. Then, substituting into the circle equation:(x - 0.5)^2 + (y - 0.5)^2 = 0.25Substitute x and y:(1 + t*cosθ - 0.5)^2 + (1 + t*sinθ - 0.5)^2 = 0.25Simplify:(0.5 + t*cosθ)^2 + (0.5 + t*sinθ)^2 = 0.25Expanding both terms:(0.25 + t*cosθ*1 + t²*cos²θ) + (0.25 + t*sinθ*1 + t²*sin²θ) = 0.25Combine like terms:0.25 + 0.25 + t*(cosθ + sinθ) + t²*(cos²θ + sin²θ) = 0.25Simplify:0.5 + t*(cosθ + sinθ) + t²*(1) = 0.25Bring all terms to left side:t² + t*(cosθ + sinθ) + 0.5 - 0.25 = 0Which simplifies to:t² + t*(cosθ + sinθ) + 0.25 = 0So, quadratic equation in t:t² + t*(cosθ + sinθ) + 0.25 = 0To solve for t, use quadratic formula:t = [ - (cosθ + sinθ) ± sqrt( (cosθ + sinθ)^2 - 4*1*0.25 ) ] / 2Compute discriminant:D = (cosθ + sinθ)^2 - 1Expand (cosθ + sinθ)^2:cos²θ + 2 sinθ cosθ + sin²θ = 1 + sin(2θ)So, D = 1 + sin(2θ) - 1 = sin(2θ)Thus, t = [ - (cosθ + sinθ) ± sqrt(sin(2θ)) ] / 2Wait, sqrt(sin(2θ)) must be real, so sin(2θ) ≥ 0. Therefore, θ must be in the range where 2θ is in [0, π], so θ in [0, π/2], considering that θ is the angle from the x-axis to the line. But since the line passes through U (1,1), perhaps θ can vary in other ranges, but the discriminant requires sin(2θ) ≥ 0. So θ between 0 and π/2 or between π and 3π/2. But since the line is passing through (1,1), which is in the first quadrant, perhaps θ is between -π/4 (pointing towards the square) to 3π/4. Hmm, maybe. Let's not get bogged down here yet.Assuming sin(2θ) ≥ 0, then the solutions are real. Therefore, t exists only when sin(2θ) ≥ 0, which would be for θ ∈ [0, π/2] ∪ [π, 3π/2], but since our square is in the first quadrant, and the line passes through (1,1), the angles θ that make sense for intersecting the circle inside the square would probably be between, say, θ = 0 (horizontal line) to θ = π/2 (vertical line). Wait, but a horizontal line through U (1,1) would be y=1, which would intersect the circle at x=1. Let's check.Wait, circle is centered at (0.5,0.5) with radius 0.5. So, the horizontal line y=1 would intersect the circle?The circle equation is (x - 0.5)^2 + (y - 0.5)^2 = 0.25. If y=1, then:(x - 0.5)^2 + (0.5)^2 = 0.25(x - 0.5)^2 = 0.25 - 0.25 = 0Therefore, x = 0.5. So, the horizontal line y=1 is tangent to the circle at (0.5, 1). But the line y=1 passes through U (1,1), which is outside the circle, so it only touches the circle at one point. Therefore, in this case, points I and E would coincide at (0.5,1). But the problem states that U, I, E are colinear, so maybe in this case, it's a tangent line, so only one intersection point. But the problem says "points I and E lie on ω such that U, I, and E are collinear". So, perhaps in the case of tangent, they just have I=E? But the problem probably requires I and E to be distinct points, so maybe the line must be a secant, intersecting the circle at two points. Therefore, we need sin(2θ) > 0, so θ in (0, π/2). Because when θ=0, the line y=1 is tangent, as above, and θ=π/2 would be the vertical line x=1, which is also tangent at (1,0.5). So, for lines between θ in (0, π/2), the lines through U (1,1) intersect the circle at two distinct points I and E.Therefore, θ ∈ (0, π/2), which gives sin(2θ) > 0, so sqrt(sin(2θ)) is real, and we have two solutions for t.Thus, t = [ - (cosθ + sinθ) ± sqrt(sin(2θ)) ] / 2But since the line is parametrized as starting at U (1,1), and moving in direction θ, the parameter t would be the distance from U to the point on the line. However, since the circle is of radius 0.5, the points I and E are located at some t from U towards the circle.Wait, but the solutions for t would correspond to the parameter values where the line intersects the circle. Let's note that in the parametrization x = 1 + t*cosθ, y = 1 + t*sinθ, t=0 corresponds to U (1,1). Then, moving along the line in the direction of θ, t increases, and in the opposite direction, t decreases.But since the circle is centered at (0.5,0.5), the line passes through U (1,1) and intersects the circle at two points I and E. Depending on the direction θ, these points could be on either side of U, but since U is outside the circle, the line must pass through U and intersect the circle at two points. Therefore, the two solutions for t will be negative (since the direction from U towards the circle is opposite to the direction of θ). Wait, maybe not necessarily. Let's compute the roots.Given the quadratic equation t² + t*(cosθ + sinθ) + 0.25 = 0, the product of the roots is 0.25 (from c/a), so both roots are positive or both negative. Since the coefficient of t is (cosθ + sinθ). If θ is in (0, π/2), cosθ + sinθ is positive, since both cos and sin are positive. Therefore, the sum of the roots is - (cosθ + sinθ), which is negative, and the product is positive (0.25). Therefore, both roots are negative. Therefore, both points I and E are in the direction opposite to θ from U. So, moving from U towards the circle in the direction opposite to θ. Therefore, the parameter t is negative for both points.Therefore, the coordinates of I and E are:x = 1 + t*cosθy = 1 + t*sinθwith t negative.Alternatively, perhaps we can parametrize the line in terms of a different parameter. Let's instead parametrize the line from U towards the circle. Let me consider a line passing through U (1,1) with direction vector (cosθ, sinθ), but pointing towards the circle. However, given the previous result, both intersection points are in the direction opposite to θ. So maybe θ is measured from the line connecting U to the center of the circle? Hmm, maybe complicating.Alternatively, perhaps parametrize the line as starting at U and moving towards the circle. Let's define a parameter s such that s=0 is at U, and s increases towards the circle. Then, the parametric equations would be:x = 1 - s*cosθy = 1 - s*sinθWhere θ is the angle from the positive x-axis to the line from U towards the circle. Then, substituting into the circle equation:(x - 0.5)^2 + (y - 0.5)^2 = 0.25(1 - s*cosθ - 0.5)^2 + (1 - s*sinθ - 0.5)^2 = 0.25Which simplifies to:(0.5 - s*cosθ)^2 + (0.5 - s*sinθ)^2 = 0.25Expanding:0.25 - s*cosθ + s²*cos²θ + 0.25 - s*sinθ + s²*sin²θ = 0.25Combine terms:0.5 - s*(cosθ + sinθ) + s²*(cos²θ + sin²θ) = 0.25Which simplifies to:s² - s*(cosθ + sinθ) + 0.25 = 0Same quadratic equation as before, but with s instead of t. Therefore, s = [ (cosθ + sinθ) ± sqrt( (cosθ + sinθ)^2 - 1 ) ] / 2Wait, no. Wait, let's redo the discriminant:s² - s*(cosθ + sinθ) + 0.25 = 0Discriminant D = (cosθ + sinθ)^2 - 4*1*0.25 = (cos²θ + 2sinθcosθ + sin²θ) - 1 = (1 + sin2θ) - 1 = sin2θTherefore, s = [ (cosθ + sinθ) ± sqrt(sin2θ) ] / 2Since sin2θ > 0 (as θ ∈ (0, π/2)), sqrt(sin2θ) is real.Therefore, the two values of s correspond to the two intersection points I and E. Since we have two points, we can denote them as s1 and s2, where s1 and s2 are the two roots.Therefore, the coordinates of I and E are:For s1: x1 = 1 - s1*cosθ, y1 = 1 - s1*sinθFor s2: x2 = 1 - s2*cosθ, y2 = 1 - s2*sinθTherefore, points I and E are (x1, y1) and (x2, y2).Now, the area of triangle PIE, with P at (0,0), can be computed using the determinant formula:Area = 0.5 * |x1*y2 - x2*y1|Let me substitute x1, y1, x2, y2:x1*y2 - x2*y1 = [1 - s1*cosθ][1 - s2*sinθ] - [1 - s2*cosθ][1 - s1*sinθ]Let me expand this expression:= [1*(1 - s2*sinθ) - s1*cosθ*(1 - s2*sinθ)] - [1*(1 - s1*sinθ) - s2*cosθ*(1 - s1*sinθ)]= [1 - s2*sinθ - s1*cosθ + s1*s2*cosθ*sinθ] - [1 - s1*sinθ - s2*cosθ + s1*s2*cosθ*sinθ]Simplify term by term:First part: 1 - s2*sinθ - s1*cosθ + s1*s2*cosθ*sinθSecond part: -1 + s1*sinθ + s2*cosθ - s1*s2*cosθ*sinθCombine them:1 - s2*sinθ - s1*cosθ + s1*s2*cosθ*sinθ -1 + s1*sinθ + s2*cosθ - s1*s2*cosθ*sinθSimplify:(1 -1) + (-s2*sinθ + s1*sinθ) + (-s1*cosθ + s2*cosθ) + (s1*s2*cosθ*sinθ - s1*s2*cosθ*sinθ)Which becomes:0 + sinθ*( -s2 + s1 ) + cosθ*( -s1 + s2 ) + 0So,(s1 - s2)(sinθ - cosθ)Therefore, the determinant x1*y2 - x2*y1 = (s1 - s2)(sinθ - cosθ)Therefore, the area is 0.5 * |(s1 - s2)(sinθ - cosθ)|Now, s1 and s2 are the roots of the quadratic equation s² - (cosθ + sinθ)s + 0.25 = 0. The difference between the roots of a quadratic equation ax² + bx + c =0 is sqrt(D)/a, where D is the discriminant. Here, a=1, D = sin2θ, so sqrt(D) = sqrt(sin2θ). Therefore, s1 - s2 = ± sqrt(sin2θ). But since s1 and s2 are two roots, the difference is either +sqrt(sin2θ) or -sqrt(sin2θ). However, since we take absolute value in the area, the sign doesn't matter. Therefore, |s1 - s2| = sqrt(sin2θ).Therefore, the area becomes:0.5 * sqrt(sin2θ) * |sinθ - cosθ|Therefore, Area = 0.5 * sqrt(sin2θ) * |sinθ - cosθ|So, our goal is to maximize this expression over θ ∈ (0, π/2)Therefore, the problem reduces to maximizing the function:A(θ) = 0.5 * sqrt(sin2θ) * |sinθ - cosθ|for θ ∈ (0, π/2)Let me denote A(θ) = 0.5 * sqrt(sin2θ) * |sinθ - cosθ|Since θ ∈ (0, π/2), 2θ ∈ (0, π), so sin2θ is positive, so sqrt(sin2θ) is real.Also, |sinθ - cosθ| is equal to either sinθ - cosθ or cosθ - sinθ depending on whether sinθ ≥ cosθ or not. The switch happens at θ = π/4, where sinθ = cosθ.Therefore, let's split the problem into two cases: θ ∈ (0, π/4] and θ ∈ [π/4, π/2)Case 1: θ ∈ (0, π/4]Here, cosθ ≥ sinθ, so |sinθ - cosθ| = cosθ - sinθCase 2: θ ∈ [π/4, π/2)Here, sinθ ≥ cosθ, so |sinθ - cosθ| = sinθ - cosθTherefore, the area function can be written as:For θ ∈ (0, π/4]:A(θ) = 0.5 * sqrt(sin2θ) * (cosθ - sinθ)For θ ∈ [π/4, π/2):A(θ) = 0.5 * sqrt(sin2θ) * (sinθ - cosθ)Our goal is to maximize A(θ) over θ ∈ (0, π/2)To find the maximum, let's compute the derivative of A(θ) with respect to θ, set it to zero, and solve for θ. However, since the expression involves absolute value, we can handle it by considering the two cases separately.First, let me work on Case 1: θ ∈ (0, π/4]A(θ) = 0.5 * sqrt(sin2θ) * (cosθ - sinθ)Let me express this in terms of trigonometric identities to simplify differentiation.Note that sqrt(sin2θ) = (sin2θ)^{1/2} = [2 sinθ cosθ]^{1/2}Also, cosθ - sinθ can be written as sqrt(2) cos(θ + π/4), but not sure if that helps.Alternatively, perhaps square the area function to make differentiation easier. Since the maximum of A(θ) will occur at the same θ as the maximum of A(θ)^2.Let’s compute A(θ)^2:A(θ)^2 = (0.5)^2 * sin2θ * (cosθ - sinθ)^2= 0.25 * sin2θ * (cosθ - sinθ)^2Let me expand (cosθ - sinθ)^2:= cos²θ - 2 sinθ cosθ + sin²θ= 1 - 2 sinθ cosθ - 2 sinθ cosθWait, no. Wait, (cosθ - sinθ)^2 = cos²θ - 2 sinθ cosθ + sin²θ = (cos²θ + sin²θ) - 2 sinθ cosθ = 1 - sin2θTherefore, A(θ)^2 = 0.25 * sin2θ * (1 - sin2θ)So, A(θ)^2 = 0.25 * sin2θ * (1 - sin2θ)Similarly, for Case 2: θ ∈ [π/4, π/2)A(θ) = 0.5 * sqrt(sin2θ) * (sinθ - cosθ)Then, A(θ)^2 = 0.25 * sin2θ * (sinθ - cosθ)^2But (sinθ - cosθ)^2 is the same as (cosθ - sinθ)^2 = 1 - sin2θTherefore, A(θ)^2 = 0.25 * sin2θ * (1 - sin2θ)Therefore, in both cases, A(θ)^2 = 0.25 * sin2θ * (1 - sin2θ)Therefore, regardless of θ being in (0, π/4] or [π/4, π/2), the square of the area is the same function of sin2θ.Therefore, A(θ)^2 = 0.25 * sin2θ * (1 - sin2θ)Let me set t = sin2θ, where θ ∈ (0, π/2), so 2θ ∈ (0, π), hence t ∈ (0, 1]Then, A^2 = 0.25 * t * (1 - t) = 0.25(t - t^2)Therefore, to maximize A^2, we need to maximize t - t^2 over t ∈ (0,1]The maximum of f(t) = t - t^2 occurs at t = 0.5, since derivative f’(t) = 1 - 2t, setting to zero gives t=0.5. The maximum value is 0.5 - 0.25 = 0.25.Therefore, maximum A^2 = 0.25 * 0.25 = 0.0625Therefore, maximum A = sqrt(0.0625) = 0.25Wait, but this seems too straightforward. The maximum area is 0.25?But let's verify. If t = sin2θ = 0.5, then sin2θ = 0.5, so 2θ = π/6 or 5π/6. Therefore, θ = π/12 or 5π/12. But θ ∈ (0, π/2), so θ = π/12 ≈ 15°, or θ = 5π/12 ≈ 75°. Wait, but θ is measured as the angle from the x-axis to the line UE. Let's check both possibilities.First, θ = π/12 (15°). Then, 2θ = π/6, sin2θ = 0.5, which is t=0.5. Then, A^2 = 0.25 * 0.5 * (1 - 0.5) = 0.25 * 0.5 * 0.5 = 0.0625, so A=0.25.Similarly, θ = 5π/12 (75°). Then, 2θ = 5π/6, sin2θ = 0.5 as well. Then, same result.But θ=π/12 is in (0, π/4] (since π/12 ≈15° < π/4 ≈45°), and θ=5π/12 is in [π/4, π/2). Therefore, both cases are covered.Therefore, the maximum area is 0.25. But wait, the unit square has area 1, so a triangle with area 0.25 is a quarter of the square. But is this possible?Wait, let's check. For instance, if θ=π/4 (45°), the line would be the diagonal from U (1,1) towards the center of the circle (0.5,0.5). Then, the points I and E would be the two intersections of that line with the circle.Let's compute the coordinates. The line from U (1,1) through the center (0.5,0.5) has direction vector (-0.5, -0.5). Parametrizing as (1 - 0.5t, 1 - 0.5t). The circle equation is (x - 0.5)^2 + (y - 0.5)^2 = 0.25. Substitute x = 1 - 0.5t, y =1 -0.5t:(0.5 -0.5t)^2 + (0.5 -0.5t)^2 = 0.252*(0.25 -0.5t +0.25t²) = 0.250.5 - t +0.5t² =0.250.5t² -t +0.25 =0Multiply by 2:t² -2t +0.5=0Solutions:t = [2 ± sqrt(4 - 2)]/2 = [2 ± sqrt(2)]/2 = 1 ± sqrt(2)/2So, points I and E are at t=1 + sqrt(2)/2 and t=1 - sqrt(2)/2. Wait, but t is the parameter in the parametrization (1 -0.5t, 1 -0.5t). Wait, but this might be different from the previous parametrization. Wait, maybe I need to check again.Alternatively, compute the distance from U (1,1) to the center (0.5,0.5) is sqrt(0.5^2 +0.5^2)=sqrt(0.5)=√2/2≈0.707. The radius of the circle is 0.5. So, the line from U to the center passes through the circle. The two intersection points would be at distances d1 and d2 from U, where d1 and d2 satisfy the secant line formula.But the maximum area in this case: when θ=π/4, which is the line towards the center. Then, points I and E are symmetric with respect to the line. But let's compute the area.If θ=π/12, then let's compute points I and E.Alternatively, since we have the maximum area as 0.25, let's see if that's correct.Wait, if the maximum area is 0.25, then the area of triangle PIE is 1/4. Let's think of a triangle with vertices at (0,0), and two points on the circle. If the maximum area is 1/4, which is the same as the area of the triangle formed by (0,0), (0.5,1), and (1,0.5). Wait, but those points might not lie on the circle or be colinear with U.Alternatively, perhaps when θ=π/12, points I and E are arranged such that the determinant x1*y2 - x2*y1 is maximized. But according to our previous calculation, the maximum area is 0.25.Wait, but let's test with θ=π/4. If the line is from U (1,1) towards the center (0.5,0.5), then the two intersection points would be along that line. The center is at (0.5,0.5), and the circle has radius 0.5, so the line from U to the center intersects the circle at two points: one closer to U, and one further away.But the distance from U to the center is sqrt(0.5), which is approximately 0.707, and the radius is 0.5, so the two intersection points are at distance sqrt(0.5) ± 0.5 from U? Wait, actually, using the power of a point: the product of the distances from U to I and U to E is equal to the power of U with respect to the circle.Power of U with respect to circle ω is |OU|² - r², where OU is the vector from center to U. The center is at (0.5,0.5), so vector OU is (0.5,0.5), |OU|² = 0.5² +0.5²=0.5, and r²=0.25. Therefore, power of U is 0.5 -0.25=0.25. Therefore, the product of the distances from U to I and U to E is 0.25.Therefore, if UI * UE =0.25. If UI and UE are the two segments from U to I and E, then their product is 0.25. But how does that relate to the area of triangle PIE?Alternatively, perhaps not directly. But if we have UI * UE =0.25, and the area of triangle PIE is related to the positions of I and E.But according to our earlier calculation, the maximum area is 0.25. Let's see if that's achievable.Suppose θ=π/12, so 2θ=π/6, sin2θ=0.5. Then, A=0.25. Let's compute coordinates.Wait, let's compute s1 and s2 from the quadratic equation:s² - (cosθ + sinθ)s +0.25=0At θ=π/12, cosθ=cos(π/12)=sqrt(6)+sqrt(2))/4≈0.9659, sinθ=sin(π/12)=sqrt(6)-sqrt(2))/4≈0.2588. Then, cosθ + sinθ≈0.9659 +0.2588≈1.2247. The quadratic equation is s² -1.2247s +0.25=0Solutions:s=(1.2247±sqrt(1.2247² -1))/2Calculate discriminant:1.2247²≈1.499, so sqrt(1.499 -1)=sqrt(0.499)≈0.706Therefore, s≈(1.2247 ±0.706)/2So, s1≈(1.2247+0.706)/2≈1.9307/2≈0.9653s2≈(1.2247-0.706)/2≈0.5187/2≈0.2593Therefore, coordinates:For s1≈0.9653:x1=1 - s1*cosθ≈1 -0.9653*0.9659≈1 -0.932≈0.068y1=1 -s1*sinθ≈1 -0.9653*0.2588≈1 -0.250≈0.750For s2≈0.2593:x2=1 -0.2593*0.9659≈1 -0.250≈0.750y2=1 -0.2593*0.2588≈1 -0.067≈0.933Therefore, points I≈(0.068, 0.750) and E≈(0.750, 0.933)Now, compute the area of triangle PIE with P=(0,0), I≈(0.068,0.750), E≈(0.750,0.933)Using determinant formula:Area =0.5*|0.068*0.933 -0.750*0.750|Calculate:0.068*0.933≈0.06340.750*0.750=0.5625Difference: 0.0634 -0.5625≈-0.4991Absolute value:0.4991Area≈0.5*0.4991≈0.2495≈0.25So, yes, that's approximately 0.25. Therefore, in this case, the area is indeed 0.25. Therefore, our previous conclusion that the maximum area is 0.25 is correct.But wait, this seems like the maximum occurs when sin2θ=0.5, so 2θ=30°, θ=15°, or 2θ=150°,θ=75°, and we achieve the maximum area of 0.25.Is there a possibility of a larger area? Let's check another angle. For example, θ=π/8 (22.5°), so 2θ=π/4, sin2θ=√2/2≈0.707.Compute A(θ)=0.5*sqrt(sin2θ)*|sinθ - cosθ|Wait, but for θ=π/8, which is 22.5°, let's compute.sinθ≈0.3827, cosθ≈0.9239, sinθ - cosθ≈-0.5412, absolute value≈0.5412sqrt(sin2θ)=sqrt(sin(π/4))=sqrt(√2/2)= (2)^{1/4}≈0.8409Therefore, A(θ)=0.5 *0.8409 *0.5412≈0.5*0.8409*0.5412≈0.5*0.455≈0.227, which is less than 0.25.Similarly, if we check θ=0, but θ=0 is the tangent case, area zero. Similarly, θ=π/2, area zero. Therefore, the maximum seems to indeed occur at θ=π/12 and θ=5π/12 with area 0.25.But let's confirm with θ=5π/12 (75°). Let's compute.θ=5π/12, 2θ=5π/6, sin2θ=0.5, so same as before.cosθ=cos(5π/12)= (sqrt(6)-sqrt(2))/4≈0.2588, sinθ=sin(5π/12)= (sqrt(6)+sqrt(2))/4≈0.9659Therefore, cosθ + sinθ≈0.2588 +0.9659≈1.2247, same as before.Therefore, quadratic equation same as before, s1≈0.9653, s2≈0.2593Therefore, coordinates:For s1≈0.9653:x1=1 - s1*cosθ≈1 -0.9653*0.2588≈1 -0.250≈0.750y1=1 -s1*sinθ≈1 -0.9653*0.9659≈1 -0.932≈0.068For s2≈0.2593:x2=1 -0.2593*0.2588≈1 -0.067≈0.933y2=1 -0.2593*0.9659≈1 -0.250≈0.750Therefore, points I≈(0.750,0.068) and E≈(0.933,0.750)Area of triangle PIE:0.5*|0.750*0.750 -0.933*0.068|Calculate:0.750*0.750=0.56250.933*0.068≈0.0634Difference:0.5625 -0.0634≈0.4991Absolute value:0.4991Area≈0.5*0.4991≈0.2495≈0.25Same result. Therefore, in both cases, the area is approximately 0.25. Therefore, the maximum area is indeed 0.25.But let's just verify once again if there's a possible configuration where the area could be larger. Suppose we take a different line, not passing through the center. For example, take a line that is closer to the corner P. Let's say a line passing through U (1,1) and intersecting the circle near P (0,0). But would that give a larger area?Wait, if points I and E are near P, then the triangle PIE would have a base near P and height near P, but maybe not. Wait, but the determinant area depends on the coordinates of I and E. If I is near (0.5,1) and E is near (1,0.5), then the area could be larger? Wait, but according to our calculation, the maximum is 0.25.Alternatively, let's consider the line tangent to the circle. If we take the tangent line at point (0.5,1), which is horizontal line y=1. Then, points I and E coincide at (0.5,1). Then, the area of triangle PIE would be the area of triangle P(0.5,1)E, but since I=E, the area is zero. Similarly, for the vertical tangent line at (1,0.5), the area is zero. Therefore, tangent lines do not contribute.Alternatively, take a line passing through U and intersecting the circle at two points such that one of them is close to (0,0.5) or (0.5,0). But the determinant area might not be larger.For example, suppose we take a line passing through U (1,1) and point (0,0.5). Let's find the intersection points.First, the line through (1,1) and (0,0.5). The slope m=(0.5 -1)/(0 -1)= (-0.5)/(-1)=0.5. The equation is y -1=0.5(x -1), so y=0.5x +0.5.Find intersection with circle (x-0.5)^2 + (y-0.5)^2=0.25.Substitute y=0.5x +0.5:(x -0.5)^2 + (0.5x +0.5 -0.5)^2=0.25Simplify:(x -0.5)^2 + (0.5x)^2=0.25Expand:x² -x +0.25 +0.25x²=0.25Combine:1.25x² -x +0.25=0.25Subtract 0.25:1.25x² -x=0Multiply by 4/5 to simplify:x² -0.8x=0x(x -0.8)=0Solutions x=0 or x=0.8x=0: y=0.5*0 +0.5=0.5, which is point (0,0.5)x=0.8: y=0.5*0.8 +0.5=0.4 +0.5=0.9, so point (0.8,0.9)Therefore, points I=(0,0.5) and E=(0.8,0.9). But wait, but (0,0.5) is not on the circle. Wait, the circle is radius 0.5 centered at (0.5,0.5). So, (0,0.5) is distance 0.5 from the center along the x-axis, but since x=0, it's 0.5 units left from center, so it is on the circle. Yes, (0,0.5) is on the circle: (0 -0.5)^2 + (0.5 -0.5)^2=0.25 +0=0.25. Correct.Therefore, points I=(0,0.5) and E=(0.8,0.9). Compute area of triangle PIE.P=(0,0), I=(0,0.5), E=(0.8,0.9)Area=0.5*|0*0.9 -0.8*0.5|=0.5*|0 -0.4|=0.5*0.4=0.2Which is less than 0.25. Therefore, even though one point is at the edge of the circle, the area is smaller.Another example: take line from U (1,1) through center (0.5,0.5), intersecting the circle at (0.5 + 0.5*cosθ, 0.5 +0.5*sinθ). Wait, but earlier calculations show the area is 0.25.Wait, but let's compute area for line through U and center.As previously, points I and E are (1 - s1*cosθ, 1 - s1*sinθ) and (1 - s2*cosθ, 1 - s2*sinθ). When the line passes through the center, the angle θ is π/4 (45°), but we saw that the area in this case is 0.25 as well. Wait, but when θ=π/4, sin2θ=1, so A(θ)=0.5 *1 * |sinθ - cosθ|At θ=π/4, sinθ=cosθ=√2/2, so |sinθ - cosθ|=0, so Area=0. Wait, contradiction?Wait, no. Wait, if θ=π/4, then the line is passing through U (1,1) and the center (0.5,0.5), so it's the line y=x. The intersection points with the circle are at (0.5 + 0.5*cosπ/4, 0.5 +0.5*sinπ/4) and (0.5 -0.5*cosπ/4, 0.5 -0.5*sinπ/4)Wait, parametrize the line y=x from U (1,1):Parametric equations x=1 - t, y=1 - t. Plug into circle:(x -0.5)^2 + (y -0.5)^2=0.25(0.5 -t)^2 + (0.5 -t)^2=0.252*(0.5 -t)^2=0.25(0.5 -t)^2=0.1250.5 -t=±sqrt(0.125)=±(√2)/4≈±0.3536Thus, t=0.5 ±(√2)/4Therefore, points:For t=0.5 +√2/4≈0.5+0.3536≈0.8536:x=1 -0.8536≈0.1464, y=0.1464For t=0.5 -√2/4≈0.5 -0.3536≈0.1464:x=1 -0.1464≈0.8536, y=0.8536Thus, points I=(0.1464,0.1464) and E=(0.8536,0.8536)Area of triangle PIE:0.5*|0.1464*0.8536 -0.8536*0.1464|=0.5*|0.125 -0.125|=0.5*0=0Wait, that's zero. Because points I and E are symmetric with respect to the line y=x, so the determinant becomes zero. Therefore, the area is zero. Which makes sense because points P, I, E are colinear? Wait, no. Points P=(0,0), I=(0.1464,0.1464), E=(0.8536,0.8536). These three points are colinear on the line y=x. Therefore, the area is zero. So, in this case, the area is zero, as calculated. But earlier when we considered θ=π/4, the area expression was A=0.5 * sqrt(sin2θ)*|sinθ - cosθ|. But at θ=π/4, sinθ - cosθ=0, so A=0. So, the formula holds. Therefore, when the line is passing through the center, the area is zero. So, the maximum area occurs at θ=π/12 and θ=5π/12, not at θ=π/4.Therefore, the maximum area is indeed 0.25, which is 1/4.But just to confirm, let's consider θ=π/12 as before, which gave area≈0.25, and the other angle θ=5π/12, same area. Therefore, the maximum area is 1/4.Therefore, the greatest possible area for triangle PIE is 1/4.Final Answerboxed{dfrac{1}{4}}

✅:To find the greatest possible area of triangle ( triangle PIE ) where circle ( omega ) is inscribed in the unit square ( PLUM ) and points ( I ) and ( E ) lie on ( omega ) such that ( U, I, ) and ( E ) are collinear, we start by considering the coordinates of the square and the circle. The unit square ( PLUM ) has vertices ( P(0,0) ), ( L(1,0) ), ( U(1,1) ), and ( M(0,1) ). The inscribed circle ( omega ) is centered at ( (0.5, 0.5) ) with radius ( 0.5 ).We parametrize the line through ( U(1,1) ) intersecting the circle ( omega ) using a parameter ( theta ) representing the angle from the x-axis. The parametric equations for the line are ( x = 1 + t cos theta ) and ( y = 1 + t sin theta ). Substituting these into the circle's equation, we solve for ( t ) and find the points ( I ) and ( E ).The area of ( triangle PIE ) is given by ( 0.5 times |x_1 y_2 - x_2 y_1| ). Using trigonometric identities and the quadratic formula, we derive the expression for the area in terms of ( theta ). After simplifying, we find that the area depends on ( sqrt{sin 2theta} ) and ( |sin theta - cos theta| ).Maximizing this expression, we find that the maximum area occurs when ( sin 2theta = 0.5 ), corresponding to ( theta = pi/12 ) or ( 5pi/12 ). This results in the maximum area of ( 0.25 ).Thus, the greatest possible area for ( triangle PIE ) is (boxed{dfrac{1}{4}}).

🔑:1. Set up the diagram in the Cartesian coordinate plane: - Place the unit square (PLUM) with (U = (0, 0)) and (P = (2, 2)). This simplifies calculations by scaling the square to a (2 times 2) square. - The circle (omega) is inscribed in the square, centered at ((1, 1)) with radius 1.2. Equation of the line (UE): - Let the slope of (overline{UE}) be (k). The equation of the line (UE) is (y = kx).3. Equation of the circle (omega): - The equation of the circle is ((x - 1)^2 + (y - 1)^2 = 1).4. Find coordinates of (I) and (E): - Substitute (y = kx) into the circle's equation: [ (x - 1)^2 + (kx - 1)^2 = 1 ] [ x^2 - 2x + 1 + k^2x^2 - 2kx + 1 = 1 ] [ (k^2 + 1)x^2 - 2(k + 1)x + 1 = 0 ] - Solve the quadratic equation using the quadratic formula: [ x = frac{2(k + 1) pm sqrt{4(k + 1)^2 - 4(k^2 + 1)}}{2(k^2 + 1)} ] [ x = frac{2(k + 1) pm sqrt{4k + 4}}{2(k^2 + 1)} ] [ x = frac{2(k + 1) pm 2sqrt{k + 1}}{2(k^2 + 1)} ] [ x = frac{k + 1 pm sqrt{k + 1}}{k^2 + 1} ] - Let (r_1 = frac{k + 1 - sqrt{k + 1}}{k^2 + 1}) and (r_2 = frac{k + 1 + sqrt{k + 1}}{k^2 + 1}).5. Coordinates of (E) and (I): - (E = (r_1, kr_1)) and (I = (r_2, kr_2)).6. Use the Shoelace formula to find the area of (triangle PIE): - Coordinates of (P = (2, 2)), (I = (r_2, kr_2)), and (E = (r_1, kr_1)). - Shoelace formula: [ [triangle PIE] = frac{1}{2} left| 2kr_1 + kr_1r_2 + 2r_2 - (2r_1 + kr_1r_2 + 2kr_2) right| ] [ = frac{1}{2} left| 2kr_1 - 2r_2 - 2r_1 - 2kr_2 right| ] [ = frac{1}{2} left| 2(k - 1)(r_1 - r_2) right| ] [ = |(k - 1)(r_1 - r_2)| ]7. Simplify (r_1 - r_2): - (r_1 - r_2 = frac{2sqrt{k + 1}}{k^2 + 1}).8. Expression for the area: - [ [triangle PIE] = (k - 1) left( frac{2sqrt{k + 1}}{k^2 + 1} right) ]9. Maximize the area: - Let (a(x) = (x - 1) left( frac{2sqrt{x + 1}}{x^2 + 1} right)). - Differentiate (a(x)) and find the critical points: [ a'(x) = frac{d}{dx} left( (x - 1) left( frac{2sqrt{x + 1}}{x^2 + 1} right) right) ] - Use the product rule and quotient rule to find (a'(x)): [ a'(x) = left( frac{2sqrt{x + 1}}{x^2 + 1} right) + (x - 1) left( frac{d}{dx} left( frac{2sqrt{x + 1}}{x^2 + 1} right) right) ] - Solve (a'(x) = 0) to find the critical points.10. Find the maximum value: - The critical points are (x = 2 pm sqrt{3}). - Since (k > 1), choose (k = 2 + sqrt{3}).11. Calculate the maximum area: - [ a(2 + sqrt{3}) = (2 + sqrt{3} - 1) left( frac{2sqrt{2(2 + sqrt{3})}}{(2 + sqrt{3})^2 + 1} right) ] - Simplify to find: [ a(2 + sqrt{3}) = 1 ] - Scale back by (frac{1}{4}) due to the initial scaling of the square: [ boxed{frac{1}{4}} ]

🧠:Alright, let's try to figure out this problem step by step. The question is about finding the number of integers n for which the number of solutions s_n to the equation a1 + a2 + a3 + a4 + b1 + b2 = n is odd. Here, each ai is from {2, 3, 5, 7} and each bi is from {1, 2, 3, 4}. So, s_n counts the number of ordered solutions, and we need to find how many n have s_n odd.First, I need to recall how to approach such problems. When dealing with the number of solutions to an equation with variables taking specific values, generating functions are usually a powerful tool. Since the variables are ordered, each variable contributes independently, so the generating function should be the product of the generating functions for each variable.Let me remember that the generating function for a single variable that can take values in a set S is the sum of x^k for each k in S. So, for each ai (which is in {2, 3, 5, 7}), the generating function would be x^2 + x^3 + x^5 + x^7. Similarly, for each bi (in {1, 2, 3, 4}), the generating function would be x^1 + x^2 + x^3 + x^4.Since there are four ai variables and two bi variables, the overall generating function G(x) would be [x^2 + x^3 + x^5 + x^7]^4 multiplied by [x + x^2 + x^3 + x^4]^2. Then, the coefficient of x^n in G(x) is exactly s_n. So, we need to find all n where this coefficient is odd.The problem now reduces to analyzing the parity of the coefficients of the generating function. The key insight here might be to consider the generating functions modulo 2. Since we're only interested in whether the coefficients are odd or even, working modulo 2 simplifies things. That is, we can compute G(x) mod 2 and find the exponents n where the coefficient is 1 (since 1 mod 2 is 1, meaning odd).So, let's define this step by step. First, compute the generating functions for ai and bi modulo 2.For ai: Each ai is in {2, 3, 5, 7}. Let's write the generating function for one ai:x^2 + x^3 + x^5 + x^7Modulo 2, this remains the same because the coefficients are all 1, and 1 mod 2 is 1. So, (x^2 + x^3 + x^5 + x^7) mod 2.Similarly, for each bi in {1, 2, 3, 4}, the generating function is x + x^2 + x^3 + x^4. Again, modulo 2, this is the same.Therefore, the entire generating function mod 2 is:[(x^2 + x^3 + x^5 + x^7)]^4 * [(x + x^2 + x^3 + x^4)]^2 mod 2Our goal is to compute this product and then find the exponents n where the coefficient is 1. The number of such n will be our answer.To proceed, perhaps we can simplify each part of the generating function modulo 2 first.Starting with the bi variables: let's compute [(x + x^2 + x^3 + x^4)]^2 mod 2.First, compute (x + x^2 + x^3 + x^4) squared. Let's do this step by step.Multiply (x + x^2 + x^3 + x^4) by itself:= x*(x + x^2 + x^3 + x^4) + x^2*(x + x^2 + x^3 + x^4) + x^3*(x + x^2 + x^3 + x^4) + x^4*(x + x^2 + x^3 + x^4)= (x^2 + x^3 + x^4 + x^5) + (x^3 + x^4 + x^5 + x^6) + (x^4 + x^5 + x^6 + x^7) + (x^5 + x^6 + x^7 + x^8)Now, combine like terms:x^2 + (x^3 + x^3) + (x^4 + x^4 + x^4) + (x^5 + x^5 + x^5 + x^5) + (x^6 + x^6 + x^6) + (x^7 + x^7) + x^8But modulo 2, coefficients are reduced mod 2:x^2 + (2x^3) + (3x^4) + (4x^5) + (3x^6) + (2x^7) + x^8 mod 2Which simplifies to:x^2 + 0x^3 + x^4 + 0x^5 + x^6 + 0x^7 + x^8So, (x + x^2 + x^3 + x^4)^2 ≡ x^2 + x^4 + x^6 + x^8 mod 2.Alternatively, we can factor the original polynomial for bi:x + x^2 + x^3 + x^4 = x(1 + x + x^2 + x^3) = x*(1 + x)*(1 + x^2) (since 1 + x + x^2 + x^3 = (1 + x)(1 + x^2))But maybe factoring isn't necessary here. Let's see.Similarly, for the ai variables: each ai has generating function x^2 + x^3 + x^5 + x^7.Let me check if this can be factored or simplified. Let's write it as x^2(1 + x + x^3 + x^5). Hmm, 1 + x + x^3 + x^5. Not sure if that factors nicely. Alternatively, maybe writing in terms of exponents:Wait, 1 + x + x^3 + x^5. Let me see if it can be expressed as a product. Let's try to factor it modulo 2.Testing possible roots: plug x=1: 1 +1 +1 +1 = 4 ≡ 0 mod 2. So, (x - 1) is a factor, but over GF(2), (x - 1) is (x + 1). Let's divide 1 + x + x^3 + x^5 by (x + 1).Using polynomial division:Divide x^5 + x^3 + x + 1 by x + 1.x^5 / x = x^4. Multiply (x + 1) by x^4: x^5 + x^4.Subtract (x^5 + x^4) from x^5 + x^3 + x +1: gives -x^4 + x^3 + x +1. But mod 2, subtraction is addition, so x^4 + x^3 + x +1.Next term: x^4 / x = x^3. Multiply (x +1) by x^3: x^4 + x^3.Subtract: (x^4 + x^3) from x^4 + x^3 + x +1: gives 0x^4 +0x^3 + x +1.Now, remainder is x +1. The degree is less than divisor, so we have:x^5 +x^3 +x +1 = (x +1)(x^4 +x^3) + (x +1)So, factor again: (x +1)(x^4 +x^3 +1). Wait, let me check that:Wait, perhaps my division was wrong. Let me try again.Divide x^5 + x^3 + x +1 by x +1.Alternatively, since over GF(2), coefficients are mod 2. Let me list the coefficients:Dividend: x^5 +x^3 +x +1 corresponds to coefficients:1 0 1 0 1 1 (for degrees 5,4,3,2,1,0)Divisor: x +1 is 1 1 (degrees 1,0)The division process:Start with 1 0 1 0 1 1Subtract (1 1) shifted by 4 positions (since leading term is x^5):1 0 1 0 1 1-1 1 0 0 0 0-------------------0 1 1 0 1 1 (mod 2: subtraction is addition)Now, next term is x^4 coefficient 1. Multiply divisor by x^3 (shift 3 positions):1 1 0 0 0Subtract from current remainder 0 1 1 0 1 1:0 1 1 0 1 1-0 1 1 0 0 0-------------------0 0 0 0 1 1Now, remainder is x +1. So, the division yields quotient x^4 +x^3 and remainder x +1. Thus,x^5 +x^3 +x +1 = (x +1)(x^4 +x^3) + (x +1) mod 2Factor out (x +1):= (x +1)(x^4 +x^3 +1)Hence, 1 +x +x^3 +x^5 = (x +1)(x^4 +x^3 +1) mod 2.Therefore, the generating function for ai is x^2*(x +1)*(x^4 +x^3 +1)Therefore, the generating function for one ai is x^2*(1 +x)*(1 +x^3 +x^4) mod 2.But perhaps this factorization might help in simplifying the overall generating function.But let's see. Since we need to raise this to the 4th power. So, [x^2*(1 +x)*(1 +x^3 +x^4)]^4 mod 2.Which is x^8*(1 +x)^4*(1 +x^3 +x^4)^4 mod 2.Similarly, we have the generating function for the bi part: (x +x^2 +x^3 +x^4)^2 = x^2*(1 +x +x^2 +x^3)^2. Which we already found to be x^2 +x^4 +x^6 +x^8 mod 2, but we also could factor it as x^2*(1 +x)^2*(1 +x^2)^2 mod 2.But perhaps it's better to compute each part separately and then multiply them.Alternatively, let's try to compute both generating functions mod 2.First, let's compute [(x^2 +x^3 +x^5 +x^7)]^4 mod 2.Given that (x^2 +x^3 +x^5 +x^7) is x^2*(1 +x +x^3 +x^5), which we factored as x^2*(1 +x)*(1 +x^3 +x^4). So, raising this to the 4th power:[x^2*(1 +x)*(1 +x^3 +x^4)]^4 = x^8*(1 +x)^4*(1 +x^3 +x^4)^4 mod 2.Now, let's compute (1 +x)^4 mod 2. Since (1 +x)^2 = 1 +x^2 mod 2, then (1 +x)^4 = (1 +x^2)^2 = 1 +x^4 mod 2.Similarly, (1 +x^3 +x^4)^4 mod 2. Let's compute (1 +x^3 +x^4)^2 first:(1 +x^3 +x^4)^2 = 1 + x^6 +x^8 + 2x^3 + 2x^4 + 2x^7. Modulo 2, this becomes 1 +x^6 +x^8.Then, squaring again: (1 +x^6 +x^8)^2 = 1 +x^{12} +x^{16} + 2x^6 + 2x^8 + 2x^{14} ≡ 1 +x^{12} +x^{16} mod 2.Therefore, (1 +x^3 +x^4)^4 ≡ 1 +x^{12} +x^{16} mod 2.Putting it all together:[x^2 +x^3 +x^5 +x^7]^4 ≡ x^8*(1 +x^4)*(1 +x^{12} +x^{16}) mod 2Multiplying out:x^8*(1 +x^4)*(1 +x^{12} +x^{16}) = x^8*(1 +x^{12} +x^{16} +x^4 +x^{16} +x^{20}) mod 2Simplify, noting that x^{16} +x^{16} = 0 mod 2:= x^8*(1 +x^4 +x^{12} +x^{20}) mod 2Therefore:= x^8 +x^{12} +x^{20} +x^{28} mod 2So, the ai part raised to the 4th power mod 2 is x^8 +x^{12} +x^{20} +x^{28}.Now, the bi part was [x +x^2 +x^3 +x^4]^2 mod 2, which we found earlier as x^2 +x^4 +x^6 +x^8.So, the entire generating function G(x) mod 2 is:(x^8 +x^{12} +x^{20} +x^{28}) * (x^2 +x^4 +x^6 +x^8) mod 2Now, we need to compute this product. Let's do this term by term.Multiply each term in the first polynomial by each term in the second polynomial and add them up modulo 2.First, take x^8 * (x^2 +x^4 +x^6 +x^8) = x^{10} +x^{12} +x^{14} +x^{16}Then, x^{12} * (x^2 +x^4 +x^6 +x^8) = x^{14} +x^{16} +x^{18} +x^{20}Then, x^{20} * (x^2 +x^4 +x^6 +x^8) = x^{22} +x^{24} +x^{26} +x^{28}Then, x^{28} * (x^2 +x^4 +x^6 +x^8) = x^{30} +x^{32} +x^{34} +x^{36}Now, combine all these terms:x^{10} +x^{12} +x^{14} +x^{16} +x^{14} +x^{16} +x^{18} +x^{20} +x^{22} +x^{24} +x^{26} +x^{28} +x^{30} +x^{32} +x^{34} +x^{36}Combine like terms:x^{10} +x^{12} + (x^{14} +x^{14}) + (x^{16} +x^{16}) +x^{18} +x^{20} +x^{22} +x^{24} +x^{26} +x^{28} +x^{30} +x^{32} +x^{34} +x^{36}Since coefficients are mod 2, terms with even coefficients cancel out:= x^{10} +x^{12} +0x^{14} +0x^{16} +x^{18} +x^{20} +x^{22} +x^{24} +x^{26} +x^{28} +x^{30} +x^{32} +x^{34} +x^{36}Thus, the generating function mod 2 is:x^{10} +x^{12} +x^{18} +x^{20} +x^{22} +x^{24} +x^{26} +x^{28} +x^{30} +x^{32} +x^{34} +x^{36}So, the exponents where the coefficient is 1 (i.e., s_n is odd) are 10, 12, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36.Therefore, the number of such n is 12.Wait, but let me check this calculation again because it's critical.Original product:First polynomial (ai part): x^8 +x^{12} +x^{20} +x^{28}Second polynomial (bi part): x^2 +x^4 +x^6 +x^8Multiplying each term:x^8 * x^2 = x^{10}x^8 * x^4 = x^{12}x^8 * x^6 = x^{14}x^8 * x^8 = x^{16}Similarly for x^{12}:x^{12}*x^2 =x^{14}x^{12}*x^4 =x^{16}x^{12}*x^6 =x^{18}x^{12}*x^8 =x^{20}For x^{20}:x^{20}*x^2 =x^{22}x^{20}*x^4 =x^{24}x^{20}*x^6 =x^{26}x^{20}*x^8 =x^{28}For x^{28}:x^{28}*x^2 =x^{30}x^{28}*x^4 =x^{32}x^{28}*x^6 =x^{34}x^{28}*x^8 =x^{36}So, the terms are:x^{10}, x^{12}, x^{14}, x^{16},x^{14}, x^{16}, x^{18}, x^{20},x^{22}, x^{24}, x^{26}, x^{28},x^{30}, x^{32}, x^{34}, x^{36}Now, combining like terms:x^{10}, x^{12}, (x^{14}+x^{14}=0), (x^{16}+x^{16}=0), x^{18}, x^{20}, x^{22}, x^{24}, x^{26}, x^{28}, x^{30}, x^{32}, x^{34}, x^{36}Hence, the remaining terms are:10,12,18,20,22,24,26,28,30,32,34,36. So 12 terms. Thus, the number of n where s_n is odd is 12. So, the answer is 12. But wait, let's double-check.But before confirming, let's consider the possible range of n. What is the minimum and maximum possible value of n?Each ai is at least 2, so 4 variables contribute at least 8. Each bi is at least 1, so 2 variables contribute at least 2. Thus, the minimum n is 8 + 2 =10.Maximum n: each ai is at most 7, so 4*7=28. Each bi is at most 4, so 2*4=8. Total maximum n is 28 +8=36.So, n ranges from 10 to 36. The exponents we found are from 10 to 36, but only specific ones. Indeed, in our generating function mod 2, we have exponents 10,12,18,20,22,24,26,28,30,32,34,36. So, every even number from 10 to 36, skipping some. Let me check the list:10,12,18,20,22,24,26,28,30,32,34,36. So, between 10 and 36, inclusive, even numbers, but missing 14,16, which got canceled out. Therefore, the total count is 12 numbers. So, the answer is 12.But wait, before finalizing, let's think if there's any mistake in the process. For example, when we computed the generating function for the ai part, we had [x^2 +x^3 +x^5 +x^7]^4 mod 2. We factored it and found that it's x^8*(1 +x^4)*(1 +x^{12} +x^{16}) mod 2. Then multiplied out to get x^8 +x^{12} +x^{20} +x^{28} mod 2. Then multiplied by the bi part x^2 +x^4 +x^6 +x^8. That multiplication gave us the exponents from 10 to 36 in steps, with some terms canceled. Then we counted 12 exponents. That seems correct.But let me verify with a small example. Suppose we consider n=10. The minimum n is 10. The number of solutions for n=10 should be the product of the number of ways to get a1+a2+a3+a4=8 (since bi's contribute 2) and b1+b2=2. Wait, no: a1+a2+a3+a4 +b1+b2=10, so a1+a2+a3+a4=10 - (b1 +b2). Since each bi is at least 1, b1 +b2 is at least 2, so a1+...+a4=8. But each ai is at least 2, so a1+...+a4=8 implies each ai=2. So, only one way for a's: all 2s. For b's, b1 +b2=2. Since bi >=1, the only way is b1=1, b2=1. But since they are ordered, there is exactly one solution for b's: (1,1). Hence, total solutions for n=10: 1*1=1, which is odd. So s_10=1, which is odd. So n=10 is indeed counted. Similarly, n=12: Let's check. For n=12, a1+...+a4 +b1+b2=12. The possible sums for a's and b's. Let's think. For example, a's can sum to 8, 9, 10, 11, 12, etc., but given that each a_i is in {2,3,5,7}, the possible sums for a's are even or odd? Let's see. Each a_i is 2,3,5,7: which are 2 (even), 3 (odd), 5 (odd), 7 (odd). So, each a_i contributes even or odd. The sum of four numbers each even or odd: the parity of the sum is the sum of the parities mod 2. Each a_i is even (2) or odd (3,5,7). So, the total parity for a1+a2+a3+a4 is even if there are an even number of odd a_i's, and odd otherwise. Similarly, b1 and b2: each bi is in {1,2,3,4}, which are 1 (odd), 2 (even), 3 (odd), 4 (even). So the sum b1+b2's parity is the sum of the parities of b1 and b2. Therefore, the total parity of n is (parity of a's sum) + (parity of b's sum) mod 2. Since n = a's sum + b's sum. So, parity of n is parity(a's sum) + parity(b's sum).But since we are working mod 2, maybe this can help. However, in our generating function approach, we didn't consider parity, but rather counted all coefficients mod 2. However, the fact that we have a generating function mod 2 which is a sparse polynomial with exponents at even positions from 10 to 36, stepping by 2, but missing some. However, according to our calculation, s_n is odd only at those 12 exponents. But let's check n=14. According to our result, s_14 should be even. Let's see.For n=14, a's can be from 8 (all 2s) up to 28 (all 7s), but since n=14, the a's sum is 14 - (b's sum). The b's sum can be from 2 to 8 (since each bi up to 4). So a's sum would be from 14 -8=6 (but a's minimum is 8) up to 14 -2=12. Wait, this seems conflicting. Wait, if n=14, then a's sum + b's sum=14. Since a's sum is at least 8 and b's sum is at least 2, the minimum n is 10. For n=14, a's sum can be from 8 to 12 (since b's sum can be from 2 to 6). Wait, but 14 - b's sum must be >=8. So b's sum <=6. Because a's sum >=8. So b's sum can be 2,3,4,5,6. Thus, for n=14, we need to compute the number of solutions where a's sum=12,11,10,9,8 and b's sum=2,3,4,5,6 respectively. But each a_i is in {2,3,5,7}, so the sums possible for a's are sums of four numbers each 2,3,5,7. The possible sums here would be even or odd. Let's see. If a's sum=8: all 2s. Then b's sum=6. The number of ordered solutions for a's sum=8 is 1 (all 2s). For b's sum=6: the number of ordered pairs (b1,b2) with b1,b2 in {1,2,3,4} and sum 6. The possible pairs are (2,4), (3,3), (4,2). So three ordered pairs. So total solutions for this part: 1*3=3.Next, a's sum=9. How many ordered quadruples (a1,a2,a3,a4) in {2,3,5,7} sum to 9. Since each a_i is at least 2, and the sum is 9. So 9=2+2+2+3. So the only way is three 2s and one 3. The number of ordered solutions is C(4,1)=4. Then, b's sum=14 -9=5. Number of ordered pairs for b's sum=5: possible pairs are (1,4), (2,3), (3,2), (4,1). So 4 ordered pairs. Total solutions: 4*4=16.Similarly, a's sum=10. Number of ordered quadruples: possible combinations. 10 can be achieved by:- Two 2s and two 3s: 2+2+3+3=10. The number of ordered solutions: C(4,2)=6.- Or one 2, one 3, and two 5s: 2+3+5+5=15, which is too big. Wait, no. Wait, 2+3+5=10, but we have four numbers. Let's see.Wait, 2+2+2+4=10, but 4 is not allowed. Wait, the possible numbers are 2,3,5,7. So to get 10:Possible combinations:- 2,2,2,4 → but 4 is not allowed.- 2,2,3,3: sum=10. That's the only way. So number of ordered solutions is 4C2=6.Then, b's sum=14 -10=4. Number of ordered pairs for sum=4: (1,3), (2,2), (3,1), (1,3) again. Wait, no: ordered pairs. So (1,3), (2,2), (3,1). So three ordered pairs. So total solutions: 6*3=18.Next, a's sum=11. Number of ordered quadruples. To get 11 with four numbers each 2,3,5,7.Possible combinations:- Three 3s and one 2: 3+3+3+2=11. Number of ordered solutions: C(4,1)=4.- One 5, two 2s, and one 3: 5+2+2+3=12, which is over.Wait, 2+2+2+5=11? 2+2+2+5=11. Yes, that works. So another combination: three 2s and one 5. Number of ordered solutions: C(4,1)=4.- Also, one 3 and one 5 and two 2s: 3+5+2+2=12, which is over.Alternatively, maybe 3+3+5 + something? 3+3+5=11, but need four numbers. 3+3+5 +0, but 0 isn't allowed. No.So possible combinations:Either three 2s and one 5, or three 3s and one 2. So total ordered solutions: 4 +4=8.Then, b's sum=14 -11=3. Number of ordered pairs for sum=3: (1,2), (2,1). So 2 ordered pairs. So total solutions:8*2=16.Finally, a's sum=12. Number of ordered quadruples. To get 12 with a_i in {2,3,5,7}.Possible combinations:- All 3s: 3*4=12. So four 3s: 1 way.- Two 2s and two 5s: 2+2+5+5=14, which is over.- One 2, two 3s, and one 5: 2+3+3+5=13, over.- Three 2s and one 6: but 6 isn't allowed.Wait, maybe other combinations:- Two 3s, one 2, and one 4: 4 not allowed.Alternatively:- One 2, one 3, one 5, one 2: 2+3+5+2=12. That works. So permutations of two 2s, one 3, one 5. Number of ordered solutions: 4!/(2!1!1!)=12.Wait, 2,2,3,5. So 12 ways.Also, one 7, and the rest 2s and 3s to sum to 12 -7=5 with three numbers. 7+2+2+1=12, but 1 is not allowed. Alternatively, 7 + 3 +2 +0=12, which isn't allowed.Wait, another approach. Let's consider possible a's:Possible numbers: 2,3,5,7.Possible combinations for sum=12:- Four 3s: 3*4=12: 1 way.- Two 2s, one 3, and one 5: 2+2+3+5=12: number of ordered arrangements: 4!/(2!1!1!)=12.- One 2, three 5s: 2 +5+5+5=17, too big.- One 7, and the rest sum to 5: 7 +2+2+1=12, but 1 not allowed. 7 +3 +2 +0=12, no.So only two types: four 3s (1 way) and two 2s, one 3, one 5 (12 ways). Total 13.Wait, but this contradicts previous thought. Wait, let's check:Four 3s: 3*4=12: 1 way.Two 2s, one 3, one 5: 2+2+3+5=12: permutations: 4! / (2!1!1!) = 12.Any other combinations?For example, three 2s and one 6: but 6 isn't allowed.How about one 5 and three 3s: 5+3+3+3=14, over.One 7 and three 2s: 7+2+2+2=13, over.One 7, one 3, and two 2s: 7+3+2+2=14, over.So no other combinations. So total ordered solutions:1 +12=13.Then, b's sum=14 -12=2. Which is (1,1): 1 ordered pair. So total solutions:13*1=13.Thus, total solutions for n=14:From a's sum=8: 1*3=3a's sum=9:4*4=16a's sum=10:6*3=18a's sum=11:8*2=16a's sum=12:13*1=13Total:3+16=19, 19+18=37, 37+16=53, 53+13=66.So s_14=66, which is even. Therefore, n=14 is not counted, which matches our generating function result. So this is correct.Similarly, let's check n=12. From our GF mod 2, s_12 should be odd. Let's compute s_12.For n=12: a's sum + b's sum=12.Possible a's sum: from 8 to 10 (since b's sum from 2 to 4). Wait, a's sum=12 - b's sum. Since b's sum can be up to 8, but a's sum must be at least 8, so b's sum <=4. Therefore, a's sum=8,9,10,12 -4=8, but a's sum must be >=8. Wait, if n=12, then a's sum=12 - b's sum. Since b's sum is at least 2, a's sum=12 -2=10, down to 12 -8=4, but a's sum >=8. Therefore, b's sum <=4. So a's sum=12 -2=10, 12 -3=9, 12 -4=8. So a's sum=10,9,8.For a's sum=10: number of solutions as before was 6. Then b's sum=2: ordered pairs (1,1):1. So 6*1=6.a's sum=9: number of solutions=4. b's sum=3: ordered pairs (1,2),(2,1). So 2. Total 4*2=8.a's sum=8: number of solutions=1. b's sum=4: ordered pairs (1,3),(2,2),(3,1). So 3. Total 1*3=3.Total solutions=6+8+3=17. Which is odd. Therefore, s_12=17, which is odd. Hence, n=12 is counted. So our generating function approach gives correct result.Similarly, check n=18. According to our GF, s_18 is odd. Let's see.For n=18: a's sum + b's sum=18.Minimum a's sum=8, so b's sum=10. But b's sum max is 8. Therefore, impossible. Wait, no, wait: b's sum can be up to 8 (4+4). Therefore, a's sum=18 -8=10, up to 18 -2=16. But a's sum can be up to 28, but here a's sum from 10 to16.So a's sum=10,11,12,13,14,15,16. Wait, but a's sum must be achievable by four numbers in {2,3,5,7}. Let's see.But maybe it's too time-consuming to compute. However, according to generating function, s_18 is odd. Since our generating function approach seems correct for n=10,12,14, maybe we can trust it for n=18.Therefore, the conclusion is that there are 12 values of n where s_n is odd: 10,12,18,20,22,24,26,28,30,32,34,36. Therefore, the answer is 12.But just to make sure, let's check the number of exponents in the final generating function mod 2. We had 12 terms:10,12,18,20,22,24,26,28,30,32,34,36. Yes, 12 terms. So the answer is 12.Final Answerboxed{12}

✅:To find the number of ( n ) for which ( s_n ) is odd, we use generating functions modulo 2. The generating function for each ( a_i ) (where ( a_i in {2, 3, 5, 7} )) is ( x^2 + x^3 + x^5 + x^7 ), and for each ( b_i ) (where ( b_i in {1, 2, 3, 4} )) is ( x + x^2 + x^3 + x^4 ).The generating function for four ( a_i ) variables is ( (x^2 + x^3 + x^5 + x^7)^4 ) and for two ( b_i ) variables is ( (x + x^2 + x^3 + x^4)^2 ). We compute these modulo 2:1. Generating function for ( b_i ) variables modulo 2: [ (x + x^2 + x^3 + x^4)^2 equiv x^2 + x^4 + x^6 + x^8 mod 2 ]2. Generating function for ( a_i ) variables modulo 2: [ (x^2 + x^3 + x^5 + x^7)^4 equiv x^8 + x^{12} + x^{20} + x^{28} mod 2 ]3. Multiplying the two results modulo 2: [ (x^8 + x^{12} + x^{20} + x^{28}) cdot (x^2 + x^4 + x^6 + x^8) mod 2 ] Expanding this product: [ x^{10} + x^{12} + x^{18} + x^{20} + x^{22} + x^{24} + x^{26} + x^{28} + x^{30} + x^{32} + x^{34} + x^{36} ]The exponents where the coefficients are 1 (indicating ( s_n ) is odd) are 10, 12, 18, 20, 22, 24, 26, 28, 30, 32, 34, and 36. Thus, the number of ( n ) for which ( s_n ) is odd is (boxed{12}).

🔑:1. Formulate the generating function: We need to find the number of solutions to the equation (a_1 + a_2 + a_3 + a_4 + b_1 + b_2 = n), where (a_i in {2, 3, 5, 7}) and (b_i in {1, 2, 3, 4}). The generating function for (a_i) is: [ (x^2 + x^3 + x^5 + x^7)^4 ] and for (b_i) is: [ (x + x^2 + x^3 + x^4)^2 ] Therefore, the generating function for the entire equation is: [ (x^2 + x^3 + x^5 + x^7)^4 (x + x^2 + x^3 + x^4)^2 ]2. Expand the generating function: We need to expand the generating function to find the coefficients of (x^n). The expansion is: [ (x^2 + x^3 + x^5 + x^7)^4 (x + x^2 + x^3 + x^4)^2 ] This expansion results in: [ x^{36} + 2x^{35} + 7x^{34} + 12x^{33} + 25x^{32} + 42x^{31} + 67x^{30} + 104x^{29} + 141x^{28} + 194x^{27} + 241x^{26} + 292x^{25} + 337x^{24} + 362x^{23} + 381x^{22} + 372x^{21} + 351x^{20} + 314x^{19} + 261x^{18} + 208x^{17} + 152x^{16} + 104x^{15} + 66x^{14} + 36x^{13} + 17x^{12} + 6x^{11} + x^{10} ]3. Identify the odd coefficients: We need to count the number of odd coefficients in the expansion. The coefficients are: [ 1, 2, 7, 12, 25, 42, 67, 104, 141, 194, 241, 292, 337, 362, 381, 372, 351, 314, 261, 208, 152, 104, 66, 36, 17, 6, 1 ] The odd coefficients are: [ 1, 7, 25, 67, 141, 241, 337, 381, 351, 261, 17, 1 ] There are 12 odd coefficients.4. Conclusion: The number of (n) for which (s_n) is odd is 12.The final answer is (boxed{12})