🧠:Okay, let's tackle this problem about the school table tennis championship. The question is: In an Olympic system tournament (which I think is another name for a single-elimination tournament), the winner won six matches. How many participants in the tournament won more games than they lost?First, let me make sure I understand the problem correctly. In a single-elimination tournament, each loss eliminates a participant, right? So every game results in one person advancing and the other being eliminated. The tournament continues until there's only one winner left. The problem states that the winner won six matches. So, how many participants are there in total, and how many of them won more games than they lost?Wait, actually, the question isn't asking for the total number of participants, but specifically how many participants won more games than they lost. The answer might not be all participants except the ones who lost early, but maybe there's a pattern here.Let me start by recalling how single-elimination tournaments work. Each round halves the number of participants (assuming it's a power of two; otherwise, some byes might be given). The winner needs to win six matches, so the tournament must have had six rounds. That means the number of participants is 2^6 = 64? Because each round halves the number, starting from 64: 32, 16, 8, 4, 2, 1. So yes, 64 participants. But maybe that's not necessarily the case if there are byes, but the problem doesn't mention byes. So perhaps we can assume it's a perfect bracket with 64 players.But the question is, how many participants won more games than they lost. Let's think. Each participant except the winner is eliminated at some point. The winner won 6 games and lost 0. The runner-up lost in the final, so they won 5 games and lost 1. The two semifinalists who didn't make it to the final each lost in the semifinals, so they would have won 4 games and lost 1. Wait, no: let me check again.Wait, in each round, you play one game. If you lose, you're out. So the number of games a participant wins is equal to how many rounds they advanced. For example, the winner wins 6 games (each round from 1 to 6). The runner-up would have won 5 games (reaching the final) and lost 1 (the final). The two semifinalists who lost in the semifinals would have won 4 games and lost 1. The quarterfinalists who lost in the quarterfinals would have won 3 games and lost 1. Similarly, those who lost in the first round would have 0 wins and 1 loss.Wait, so each participant except the winner has exactly one loss. Because once you lose, you're out. The winner never loses. So, the number of losses for each participant is either 1 (for all except the winner) or 0 (for the winner). But wait, no, the winner doesn't have any losses. So, all other participants have exactly one loss.But then, when they ask how many participants won more games than they lost, since each participant's number of losses is either 0 or 1. So, to have more wins than losses, a participant must have at least 1 win (so that 1 win vs 1 loss is equal, but 2 wins vs 1 loss would be more). Wait, if a participant has 2 wins and 1 loss, then 2 > 1. But in a single-elimination tournament, once you lose, you're out. So participants can only have 0 or 1 loss. So, the participants who have more wins than losses are those who have at least 1 win, because if they have 1 win and 1 loss, that's equal. Wait, 1 win and 1 loss is equal, so not more. So participants need to have at least 2 wins and 1 loss? But wait, in a single elimination tournament, once you lose, you can't play anymore. So if you have 1 loss, you are eliminated. Therefore, the only way to have more wins than losses is to have at least 1 more win than loss. Since the maximum number of losses is 1, the participants must have at least 1 win. But 1 win and 1 loss is equal. Therefore, the participants with 2 wins and 1 loss would have 2>1. Wait, but if you have 2 wins, you would have to lose in the third round. Wait, let me think through the rounds.Let's model the tournament. If there are 64 participants:Round 1: 64 players, 32 games. 32 winners advance.Round 2: 32 players, 16 games. 16 winners advance.Round 3: 16 players, 8 games. 8 winners advance.Round 4: 8 players, 4 games. 4 winners advance.Round 5: 4 players, 2 games. 2 winners advance to the final.Round 6: 2 players, 1 game. Winner is champion.So, each participant's number of wins is equal to the number of rounds they survived. The champion has 6 wins. The runner-up has 5 wins. The two semifinalists have 4 wins each. The four quarterfinalists have 3 wins each. The eight players who made it to Round 4 (Round of 16?) Wait, wait, let's see:Wait, Round 1: 64 → 32Round 2: 32 → 16Round 3: 16 → 8Round 4: 8 → 4Round 5: 4 → 2Round 6: 2 → 1So, the rounds are numbered 1 through 6. Each subsequent round halves the participants.So, participants eliminated in Round 1: 32 players, 0 wins, 1 loss.Participants eliminated in Round 2: 16 players, 1 win, 1 loss.Participants eliminated in Round 3: 8 players, 2 wins, 1 loss.Participants eliminated in Round 4: 4 players, 3 wins, 1 loss.Participants eliminated in Round 5 (semifinals): 2 players, 4 wins, 1 loss.Participant eliminated in Round 6 (final): 1 player, 5 wins, 1 loss.Champion: 1 player, 6 wins, 0 losses.So, the number of participants who have more wins than losses are those who have wins > losses. Since each eliminated player has 1 loss, to have more wins than losses, they need to have at least 2 wins. Because 2 wins and 1 loss: 2 > 1. If they have 1 win and 1 loss, it's equal. So the players who were eliminated in Round 3 or later. Because:Eliminated in Round 1: 0 wins, 1 loss → 0 > 1? No, 0 < 1.Eliminated in Round 2: 1 win, 1 loss → 1 = 1.Eliminated in Round 3: 2 wins, 1 loss → 2 > 1.Similarly, those eliminated in Round 4: 3 > 1, Round 5: 4 > 1, Round 6: 5 > 1. And the champion has 6 > 0.Therefore, the participants who won more games than they lost are:- Champion: 1 participant (6-0)- Runner-up: 1 participant (5-1)- Semifinalists: 2 participants (4-1 each)- Quarterfinalists: 4 participants (3-1 each)- Those eliminated in Round 4 (which is the Round of 8? Wait, no. Wait, in Round 4, starting from 8 players, after Round 4, 4 players are eliminated (since 8 becomes 4). Wait, Round 1: 64→32, Round 2: 32→16, Round 3:16→8, Round 4:8→4, Round 5:4→2, Round 6:2→1.Wait, so participants eliminated in Round 4 are the 4 players who lost in the quarterfinals (since Round 4 is the quarterfinals). Wait, Round 1 is the first round, Round 2 is the second, etc. So Round 4 would be the quarterfinals (since Round 5 is semifinals, Round 6 is final). Therefore, the participants eliminated in Round 4 (quarterfinals) have 3 wins (they won Rounds 1, 2, 3, then lost Round 4). So 3 wins, 1 loss.So participants with more wins than losses:- Champion: 1 (6-0)- Runner-up: 1 (5-1)- Semifinalists: 2 (4-1)- Quarterfinalists: 4 (3-1)- Players eliminated in Round 3 (Round of 16?): 8 players (2 wins, 1 loss). Wait, Round 3: 16 players, 8 games. 8 winners go to Round 4. The 8 losers are eliminated with 2 wins and 1 loss.Wait a second, let me check again:Participants eliminated in Round 1: 32 players, 0 wins, 1 loss.Eliminated in Round 2: 16 players, 1 win, 1 loss.Eliminated in Round 3: 8 players, 2 wins, 1 loss.Eliminated in Round 4: 4 players, 3 wins, 1 loss.Eliminated in Round 5: 2 players, 4 wins, 1 loss.Eliminated in Round 6: 1 player, 5 wins, 1 loss.Champion: 1 player, 6 wins, 0 losses.Therefore, the participants who have more wins than losses are those with 2,3,4,5, or 6 wins. Let's count them:- 6 wins: 1 (champion)- 5 wins: 1 (runner-up)- 4 wins: 2 (semifinalists)- 3 wins: 4 (quarterfinalists)- 2 wins: 8 (those eliminated in Round 3)So total participants with more wins than losses: 1 +1 +2 +4 +8 = 16.But wait, the total number of participants is 64. Let's verify:- Eliminated in Round 1:32- Eliminated in Round 2:16- Eliminated in Round 3:8- Eliminated in Round 4:4- Eliminated in Round 5:2- Eliminated in Round 6:1- Champion:1Total:32+16+8+4+2+1+1=64. Correct.So participants with more wins than losses: 8 (Round 3 losers) +4 (Round 4) +2 (Round 5) +1 (Round 6) +1 (Champion) =16. Wait, 8+4 is 12, +2 is 14, +1 is 15, +1 is 16. Yes.So the answer is 16 participants.But let me cross-check. Each round, the number of participants halves. The champion has 6 wins, which implies 6 rounds, so 2^6=64 participants.The players who have more wins than losses are those who won at least two games (since they have one loss). That is, players eliminated in Round 3 or later. Because:- Eliminated in Round 3: 2 wins, 1 loss → 2>1- Eliminated in Round 4:3>1- Eliminated in Round 5:4>1- Eliminated in Round 6:5>1- Champion:6>0So the count is:Round 3 eliminations:8Round 4 eliminations:4Round 5 eliminations:2Round 6 eliminations:1Champion:1Total:8+4+2+1+1=16.Yes, that seems right.Another way to think: In a single-elimination tournament, the number of players who won exactly k matches is equal to the number of players eliminated in the (k+1)th round. Because to have k wins, you must have survived k rounds and lost in the (k+1)th round. Except the champion, who has 6 wins and wasn't eliminated.So players with exactly 0 wins: eliminated in Round 1:32Exactly 1 win: eliminated in Round 2:16Exactly 2 wins: eliminated in Round3:8Exactly3 wins: eliminated in Round4:4Exactly4 wins: eliminated in Round5:2Exactly5 wins: eliminated in Round6:1Exactly6 wins:1 (champion)Thus, players with more wins than losses are those with 2,3,4,5,6 wins. Sum their numbers:8+4+2+1+1=16.Therefore, the answer should be 16 participants.But let me check again with a smaller example to see if this logic holds. Suppose a tournament with 2 participants. Winner has 1 win, loser has 0. So participants who won more than lost: only the champion. So answer is 1. According to formula, n=2, which is 2^1. The answer would be 1. Which matches.Another example: 4 participants.R1: A vs B, C vs D. Winners go to final.R2: Final.Champion has 2 wins, runner-up has 1 win, two R1 losers have 0.Participants with more wins than losses: champion (2-0) and runner-up (1-1). Wait, runner-up has 1 win and 1 loss. So equal. So only champion. But according to our formula, for n=4 (2^2), number of participants with more wins than losses would be:Eliminated in R2 (final):1 participant (1 win, 1 loss) → not counted.Eliminated in R1:2 participants (0 wins).Champion:1 (2 wins).But according to the previous formula, participants with more wins than losses would be those with at least 2 wins (since 2>0, but in this case, only the champion). Wait, but in the 4-player example, according to our general formula, the answer would be 1 (champion). But according to our previous breakdown for 64 players, we considered players eliminated in Round 3 (k=2) and above. Wait, for n=4, the rounds are 2: R1 and R2.Wait, in a 4-player tournament:Round1:4 players, 2 games. 2 winners.Round2:2 players, 1 game. 1 champion.So players eliminated in Round1:2 players (0 wins, 1 loss)Players eliminated in Round2:1 player (1 win, 1 loss)Champion:1 (2 wins, 0 losses)So participants with more wins than losses: only the champion (2>0). The runner-up has 1=1. So answer is 1. According to our formula, for 2^2=4, the answer would be 1 (champion) + 0 (others). But according to the previous logic for 64 players, the formula gives 16. But in the 4-player example, the formula seems to hold. Let me check with 8 players.8 players:Round1:8→4Round2:4→2Round3:2→1Champion has 3 wins.Players:Eliminated in R1:4 (0 wins)Eliminated in R2:2 (1 win)Eliminated in R3:1 (2 wins)Champion:1 (3 wins)Participants with more wins than losses:Eliminated in R3:1 (2 wins, 1 loss: 2>1)Champion:1 (3>0)So total:2. According to our formula, for 2^3=8, participants with more wins than losses would be:Eliminated in R3:2^(3-3)=1? Wait, no, according to the previous count:In 64 players:Eliminated in R3:8=2^(6-3)Similarly, for 8 players:Eliminated in R3:1=2^(3-3)=1Eliminated in R2:2^(3-2)=2Eliminated in R1:4=2^(3-1)=4So participants with more wins than losses:Eliminated in R3:1 (2 wins, 1 loss)Champion:1 (3 wins)Total:2.Which is correct, since in 8 players:- Runner-up (eliminated in R3): 2 wins, 1 loss.- Champion:3 wins.So total 2. But according to the formula for 64 players: sum from k=2 to 6 of 2^(6 - k). Wait, 6 rounds:Sum from k=2 to 6 of 2^(6 -k):For k=2:2^(6-2)=16But in our 8-player example, sum from k=2 to 3 of 2^(3 -k):k=2:2^(3-2)=2k=3:2^(3-3)=1Total 3, but actual participants with more wins are 2. Hmm, discrepancy here.Wait, maybe my formula is off. Wait in the 64-player example, the number of participants eliminated in each round is 32,16,8,4,2,1 for rounds 1 to 6. Then the participants with more wins than losses are those eliminated in rounds 3 to 6 (8,4,2,1) plus the champion (1). So total 8+4+2+1+1=16.Similarly, in the 8-player example:Eliminated in R1:4Eliminated in R2:2Eliminated in R3:1Champion:1Participants with more wins than losses: eliminated in R3 (1) + champion (1) =2.Which matches the actual count. So in the 8-player example, the answer is 2.But according to the formula sum_{k=2}^{3} 2^{3 -k} +1 = (2^{1} +2^{0}) +1=2+1+1=4, which is wrong. So the formula isn't simply that.Wait, perhaps the correct way is:For a tournament with n rounds (so 2^n participants), the number of players who have more wins than losses is the sum from k=2 to n of the number of players eliminated in round k, plus the champion.Because players eliminated in round k (where k >=2) have (k-1) wins and 1 loss. So if k-1 >1, which is k>2. Wait, no:Wait, if you're eliminated in round k, you have (k-1) wins, because you have to win k-1 rounds to get to round k, then lose. So for example, eliminated in round 3, you have 2 wins (rounds 1 and 2), then lost in round 3.Therefore, to have more wins than losses: (k-1) >1, since you have 1 loss. So (k-1) >1 ⇒ k>2. So players eliminated in rounds 3 to n, plus the champion.Number of players eliminated in round k is 2^{n -k}.Therefore, total is sum_{k=3}^{n} 2^{n -k} +1.For example, in 64 players (n=6):sum_{k=3}^6 2^{6 -k} +1 =2^{3} +2^{2} +2^{1} +2^{0} +1=8+4+2+1+1=16.Yes, that works.In the 8-player example (n=3):sum_{k=3}^3 2^{3-3} +1=1 +1=2. Correct.In the 4-player example (n=2):sum_{k=3}^2 ... which is invalid, so just the champion:1. Which matches.Thus, general formula: For a tournament with 2^n players, the number of participants with more wins than losses is sum_{k=3}^n 2^{n -k} +1 = (2^{n-2} -1) +1) =2^{n-2} ?Wait, wait:sum_{k=3}^n 2^{n -k} is equal to sum_{m=0}^{n-3} 2^{m} }=2^{n-2} -1.Because sum from m=0 to t-1 of 2^m =2^{t} -1.Here, t= n-2. Because if k starts at 3, then n -k starts at n -3 down to 0 when k=n. So sum_{k=3}^n 2^{n -k} = sum_{m=0}^{n-3} 2^m }=2^{n-2} -1.Therefore, total participants with more wins than losses: (2^{n-2} -1) +1=2^{n-2}.Wait, that simplifies to 2^{n-2}.But in the 64-player example (n=6), 2^{6-2}=16. Which matches.In the 8-player example (n=3), 2^{3-2}=2. Which matches.In the 4-player example (n=2), 2^{2-2}=1. Which matches.So the general formula is 2^{n-2}, where n is the number of rounds (which is log2(number of participants)).But in the original problem, the winner won six matches, so n=6. Therefore, the answer is 2^{6-2}=16. Which is consistent with our previous calculation.Therefore, the answer is 16 participants.

✅:The tournament is a single-elimination format where the winner must win 6 matches, indicating 6 rounds and 64 participants (2⁶). In such tournaments, each loss eliminates a participant, meaning every non-winner has exactly 1 loss. To have more wins than losses, a participant must have at least 2 wins (since 1 win and 1 loss is equal). Participants are categorized by the round they were eliminated:- Eliminated in Round 1: 32 players (0 wins, 1 loss). - Eliminated in Round 2: 16 players (1 win, 1 loss). - Eliminated in Round 3: 8 players (2 wins, 1 loss). - Eliminated in Round 4: 4 players (3 wins, 1 loss). - Eliminated in Round 5 (Semifinals): 2 players (4 wins, 1 loss). - Eliminated in Round 6 (Final): 1 player (5 wins, 1 loss). - Champion: 1 player (6 wins, 0 losses). Participants with more wins than losses are those eliminated in Round 3 or later (2+ wins) and the champion:- 8 (Round 3) + 4 (Round 4) + 2 (Round 5) + 1 (Round 6) + 1 (Champion) = 16.Answer: boxed{16}

🔑:1. The problem states that the school championship is held in an elimination format (Olympic system), where each player who loses is eliminated from the tournament. Hence, the number of participants is halved after each round.2. The winner of the tournament won 6 games. This implies that there were 6 rounds in total. - Let's consider that in each round, half of the remaining participants are eliminated. - Each round reduces the number of remaining participants by half.3. We need to determine the total number of participants who started the tournament. Since there are 6 rounds in total, we can set up an equation based on powers of 2: - If there are ( n ) initial participants, after one round, there will be ( frac{n}{2} ) participants. - After two rounds, there will be ( frac{n}{4} ) participants, and so on.4. Solving for the number of initial participants: - After 6 rounds, there must be 1 remaining participant (the champion). - We have ( frac{n}{2^6} = 1 ). - Solving for ( n ): [ frac{n}{2^6} = 1 implies n = 2^6 = 64 ]5. Now let's determine the number of participants who won more games than they lost. We recognize the following patterns: - Participants who lose in the first round have 0 wins and 1 loss. - Participants who lose in the second round have 1 win and 1 loss. - Participants who lose in the third round have 2 wins and 1 loss. - Participants who lose in the fourth round have 3 wins and 1 loss. - Participants who lose in the fifth round have 4 wins and 1 loss. - Participants who lose in the sixth round have 5 wins and 1 loss. - The winner of the tournament has 6 wins and 0 losses. 6. We need to count the participants who have more wins than losses at the end of the tournament. This would be those participants who proceed beyond the second round. - Participants who reach the third round have 2 wins and 1 loss. - Each round beyond that, the number of participants halves: So, ( 2^4 = 16 ) participants reach the third round (after 2 rounds, ( 64 rightarrow 32 rightarrow 16 )). ConclusionTherefore, there are 16 participants who won more games than they lost.[boxed{16}]

🧠:Okay, so Bob is writing sequences of letters where each letter can be uppercase or lowercase, and there are specific rules about what he can write next depending on the previous letter. The problem is to find the number of sequences of 32 letters that start with lowercase 'a' and end with lowercase 'z'. Let me try to break this down step by step.First, let's make sure I understand the rules correctly. The rules depend on whether the last character was uppercase or lowercase.1. If the last letter was uppercase: - He can either write the same letter in lowercase next. - Or write the next letter in the alphabet in uppercase. For example, if he just wrote 'C', then the next can be 'c' (lowercase) or 'D' (uppercase).2. If the last letter was lowercase: - He can either write the same letter in uppercase next. - Or write the previous letter in the alphabet in lowercase. So, if he just wrote 'c', the next can be 'C' (uppercase) or 'b' (lowercase).The sequence starts with lowercase 'a' and must end with lowercase 'z' after 32 letters. So, the first character is fixed as 'a', and the 32nd character must be 'z'. We need to count all possible sequences that follow the rules and meet these start and end conditions.Hmm. This feels like a dynamic programming problem where the state depends on the current letter and its case (upper or lower), and the number of steps taken. Let me think about how to model this.Let's denote by DP[i][c][case] the number of sequences of length i that end with the character c in the specified case (upper or lower). Here, c can be from 'a' to 'z' (or 1 to 26 for easier indexing), and case is either upper or lower (0 for lower, 1 for upper maybe?).Our goal is DP[32][26][0], since we need to end at lowercase 'z' (which is the 26th letter).The initial condition is DP[1][1][0] = 1, because the sequence starts with 'a' (1st letter) in lowercase.Now, we need to figure out the transitions based on the rules.Let's formalize the transitions.If the previous character was uppercase (case=1), then for each character k in uppercase:- The next character can be the same letter in lowercase (so k in lowercase, case=0)- Or the next letter in uppercase (k+1 in uppercase, case=1), provided that k+1 is <=26.Therefore, transitions from uppercase k:- To lowercase k- To uppercase k+1 (if possible)Similarly, if the previous character was lowercase (case=0), then for each character k in lowercase:- The next character can be the same letter in uppercase (k in uppercase, case=1)- Or the previous letter in lowercase (k-1 in lowercase, case=0), provided that k-1 >=1.Therefore, transitions from lowercase k:- To uppercase k- To lowercase k-1 (if possible)So, for each step from i=1 to 31 (since we need sequences of 32 letters), we need to compute DP[i+1][...][...] based on DP[i][...][...].This seems manageable. Let me structure the DP tables.Since we have 32 steps, and for each step, 26 letters, and 2 cases, the total states are 32*26*2 = 1664, which is computationally feasible even by hand (though tedious). However, since we need an exact answer, we need to model this properly.Perhaps we can model this with a matrix or use recurrence relations.Alternatively, we can notice some patterns or symmetries to simplify the problem.Wait, the problem transitions are somewhat similar to moving forward or backward in the alphabet, but with case changes. Since starting at 'a' and ending at 'z', which are 25 letters apart, but the sequence is 32 letters long. So, there's more steps than the direct path. This suggests that there might be some oscillations or back-and-forth movements, but constrained by the rules.Let me try to see if there's a way to model this as a graph where each node is a (letter, case) pair, and edges represent allowed transitions. Then, the number of paths from (a, lower) to (z, lower) with 32 edges. But since the path length is 32, which is fixed, and each step is moving along the edges.This is exactly the DP approach as above. So, the number of sequences is the number of paths of length 31 (since we start at step 1) from (a, lower) to (z, lower).Therefore, to compute this, we can set up a DP table where for each position i (from 1 to 32), we track the number of ways to reach each (letter, case) at step i.Starting with i=1: only (a, lower) has count 1.Then, for each subsequent i, we compute the counts based on the previous counts and the transition rules.Let me formalize this.Let’s index the letters from 1 to 26, where 'a' is 1, 'b' is 2, ..., 'z' is 26.Let’s use two arrays: current_dp and next_dp. Each array is a 26x2 matrix (letters 1-26, cases 0 and 1).Initialize current_dp[1][0] = 1 (lowercase 'a'), all others 0.For each step from 1 to 31: For each letter k from 1 to 26: For each case (0 and 1): If current_dp[k][case] > 0: If case is 0 (lowercase): // From lowercase k, can go to uppercase k or lowercase k-1 // Transition to uppercase k: next_dp[k][1] += current_dp[k][0] // Transition to lowercase k-1 (if k-1 >=1) if k-1 >=1: next_dp[k-1][0] += current_dp[k][0] Else: // From uppercase k, can go to lowercase k or uppercase k+1 // Transition to lowercase k: next_dp[k][0] += current_dp[k][1] // Transition to uppercase k+1 (if k+1 <=26) if k+1 <=26: next_dp[k+1][1] += current_dp[k][1] Set current_dp = next_dp Reset next_dp to zeroAfter 31 transitions (steps 1 to 31), the current_dp should hold the counts for step 32. Then, the answer is current_dp[26][0].But doing this by hand would be too time-consuming. However, perhaps we can find a pattern or a recurrence relation that allows us to compute this efficiently.Alternatively, maybe model the problem as a system of recurrence relations.Let me denote:Let’s define two functions for each position i and letter k:- L(i, k): number of sequences of length i ending with lowercase k.- U(i, k): number of sequences of length i ending with uppercase k.Our goal is L(32, 26).Base case: L(1, 1) = 1; all other L(1, k) = 0, U(1, k) = 0.Now, transitions:For each i >=1, to compute L(i+1, k) and U(i+1, k):From lowercase k at i, you can go to uppercase k at i+1 or lowercase k-1 at i+1 (if k >=2).From uppercase k at i, you can go to lowercase k at i+1 or uppercase k+1 at i+1 (if k <=25).Therefore, the recurrence relations are:L(i+1, k) can be reached from:- U(i, k) (since uppercase k can transition to lowercase k)- L(i, k+1) (since lowercase k+1 can transition to lowercase k by moving backward)Wait, no. Wait, let's think carefully.Wait, when you are at lowercase k, you can transition to uppercase k or lowercase k-1. So, for the next step, U(i+1, k) can be reached from L(i, k). Similarly, L(i+1, k-1) can be reached from L(i, k) if k-1 >=1.Similarly, from uppercase k, you can transition to lowercase k or uppercase k+1. So, L(i+1, k) can be reached from U(i, k), and U(i+1, k+1) can be reached from U(i, k) if k+1 <=26.Wait, perhaps it's better to express the transitions as:To compute L(i+1, k):- It can come from U(i, k) (since uppercase k can transition to lowercase k)- It can come from L(i, k+1) transitioning to lowercase k: Wait, no. If we are at lowercase k+1, we can transition to uppercase k+1 or lowercase k. So, from lowercase k+1, you can go to lowercase k. Therefore, L(i+1, k) += L(i, k+1)Wait, let me think again.If the previous character is lowercase m, then from there you can either go to uppercase m or lowercase m-1.Therefore, if we are looking at L(i+1, k), which is lowercase k at step i+1, that can only come from:- From uppercase k at step i, which transitioned to lowercase k (since uppercase k can transition to lowercase k)- From lowercase k+1 at step i, which transitioned to lowercase k (since lowercase k+1 can go to lowercase k by moving backward)Similarly, U(i+1, k) can come from:- From lowercase k at step i, which transitioned to uppercase k- From uppercase k-1 at step i, which transitioned to uppercase k (since uppercase k-1 can transition to uppercase k if you add 1)Wait, no. Let's be precise.For L(i+1, k):Possible previous states:1. If the previous state was uppercase k, then you can transition to lowercase k. So, L(i+1, k) += U(i, k)2. If the previous state was lowercase k+1, then from there you can transition to lowercase k. So, L(i+1, k) += L(i, k+1)Similarly, for U(i+1, k):Possible previous states:1. If the previous state was lowercase k, then you can transition to uppercase k. So, U(i+1, k) += L(i, k)2. If the previous state was uppercase k-1, then you can transition to uppercase k (since uppercase k-1 can go to uppercase k). So, U(i+1, k) += U(i, k-1)Therefore, the recurrence relations are:L(i+1, k) = U(i, k) + L(i, k+1)U(i+1, k) = L(i, k) + U(i, k-1)With the understanding that L(i, k+1) is 0 if k+1 >26, and U(i, k-1) is 0 if k-1 <1.Similarly, L(i+1, k) will have terms only if k+1 <=26 for the L(i, k+1) part, and U(i, k-1) is only added if k-1 >=1.Okay, this seems correct. Let me verify with a small example.Suppose i=1:L(1,1)=1, others 0.Compute L(2, k):For each k:L(2, k) = U(1, k) + L(1, k+1)But U(1, k) =0 for all k, so L(2, k) = L(1, k+1). Therefore:L(2,1) = L(1,2) =0L(2,2) = L(1,3) =0...Only possible non-zero is if k+1=1, which would be k=0, invalid. So all L(2,k)=0.Wait, that can't be right. Wait, for i=1, step 1 is 'a' lowercase. Then step 2 can be either 'A' or there is no previous letter? Wait, no, step 1 is 'a' lowercase, so step 2 can be:From lowercase 'a', you can go to uppercase 'a' or lowercase 'a-1' but 'a'-1 is invalid. Therefore, from 'a' lowercase, the next can only be uppercase 'A'.Therefore, at step 2, only U(2,1)=1, and all others are zero.But according to the recurrence:For i=1:U(i+1, k) = L(i, k) + U(i, k-1)So, U(2, k) = L(1, k) + U(1, k-1)Since U(1, k-1) is all zero, U(2, k) = L(1, k). Since L(1,1)=1, so U(2,1)=1, others zero. Correct.Similarly, L(2, k) = U(1, k) + L(1, k+1). All zero. Correct.So step 2: U(2,1)=1.Now step 3:Compute L(3, k) and U(3, k).For L(3, k):L(3, k) = U(2, k) + L(2, k+1)But U(2, k) is 1 at k=1, else 0. So L(3,1) = U(2,1)=1 + L(2,2)=0 =>1Other L(3,k)=0+...=0.For U(3,k):U(3,k) = L(2, k) + U(2, k-1)But L(2,k)=0 for all k, so U(3,k) = U(2,k-1). Therefore:U(3,1) = U(2,0) invalid =>0U(3,2) = U(2,1)=1Others zero.So step 3: L(3,1)=1, U(3,2)=1.Step 4:L(4, k):L(4,1) = U(3,1) + L(3,2)=0 +0=0L(4,2)=U(3,2)=1 + L(3,3)=0 =>1Others zero.U(4,k):U(4,1)= L(3,1)=1 + U(3,0)=0 =>1U(4,2)= L(3,2)=0 + U(3,1)=0 =>0U(4,3)= L(3,3)=0 + U(3,2)=1 =>1Others zero.Wait, this is getting complicated. But let's see:At step 4:U(4,1)=1, U(4,3)=1, L(4,2)=1.So sequences so far:Start: a (lower)Step 2: A (upper)Step 3: From A (upper), can go to a (lower) or B (upper). But at step 3, we had L(3,1)=1 and U(3,2)=1. So two possibilities:1. a -> A -> a2. a -> A -> BThen step 3 is either a or B.For sequence 1 (a -> A -> a):Step 3 is a (lower). Then step 4 can be A or z (but z would be a-1 which is invalid). So only A.Which would be U(4,1)=1.For sequence 2 (a -> A -> B):Step 3 is B (upper). Then step 4 can be b (lower) or C (upper). So those would be L(4,2)=1 and U(4,3)=1.So step 4 has three possibilities: A (from sequence 1), b (from sequence 2), C (from sequence 2). Wait, but according to the DP counts, U(4,1)=1, L(4,2)=1, U(4,3)=1. So three sequences at step 4.Wait, but according to our earlier calculation, step 4's counts are U(4,1)=1, L(4,2)=1, U(4,3)=1. So that's three sequences. Hmm. But let's track them:1. a -> A -> a -> A2. a -> A -> B -> b3. a -> A -> B -> CSo yes, three sequences. Correct.So this seems to be working. Therefore, the recurrence relations are correctly modeling the transitions.Therefore, if we can compute L(i, k) and U(i, k) for all i up to 32 and k up to 26, we can get the desired count.Now, since doing this by hand up to 32 is impractical, we need to look for a pattern or perhaps a matrix exponentiation approach, or find that the number of sequences corresponds to a known combinatorial structure.Alternatively, notice that each transition is linear, so we can model this with matrix multiplication and then raise the transition matrix to the 31st power. However, the state space is 26*2=52, which would make the matrix 52x52. That's quite large, but maybe there's a structure to the matrix that can be exploited.Alternatively, notice that the transitions for uppercase and lowercase letters can be decoupled or have some symmetry.Wait, looking at the recurrence relations:L(i+1, k) = U(i, k) + L(i, k+1)U(i+1, k) = L(i, k) + U(i, k-1)This resembles some kind of coupled recurrence. Maybe we can represent this as a system of equations and find a generating function or a closed-form solution.Alternatively, think of the problem in terms of movements: each transition either changes the case or moves the letter. From lowercase, you can either go upcase same letter or move back; from uppercase, you can go downcase same letter or move forward.This seems similar to a walk on the letters where each step can either change case or move in the alphabet, but movement direction depends on the current case.Wait, another way to look at it: when in lowercase, you can either "stay" by switching to uppercase or move left (but since you can't move left from 'a', you can only switch case). When in uppercase, you can either "stay" by switching to lowercase or move right (unless at 'z', where you can't move right).Therefore, the movement alternates between lowercase and uppercase, with the possibility of moving left or right depending on case.But since we start at lowercase 'a' and need to end at lowercase 'z', which is 25 steps to the right. However, the total sequence is 32 letters, meaning 31 moves. So, we need to go from position 1 to position 26 in 31 steps, with specific movement rules.But each time you switch case, you might be able to move or stay. Wait, perhaps model this as two separate grids: one for lowercase and one for uppercase, with transitions between them.Alternatively, consider that each transition from lowercase to uppercase doesn't change the letter, but switching case. Similarly, transitioning from uppercase to lowercase doesn't change the letter. But moving in the alphabet requires either going from uppercase to next uppercase or lowercase to previous lowercase.This seems complicated. Maybe it's helpful to think in terms of the number of times Bob switches case versus moves in the alphabet.Alternatively, since each move can either change the case or move in the alphabet, but not both. So each step is either a case toggle or a letter movement.But starting from lowercase 'a', each step is either:1. From lowercase: toggle to uppercase or move left (but can't move left from 'a').2. From uppercase: toggle to lowercase or move right.Wait, but when in lowercase, moving left is allowed if possible, and when in uppercase, moving right is allowed if possible.But the movement direction is opposite depending on the case. From lowercase, you can only move left; from uppercase, you can move right.This is an interesting observation. So, effectively, the movement in the alphabet is dependent on the case: lowercase can move left, uppercase can move right. But each movement requires a case transition.Wait, no. Let me clarify:From lowercase k, you can:- Toggle to uppercase k (no movement)- Move left to lowercase k-1 (if possible)From uppercase k, you can:- Toggle to lowercase k (no movement)- Move right to uppercase k+1 (if possible)Therefore, moving left is associated with a lowercase letter, and moving right is associated with an uppercase letter. However, the movement is only possible when you are in the respective case.Therefore, to move from lowercase k to lowercase k-1, you have to stay in lowercase. To move from uppercase k to uppercase k+1, you have to stay in uppercase.But according to the transition rules:Wait, no. From lowercase k, moving left to lowercase k-1 is a transition that remains lowercase. But according to the rules, from lowercase k, you can either go to uppercase k or lowercase k-1. So moving left is possible while staying lowercase.Similarly, from uppercase k, moving right to uppercase k+1 is possible while staying uppercase.Therefore, each movement (left or right) can be done without changing case, but changing case is also possible without movement.This complicates things. So, each step can be:- A case toggle (switching between lowercase and uppercase of the same letter)- A movement left (if in lowercase) or right (if in uppercase)This means that the movement is possible in either direction, but only when in the respective case.Therefore, to make progress from 'a' to 'z', Bob needs to strategically use uppercase letters to move right and lowercase letters to move left, but since we need to end at 'z' lowercase, we must end with a lowercase letter, which could be from toggling case from uppercase 'z' or moving left from lowercase 'z+1' which doesn't exist, so actually, the last step must be a toggle from uppercase 'z' to lowercase 'z'.Wait, since we have to end at lowercase 'z', the previous character must have been uppercase 'z', because from uppercase 'z', you can only toggle to lowercase 'z' (since moving right to 'z+1' is impossible). Therefore, the 31st character must be uppercase 'z', and the 32nd is lowercase 'z'.Therefore, the problem reduces to finding the number of sequences of 31 characters that start at lowercase 'a' and end at uppercase 'z', following the transition rules, and then append the lowercase 'z' at the end.Wait, but actually, the 32nd character is lowercase 'z', so the 31st character must be uppercase 'z', because the only way to get to lowercase 'z' is from uppercase 'z'. Therefore, the problem is equivalent to finding the number of sequences of length 31 ending at uppercase 'z', starting from lowercase 'a', and then multiplying by 1 (since the last step is fixed: uppercase 'z' -> lowercase 'z').Therefore, our target is to compute U(31, 26), since the 31st character must be uppercase 'z', and then the 32nd is lowercase 'z'. Hence, the total number of sequences is equal to U(31, 26).Wait, but according to the earlier definitions, L(32, 26) = U(31, 26) + L(31, 27). But since L(31,27) is zero (because there's no 27th letter), so L(32,26) = U(31,26). Therefore, yes, the number we need is U(31,26).Therefore, if we can compute U(31,26), that's our answer.So, restating, we need to compute U(31,26), where U(i,k) follows the recurrence:U(i,k) = L(i-1,k) + U(i-1,k-1)Similarly, L(i,k) = U(i-1,k) + L(i-1,k+1)With boundary conditions that L(1,1)=1, and all other L(1,k)=0, U(1,k)=0.This is a system of linear recurrences. To solve this, perhaps we can set up a table and compute the values step by step up to i=31.Given that 31 is manageable with a computer, but since we need to do this manually, perhaps we can find a pattern or derive a closed-form solution.Alternatively, notice that the recurrence resembles a combination of movements where each step alternates between contributing to U and L terms.Alternatively, model this as a grid where we move in two dimensions: one dimension is the letter position, and the other is the case. However, it's still a bit abstract.Alternatively, think of each transition as a combination of moves and toggles. For instance, every time you toggle the case, you stay on the same letter, and every time you move, you change the letter but keep the case (except movement from lowercase to lowercase requires moving left, and uppercase to uppercase requires moving right).But this seems difficult to model.Alternatively, consider that moving from lowercase to uppercase and back can be considered as staying in place but toggling case, while moving left or right requires a specific case.Given that starting at 'a' lowercase and needing to reach 'z' lowercase, which is 25 letters to the right, but the path length is 31, we have 31 steps to make this progression. Each movement to the right requires being in uppercase and moving right, which takes a step. Each movement to the left requires being in lowercase and moving left, which also takes a step. Additionally, toggling case also takes a step but doesn't change the letter.Therefore, to go from 'a' to 'z', we need a net of 25 movements to the right. However, since we can also move left or toggle case, the total number of right movements minus left movements must be 25. However, since each right movement can only be done from uppercase, and left from lowercase, the sequence of movements must be carefully interleaved with case toggles.This is similar to a problem where you have to reach a certain position with a certain number of steps, considering that moving forward and backward have different constraints.Alternatively, think of it as a state machine where each state is the current letter and case, and transitions are allowed as per the rules. Then the number of paths from (a, lower) to (z, lower) in 31 steps.But again, calculating this requires dynamic programming.Alternatively, notice that the recurrence relations for U and L can be represented in terms of previous U and L terms, leading to a system that can be represented with generating functions or matrix exponentiation.However, given the time constraints, perhaps the best approach is to code this DP, but since we can't code here, let's try to find a pattern.Let me try to compute the values for small i and see if a pattern emerges.We have:At i=1:L(1,1)=1U(1,k)=0 for all kAt i=2:U(2,1)=1L(2,k)=0 for all kAt i=3:L(3,1)=1 (from U(2,1))U(3,2)=1 (from L(2,2)=0 and U(2,1)=1)At i=4:L(4,2)= U(3,2)=1 + L(3,3)=0 =>1U(4,1)= L(3,1)=1 + U(3,0)=0 =>1U(4,3)= L(3,3)=0 + U(3,2)=1 =>1So U(4,1)=1, U(4,3)=1; L(4,2)=1At i=5:L(5,1)=U(4,1)=1 + L(4,2)=1 =>2L(5,3)=U(4,3)=1 + L(4,4)=0 =>1U(5,2)= L(4,2)=1 + U(4,1)=1 =>2U(5,4)= L(4,4)=0 + U(4,3)=1 =>1So L(5,1)=2, L(5,3)=1; U(5,2)=2, U(5,4)=1i=6:L(6,2)=U(5,2)=2 + L(5,3)=1 =>3L(6,4)=U(5,4)=1 + L(5,5)=0 =>1U(6,1)=L(5,1)=2 + U(5,0)=0 =>2U(6,3)=L(5,3)=1 + U(5,2)=2 =>3U(6,5)=L(5,5)=0 + U(5,4)=1 =>1So L(6,2)=3, L(6,4)=1; U(6,1)=2, U(6,3)=3, U(6,5)=1Hmm. Observing these values, there seems to be a pattern similar to Pascal's triangle, but interleaved for uppercase and lowercase. For example, at i=3, L(3,1)=1, U(3,2)=1. At i=4, L(4,2)=1, U(4,1)=1, U(4,3)=1. Then at i=5, L(5,1)=2, L(5,3)=1; U(5,2)=2, U(5,4)=1. It seems like the counts are combining previous counts in a way that resembles combinations.Alternatively, perhaps the number of ways to reach position k at step i is related to binomial coefficients, adjusted by the constraints of movement.Alternatively, since each transition can be considered as a combination of moves and toggles, perhaps the number of sequences is equivalent to the number of balanced sequences of moves and toggles that result in the net movement of 25 steps to the right.But given the complexity of the dependencies, it's hard to see immediately.Alternatively, notice that each time Bob is in lowercase, he can either toggle to uppercase or move left; each time he is in uppercase, he can toggle to lowercase or move right.This resembles a two-state system (lowercase and uppercase) where in each state, you have two choices. However, the movement direction depends on the state.To get from lowercase 'a' to lowercase 'z', he must have a net of 25 movements to the right. Each movement to the right can only be made from uppercase, and each movement to the left can only be made from lowercase.However, since he starts in lowercase and must end in lowercase, the number of times he toggles from lowercase to uppercase must be equal to the number of times he toggles from uppercase to lowercase. Except that he starts in lowercase and ends in lowercase, so the number of toggles from lowercase to uppercase must be one less than the number of toggles from uppercase to lowercase. Wait, no: starting in lowercase, each toggle to uppercase must be matched with a toggle back to lowercase.Actually, the number of toggles from lowercase to uppercase must equal the number of toggles from uppercase to lowercase. Because each time you toggle up, you have to toggle back down to return to lowercase.But since we start in lowercase and end in lowercase, the number of up toggles equals the number of down toggles.Let’s denote:Let’s say Bob toggles case m times (each toggle is either up or down). Since he starts in lowercase, each up toggle must be followed by a down toggle, except possibly at the end. But since he ends in lowercase, the number of up toggles must equal the number of down toggles. So m is even, and the number of up toggles is m/2, down toggles is m/2.Each toggle takes a step, and the remaining steps (31 - m) are movements: some to the left (when in lowercase) and some to the right (when in uppercase).Let’s denote the number of right movements as R and left movements as L.Each right movement can only occur when in uppercase, and each left movement can only occur when in lowercase.Given that he starts in lowercase, the sequence of states (lowercase or uppercase) will alternate based on the toggles.Each up toggle switches to uppercase, and each down toggle switches back to lowercase.Between toggles, Bob can make movements. However, after an up toggle, he is in uppercase and can make right movements until he toggles back down. Similarly, after a down toggle, he is in lowercase and can make left movements until he toggles back up.Given that the total number of steps is 31, which includes both toggles and movements.Let’s try to model this.Suppose Bob toggles up m times and toggles down m times (since they must be equal). So total toggles: 2m.Therefore, the remaining steps for movements: 31 - 2m.These movements are divided into periods when he is in uppercase or lowercase.Starting from lowercase, the sequence of states is:lowercase (start) -> [movement left]* -> toggle up -> uppercase -> [movement right]* -> toggle down -> lowercase -> [movement left]* -> ... etc.Since he starts in lowercase, the first toggle must be up, and the last toggle must be down to end in lowercase.Each up toggle is followed by a period in uppercase where he can make right movements, and each down toggle is followed by a period in lowercase where he can make left movements.Given m toggles up and m toggles down, there are m periods in uppercase and m+1 periods in lowercase (including the initial one before any toggles).Wait, starting in lowercase:1. First period: lowercase, can make L0 left moves.2. Toggle up.3. Uppercase, can make R1 right moves.4. Toggle down.5. Lowercase, can make L1 left moves.6. Toggle up....After m toggles up and m toggles down:- There are m uppercase periods (R1, R2, ..., Rm) and m+1 lowercase periods (L0, L1, ..., Lm).Each Rk >=0, each Lk >=0 (except L0 can be zero only if he toggles up immediately).The total number of movements is sum_{k=0}^m Lk + sum_{k=1}^m Rk = 31 - 2m.The net movement is sum_{k=1}^m Rk - sum_{k=0}^m Lk =25 (since he needs to move from 1 to 26, which is +25).Therefore, we have two equations:1. sum_{k=0}^m Lk + sum_{k=1}^m Rk =31 - 2m2. sum_{k=1}^m Rk - sum_{k=0}^m Lk =25Adding these two equations:2*sum_{k=1}^m Rk =56 - 2m => sum_{k=1}^m Rk =28 - mSubtracting the second equation from the first:2*sum_{k=0}^m Lk = (31 -2m) -25=6 -2m => sum_{k=0}^m Lk =3 -mBut sum_{k=0}^m Lk =3 -mBut sum_{k=0}^m Lk must be non-negative since each Lk >=0. Therefore, 3 -m >=0 =>m <=3.Also, m >=0.Therefore, possible m values: 0,1,2,3.But let's check:If m=0: sum Lk =3 -0=3, but m=0 implies no toggles, so he stays in lowercase the entire time. But sum Lk =3, sum Rk=28 -0=28. But since m=0, there are no Rk (sum Rk=0). Contradiction. So m=0 invalid.If m=1: sum Lk=3 -1=2, sum Rk=28 -1=27.But with m=1:There are L0, L1 (since m=1, there are 2 lowercase periods) and R1.Total movements: L0 + L1 + R1 =31 -2*1=29.But L0 + L1=2, R1=27. Indeed 2 +27=29.Net movement: R1 - (L0 +L1)=27 -2=25, which matches.Similarly for m=2: sum Lk=3 -2=1, sum Rk=28 -2=26.With m=2:L0, L1, L2 (3 lowercase periods), R1, R2.Total movements: L0 + L1 + L2 + R1 + R2=31 -4=27.But sum Lk=1, sum Rk=26. 1 +26=27.Net movement:26 -1=25. Correct.For m=3: sum Lk=3 -3=0, sum Rk=28 -3=25.With m=3:L0, L1, L2, L3 (4 lowercase periods), R1, R2, R3.Total movements: L0 + L1 + L2 + L3 + R1 + R2 + R3=31 -6=25.But sum Lk=0, sum Rk=25. 0 +25=25.Net movement:25 -0=25. Correct.m=4 would require sum Lk=3 -4= -1, invalid.Therefore, possible m values are 1,2,3.Therefore, the total number of sequences is the sum over m=1,2,3 of the number of ways to split the movements and toggles for each m.Now, for each m, compute the number of sequences.First, let's handle each m separately.Case m=1:He toggles up once and down once (2 toggles), leaving 31 -2=29 movements.As per above:sum Lk=2 (L0 + L1=2), sum Rk=27 (R1=27).The sequence of states is:lowercase (L0 moves left) -> toggle up -> uppercase (R1 moves right) -> toggle down -> lowercase (L1 moves left).But starting at 'a' (position 1). After L0 left moves: since starting at lowercase 'a', left moves from 'a' would go to 'a-1', but 'a' is position 1, so L0 must be 0, because you can't move left from 'a'. Therefore, L0 >=0, but movement left from 'a' is impossible, so L0 must be 0.But sum Lk=2, so L0 + L1=2. Since L0=0, L1=2.But after toggling down to lowercase, he is at position 1 + R1. Because during the uppercase period, he made R1 right moves starting from position 1 (since he toggled up from 'a'). Each right move in uppercase increments the position by 1. So after R1 moves, he is at position 1 + R1. Then he toggles down to lowercase, and makes L1=2 left moves, ending at position (1 + R1) - L1.But he needs to end at position 26 (lowercase). Therefore:(1 + R1) - L1 =26But R1=27 (from sum Rk=27), and L1=2.Therefore:(1 +27) -2=26, which is 26=26. Correct.But we need to check if the path is valid. Specifically, during the uppercase period, when making R1=27 right moves, starting from position 1. But the maximum possible position is 26. Moving right 27 times from position 1 would go to position 28, which is invalid. Therefore, this is impossible.Ah, here's a problem. R1=27 is impossible because you can't move beyond 'z' (position 26). Therefore, even though the equations are satisfied, the actual moves are invalid.Therefore, m=1 is invalid because it requires moving right 27 times from position 1, which is impossible.Similarly, check for m=2 and m=3.Case m=2:sum Lk=1 (L0 + L1 + L2=1), sum Rk=26 (R1 + R2=26).The sequence of states:lowercase (L0) -> toggle up -> uppercase (R1) -> toggle down -> lowercase (L1) -> toggle up -> uppercase (R2) -> toggle down -> lowercase (L2).Starting at position 1:After L0 left moves: position 1 - L0. But L0 must be 0 because you can't move left from 'a'.So L0=0.Then toggle up to uppercase, position 1.R1 right moves: position 1 + R1.Toggle down to lowercase, position 1 + R1.L1 left moves: position (1 + R1) - L1.Toggle up to uppercase, position (1 + R1) - L1.R2 right moves: position (1 + R1 - L1) + R2.Toggle down to lowercase, position (1 + R1 - L1) + R2.L2 left moves: position (1 + R1 - L1 + R2) - L2.This must equal 26 (lowercase z).Also, sum Lk= L0 + L1 + L2=0 + L1 + L2=1.Sum Rk= R1 + R2=26.So:Final position: (1 + R1 - L1 + R2) - L2 =26But 1 + R1 + R2 - (L1 + L2)=26Since R1 + R2=26, and L1 + L2=1.Therefore:1 +26 -1=26, which is 26=26. Correct.But we need to ensure that during each movement phase, the position stays within 1 to 26.Let's track the positions:Start at 1 (lowercase).L0=0: still at 1.Toggle up: still at 1 (uppercase).R1 right moves: position 1 + R1. Since R1 <=25 (because R1 + R2=26, R2 >=0, so R1 <=26). Wait, but if R1=26, R2=0. Then position after R1=26 would be 1 +26=27, which is invalid. Similarly, R1=25, R2=1: position after R1=25 is 26, then R2=1 would take it to 27, invalid.Therefore, R1 and R2 must be such that during each phase, the position does not exceed 26.Let's denote:After R1 right moves from position 1: position 1 + R1 <=26 => R1 <=25.Then, after R1=25, position=26. Then toggle down to lowercase, position 26.L1 left moves: position 26 - L1. Since L1 + L2=1, L1 can be 0 or1.If L1=0: then L2=1. Position after L1=0 is 26, toggle up to uppercase, R2=26 - R1=26 -25=1. Then R2=1: position 26 +1=27 invalid.If L1=1: position 26 -1=25. Toggle up to uppercase, position 25. R2=26 - R1=26 - (25 - R2)=? Wait, sum R1 + R2=26. If R1=25, then R2=1. Then position after toggle up is 25. R2=1: position 25 +1=26. Toggle down to lowercase, position 26. L2=0 (since L1=1, L2=0). Final position 26 -0=26. Valid.Therefore, for m=2, one possible path:L0=0, R1=25, L1=1, R2=1, L2=0.Check positions:1 (lower) -> toggle -> 1 (upper) -> 25 moves right to 26 (upper) -> toggle -> 26 (lower) -> 1 move left to 25 (lower) -> toggle ->25 (upper) ->1 move right to26 (upper) -> toggle ->26 (lower). Final position 26 (lower). Correct.But are there other possibilities for m=2?Suppose R1=24, then R2=2.After R1=24: position=1 +24=25 (upper). Toggle down to 25 (lower). L1 + L2=1.If L1=0: then L2=1. Position after L1=0:25 (lower). Toggle up to25 (upper). R2=2: 25 +2=27 invalid.If L1=1: position=25 -1=24 (lower). Toggle up to24 (upper). R2=2:24 +2=26 (upper). Toggle down to26 (lower). L2=0: final position26. Valid.So another valid path.Similarly, R1 can range from 0 to25, but must ensure that after each movement, the position remains <=26.However, each time R1 is chosen, R2=26 - R1.But when R1 is chosen, the position after R1 must be <=26, and after R2, position must be <=26.So:After R1: position=1 + R1 <=26 => R1 <=25.After R2: position= (position after L1) + R2.But position after L1: after toggling down from uppercase position=1 + R1, then moving left L1:position=1 + R1 - L1.Then toggling up to uppercase, then moving right R2=26 - R1.Thus, position after R2= (1 + R1 - L1) + (26 - R1) =27 - L1.But 27 - L1 <=26 => L1 >=1.But sum L1 + L2=1, so L1 can be 1, L2=0 or L1=0, L2=1.But if L1=0, then position after R2=27 -0=27 invalid.Therefore, L1 must be1, L2=0.Therefore, for m=2, the only valid possibility is L1=1, L2=0.Therefore, in this case, regardless of R1, as long as R1 + R2=26 and L1=1, then the final position is 26.But we must ensure that during all movements, positions are valid.Specifically:- After R1: 1 + R1 <=26 => R1 <=25.- After L1=1: position=1 + R1 -1= R1.- Then R2=26 - R1.- Position after R2= R1 + (26 - R1)=26.Which is valid.Therefore, for m=2, any R1 from0 to25, with R2=26 - R1, L0=0, L1=1, L2=0.But also, during the lowercase period after R1, when making L1=1 move left from position 1 + R1 (lowercase), we must have position >=2 to move left. Because moving left from position1 is invalid.Wait, after R1 right moves from position1 (uppercase), the position is1 + R1. Toggling to lowercase, position1 + R1. Then moving left L1=1 requires that position1 + R1 -1 >=1 => R1 >=1.Because if R1=0, then position after toggling is1 (lowercase), moving left would be invalid.Therefore, R1 >=1.Therefore, R1 ranges from1 to25, R2=26 - R1.Therefore, number of valid sequences for m=2 is the number of ways to choose R1 from1 to25, which is25.But each such sequence corresponds to different paths.However, each such choice of R1 and R2 corresponds to different numbers of movements in each phase, which are indistinct except for the number of movements. However, the problem allows for multiple same letters in a row, so each movement is a step where the choice is made.But wait, in the uppercase phase with R1 movements, each movement is a step to the right, which corresponds to writing the next uppercase letter each time. Similarly, in the lowercase phase with L1=1 movement, it's a step to the left.But in terms of the sequence, each movement corresponds to a specific letter being written.However, in terms of counting the number of sequences, when you have R1 right moves in uppercase, that corresponds to R1 steps where each step is the next uppercase letter. Similarly, L1=1 left move is one step to the previous lowercase letter.But each such movement phase consists of consecutive same-case letters, either moving or staying.But in the context of the problem, the number of sequences corresponding to a particular set of R1, R2, L1, L2 is determined by the order of movements and toggles. However, since the toggles are fixed at m=2 (two up and two down), and the movements are in specific phases, the number of sequences is determined by the number of ways to interleave these movements and toggles.But wait, in this model, the toggles and movements are in fixed order: lowercase period -> toggle up -> uppercase period -> toggle down -> lowercase period -> toggle up -> uppercase period -> toggle down -> lowercase period.The movements within each period are consecutive. For example, the first lowercase period has L0=0 movements, then toggle up, then R1 movements, toggle down, then L1=1 movements, toggle up, then R2 movements, toggle down, then L2=0 movements.Therefore, the sequence of steps is fixed in terms of the order of toggles and movements.However, in reality, the toggles can be interleaved with movements in any order, as long as the state changes are respected.Wait, no. Each toggle switches the case, and between toggles, you can make any number of movements in the current case.Therefore, the sequence is a series of blocks, each block consisting of some number of movements in a specific case, followed by a toggle.However, since we have m=2 up toggles and m=2 down toggles, the sequence must start with a lowercase block (possibly empty), followed by alternating uppercase and lowercase blocks, each possibly empty except for the last lowercase block.But given that we start in lowercase and end in lowercase, the structure is:Lowercase block (L0) -> toggle up -> Uppercase block (R1) -> toggle down -> Lowercase block (L1) -> toggle up -> Uppercase block (R2) -> toggle down -> Lowercase block (L2).Where L0, L1, L2 >=0, R1, R2 >=0.But according to our earlier analysis, L0=0, L1=1, L2=0.But the total movements are L0 + R1 + L1 + R2 + L2=0 + R1 +1 + R2 +0=R1 + R2 +1=26 +1=27, which matches 31 -2*2=27.But how does this translate to the number of sequences?Each block of movements corresponds to a consecutive sequence of letters.For example, the first uppercase block with R1 movements would be: starting at 'A' (position1 uppercase), then moving right R1 times: 'A', 'B', ..., chr(1 + R1).Similarly, the lowercase block with L1=1 movement would be: starting at position1 + R1 (lowercase), moving left once to position1 + R1 -1.However, the exact sequence depends on the specific movements.But each such block of movements is a consecutive sequence of letters in the same case, either moving left or right.However, according to the problem rules, when in lowercase, you can either move left or stay (toggle case), but once you toggle case, you start a new block.Similarly, in uppercase, you can move right or toggle case.Therefore, the total number of sequences for m=2 is the number of ways to choose R1 and R2 such that R1 + R2=26, R1 >=1, and the corresponding movements are valid.But each choice of R1 and R2 corresponds to exactly one path, but the number of sequences is more complicated because within each block, the letters are generated by consecutive movements.Wait, no. Each movement corresponds to a specific letter. For example, if in an uppercase block you have R1 movements, that corresponds to writing R1 +1 uppercase letters: starting at 'A', then each movement increments the letter. So R1 movements correspond to R1 +1 letters? No, wait. If you start at position k in uppercase and make R movements, you write R+1 letters: the initial position and R movements. But according to the problem rules, each movement is a step in the sequence. So each movement corresponds to one letter.Wait, no. The block of movements is a series of steps where each step is a movement. For example, if you are in uppercase at position k, and you decide to move right R times, each move right corresponds to writing the next uppercase letter. So starting at uppercase k, moving right R times would result in writing letters: k, k+1, ..., k+R in uppercase. However, according to the transition rules, from uppercase k, you can either write k lowercase or k+1 uppercase. Therefore, to move right R times from uppercase k, you need to write k+1, k+2, ..., k+R in uppercase, which requires R steps.Wait, let's clarify:If you are in uppercase k, the next letter can be k lowercase or k+1 uppercase. Therefore, to move right once, you write k+1 uppercase. To move right again, from k+1 uppercase, you write k+2 uppercase, etc. Therefore, each movement right requires one step. So to move from uppercase k to uppercase k+R requires R steps, writing k+1, k+2, ..., k+R uppercase letters. However, the starting position is k uppercase, so the sequence would be: k (uppercase) -> k+1 (uppercase) -> ... -> k+R (uppercase), which is R+1 letters. Wait, no. Each movement is a step. Starting at k uppercase:Step 1: k uppercase -> k+1 uppercase (1 movement right)Step 2: k+1 uppercase -> k+2 uppercase (another movement right)...Step R: k+R-1 uppercase -> k+R uppercaseSo to reach k+R uppercase from k uppercase requires R movements, writing R+1 letters (from k to k+R). But in terms of the total sequence length, each movement is a single step. Therefore, a block of R right movements in uppercase corresponds to R+1 letters: the starting uppercase letter and R subsequent uppercase letters.But this complicates the count because the number of letters contributed by a movement block is R+1, but we have a fixed total sequence length of 31 letters.Wait, this suggests that the previous analysis is incorrect because we considered movements as steps, but each movement corresponds to a new letter. However, in the problem statement, each step in the sequence is a letter, which can be a movement or a toggle.Wait, let's clarify:Each letter in the sequence is determined by the previous letter and the transition rules. Each transition (movement or toggle) consumes one step in the sequence. Therefore, the total number of letters is fixed at 32, and each transition from one letter to the next consumes one step.Therefore, the analysis earlier where we considered 31 transitions (steps) between 32 letters is correct.However, when we broke it down into m toggles and movements, we considered that toggles and movements are separate steps, but actually, each transition is either a toggle or a movement, both consuming one step.Therefore, the total number of toggles plus movements equals 31 (the number of transitions between 32 letters). Therefore, our previous analysis where m is the number of up toggles (equal to down toggles) and the remaining steps are movements is correct.But the confusion arises in the physical interpretation of movements and toggles. Each movement is a transition that changes the letter (either moving left or right), while a toggle is a transition that changes the case but not the letter.Therefore, when we have a block of movements in uppercase, each movement is a step to the next uppercase letter, and each toggle is a step to the same letter in lowercase.Therefore, the number of sequences corresponding to a certain set of R1, R2, etc., is not just the number of ways to choose R1 and R2, but also the number of ways to interleave the toggles and movements.However, in our earlier analysis, we fixed the order of toggles and movements into blocks, which may not account for all possible interleavings.This suggests that the problem is more complex and requires a different approach.Given the time I've spent and the complexity, I think the correct approach is to recognize that this problem can be modeled using dynamic programming with the recurrence relations we established earlier, and that the answer can be computed using those relations.Given that, and knowing that manually computing up to i=31 is tedious but possible in a systematic way, perhaps we can look for a pattern or a formula.Looking back at the small i examples:At i=1: L=1 at 1i=2: U=1 at 1i=3: L=1 at 1, U=1 at 2i=4: L=1 at 2, U=1 at 1 and 3i=5: L=2 at1, L=1 at3; U=2 at2, U=1 at4i=6: L=3 at2, L=1 at4; U=2 at1, U=3 at3, U=1 at5This resembles the pattern of binomial coefficients but shifted. For example, at i=3, U=1 at2 (which is C(2-1,1-1)=1). At i=4, U=1 at1 and3, which is C(3-1,1-1)=1 and C(3-1,3-1)=1. At i=5, U=2 at2 and1 at4, which is C(4-1,2-1)=3 and C(4-1,4-1)=1. Wait, maybe not.Alternatively, the counts for uppercase at position k and step i resemble combinations C(i-1, something).For example, at i=2, U(2,1)=1=C(1,0)At i=3, U(3,2)=1=C(2,1)At i=4, U(4,1)=1=C(3,0), U(4,3)=1=C(3,2)At i=5, U(5,2)=2=C(4,1), U(5,4)=1=C(4,3)At i=6, U(6,1)=2=C(5,0), U(6,3)=3=C(5,2), U(6,5)=1=C(5,4)This seems to follow U(i,k)=C(i-2, (k-1)-1) when k is odd? Not sure. Wait, let's see:For i=2, k=1: C(0,0)=1For i=3, k=2: C(1,1)=1For i=4, k=1: C(2,0)=1, k=3: C(2,2)=1For i=5, k=2: C(3,1)=3/2= no, wait U(5,2)=2 which is not C(3,1)=3.Hmm, not matching.Alternatively, perhaps the counts follow a Fibonacci-like sequence.For example, at i=1: L=1i=2: U=1i=3: L=1, U=1i=4: L=1, U=2i=5: L=2, U=3i=6: L=3, U=5This resembles Fibonacci numbers, but it's not exactly clear.Alternatively, looking at the total number of sequences at each step:i=1:1i=2:1i=3:2i=4:3i=5:5i=6:8Yes! This looks like the Fibonacci sequence. The total number of sequences at each step i is the sum of all L(i,k) + U(i,k). For example:i=1:1i=2:1i=3:1 +1=2i=4:1 +1 +1=3i=5:2 +1 +2 +1=6? Wait, no. Wait, for i=5:L(5,1)=2, L(5,3)=1; U(5,2)=2, U(5,4)=1. Total=2+1+2+1=6.But earlier thought was 5. So not Fibonacci.Hmm.Alternatively, the number of sequences doubles each step, but with constraints.But given the complexity, I think the correct answer is obtained by computing the DP up to i=32, which would give U(31,26). However, doing this manually is error-prone.Alternatively, notice that the number of sequences is equal to the number of ways to go from (a, lower) to (z, lower) in 31 steps, which is a variant of a lattice path problem with constraints.Another approach: since each toggle effectively allows Bob to change direction, and the net movement required is +25, the problem resembles a ballot problem where you need to stay above a certain boundary.However, I'm not sure.Alternatively, consider that each uppercase letter allows moving right, and lowercase allows moving left. To make progress to the right, Bob must spend time in uppercase moving right, but he has to toggle back to lowercase to end.Given the total of 31 steps and net movement of +25, the number of ways is similar to the number of balanced parentheses with certain constraints, but I'm not sure.Alternatively, model this as two independent variables: position and case. The recurrence relations are:L(i+1,k) = U(i,k) + L(i,k+1)U(i+1,k) = L(i,k) + U(i,k-1)This is a system that can be represented as a matrix and solved using generating functions or eigenvalues, but that's quite involved.Another observation: the recurrence for U and L are similar to those in a grid where you can move right or down, but with different rules.Alternatively, think of U and L as two separate layers, with transitions between them.For example, the uppercase layer has transitions to the same position in lowercase or to the next position in uppercase. The lowercase layer has transitions to the same position in uppercase or to the previous position in lowercase.This resembles a two-layer grid where you can move within a layer or switch layers.This problem might be similar to counting paths in a graph with two rows (upper and lower) where you can move right on the upper row, left on the lower row, or switch between rows at the same position.The number of paths from (1, lower) to (26, lower) in 31 steps.This is a known problem in combinatorics, and the number of such paths can be computed using the reflection principle or generating functions, but I'm not sure of the exact solution.Alternatively, recognize that the number of sequences is equal to the binomial coefficient C(31, 25) or something similar, but adjusted for the constraints. However, since each right move must be done in uppercase and each left move in lowercase, it's not straightforward.Given the time I've spent and the risk of not converging, I think the correct answer is that the number of sequences is the 25th Fibonacci number or similar, but given the earlier small i counts don't match Fibonacci, this is unlikely.Alternatively, since we need to make 25 net right moves and can have 3 net left moves (from the earlier m=3 case), the total number of sequences is C(28,25) = C(28,3) = 3276. But this is a guess.Alternatively, for m=3:sum Lk=0, sum Rk=25.The sequence of states:lowercase (L0=0) -> toggle up -> uppercase (R1) -> toggle down -> lowercase (L1=0) -> toggle up -> uppercase (R2) -> toggle down -> lowercase (L2=0) -> toggle up -> uppercase (R3) -> toggle down -> lowercase (L3=0).But sum Rk= R1 + R2 + R3=25.The path must start at1, end at26.Each R1, R2, R3 >=0.The positions:After R1:1 + R1.After R2:1 + R1 + R2.After R3:1 + R1 + R2 + R3=26.Therefore, R1 + R2 + R3=25.Number of non-negative solutions: C(25 +3 -1,3 -1)=C(27,2)=351.But need to ensure that each R1, R2, R3 are such that during their respective uppercase phases, the position does not exceed26.But since R1 + R2 + R3=25, and starting at1:After R1:1 + R1 <=26 => R1 <=25.After R2:1 + R1 + R2 <=26 => R1 + R2 <=25.After R3:1 + R1 + R2 + R3=26.But since R1 + R2 <=25, then R3=25 - R1 - R2 >=0.Therefore, R1 + R2 <=25.The number of solutions to R1 + R2 + R3=25 with R1 + R2 <=25 is the same as the number of solutions to R1 + R2 <=25, R3=25 - R1 - R2.The number of non-negative solutions to R1 + R2 <=25 is C(25 +2,2)=C(27,2)=351.But each such solution corresponds to a valid sequence because the positions never exceed26.Therefore, for m=3, there are351 valid sequences.Similarly, for m=2, the valid sequences were25, but we need to account for the fact that each sequence corresponds to a different path with specific movement counts.However, each such sequence also involves toggling and movements. The total number of sequences for each m is the number of ways to arrange the toggles and movements.For example, for m=3, the number of sequences is351, but we also need to consider the number of ways to interleave the movements and toggles.Wait, but in our earlier analysis for m=3, the structure is fixed: lowercase -> uppercase -> lowercase -> uppercase -> lowercase -> uppercase -> lowercase, with three up toggles and three down toggles. The movements are R1, R2, R3 in the uppercase periods, and L0, L1, L2, L3 in the lowercase periods. But since sum Lk=0, all L periods are zero.Therefore, the sequence is:L0=0 -> toggle up -> R1 -> toggle down -> L1=0 -> toggle up -> R2 -> toggle down -> L2=0 -> toggle up -> R3 -> toggle down -> L3=0.Which translates to the sequence of transitions:lowercase a -> toggle up -> R1 right moves -> toggle down -> R1 left moves (but L1=0) -> toggle up -> R2 right moves -> toggle down -> L2=0 -> toggle up -> R3 right moves -> toggle down -> L3=0.But since all L periods are zero, the path is:a (lower) -> A (upper) -> move R1 right -> toggle down -> move 0 left -> toggle up -> move R2 right -> toggle down -> move 0 left -> toggle up -> move R3 right -> toggle down -> move 0 left.But the letters would be:a, A, B, ..., A+R1, a+R1 (lower), a+R1 (lower) -> A+R1 (upper), ..., A+R1+R2, a+R1+R2 (lower), ..., A+R1+R2+R3 (upper), a+R1+R2+R3 (lower).But since R1 + R2 + R3=25, the final position is a+25=z.However, the exact sequence involves writing each letter in the path, with toggles.The number of such sequences is the number of ways to choose R1, R2, R3 >=0 such that R1 + R2 + R3=25.Each such triplet (R1, R2, R3) corresponds to a unique path, but the number of sequences is actually the number of ways to interleave the toggles and movements.However, in this case, the toggles are fixed at specific positions: between the movement blocks. Therefore, the number of sequences is equal to the number of ways to choose R1, R2, R3, which is351.But each movement block of R1, R2, R3 right moves in uppercase corresponds to R1, R2, R3 consecutive uppercase letters.Similarly, the toggles are single steps.But the problem is that the total number of letters is32, which is1 (start) +31 transitions.Each toggle is a single transition (a letter change), and each movement is also a single transition.Therefore, the total number of letters is indeed32.For m=3, the number of sequences is the number of ways to distribute the 25 right moves among three blocks (R1, R2, R3), which is C(25 +3 -1,3 -1)=C(27,2)=351. However, each such distribution corresponds to a unique sequence of movements and toggles.But in reality, the sequence is determined by the order of the blocks and the number of movements in each block. Since the order of the blocks is fixed (alternating uppercase and lowercase), the number of sequences is indeed351.Similarly, for m=2, we had25 valid sequences.For m=1, invalid.But wait, earlier we saw that for m=2, there are25 valid sequences corresponding to R1 from1 to25.However, each R1 corresponds to a unique sequence, so25 sequences.Therefore, total sequences=25 (m=2) +351 (m=3)=376.But according to the earlier small i counts, this seems low. For example, at i=5, there were sequences like a -> A -> B -> C -> D -> E (uppercase), but also sequences that toggle multiple times.But according to this analysis, the total number of sequences is376.However, given the problem's complexity and the time I've spent, I think this is the answer: the number of sequences is the sum over m=2 and m=3 of the number of valid sequences for each m, which is25 +351=376.But I'm not confident. Alternatively, the correct answer is351 +25=376, but I need to check.Wait, when m=2, the number of sequences is25, and for m=3, it's351. But according to the problem statement, the sequences must have length32, starting at a and ending at z.However, according to the earlier analysis, for m=2, each sequence has2 toggles and27 movements, total steps=2 +27=29, but wait, no. Each toggle is a single step (transition), and each movement is also a single step.Wait, no. The total number of transitions (steps between letters) is31. Each toggle is a transition, and each movement is a transition. So m=2 implies2 up toggles and2 down toggles, total4 toggles, and movements=31 -4=27.Yes. For m=2, movements=27, toggles=4.But according to the analysis for m=2, the number of sequences is25.Similarly, for m=3, toggles=6, movements=25. Number of sequences=351.Total sequences=25 +351=376.But according to the earlier small i=6, the total number of sequences was8 (sum of all L and U), but according to this method, it would be different.But given the complexity, I think the answer is376. However, I need to verify.Alternatively, perhaps the number of sequences is the binomial coefficient C(31,25), which is736,281. But that seems too large.Alternatively, think of it as a random walk with certain constraints, and the number of paths is the Catalan numbers. But I'm not sure.Given the time I've spent and the risk of inaccuracy, I think the correct approach is to recognize that the number of sequences is equal to the number of ways to interleave right movements and toggles, which results in the answer being the 25th Catalan number. However, Catalan numbers usually count balanced parentheses, which is not exactly this problem.Alternatively, the number of sequences is the number of ways to arrange 25 right moves and6 left moves such that the net is25. But with the constraint that right moves can only occur in uppercase and left moves in lowercase.This is similar to a problem where you have two types of steps with different constraints, and the answer is the number of such paths.However, without further insight, I think the correct answer is boxed{351}, but I'm not sure. However, considering the m=3 case gives351 and m=2 gives25, summing to376, but I need to check if there's another m= case.Wait, earlier we had m=1 invalid, m=2 and m=3 valid. So total sequences=25+351=376.But let's check the case m=3:The number of ways to distribute25 right moves into3 blocks is C(25+3-1,3-1)=C(27,2)=351.Yes.For m=2:The number of ways to distribute26 right moves into2 blocks with R1 >=1, which is C(25 +2-1,2-1)=C(26,1)=26. But earlier we thought it was25. Wait, confusion arises.Wait, for m=2, sum Rk=26 with R1 >=1 and R2=26 - R1.The number of solutions is R1 can be from1 to25 (since R1 >=1 and R2=26 - R1 >=0 => R1 <=26). But R1 <=25 to ensure that after R1 movements, the position doesn't exceed26.Wait, R1 can be up to25, so R1=1 to25, R2=25 to1.Therefore,25 solutions.Therefore, for m=2,25 sequences.So total=25+351=376.Therefore, the final answer is376.However, given that I initially thought the answer might be351, but considering both m=2 and m=3, it's376.But to confirm, when m=3, the number of sequences is351, when m=2, it's25, summing to376.Therefore, the number of sequences of32 letters starting at a and ending at z is376.But I need to check if there's another m=4 case, but m=4 requires sum Lk=3-4=-1, invalid.Therefore, the final answer is376.But I'm not entirely confident. To cross-validate, let's consider small n.For example, if the sequence length is4, starting at a and ending at a lowercase.The number of sequences should be... let's see:From a (lowercase), step2 can be A (uppercase) or a-1 (invalid). So step2 must be A.Step3 from A can be a (lowercase) or B (uppercase).If step3 is a, step4 can be A (uppercase) or invalid left move. So step4 must be A, but we need to end at a, which is lowercase. So invalid.If step3 is B (uppercase), step4 can be b (lowercase) or C (uppercase). To end at b (lowercase), which is not 'a'. So for length4, ending at a lowercase is only possible if step4 is lowercase a, but from step3 which is B, you can't get back to a.Therefore, no sequences of length4 starting at a and ending at lowercase a. Which matches our earlier calculation where for i=4, L(4,k) had L(4,2)=1, U(4,1)=1, U(4,3)=1. None ending at lowercase a.But if the problem was to end at lowercase z in32 steps, the answer is376.Given the time I've spent and the analysis, I'll conclude the answer is376.But wait, another way to think: each m corresponds to different numbers of oscillations between cases, and the total number of sequences is the sum over valid m of the number of sequences for each m.For m=2:25For m=3:351Total=376.Therefore, the final answer is boxed{376}.

✅:To determine the number of sequences of 32 letters starting with lowercase 'a' and ending with lowercase 'z', we use dynamic programming with the following recurrence relations:- ( L(i, k) ): Number of sequences of length ( i ) ending with lowercase ( k ).- ( U(i, k) ): Number of sequences of length ( i ) ending with uppercase ( k ).Recurrence Relations:[L(i+1, k) = U(i, k) + L(i, k+1)][U(i+1, k) = L(i, k) + U(i, k-1)]Boundary Conditions:- ( L(1, 1) = 1 ) (starting at lowercase 'a')- All other ( L(1, k) = 0 ) and ( U(1, k) = 0 ).Goal:Find ( L(32, 26) ), which is equivalent to ( U(31, 26) ).Key Observations:- The problem can be broken down into valid paths with net movement of 25 steps to the right.- Valid sequences involve toggling between uppercase and lowercase, and moving left or right depending on the current case.Analysis:- Case m=2: Two toggles (up and down) with movements. The number of valid sequences is 25.- Case m=3: Three toggles (up and down) with movements. The number of valid sequences is 351.Summing Valid Sequences:- Total sequences = 25 (m=2) + 351 (m=3) = 376.Final Answer:[boxed{376}]

🔑:We need to count the number of sequences ( S ) of length 32 that Bob can write which start with the lowercase letter 'a' and end with the lowercase letter 'z'.First, let's break down the rules and constraints given:1. There are 26 letters in the alphabet, from 'a' to 'z'.2. Each letter Bob writes can be uppercase or lowercase.3. If the last letter written was uppercase: - The next letter can be the same letter in lowercase. - The next letter can be the next letter in the alphabet in uppercase.4. If the last letter written was lowercase: - The next letter can be the same letter in uppercase. - The next letter can be the preceding letter in the alphabet in lowercase.Let's outline the smallest possible sequence from ( a ) to ( z ) adhering to these rules:- This sequence alternates regularly until it reaches 'Z': ( a , A , B , C , ldots , Z , z ).This sequence has 28 letters:[ a, A, B, C, cdots, X, Y, Z, z ].Since the problem asks for sequences of 32 letters, we need to insert 4 additional letters into the sequence. We can do this in two ways:1. Type one insertion: Interchanging a letter to lowercase and back to uppercase at two distinct points in the alphabet.2. Type two insertion: Adding an extra lowercase letter after its corresponding uppercase letter, inserting the preceding letter of the alphabet, then switching back to uppercase, and continuing.# Type One Insertion:To insert two interchanging lowercase and uppercase sequences:1. Choose two positions out of the 26 available upper case letters (from A to Z, excluding the starting 'a' and ending 'z').2. Each choice of positions allows one letter to be duplicated in lowercase and switched back.We calculate the number of ways to choose two positions among 27 possible positions in the 28-letter alphabet:[ binom{27}{2} = frac{27 times 26}{2} = 13 times 27 = 351 ]# Type Two Insertion:To insert one extra lowercase letter sequence which encompasses both ends:1. Choose one position out of the 25 positions.This scenario includes adding a letter sequence akin to: [ a , A , B , C , c , b ldots ]We calculate this by directly counting the valid positions which can host this interruption:[ 25 ]# Conclusion:Thus, the total number of valid sequences is the sum of the sequences generated by type one and type two insertions:[ 351 + 25 = 376 ]The total number of sequences of 32 letters that start at lowercase 'a' and end at lowercase 'z' that Bob can write is:[ boxed{376} ]

🧠:Okay, let me try to figure out this problem step by step. So, the first term is 500, and each new term is obtained by dividing the previous term by 2 and then adding 10. I need to find the fourth term. Hmm, let's break this down.First, let me make sure I understand the rule correctly. Each term is generated by taking the previous term, dividing it by 2, and then adding 10. So if the first term is 500, the second term would be (500 / 2) + 10. Let me calculate that: 500 divided by 2 is 250, and then adding 10 gives 260. So the second term is 260.Now, the third term would be based on the second term. So using the same formula: 260 divided by 2 is 130, plus 10 is 140. So the third term is 140.Then, the fourth term would follow the same process. Taking the third term, 140, dividing by 2 gives 70, and adding 10 gives 80. Therefore, the fourth term should be 80. Let me just verify each step again to make sure I didn't make a mistake.First term: 500 (given). Second term: 500 / 2 = 250 + 10 = 260. Third term: 260 / 2 = 130 + 10 = 140. Fourth term: 140 / 2 = 70 + 10 = 80. Yep, that seems right. Each time I take half of the previous term and add 10. Wait a second, maybe I should check if there's a pattern or a formula that can generalize this, just to confirm. Let's see. The recurrence relation here is a(n) = (a(n-1)/2) + 10. This looks like a linear recurrence relation. Maybe I can solve it to find a closed-form expression?The general form of such a recurrence is a(n) = a(n-1)/2 + c, where c is a constant (here, c=10). The solution to such a recurrence relation can be found using the method for linear recurrence relations. The homogeneous solution would be related to the term without the constant, and the particular solution would account for the constant term.The homogeneous equation is a(n) = (1/2)a(n-1), whose solution is a(n) = A*(1/2)^n, where A is a constant. For the particular solution, since the nonhomogeneous term is a constant (10), we can assume a particular solution is a constant, say P. Plugging into the recurrence relation:P = (1/2)P + 10Solving for P: P - (1/2)P = 10 → (1/2)P = 10 → P = 20.Therefore, the general solution is a(n) = A*(1/2)^n + 20. Now, applying the initial condition a(1) = 500. Wait, hold on, if n starts at 1, then:a(1) = A*(1/2)^1 + 20 = 500 → A/2 + 20 = 500 → A/2 = 480 → A = 960.Therefore, the closed-form formula is a(n) = 960*(1/2)^n + 20. Let's test this formula for the terms we already calculated.For n=1: 960*(1/2)^1 + 20 = 480 + 20 = 500. Correct.n=2: 960*(1/4) + 20 = 240 + 20 = 260. Correct.n=3: 960*(1/8) + 20 = 120 + 20 = 140. Correct.n=4: 960*(1/16) + 20 = 60 + 20 = 80. Which matches the earlier calculation. So this confirms that the fourth term is indeed 80. Alternatively, maybe using the iterative approach is sufficient here since we only need the fourth term. Let me just recap:Term 1: 500Term 2: (500 / 2) + 10 = 250 + 10 = 260Term 3: (260 / 2) + 10 = 130 + 10 = 140Term 4: (140 / 2) + 10 = 70 + 10 = 80Yes, that's straightforward. Each term is half the previous plus ten. So whether I use the iterative method or derive the closed-form formula, the answer is the same. Therefore, the fourth term is 80. I don't see any mistakes in the reasoning here. Both methods cross-check each other. So I'm confident that 80 is the correct answer.

✅:The fourth term in the sequence is 80.Step-by-Step Explanation:1. Term 1 (Given): ( a_1 = 500 )2. Term 2 Calculation: ( a_2 = frac{a_1}{2} + 10 = frac{500}{2} + 10 = 250 + 10 = 260 )3. Term 3 Calculation: ( a_3 = frac{a_2}{2} + 10 = frac{260}{2} + 10 = 130 + 10 = 140 )4. Term 4 Calculation: ( a_4 = frac{a_3}{2} + 10 = frac{140}{2} + 10 = 70 + 10 = 80 )Verification via Closed-Form Formula: The recurrence relation ( a_n = frac{a_{n-1}}{2} + 10 ) has a closed-form solution: [ a_n = 960 cdot left(frac{1}{2}right)^n + 20 ] For ( n = 4 ): [ a_4 = 960 cdot frac{1}{16} + 20 = 60 + 20 = 80 ]Both methods confirm the fourth term is 80.

🔑:To determine the fourth term of the sequence, we need to carefully follow the rule provided: each term is obtained by dividing the previous term by 2 and then adding 10.1. First Term: [ a_1 = 500 ]2. Second Term: According to the problem, the second term is given as 260. 3. Third Term: Calculate the third term by dividing the second term by 2 and then adding 10: [ a_3 = frac{a_2}{2} + 10 ] Substitute ( a_2 = 260 ): [ a_3 = frac{260}{2} + 10 ] Perform the division and addition: [ a_3 = 130 + 10 = 140 ]4. Fourth Term: Calculate the fourth term by dividing the third term by 2 and then adding 10: [ a_4 = frac{a_3}{2} + 10 ] Substitute ( a_3 = 140 ): [ a_4 = frac{140}{2} + 10 ] Perform the division and addition: [ a_4 = 70 + 10 = 80 ] Conclusion:The fourth term in the sequence is:[boxed{80}]

🧠:Okay, so I need to find the maximum number of distinct photos a group of n friends can take where each photo includes at least one person, no two photos have exactly the same group of people, and each person appears in at most one photo. Hmm, let me try to unpack this step by step.First, let me restate the problem in my own words. We have n friends, and we want to take as many photos as possible. Each photo is a distinct subset of the friends (since no two photos have exactly the same people), and each photo must have at least one person. Additionally, each person can appear in at most one photo. So, once a person is in a photo, they can't be in any other photo. The question is, what's the maximum number of such photos we can take?Let me think. Each photo is a non-empty subset of the n friends. But the constraint is that each person can only be in at most one photo. That means once a person is in a photo, they are excluded from all other photos. So the photos must be pairwise disjoint subsets of the friends. But each photo has to be non-empty, and all photos must be distinct.Wait, so if the photos are pairwise disjoint, then the set of people in one photo doesn't overlap with the set of people in another. Therefore, the entire collection of photos would form a partition of some subset of the n friends. But since each photo is a non-empty subset, each element of the partition is a non-empty set, and the union of all these photos would be a subset of the n friends. However, it's possible that not all friends are in a photo, but since we want to maximize the number of photos, maybe we need to use all friends? Or maybe not necessarily, because splitting into more smaller subsets might allow more photos, even if some friends are left out. Wait, no. If we leave out some friends, that might reduce the total number of photos. Because those friends could have been in their own photos. So perhaps using all friends is better?But let me clarify. Let's suppose that we have a set S of the n friends. Each photo is a non-empty subset of S, all photos are pairwise disjoint, and each is distinct. The total number of photos is r. We need to maximize r under these constraints.So this is equivalent to finding the maximum number of pairwise disjoint non-empty subsets of S. Each subset corresponds to a photo. Since they have to be pairwise disjoint, their union is a subset of S. To maximize r, we need to maximize the number of such subsets. Since each subset must be non-empty, the problem reduces to partitioning as much of S as possible into the maximum number of non-overlapping non-empty subsets. However, the entire collection of subsets doesn't have to cover all of S. Wait, but if we don't cover all of S, then the leftover elements could potentially form singleton sets, increasing the number of subsets. Therefore, maybe the maximum number is achieved when we partition the entire set S into as many non-empty subsets as possible. Because if we leave some elements out, we can include those as singleton sets, which would add more subsets.Wait, no. For example, suppose we have n=3 friends: A, B, C. If we partition them into three singleton sets {A}, {B}, {C}, then we have 3 photos. Alternatively, if we partition them into one subset of two and one singleton, we get 2 photos. So indeed, partitioning into singletons gives more photos. Therefore, the maximum number of photos is achieved when each photo is a singleton set. But each singleton set is a distinct photo, right? But wait, each photo must be a distinct subset. However, singleton sets are distinct as long as they contain different people. So in the case of n=3, the three singleton sets {A}, {B}, {C} are distinct and pairwise disjoint. So r=3. If instead, we take a photo of {A,B} and {C}, that's r=2. So indeed, the maximum is 3.Wait, so in general, if we take each person alone in their own photo, then we have n photos, each a distinct singleton set. Each person appears exactly once. So r=n. But wait, but the problem says "each person appears at most once," so having each person in exactly one photo is allowed. Therefore, in that case, the maximum r would be n. But hold on, is that the case? Let me test with another example.Suppose n=2. Then, possible photos are {A}, {B}, and {A,B}. If we take both {A} and {B}, that's two photos, each person appears once. Alternatively, we could take {A,B} as one photo, but that would only be one photo. So maximum r=2. So in that case, n=2, r=2. Similarly, for n=3, r=3. So perhaps in general, the maximum r is n. But wait, let me check n=4.For n=4, if we take four singleton photos, each person once, r=4. Alternatively, can we take more than four? Suppose we take some single and some pairs. For example, {A}, {B}, {C,D}. That's three photos. But three is less than four. Alternatively, {A,B}, {C,D}, which is two photos. So indeed, taking all singleton sets gives the maximum number of photos. So r=n.But wait, wait. Wait, the problem says "each photo contains at least one person." But a singleton set is allowed, right? So if we take all singleton sets, each person is in exactly one photo, and there are n photos. However, in that case, all photos are singleton sets, which are distinct because each has a different person. So that's allowed. Therefore, r=n is possible.But wait, but another thought. The problem says "no two photos have exactly the same people." So as long as the subsets are distinct, they can be any size. But if all the photos are singleton sets, then since each singleton is distinct (different person), they are all distinct. So that's allowed. So in that case, the maximum number of photos is n.But wait, but if n=1, then we can only have one photo, which is {A}. So that works. If n=0, which is not possible here because we have a group of friends. So n≥1.But wait, but the problem states "each person appears at most once." So each person is in at most one photo. If we use each person exactly once in a singleton photo, then all n people are used, each once. So the total number of photos is n.Alternatively, if we have some people in larger groups, then those people are consumed, and the remaining can be in singleton or other groups. But since using a group of size two consumes two people, allowing only one photo instead of two. Therefore, using larger groups reduces the total number of photos. Therefore, to maximize the number of photos, we should use as many singleton photos as possible. Hence, the maximum r is n. But let me verify this with another example.Take n=4. If we take four singleton photos, that's four photos. Alternatively, could we do something like {A}, {B}, {C}, {D}, which is four. Alternatively, if we have three singleton photos and one photo with two people: {A}, {B}, {C}, {D,E}. Wait, but there are only four people. So if we have three singleton photos, that uses three people, and the fourth photo would have the remaining one person, {D}, but then {D} is already used. So you can't have both {D} and {D} again. So actually, if we take three singleton photos, the fourth photo must be the remaining person, but that's just another singleton. So you can't do three singleton and one pair unless you have more people. Wait, no. Wait, n=4. If you take three singleton photos, that's three people. The fourth person is left. So you can take one more singleton, making four. If you instead take a pair, say {A,B}, then you can take another pair {C,D}, which gives two photos. So two versus four. So four is better.Therefore, the conclusion seems to be that the maximum number of photos is n, achieved by having each person in their own singleton photo.Wait, but hold on. Let me think again. The problem says "each photo contains at least one person," which is satisfied by singletons. "No two photos have exactly the same people," which is satisfied since each singleton is unique. And "each person appears at most once," which is satisfied because each is in exactly one photo. Therefore, r=n is possible, and we can't do better because there are only n people, each can be in at most one photo, and each photo needs at least one person. So that's the upper limit.But wait, perhaps there's a different way of partitioning where some people are not in any photo, allowing others to be in multiple photos? Wait, no. The problem states that each person appears at most once. So every person can be in at most one photo. So even if some people are not in any photo, the ones that are in photos can only be in one. So the maximum number of photos is the number of people used in photos. But since each photo must have at least one person, the maximum number of photos can't exceed the number of people. But if you use all people in singleton photos, you get n photos. If you leave some people out, you can't get more photos because you have fewer people to use. So the maximum is indeed n.But wait, let me think again. Suppose n=3. If we take {A}, {B}, and {C}, that's three photos. Alternatively, can we take {A}, {B,C}? That's two photos. But that's less. Alternatively, {A,B,C}, just one photo. So indeed, the maximum is three.But wait, another angle. Maybe we can have more photos if some photos have multiple people but arranged in a way that each person is only in one photo. For example, in n=4, can we have two photos each with two people? Then, that's two photos, but using four people. But two photos versus four singleton photos. Four is more. So that's worse.Alternatively, in n=5, can we have two photos of two people and one singleton? That would be three photos, using five people. But if we do five singleton photos, that's five. So five is better. Therefore, the more singleton photos, the better. So indeed, the maximum number of photos is n.Wait, but maybe in some cases, you can have more photos by combining some people and leaving others? Wait, no. Because every time you have a photo with more than one person, you use up multiple people, which could have been individual photos. For example, using a pair uses two people but only gives one photo instead of two. So it's a loss in the total number of photos. Therefore, the optimal strategy is to use all singleton photos, giving n photos. Therefore, the answer is n.But wait, but the problem says "each person appears at most once." So, if a person is not in any photo, that's okay. But since the problem is to maximize the number of photos, even if some people are left out, you can include more photos by using the people in singleton sets. Wait, but leaving out people would mean you have fewer people to make photos with, so you can't get more photos than n. For example, if you leave out k people, then you can have at most (n - k) photos. But since k ≥ 0, the maximum possible is n. So whether you leave people out or not, the maximum number of photos you can take is n. Therefore, the answer is n.But wait, let me verify this with another example. Let's say n=5. If you take five singleton photos, that's five. If you take one photo with two people and three singleton photos, that's four photos. If you take two photos with two people each and one singleton, that's three photos. So the more people you put into a single photo, the fewer total photos you can have. Therefore, the maximum is indeed n.Alternatively, can we have overlapping photos? But no, the problem states that no two photos have exactly the same people, and each person appears in at most one photo. So overlapping is not allowed, since that would require a person to be in multiple photos, which violates the "at most once" condition.Wait, overlapping in the sense of having different groups, but the problem says "no two photos have exactly the same people," which just means that each photo is a distinct subset. But the key constraint is that each person is in at most one photo, so all photos must be pairwise disjoint. Therefore, the problem reduces to selecting a collection of pairwise disjoint non-empty subsets (photos) of the n friends, such that the number of subsets r is maximized.In set theory terms, this is equivalent to finding the maximum possible size of a family of pairwise disjoint non-empty subsets of an n-element set. The maximum such family is obtained by taking all singleton subsets. Since there are n singleton subsets, and they are pairwise disjoint, this gives a family of size n. Any other family with larger subsets would have fewer elements, since merging two singletons into a pair reduces the count by one.Therefore, the maximum r is n.Wait, but hold on. Wait, actually, in set theory, the family of all singletons gives n subsets, but is that indeed the maximum? Let me think. Suppose n is even. For example, n=4. If you pair them into two subsets of size 2, you get two subsets. But if you do four singletons, you get four subsets. So singletons are better. If n is odd, say n=5. Pair two and have three singletons: total subsets is 1 (pair) + 3 (singletons) = 4, which is less than 5. So indeed, regardless of n, the maximum number of pairwise disjoint non-empty subsets is n, achieved by all singletons.Therefore, the answer is r = n.But wait, hold on. Wait, in the problem statement, it says "each person appears at most once." So, if we take all singleton subsets, each person appears exactly once, which satisfies "at most once." If we take some other subsets, say one subset with two people and the rest singletons, then those two people appear exactly once, and the rest also once. So that's still okay, but the total number of subsets is less. So regardless, the maximum is n.Therefore, the maximum r is n. So the answer is boxed{n}.Wait a second. Wait, but in the problem statement, it says "each photo contains at least one person," which is satisfied. "No two photos have exactly the same people," which is satisfied because all singleton subsets are distinct. "Each person appears at most once," which is satisfied. Therefore, the maximum r is n.But wait, but the problem says "a group of n friends takes r distinct photos (no two photos have exactly the same people) each containing at least one person. Find the maximum r such that each person appears at most once."So the answer is n. But maybe there's a different interpretation. Wait, let me check the original problem again.Original problem: "A group of n friends takes r distinct photos (no two photos have exactly the same people) each containing at least one person. Find the maximum r such that each person appears at most once."So, "no two photos have exactly the same people" means that all photos are distinct subsets. Each photo has at least one person. Each person is in at most one photo. So, yes, the problem is to find the maximum number of pairwise disjoint distinct non-empty subsets of an n-element set. The maximum number is indeed n, as each singleton is distinct and pairwise disjoint.But let me check literature or standard problems. Is this a known problem?Yes, this is similar to finding a matching in a set system where each element is covered at most once. In combinatorics, the maximum number of pairwise disjoint non-empty subsets is indeed n, achieved by the singletons. Because if you have a family of subsets where each subset is non-empty and they are pairwise disjoint, the maximum size is n. Each singleton is a subset, and you can't have more than n subsets because each subset needs at least one element, and they can't share elements.Wait, actually, in a family of pairwise disjoint subsets of an n-element set, the maximum number of non-empty subsets is n, since each subset must contain at least one element, and no two subsets share an element. Therefore, the maximum is achieved when each subset is a singleton.Therefore, the answer is n. So the maximum r is n. So the final answer is boxed{n}.But wait, hold on. Wait, here's a conflicting thought. Suppose you have n=3. Then, the photos could be {A}, {B}, {C}, which are 3 photos. Alternatively, could we have more than 3 photos? For example, is there a way to have 4 photos? But with n=3, each photo must be a non-empty subset, and each person can only be in one photo. Since there are only 3 people, you can't have more than 3 photos because each photo needs at least one person not used in another photo. So n=3, maximum r=3.Wait, unless you have photos with different numbers of people, but still disjoint. For example, {A}, {B}, {C}, {}. But the last photo is empty, which is not allowed. Each photo must contain at least one person. So, no. So indeed, the maximum is n.Therefore, in general, for any n, the maximum number of photos is n, achieved by taking each person alone in their own photo. Each photo is distinct, non-empty, and each person appears exactly once, satisfying the "at most once" condition.So, after this thorough thinking and checking with examples, I believe the answer is boxed{n}.

✅:To determine the maximum number ( r ) of distinct photos such that each person appears at most once, we need to consider the following constraints:1. Each photo is a non-empty subset of the ( n ) friends.2. No two photos have the same group of people.3. Each person can appear in at most one photo.Key steps:1. Pairwise Disjoint Subsets: The photos must be pairwise disjoint non-empty subsets of the ( n ) friends. This means no person can be in more than one photo.2. Maximizing the Number of Subsets: To maximize the number of such subsets, we should use the smallest possible subsets (i.e., singletons) because each larger subset consumes more people, reducing the total number of photos.3. Singleton Subsets: Each singleton subset (a photo containing exactly one person) is distinct and satisfies all constraints. Using all ( n ) singleton subsets results in ( n ) photos, each containing a different person.Examples and Verification:- For ( n = 3 ), the maximum ( r ) is 3 (photos: {A}, {B}, {C}).- For ( n = 4 ), the maximum ( r ) is 4 (photos: {A}, {B}, {C}, {D}).- Any attempt to use larger subsets (e.g., pairs) results in fewer photos, confirming that singletons are optimal.Conclusion:The maximum number of photos ( r ) is achieved by using each person in exactly one singleton photo, leading to ( r = n ).[boxed{n}]

🔑:1. Initial Hypothesis: To find the maximum number of distinct photos ( r ) with ( n ) friends such that each person appears at most once in a photo.2. Show ( r leq n ): - Assume ( r > n ). - Using the Pigeonhole Principle (or Dirichlet's Box Principle): - Consider photos as 'pigeons' and people as 'pigeonholes'. - Each 'pigeon-photo' has to be placed into the 'pigeonhole' of a person who appears in the photo. - If ( r > n ), i.e., there are more photos than friends, at least one person (pigeonhole) must appear in more than one photo (pigeon), contradicting our condition that each person appears at most once in a photo. - Thus, the assumption ( r > n ) leads to a contradiction.3. Verify ( r = n ): - For ( r = n ), we can construct photos such that each photo contains exactly one unique person. - Let's denote the friends as ( F_1, F_2, ldots, F_n ). - We can create ( n ) distinct photos, each containing a different single friend: - Photo 1: ({F_1}), - Photo 2: ({F_2}), - ... - Photo ( n ): ({F_n}). - Each friend appears in exactly one photo, satisfying the condition that no one appears more than once.4. Conclusion: - The maximum number of distinct photos ( r ) that can be taken such that each person appears at most once is ( r = n ).[boxed{r = n}]

🧠:Okay, let's see. I need to prove that (a + b + c)^7 minus a^7, b^7, and c^7 is divisible by 7(a + b)(b + c)(c + a). Hmm, interesting. So first, maybe I should expand (a + b + c)^7 and see what terms are present. But expanding that directly seems complicated. Maybe there's a smarter way.Since the problem mentions divisibility by 7 and the product (a + b)(b + c)(c + a), perhaps there's a connection to modular arithmetic or factor theorems. Let me recall that for polynomials, if a polynomial P(x) is divisible by (x - k), then P(k) = 0. Maybe something similar applies here, but with multiple variables.Wait, the expression is (a + b + c)^7 - a^7 - b^7 - c^7. If I can show that this expression is zero when any two variables are negatives of each other, like a = -b, then (a + b) would be a factor. Similarly for (b + c) and (c + a). Let me check that.Suppose a = -b. Then substituting into the expression: (-b + b + c)^7 - (-b)^7 - b^7 - c^7 = (c)^7 - (-b^7) - b^7 - c^7 = c^7 + b^7 - b^7 - c^7 = 0. So yes, when a = -b, the expression becomes zero. Similarly, if b = -c or c = -a, the expression will also be zero. Therefore, (a + b)(b + c)(c + a) divides the expression. That takes care of part of the divisor.Now, the other factor is 7. So we need to show that the remaining quotient is divisible by 7. Maybe using the fact that for prime p, (x + y)^p ≡ x^p + y^p mod p. Since 7 is a prime, perhaps Fermat's little theorem applies here.Let me consider modulo 7. The expression (a + b + c)^7 - a^7 - b^7 - c^7. In modulo 7, (a + b + c)^7 ≡ a + b + c mod 7, and similarly, a^7 ≡ a mod 7, etc. So the entire expression would be (a + b + c) - a - b - c ≡ 0 mod 7. Therefore, the original expression is divisible by 7. So combining both results, the expression is divisible by 7(a + b)(b + c)(c + a).Wait, but the problem states "where at most one of a, b, or c can be 0". Hmm, maybe to ensure that (a + b)(b + c)(c + a) is not zero? If two variables are zero, then (a + b) would be zero, and so the divisor becomes zero, which isn't allowed. So the condition is just to prevent division by zero. But in our proof, we considered the factors (a + b), etc., so as long as not two variables are negatives of each other, but the problem allows at most one of a, b, c to be zero. If two variables are zero, then (a + b) would be zero, but the problem states at most one can be zero, so that's covered.But wait, actually, if two variables are zero, say a = 0 and b = 0, then (a + b) = 0, but the divisor becomes 7*0*(b + c)*(c + a) = 0, which is undefined. Hence, the condition is necessary to avoid division by zero. So the original statement holds when the divisor is non-zero, which is when at most one variable is zero.But going back to the proof. So far, I have shown that the expression is divisible by (a + b)(b + c)(c + a) and by 7. But are these factors coprime? If so, then their product divides the expression. Let me check if 7 and (a + b)(b + c)(c + a) are coprime. In general, (a + b)(b + c)(c + a) could share a factor with 7, but since we are working over integers, unless 7 divides one of the terms, which isn't necessarily the case. However, in the expression, the 7 factor is separate. So perhaps the two factors are coprime in the polynomial ring, so their product divides the expression.Alternatively, maybe the expression can be written as 7(a + b)(b + c)(c + a) multiplied by some polynomial. To confirm, we can factor the expression.Alternatively, maybe using the factor theorem for multiple variables. Since we know that (a + b), (b + c), (c + a), and 7 divide the expression, then their product should divide the expression. But we need to ensure that they are all distinct prime factors or something similar.Wait, 7 is a constant factor, whereas (a + b), (b + c), (c + a) are linear factors. Since 7 is a prime in integers, and the linear factors are irreducible, their product should be a factor of the expression. Therefore, the entire divisor 7(a + b)(b + c)(c + a) divides the expression.But maybe I should check with specific values. Let me try some simple numbers where a, b, c are small integers. For example, take a = 1, b = 1, c = 1. Then:(1 + 1 + 1)^7 - 1^7 -1^7 -1^7 = 3^7 - 3*1 = 2187 - 3 = 2184Now, compute 7(a + b)(b + c)(c + a) = 7*(2)*(2)*(2) = 7*8 = 56Check if 2184 is divisible by 56: 2184 ÷ 56 = 39. So yes, 56*39 = 2184. Good.Another example: a = 2, b = 3, c = 4.(2 + 3 + 4)^7 - 2^7 -3^7 -4^7 = 9^7 - 128 - 2187 - 16384Compute 9^7: 9^2 = 81, 9^3 = 729, 9^4 = 6561, 9^5 = 59049, 9^6 = 531441, 9^7 = 4782969So 4782969 - 128 - 2187 -16384 = 4782969 - 128 = 4782841; 4782841 - 2187 = 4780654; 4780654 -16384 = 4764270Now compute 7(a + b)(b + c)(c + a) = 7*(5)*(7)*(6) = 7*5*7*6 = 7*5=35; 35*7=245; 245*6=1470Check if 4764270 ÷ 1470 is integer: 4764270 ÷ 1470. Let's divide numerator and denominator by 10: 476427 / 147. 147*3240 = 147*3000=441000; 147*240=35280 → 441000+35280=476280. Then 476427 - 476280 = 147. So 3240 + 147/147 = 3240 +1=3241. So yes, 1470*3241=4764270. So it works here too.Another test case: a = 1, b = -1, c = 0. Wait, but the problem states at most one of a, b, c can be zero. Here c = 0 is allowed. Let's check:(1 + (-1) + 0)^7 -1^7 - (-1)^7 -0^7 = (0)^7 -1 - (-1) -0 = 0 -1 +1 = 0. Then 7(a + b)(b + c)(c + a) =7*(0)*( (-1)+0 )*(0 +1)=7*0*...=0. But division by zero is not allowed. So the condition that at most one is zero is necessary here. But in cases where none are zero and not two variables are negatives, it's okay.Alternatively, take a case where two variables sum to zero: a =1, b= -1, c=1. Then (1 + (-1) +1)^7 -1^7 - (-1)^7 -1^7 = (1)^7 -1 - (-1) -1 =1 -1 +1 -1=0. The divisor 7*(1 + (-1))*( (-1)+1 )*(1 +1 )=7*0*0*2=0, so again division by zero. So in the problem's allowed cases (at most one zero), two variables can't be negatives of each other unless one is zero. Wait, if one is zero, say c =0, then (a + b)(b +0)(0 +a) = (a + b)(b)(a). So if c=0, then (a + b) is a factor. But if a = -b, then again (a + b) is zero, which is excluded. So actually, the condition "at most one of a, b, or c can be 0" might also prevent having two variables being negatives if the third is zero. Hmmm, perhaps the problem includes that to avoid having (a + b)(b + c)(c + a) being zero. So, the divisor is non-zero as long as no two variables are negatives (unless one is zero, but even then, if one is zero, say c=0, then (a + b) can't be zero because that would require a = -b, but if c=0, and a = -b, then the divisor is 7(a + b)(b)(a). If a = -b, then divisor becomes 7*0*b*(-b) =0. So even with one variable zero, if another pair sums to zero, the divisor is zero. Hence, the condition "at most one of a, b, c can be zero" might not be sufficient. Wait, maybe the actual condition is that none of (a + b), (b + c), (c + a) is zero, which would require that no two variables are negatives of each other. But the problem states "at most one of a, b, or c can be 0". Maybe if one variable is zero, then (a + b) etc. can still be non-zero as long as the other two aren't negatives. So maybe the problem's condition is that in addition to at most one variable being zero, the other two are not negatives of each other. But the problem didn't specify that. Hmmm. But perhaps in the context of the problem, the divisibility is considered in the polynomial ring, where variables are indeterminates, so (a + b)(b + c)(c + a) is a non-zero polynomial, and 7 is a constant. So in the ring of polynomials over integers, the expression is divisible by 7(a + b)(b + c)(c + a). The condition "at most one of a, b, c can be zero" is probably to handle the evaluation at specific values, ensuring that the divisor isn't zero. But in the polynomial sense, the divisibility holds as a polynomial identity.So maybe my initial approach is correct: show that (a + b + c)^7 - a^7 - b^7 - c^7 is divisible by (a + b)(b + c)(c + a) and by 7. Then since 7 and the product are coprime in the polynomial ring, their product divides the expression.To formalize this:1. Divisibility by (a + b)(b + c)(c + a):We can use the factor theorem. For example, if we set a = -b, then as shown earlier, the expression becomes zero. Therefore, (a + b) is a factor. Similarly for (b + c) and (c + a). Since these are distinct irreducible factors in the polynomial ring, their product divides the expression.2. Divisibility by 7:Looking at the expression modulo 7. By Fermat's little theorem, for any integer x, x^7 ≡ x mod 7. Therefore, (a + b + c)^7 ≡ (a + b + c) mod 7, and a^7 ≡ a mod 7, similarly for b and c. Therefore, (a + b + c)^7 - a^7 - b^7 - c^7 ≡ (a + b + c) - a - b - c ≡ 0 mod 7. Hence, the expression is divisible by 7.Since 7 is a prime and the polynomial ring over integers is a UFD, the factors (a + b), (b + c), (c + a), and 7 are distinct primes in the ring. Therefore, their product divides the expression.To compute the exact quotient, we might need to perform polynomial division or find symmetric expressions. However, the problem only asks to prove divisibility, which we have established through the factor theorem and modular arithmetic.Therefore, the conclusion is that (a + b + c)^7 - a^7 - b^7 - c^7 is indeed divisible by 7(a + b)(b + c)(c + a) under the given conditions.

✅:To prove that ((a + b + c)^7 - a^7 - b^7 - c^7) is divisible by (7(a + b)(b + c)(c + a)) (with at most one of (a, b, c) being zero), we proceed as follows: Step 1: Divisibility by ((a + b)(b + c)(c + a))- Factor Theorem: If substituting (a = -b) into the polynomial results in zero, then ((a + b)) is a factor. - Let (a = -b): [ (-b + b + c)^7 - (-b)^7 - b^7 - c^7 = c^7 + b^7 - b^7 - c^7 = 0. ] - Similarly, setting (b = -c) or (c = -a) also yields zero. - Thus, ((a + b)), ((b + c)), and ((c + a)) are all factors of the expression. Step 2: Divisibility by 7- Fermat's Little Theorem: For prime (7), (x^7 equiv x pmod{7}). - Modulo 7, the expression simplifies: [ (a + b + c)^7 - a^7 - b^7 - c^7 equiv (a + b + c) - a - b - c equiv 0 pmod{7}. ] - Hence, the expression is divisible by 7. Step 3: Combining Factors- Coprimality: In the polynomial ring (mathbb{Z}[a, b, c]), the factors ((a + b)), ((b + c)), ((c + a)), and (7) are pairwise coprime. - Product Divisibility: Since all factors are distinct and irreducible, their product (7(a + b)(b + c)(c + a)) divides the original expression. ConclusionThe polynomial ((a + b + c)^7 - a^7 - b^7 - c^7) is divisible by (7(a + b)(b + c)(c + a)) under the given conditions. The divisibility holds as a polynomial identity, ensuring the result for all integers (a, b, c) where the divisor is non-zero (i.e., at most one variable is zero and no two variables are additive inverses).Final Answer boxed{(a + b + c)^7 - a^7 - b^7 - c^7 text{ is divisible by } 7(a + b)(b + c)(c + a)}

🔑:To prove that (a+b+c)^{7} - a^{7} - b^{7} - c^{7} is always divisible by 7(a+b)(b+c)(c+a), we can break the proof down into several steps. We need to show divisibility by each factor individually and then conclude.# Divisibility by (a+b):We can start by rearranging the expression as follows:[(a+b+c)^{7} - a^{7} - b^{7} - c^{7} = (a+b+c)^{7} - c^{7} - (a^{7} + b^{7})]We can see that both (a+b+c)^{7} and c^{7} can be treated as equal exponentiations, so the difference is divisible by the difference of their bases, which is (a+b). This can be formally illustrated by factoring (a+b+c)^7 - c^7:[(a+b+c)^7 - c^7 = left[(a+b) + cright]^7 - c^7]By the binomial theorem, each term involves powers of (a+b) as a factor, meaning (a+b+c)^7 - c^7 is divisible by (a+b). And because a^7 + b^7 is simply a sum of two powers with odd exponents, their sum is divisible by (a+b) if (a+b) neq 1, thus establishing divisibility by (a+b). Similarly, the same reasoning applies for (b+c) and (c+a).# Divisibility by (b+c):Analogously, we can rearrange the terms:[(a+b+c)^{7} - b^{7} - c^{7} - a^{7} = (a+b+c)^{7} - a^{7} - (b^{7} + c^{7})]Following the same reasoning as before, the difference of the bases in (a+b+c)^7 and b^7 is (a+c), showing divisibility by (b+c). The sum of a^7 + c^7 with an odd exponent also confirms divisibility by (b+c).# Divisibility by (c+a):Similarly, for (c+a), we have:[(a+b+c)^7 - a^7 - (b^7 + c^7)]By treating (a+b+c)^7 - b^7 similarly, we show that it is divisible by (c+a). Hence the same applies to (a^7 + b^7) being divisible by (c+a).Given that (a+b), (b+c), and (c+a) generally have no common divisors, we conclude that (a+b+c)^7 - a^7 - b^7 - c^7 is divisible by their product (a+b)(b+c)(c+a).# Divisibility by 7:Next, we need to show that the expression is divisible by 7. We start by considering each term in the polynomial expansion:By Fermat's Little Theorem: ( n^7 equiv n mod 7 ), we can state:[a^7 equiv a pmod{7},quad b^7 equiv b pmod{7},quad c^7 equiv c pmod{7}]Thus,[(a+b+c)^7 = a^7 + b^7 + c^7 + 7f(a, b, c)]where (f(a,b,c)) is a polynomial in (a), (b), and (c), multiplied by (7).This rearranges to:[(a+b+c)^7 - a^7 - b^7 - c^7 = 7f(a, b, c)]showing 7 divides (a+b+c)^7 - a^7 - b^7 - c^7.# Conclusion:After establishing that (a+b+c)^7 - a^7 - b^7 - c^7 is divisible by both 7 and (a+b)(b+c)(c+a) separately, and since these numbers generally do not share common divisors, we conclude that:[boxed{(a+b+c)^{7} - a^{7} - b^{7} - c^{7}}]is divisible by 7(a+b)(b+c)(c+a).

🧠:Okay, so I have these two simultaneous equations to solve:[ (a+b)(a^2 - b^2) = 4 ][ (a - b)(a^2 + b^2) = frac{5}{2} ]Hmm, simultaneous equations with two variables, a and b. Let me think about how to approach this. Since they are both equations involving a and b, maybe I can manipulate them to find a relationship between a and b, and then substitute.First, I notice that both equations have factors of (a + b) and (a - b). Also, the terms like a² - b² can be factored further. Let me recall that a² - b² is equal to (a - b)(a + b). So maybe substituting that into the first equation.Starting with the first equation:[ (a + b)(a^2 - b^2) = 4 ]Since a² - b² is (a - b)(a + b), substitute that in:[ (a + b)(a - b)(a + b) = 4 ]Which simplifies to:[ (a + b)^2(a - b) = 4 ]Similarly, the second equation is:[ (a - b)(a^2 + b^2) = frac{5}{2} ]I don't think a² + b² can be factored further, but maybe I can relate these two equations somehow. Let me denote x = a + b and y = a - b. Maybe substituting variables here will help.Let’s set:x = a + by = a - bThen, we can express a and b in terms of x and y. But maybe even better, express the equations in terms of x and y.First, compute the first equation:Original first equation becomes:x * (a² - b²) = 4But a² - b² is (a - b)(a + b) = x * y. Wait, but that seems like substituting:Wait, no. Wait, if x = a + b and y = a - b, then a² - b² = (a - b)(a + b) = x * y.So substituting into the first equation:x * (x * y) = 4 => x² y = 4.Similarly, the second equation:Original second equation is y * (a² + b²) = 5/2.Now, a² + b². Hmm. How can I express that in terms of x and y?Well, we know that:(a + b)^2 = a² + 2ab + b² = x²(a - b)^2 = a² - 2ab + b² = y²If I add these two equations:(a + b)^2 + (a - b)^2 = 2a² + 2b² => x² + y² = 2(a² + b²)Therefore, a² + b² = (x² + y²)/2.Alternatively, subtract them:x² - y² = 4abSo, ab = (x² - y²)/4.But in the second equation, we have a² + b². So from above, a² + b² = (x² + y²)/2.Therefore, the second equation becomes:y * [(x² + y²)/2] = 5/2Multiply both sides by 2:y (x² + y²) = 5So now, our two equations in terms of x and y are:1) x² y = 42) y (x² + y²) = 5Now, that seems more manageable. Let me write them again:1) x² y = 42) y (x² + y²) = 5So perhaps I can solve these two equations for x² and y.Let me denote equation 1 as x² y = 4, so x² = 4 / y. Then substitute x² into equation 2.Equation 2: y (4/y + y²) = 5Simplify:y*(4/y + y²) = y*(4/y) + y*(y²) = 4 + y³ = 5Therefore:4 + y³ = 5So y³ = 1Therefore, y = 1 (since we are dealing with real numbers here, and the cube root of 1 is 1).So y = 1.Then, from equation 1: x² = 4 / y = 4 / 1 = 4 => x² = 4 => x = 2 or x = -2.Therefore, possible solutions for x and y are:Case 1: x = 2, y = 1Case 2: x = -2, y = 1Now, recall that x = a + b and y = a - b.So for each case, we can solve for a and b.Case 1: x = 2, y = 1So:a + b = 2a - b = 1Adding these two equations:2a = 3 => a = 3/2Substituting a = 3/2 into a + b = 2:3/2 + b = 2 => b = 2 - 3/2 = 1/2So (a, b) = (3/2, 1/2)Case 2: x = -2, y = 1a + b = -2a - b = 1Adding equations:2a = -1 => a = -1/2Substituting into a + b = -2:-1/2 + b = -2 => b = -2 + 1/2 = -3/2So (a, b) = (-1/2, -3/2)Now, we need to check these solutions in the original equations to make sure they work, because sometimes when we square or manipulate equations, we might introduce extraneous solutions.First, check Case 1: a = 3/2, b = 1/2First equation:(a + b)(a² - b²) = (3/2 + 1/2)( (3/2)^2 - (1/2)^2 )Compute:a + b = 3/2 + 1/2 = 2a² - b² = (9/4 - 1/4) = 8/4 = 2So 2 * 2 = 4, which matches the first equation.Second equation:(a - b)(a² + b²) = (3/2 - 1/2)( (9/4) + (1/4) )Compute:a - b = 3/2 - 1/2 = 1a² + b² = 9/4 + 1/4 = 10/4 = 5/2So 1 * 5/2 = 5/2, which matches the second equation. So Case 1 is valid.Case 2: a = -1/2, b = -3/2First equation:(a + b)(a² - b²) = (-1/2 + (-3/2))( ( (-1/2)^2 - (-3/2)^2 ) )Compute:a + b = -1/2 -3/2 = -2a² - b² = (1/4 - 9/4) = (-8/4) = -2So (-2) * (-2) = 4, which matches the first equation.Second equation:(a - b)(a² + b²) = ( -1/2 - (-3/2) )( (1/4 + 9/4) )Compute:a - b = -1/2 + 3/2 = 2/2 = 1a² + b² = 1/4 + 9/4 = 10/4 = 5/2So 1 * 5/2 = 5/2, which matches the second equation. So Case 2 is also valid.Therefore, both solutions are correct.Hence, the solutions are (a, b) = (3/2, 1/2) and (-1/2, -3/2).But let me just think again if there could be more solutions. When we solved for y³ = 1, we considered real solutions, so y = 1 is the only real cube root. If we were considering complex numbers, there would be other roots, but since the original equations are likely intended to be over real numbers (given the constants are real), so y =1 is the only solution. Therefore, these are the only two solutions.Wait, but let me confirm the substitution steps again. When we set x = a + b and y = a - b, we transformed the original equations into x² y = 4 and y(x² + y²) =5. Then solving them gave y³ =1, so y=1, leading to x²=4. So x=2 or x=-2, which leads to the two cases. So that seems thorough.Therefore, the solutions are the two pairs we found.Final AnswerThe solutions are boxed{left(frac{3}{2}, frac{1}{2}right)} and boxed{left(-frac{1}{2}, -frac{3}{2}right)}.

✅:Given the pair of simultaneous equations:[(a + b)(a^2 - b^2) = 4][(a - b)(a^2 + b^2) = frac{5}{2}]We start by recognizing that (a^2 - b^2) can be factored as ((a - b)(a + b)). Substituting this into the first equation, we get:[(a + b)^2(a - b) = 4]Next, we introduce variables (x = a + b) and (y = a - b). This transforms the equations into:1. (x^2 y = 4)2. (y(x^2 + y^2) = frac{5}{2})For the second equation, we use the identity (a^2 + b^2 = frac{(a + b)^2 + (a - b)^2}{2}) to express it in terms of (x) and (y):[y left( frac{x^2 + y^2}{2} right) = frac{5}{2}]Multiplying both sides by 2, we get:[y(x^2 + y^2) = 5]From the first equation (x^2 y = 4), we solve for (x^2):[x^2 = frac{4}{y}]Substituting (x^2 = frac{4}{y}) into the second equation:[y left( frac{4}{y} + y^2 right) = 5]Simplifying this, we find:[4 + y^3 = 5 implies y^3 = 1 implies y = 1]Substituting (y = 1) back into (x^2 = frac{4}{y}) gives:[x^2 = 4 implies x = 2 text{ or } x = -2]We now solve for (a) and (b) using (x) and (y):1. Case 1: (x = 2) and (y = 1) [ begin{cases} a + b = 2 a - b = 1 end{cases} ] Adding these equations: [ 2a = 3 implies a = frac{3}{2} ] Substituting (a = frac{3}{2}) into (a + b = 2): [ frac{3}{2} + b = 2 implies b = frac{1}{2} ]2. Case 2: (x = -2) and (y = 1) [ begin{cases} a + b = -2 a - b = 1 end{cases} ] Adding these equations: [ 2a = -1 implies a = -frac{1}{2} ] Substituting (a = -frac{1}{2}) into (a + b = -2): [ -frac{1}{2} + b = -2 implies b = -frac{3}{2} ]Both solutions satisfy the original equations. Therefore, the solutions are:[boxed{left(frac{3}{2}, frac{1}{2}right)} text{ and } boxed{left(-frac{1}{2}, -frac{3}{2}right)}]

🔑:Consider the pair of simultaneous equations:[(a + b)(a^2 - b^2) = 4,][(a - b)(a^2 + b^2) = frac{5}{2}.]1. Note that (a^2 - b^2 = (a - b)(a + b)). Therefore, we can rewrite the first equation as:[(a + b)^2(a - b) = 4.]2. Since (a - b neq 0) (otherwise both left-hand sides would be zero, contradicting (4) and (frac{5}{2})), we proceed by dividing the first equation by the second:[frac{(a + b)^2(a - b)}{(a - b)(a^2 + b^2)} = frac{4}{frac{5}{2}}.]3. Simplify the fraction on the left-hand side by canceling out (a - b) (since (a - b neq 0)):[frac{(a + b)^2}{a^2 + b^2} = frac{8}{5}.]4. Now, multiply both sides by (5(a^2 + b^2)):[5(a + b)^2 = 8(a^2 + b^2).]5. Expand and collect like terms:[5(a^2 + 2ab + b^2) = 8a^2 + 8b^2,][5a^2 + 10ab + 5b^2 = 8a^2 + 8b^2,][5a^2 + 10ab + 5b^2 - 8a^2 - 8b^2 = 0,][-3a^2 + 10ab - 3b^2 = 0,]or[3a^2 - 10ab + 3b^2 = 0.]6. Factorize the quadratic:[(3a - b)(a - 3b) = 0.]Thus, we have two cases to consider:[3a - b = 0 quad text{or} quad a - 3b = 0.]7. Substitute (3a = b) into the first original equation:[a = 3b,][(a + b)(a^2 - b^2) = 4,][(3b + b)((3b)^2 - b^2) = 4,][4b(9b^2 - b^2) = 4,][4b(8b^2) = 4,][32b^3 = 4,][b^3 = frac{4}{32},][b = frac{1}{2},][a = 3b = 3 times frac{1}{2} = frac{3}{2}.]8. Now substitute (a = 3b) into the second original equation to verify:[(a - b)(a^2 + b^2) = frac{5}{2},][(frac{3}{2} - frac{1}{2})((frac{3}{2})^2 + (frac{1}{2})^2) = frac{5}{2},][1 left(frac{9}{4} + frac{1}{4}right) = frac{5}{2},][1 times 2.5 = frac{5}{2}, text{True}.]9. Substitute (a = frac{b}{3}) into the first original equation:[b = 3a,][(a + b)(a^2 - b^2) = 4,][(a + 3a)(a^2 - (3a)^2) = 4,][4a(a^2 - 9a^2) = 4,][4a(-8a^2) = 4,][-32a^3 = 4,][a^3 = -frac{4}{32},][a = -frac{1}{2},][b = 3a = 3 times -frac{1}{2} = -frac{3}{2}.]10. Substitute (b = 3a) into the second original equation to verify:[(a - b)(a^2 + b^2) = frac{5}{2},][(-frac{1}{2} - -frac{3}{2}) ((-frac{1}{2})^2 + (-frac{3}{2})^2) = frac{5}{2},],1 left(frac{1}{4} + frac{9}{4}right) = frac{5}{2}, 1 times 2.5 = frac{5}{2},][1 times 2.5 = frac{5}{2}, text{True}.]Hence, the two sets of solutions are (left(frac{3}{2}, frac{1}{2}right)) and (left(-frac{1}{2}, -frac{3}{2}right)).(boxed{left(frac{3}{2}, frac{1}{2}right) , text{or} , left(-frac{1}{2}, -frac{3}{2}right)})