🧠:[<think>Okay, let's try to figure out this problem step by step. So, there are three runners: Primus (1), Secundus (2), and Tertius (3). On Saturday, they finished in the order 1, 2, 3, which is recorded as 123. On Sunday, the race starts with the same initial order: 1 leading, then 2, then 3. But during the race, they switch places multiple times. Specifically, Primus and Secundus switch 9 times, Secundus and Tertius switch 10 times, and Primus and Tertius switch 11 times. We need to determine the final order on Sunday.First, let's make sure I understand the problem correctly. Each time two runners switch places, their positions are exchanged. For example, if Primus and Secundus switch once, the order goes from 1,2,3 to 2,1,3. If they switch again, it goes back to 1,2,3. So, each pair's number of switches will affect their final positions relative to each other.Now, the key here is to figure out the net effect of each number of switches on the order. Since each switch between a pair flips their positions, an even number of switches would result in them ending up in their original order, while an odd number of switches would leave them in the reversed order. Wait, is that right?Let me think. If two runners switch places once, they reverse positions. If they switch again, they go back to the original. So, yes, even number of switches cancel out, odd number results in a single switch. So, for each pair, if the number of switches is even, their relative order remains the same as the start; if odd, their order is reversed.But hold on, the problem states that they start in the order 1,2,3. Then, the switches happen between different pairs. But here's the catch: when two runners switch, it affects not only their positions but potentially the positions relative to the third runner as well. For example, if 1 and 2 switch, they swap places, but 3 remains third. Similarly, if 2 and 3 switch, they swap, while 1 remains first. But when 1 and 3 switch, they have to pass each other, which might involve moving through the second position. Wait, but in a race, runners can overtake each other directly. Hmm, maybe I need to model this more accurately.Wait, maybe the problem is assuming that when two runners switch places, they exchange their positions regardless of the third runner. So, for example, if the order is 1,2,3, and 1 and 3 switch places, do they just swap positions, making it 3,2,1? That seems a bit unrealistic in a race, but maybe that's how the problem is abstracting it. Because otherwise, overtaking would require moving past each other, which might involve intermediate positions.But the problem states that Primus and Tertius change places exactly 11 times. If each change is a direct swap, then 11 swaps would result in them ending up in the same relative order as the start if even, or reversed if odd. Wait, but 11 is odd, so swapping 11 times would reverse their order. Similarly, swaps between 1 and 2: 9 times (odd) would reverse their order. Swaps between 2 and 3: 10 times (even) would keep their original order.But hold on, this approach assumes that each pair's swaps are independent, which might not be the case. Because swapping two runners affects their positions relative to all others. For example, if 1 and 2 swap, their positions relative to each other change, but their positions relative to 3 stay the same. Similarly, if 2 and 3 swap, their positions relative to each other change, but 1's position relative to both 2 and 3 remains the same. But if 1 and 3 swap, how does that affect their positions relative to 2?This is where the problem might be getting tricky. If 1 and 3 swap, do they just exchange their positions regardless of 2? For example, if the order is 1,2,3 and 1 and 3 swap, does it become 3,2,1? If so, then each swap between 1 and 3 inverts their order relative to each other and also relative to 2. But in a real race, overtaking someone would require moving past them one by one. So, for 1 to overtake 3, 1 would have to first overtake 2, then overtake 3, but maybe the problem is abstracting this as a direct swap.Alternatively, maybe the problem is considering that a swap between 1 and 3 can only happen if they are adjacent? But in the starting order, 1 and 3 are not adjacent. So, maybe each time they swap, they have to pass through each other via intermediate steps. Hmm, this complicates things.Wait, the problem says "change places exactly 9 times," "change places exactly 10 times," etc. The term "change places" might imply that each time they overtake each other, their positions are swapped. So regardless of whether they are adjacent or not, each overtake counts as a swap. But in reality, overtaking non-adjacent runners would require passing through the runners in between. But maybe in the problem's context, each time two runners overtake each other, it's considered a swap, regardless of their positions. For example, if 1 is first, 2 is second, 3 is third. If 3 overtakes 1, then the new order is 3,2,1. But in reality, 3 would have to overtake 2 first, then overtake 1, resulting in two swaps (3 and 2, then 3 and 1). But perhaps the problem counts a direct overtake as a single swap even if non-adjacent.This is a crucial point. If overtaking a non-adjacent runner counts as a single swap, then the total number of swaps between each pair can be considered independent. Otherwise, if overtaking a non-adjacent runner requires multiple swaps (with the intermediate runners), then the numbers given (9, 10, 11) would be affected by each other.Given that the problem states the number of swaps between each pair explicitly (Primus and Secundus: 9, Secundus and Tertius: 10, Primus and Tertius: 11), it's likely that these swaps are considered independent. That is, each time two runners swap, regardless of their current positions, it counts as one swap. So even if they are not adjacent, swapping them counts as one and affects their order directly.Therefore, we can model each pair's swaps independently. So, for example, each swap between Primus and Secundus flips their order, each swap between Secundus and Tertius flips their order, and each swap between Primus and Tertius flips their order.However, this seems conflicting because the order of the runners affects which swaps are possible. For example, if Primus is first and Tertius is third, they can't directly swap unless they pass Secundus. But the problem might be simplifying this by allowing direct swaps regardless of position, tallying each pair's total swaps, and then figuring out the net effect.Alternatively, maybe each swap between two runners only occurs when they are adjacent. So, in order for Primus and Tertius to swap, they would have to first be adjacent, which would require Secundus to have been overtaken by one of them. But if that's the case, the number of swaps between Primus and Tertius would be dependent on the number of times they pass each other, which might involve intermediate passes through Secundus.This is getting complicated. Let me think again.Perhaps the key is that each time two runners swap, their relative order is reversed. So regardless of their current positions, a swap between them just flips their order. So, for example, if the current order is 1,2,3 and Primus and Tertius swap, the new order is 3,2,1. Similarly, if the order is 2,1,3 and Primus and Secundus swap, it becomes 1,2,3.If this is the case, then each swap between two runners is a simple inversion of their relative positions, regardless of where they are. Then, the total number of swaps between each pair will determine their final relative order.So, starting from 1,2,3:- Primus (1) and Secundus (2) swap 9 times. Since each swap flips their order, an odd number of swaps (9) means their final order is reversed. So starting as 1 before 2, after 9 swaps, 2 before 1.- Secundus (2) and Tertius (3) swap 10 times. Even number of swaps, so their order remains the same. So 2 stays before 3.- Primus (1) and Tertius (3) swap 11 times. Odd number of swaps, so their order is reversed. Starting with 1 before 3, after 11 swaps, 3 before 1.But wait, this approach might be conflicting because the swaps between different pairs affect each other. For example, swapping Primus and Secundus affects the positions of both relative to Tertius. Similarly, swapping Primus and Tertius affects their positions relative to Secundus.Therefore, we can't consider each pair's swaps in isolation. We need to model the overall permutation.Let me recall that each swap between two elements in a permutation changes the permutation's parity. The total number of swaps (transpositions) determines the parity of the permutation. However, in this case, each transposition is between a specific pair, and we have counts for each pair.But permutations can be decomposed into transpositions, but the order matters. However, if all transpositions are commutative, which they are not. For example, swapping 1 and 2 then swapping 2 and 3 is not the same as swapping 2 and 3 then swapping 1 and 2.Therefore, the order of swaps affects the final permutation. But the problem doesn't specify the order of the swaps, only the total number between each pair. Therefore, perhaps we need to find the net effect of these swaps regardless of the order.Wait, but in group theory, the symmetric group S₃ is generated by transpositions, and each element can be expressed as a product of transpositions. However, the number of transpositions between each pair affects the overall permutation.But if we consider the net effect, each transposition between 1 and 2 is a swap of those two, each transposition between 2 and 3 is a swap of those two, and each transposition between 1 and 3 is equivalent to a swap of those two. However, in S₃, swapping 1 and 3 can be achieved by swapping 1 and 2, then 2 and 3, then 1 and 2 again, which is three transpositions. So, swapping non-adjacent elements requires multiple adjacent swaps.But in the problem, the swaps between 1 and 3 are counted as direct swaps. So perhaps in this problem, a swap between 1 and 3 is considered a single transposition, even if they are not adjacent. If that's the case, then the total number of transpositions for each pair can be considered separately, and the overall permutation is the product of these transpositions.However, the problem is that transpositions don't commute. For example, swapping 1 and 2 then swapping 1 and 3 is different from swapping 1 and 3 then swapping 1 and 2.But since the problem doesn't specify the order of the swaps, we have to assume that the order doesn't matter, or that the net effect is the same regardless of order. But in reality, the order does matter.This seems like a problem. Maybe there's another way to approach it.Alternatively, think of each pair's number of swaps modulo 2, since each swap toggles their order. So, for each pair, if the number of swaps is odd, their order is reversed; if even, it remains the same.But again, the problem is that swapping one pair affects the positions of other pairs. For example, if you swap 1 and 2 an odd number of times, their order is reversed. Then swapping 1 and 3 an odd number of times would reverse their order, but their order depends on the current positions.Wait, maybe we can model the final permutation as follows:Start with the identity permutation [1, 2, 3].Each swap between 1 and 2 toggles their order.Each swap between 2 and 3 toggles their order.Each swap between 1 and 3 toggles their order.But the key is that these toggles are independent? Or do they interfere?Alternatively, we can think of the overall permutation as the composition of three separate permutation cycles: (1 2) done 9 times, (2 3) done 10 times, and (1 3) done 11 times.But in permutation algebra, applying a transposition multiple times is equivalent to applying it (number of times modulo 2). So, if you apply (1 2) 9 times, it's the same as applying it once. Similarly, (2 3) 10 times is the same as not applying it (since 10 mod 2 is 0), and (1 3) 11 times is the same as applying it once.Therefore, the total permutation would be the composition of (1 2), (1 3). So, let's compute that.First, start with the identity permutation [1, 2, 3].Apply (1 2): swaps 1 and 2 => [2, 1, 3].Then apply (1 3): swaps 1 and 3. But in the current permutation, 1 is in position 2 and 3 is in position 3. Swapping 1 and 3 would swap the elements in positions 2 and 3. So, [2, 3, 1].Therefore, the composition (1 2) followed by (1 3) results in the permutation [2, 3, 1].But is this the correct approach? Let's check.Alternatively, the composition can depend on the order. If we apply (1 3) first, then (1 2):Start with [1, 2, 3].Apply (1 3): [3, 2, 1].Then apply (1 2): swaps 3 and 2 => [2, 3, 1].Same result. Wait, so maybe in this case, the composition order doesn't matter? Wait, no, in permutation composition, the order usually matters. Let me check:Wait, in permutation notation, when we write the composition, we usually apply the rightmost permutation first. So, (1 2) followed by (1 3) would be (1 3)(1 2). Let's compute that.First, let's compute (1 2)(1 3):Start with (1 3): 1→3, 3→1, 2→2.Then (1 2): 1→2, 2→1, 3→3.So, composing (1 2) after (1 3):First apply (1 3):1→3, 3→1, 2→2.Then apply (1 2):1→2, 2→1, 3→3.So overall:1: (1→3) then (3→3) => 3.2: (2→2) then (2→1) => 1.3: (3→1) then (1→2) => 2.So the composition (1 2)(1 3) results in the permutation (1 3 2). So mapping 1→3, 3→2, 2→1.Which is the permutation [3, 1, 2].Wait, but earlier when we did it step by step, we got [2, 3, 1]. Hmm, there's a discrepancy here. Which is correct?Wait, maybe I made a mistake in the step-by-step. Let's do it again.Start with [1,2,3].Apply (1 2): swap 1 and 2 => [2,1,3].Then apply (1 3): swap 1 and 3. In the current permutation, positions are 2,1,3. Swapping 1 and 3 would swap the second and third elements. So, [2,3,1].But according to the permutation composition, (1 2)(1 3) is equivalent to (1 3 2). But according to the step-by-step application, we end up with [2,3,1], which is the same as the permutation (1 2 3) → (2,3,1). Wait, the permutation (1 3 2) maps 1→3, 3→2, 2→1. So starting from [1,2,3], applying permutation (1 3 2) would result in [3,1,2], not [2,3,1]. So there's a contradiction here.This suggests that my step-by-step approach might be wrong. Let me check again.Wait, maybe the confusion arises from active vs passive permutations. When we talk about swapping elements, are we swapping the positions or the elements? Let me clarify.In permutation notation, the cycle (1 2 3) means that 1 goes to 2, 2 goes to 3, 3 goes to 1. So applying (1 2 3) to [1,2,3] gives [2,3,1].Similarly, the transposition (1 2) swaps 1 and 2. Applying (1 2) to [1,2,3] gives [2,1,3].Then, applying (1 3) to that result swaps the first and third elements. Wait, in permutation terms, (1 3) swaps 1 and 3 regardless of their current positions. So if the current permutation is [2,1,3], applying (1 3) would swap the first and third elements: [3,1,2].But that's different from swapping the numbers 1 and 3 in the permutation. Wait, this is a key point. If we consider the transpositions as swapping the positions (indices) or the elements (labels).In standard permutation notation, when we apply a transposition (a b), we swap the elements a and b. So, in the permutation [2,1,3], applying (1 3) would swap the elements 1 and 3. But in the permutation [2,1,3], the element 1 is in position 2, and the element 3 is in position 3. So swapping elements 1 and 3 would result in [2,3,1].Wait, no. If we have the permutation [2,1,3], which means position 1: 2, position 2: 1, position 3: 3. Applying transposition (1 3), which swaps elements 1 and 3 wherever they are. So element 1 is in position 2, element 3 is in position 3. Swapping them would put 3 in position 2 and 1 in position 3. So the new permutation would be [2,3,1].But in standard permutation notation, applying the transposition (1 3) to the permutation [2,1,3] would result in swapping the first and third positions. Wait, now I'm confused.Let me clarify:There are two interpretations here.1. Active permutation: permuting the positions. For example, transposition (1 3) swaps the first and third positions.2. Passive permutation: permuting the elements. For example, transposition (1 3) swaps the elements 1 and 3 wherever they are.In the problem, the swaps are between the runners (elements), not the positions. So when Primus and Secundus swap, their positions are exchanged, regardless of where they are. So if Primus (1) is in position 1 and Secundus (2) is in position 2, swapping them results in 2,1,3. If Primus is in position 2 and Secundus is in position 1, swapping them results in 1,2,3 again.Similarly, swapping Primus (1) and Tertius (3) would exchange their positions. So if 1 is in position 1 and 3 is in position 3, swapping them results in 3,2,1. If 1 is in position 3 and 3 is in position 1, swapping them results in 1,2,3 again.Therefore, in this context, each transposition is swapping the elements (runners), not the positions. Therefore, when we apply a transposition (1 2), we swap the runners 1 and 2 in whatever positions they are. So, the permutation is affected accordingly.Therefore, perhaps the composition of swaps is better modeled as follows.Starting permutation: [1,2,3].Each swap between 1 and 2 is a transposition (1 2).Each swap between 2 and 3 is a transposition (2 3).Each swap between 1 and 3 is a transposition (1 3).Since these transpositions are applied multiple times, and in permutation group theory, applying a transposition twice cancels it out. Therefore, the total effect is equivalent to applying each transposition a number of times modulo 2.Therefore, 9 swaps between 1 and 2 is equivalent to 1 transposition (1 2).10 swaps between 2 and 3 is equivalent to 0 transpositions.11 swaps between 1 and 3 is equivalent to 1 transposition (1 3).Therefore, the total permutation is the composition of (1 2) and (1 3). The order of composition matters here.However, since the problem doesn't specify the order in which the swaps occur, we might need to consider that the composition could be in any order. However, in permutation groups, the composition of (1 2) and (1 3) can result in different permutations depending on the order.Wait, let's compute both possibilities.First, apply (1 2) then (1 3):Start with [1,2,3].Apply (1 2): [2,1,3].Apply (1 3): Swap 1 and 3. In [2,1,3], 1 is in position 2 and 3 is in position 3. Swapping them gives [2,3,1].Second, apply (1 3) then (1 2):Start with [1,2,3].Apply (1 3): [3,2,1].Apply (1 2): Swap 1 and 2. In [3,2,1], 1 is in position 3 and 2 is in position 2. Swapping them gives [3,1,2].So depending on the order, we get different results. But the problem doesn't specify the order of the swaps, only the total number between each pair.This is a problem because without knowing the order, we can't determine the exact permutation. However, perhaps there is a clever way to figure out the result regardless of the order.Alternatively, maybe the problem assumes that all swaps happen in a particular order, such as all (1 2) swaps first, then (2 3), then (1 3). But since the problem doesn't specify, we can't assume that.Wait, but in the problem statement, it says "During Sunday's race Primus and Secundus change places exactly 9 times, Secundus and Tertius change places exactly 10 times while Primus and Tertius change places exactly 11 times."The word "while" might imply that these swaps are happening concurrently throughout the race, not in a specific sequence. Therefore, the order of swaps is arbitrary and could be interleaved in any way. Therefore, the composition of the permutations would be ambiguous unless the swaps commute, which they don't.This suggests that the problem might have a unique answer despite the order, which would mean that perhaps the number of swaps is designed such that the final permutation is the same regardless of the order in which the swaps are applied.Alternatively, maybe there's a different way to model this.Let me think differently. Let's model the positions of the runners as a permutation and track the number of inversions.Each time two runners swap, it changes the permutation by a transposition. Each transposition either increases or decreases the number of inversions by 1. However, this might not be helpful here.Alternatively, let's assign a parity to each runner's position.Wait, perhaps we can model the final order by considering the number of times each runner has been swapped with others.For each runner, the number of times they swap with others affects their position. Let's see.Primus (1) swaps 9 times with Secundus and 11 times with Tertius. So, in total, Primus is involved in 9 + 11 = 20 swaps.Similarly, Secundus (2) swaps 9 times with Primus and 10 times with Tertius: 19 swaps.Tertius (3) swaps 10 times with Secundus and 11 times with Primus: 21 swaps.But I don't see how this helps. The number of swaps a runner is involved in doesn't directly translate to their position.Alternatively, perhaps we can model the net displacement of each runner.Each swap with a higher-numbered runner moves them forward or backward?Wait, in a race, overtaking someone would move you forward, and being overtaken would move you backward. But in this problem, each swap is a mutual exchange, so it's not clear if it's an overtake or being overtaken.Alternatively, think of each swap between two runners as a change in their relative order. If two runners swap an odd number of times, their relative order is reversed; if even, it remains the same.But we have three pairs:- 1 and 2 swapped 9 times (odd): reversed.- 2 and 3 swapped 10 times (even): same.- 1 and 3 swapped 11 times (odd): reversed.But how do these interact? Let's see:Starting order: 1, 2, 3.After swaps:1 and 2 have swapped an odd number of times, so their order is reversed: 2, 1, 3.2 and 3 have swapped even number of times, so their order remains: 2, 1, 3.1 and 3 have swapped an odd number of times, so their order is reversed. But in the current order, 1 is second and 3 is third. So swapping them would result in 2, 3, 1.But wait, does swapping 1 and 3 in the order 2,1,3 result in 2,3,1? Yes. So if we consider each pair's swaps independently, reversing their order if odd, the final result would be 2,3,1.But is this the correct approach? Because in reality, swapping 1 and 3 affects the order between 1 and 2 as well. Wait, no. If we first reverse 1 and 2, resulting in 2,1,3, and then reverse 1 and 3, resulting in 2,3,1, that's a possible outcome. However, if we had reversed 1 and 3 first, we would have 3,2,1, then reversing 1 and 2 would give 3,1,2. So depending on the order, we get different results. Therefore, this approach is invalid because it doesn't consider the order of swaps.Therefore, this line of reasoning is flawed. The problem is that considering each pair's swaps independently ignores the interdependency between the swaps. The order in which swaps occur affects the final result, and since the problem doesn't specify the order, we need another way to resolve this.Alternatively, perhaps the key is to recognize that the number of swaps between each pair affects the parity of their relative orderings. Let's consider all possible relative orderings.There are three pairs:1. 1 vs 2: Swapped 9 times (odd) => reversed.2. 2 vs 3: Swapped 10 times (even) => same.3. 1 vs 3: Swapped 11 times (odd) => reversed.If these were independent, then the final order would have 1 and 2 reversed, 2 and 3 same, 1 and 3 reversed. Let's see:Original order: 1, 2, 3.1 and 2 reversed: 2, 1, 3.2 and 3 same: still 2, 1, 3.1 and 3 reversed: 2, 3, 1.Therefore, the final order would be 2,3,1.But wait, let's check if this is consistent.In the final order 2,3,1:- 1 and 2: 2 comes before 1 (reversed from original).- 2 and 3: 2 comes before 3 (same as original).- 1 and 3: 3 comes before 1 (reversed from original).Which matches the parity of the swaps: 1-2 reversed (9 odd), 2-3 same (10 even), 1-3 reversed (11 odd). So regardless of the order of swaps, the final relative orderings must satisfy these parities.Therefore, the final permutation must be such that:- 1 and 2 are in reversed order,- 2 and 3 are in original order,- 1 and 3 are in reversed order.Let's see which permutation satisfies this.Possible permutations of 1,2,3:1. 123: original, doesn't satisfy reversals.2. 132: 1 vs 3 reversed, 2 vs 3 reversed, 1 vs 2 original. Doesn't fit.3. 213: 1 vs 2 reversed, 2 vs 3 original, 1 vs 3 original. Doesn't fit.4. 231: 1 vs 2 reversed (2 before 1), 2 vs 3 reversed (3 before 2), 1 vs 3 original (1 before 3). Doesn't fit.5. 312: 1 vs 2 original (3 before 1 and 2), 2 vs 3 reversed (1 before 3), 1 vs 3 reversed. Doesn't fit.6. 321: All reversed.Wait, none of these permutations satisfy exactly:- 1 vs 2 reversed,- 2 vs 3 original,- 1 vs 3 reversed.Wait, let's check permutation 231:In 231:- 2 vs 3: 2 comes before 3? No, 3 is after 2 in 231? Wait, 231 is 2, 3, 1. So:- 2 comes before 3 (original order for 2 vs 3).- 2 comes before 1 (original order for 2 vs 1, but 1 vs 2 is reversed from original). Wait, 2 comes before 1 in 231, but in the original 1 comes before 2. So 1 vs 2 is reversed here.Similarly, 3 comes before 1 (original order was 1 before 3, so reversed). So in 231:- 1 vs 2: reversed (2 before 1),- 2 vs 3: original (2 before 3),- 1 vs 3: reversed (3 before 1).Yes, that's exactly the permutation 231, which is 2,3,1. So this permutation satisfies all three conditions. Therefore, the final order should be 231.Therefore, the result is 231.But let's check another way. Let's assume that each pair's relative order is determined by the parity of swaps, independent of the others. Then, as we saw, 231 is the only permutation that satisfies all three conditions. Therefore, regardless of the order in which the swaps were performed, the final permutation must satisfy those parity conditions, so the answer is 231.Alternatively, let's think of it as the permutation parity. Each transposition changes the parity of the permutation. The identity permutation is even. Each transposition (swap) flips the parity. The total number of transpositions is 9 + 10 + 11 = 30, which is even, so the final permutation should be even. However, the permutation 231 is equivalent to the cycle (1 2 3), which is an even permutation (since it can be expressed as two transpositions: (1 2)(1 3)). Wait, actually, a 3-cycle is an even permutation. Let me confirm:A 3-cycle (1 2 3) can be written as (1 3)(1 2), which is two transpositions, hence even. Therefore, the permutation 231 is even. Since the total number of transpositions is 30 (even), the final permutation must be even. 231 is even, which matches. The other permutations:- 123: identity (even)- 132: (2 3) transposition (odd)- 213: (1 2) transposition (odd)- 231: (1 2 3) cycle (even)- 312: (1 3 2) cycle (even)- 321: (1 3) transposition (odd)Therefore, the possible even permutations are 123, 231, 312. Since we started with 123 and performed an even number of transpositions, the final permutation must be even. But we already deduced that it's 231 based on the pair parities, which is indeed even. So that checks out.Therefore, the final result is 231, recorded as '231'.Final AnswerThe result of Sunday's race is recorded as boxed{231}.</think>To determine the final order of the runners Primus (1), Secundus (2), and Tertius (3) on Sunday, we need to consider the effect of the number of times each pair swaps places. 1. Primus and Secundus swap 9 times: An odd number of swaps means their relative order is reversed.2. Secundus and Tertius swap 10 times: An even number of swaps means their relative order remains the same.3. Primus and Tertius swap 11 times: An odd number of swaps means their relative order is reversed.Starting from the initial order 1, 2, 3:- After reversing Primus and Secundus (9 times), the order becomes 2, 1, 3.- The order of Secundus and Tertius remains the same (10 times), so it stays 2, 1, 3.- After reversing Primus and Tertius (11 times), the order becomes 2, 3, 1.This final order satisfies all the conditions:- Primus (1) and Secundus (2) are reversed.- Secundus (2) and Tertius (3) remain in their original order.- Primus (1) and Tertius (3) are reversed.Thus, the result of Sunday's race is recorded as boxed{231}.,<think>Okay, let's try to figure out the order of the sportsmen after Sunday's race. So, we know that on Saturday, the result was 123, meaning Primus first, Secundus second, Tertius third. Now, on Sunday, there are a bunch of position changes between them. Specifically, Primus and Secundus swap places 9 times, Secundus and Tertius swap 10 times, and Primus and Tertius swap 11 times. We need to figure out the final order after all these swaps.First, let me make sure I understand what a position change means. If two runners swap places, their positions are exchanged. So, for example, if Primus and Secundus are in positions 1 and 2, after one swap, Secundus is first and Primus is second. If they swap again, they go back to the original positions. So, each pair of swaps between two runners can flip their order and then flip it back.Now, the key here is to figure out how each number of swaps affects the parity (whether it's even or odd) of their order. Because each swap between two runners changes their relative positions. If they swap an even number of times, their original order is preserved. If they swap an odd number of times, their order is reversed.So, let's break it down:1. Primus and Secundus (1 and 2) swap 9 times. Since 9 is odd, their order will be reversed compared to the original. Originally, Primus was ahead of Secundus. After 9 swaps, Secundus would be ahead of Primus.2. Secundus and Tertius (2 and 3) swap 10 times. 10 is even, so their order remains the same. Originally, Secundus was ahead of Tertius, so they should still be in that order.3. Primus and Tertius (1 and 3) swap 11 times. 11 is odd, so their order is reversed. Originally, Primus was ahead of Tertius. After 11 swaps, Tertius would be ahead of Primus.But wait, this seems conflicting. Let's see. If we consider each pair separately, maybe there's a way to combine these results.But hold on, swapping between 1 and 2, and 2 and 3, and 1 and 3. These swaps are not independent. Because swapping 1 and 2 and then 2 and 3 can result in a different overall order than swapping 1 and 3 directly. So maybe I need to model the permutations here.Each swap can be considered as a transposition, which is a permutation that exchanges two elements. The composition of multiple transpositions can be combined into a single permutation. The total permutation after all swaps would determine the final order.But since the order in which the swaps occur matters, the problem might be more complex. However, maybe the parity of the number of swaps can help. Wait, but even if we know the number of transpositions, since transpositions can involve different pairs, the overall permutation's parity might not just depend on the counts of each pair.Alternatively, maybe we can model each pair's swaps as affecting their relative order, but we need to consider how the swaps interact.Wait, perhaps another approach. Let's track the position of each runner step by step. But since we don't know the exact sequence of swaps, just the total number between each pair. So, maybe we can model the net effect of these swaps on the initial positions.Initially, on Sunday, the starting order is the same as Saturday: 1 (Primus), 2 (Secundus), 3 (Tertius). Then, they start swapping.Let me consider that every time two adjacent runners swap, their positions are exchanged. But in this problem, when they say "change places exactly 9 times," does that mean they swapped positions 9 times, regardless of direction? So, for example, if Primus and Secundus swap 9 times, each swap is a single exchange. So, regardless of whether they swap back and forth, each individual exchange counts towards the total.But in reality, if two runners swap positions multiple times, their final order depends on whether the number of swaps is even or odd. For example, swapping once changes their order, swapping again changes it back, so even number of swaps reverts to original, odd number inverts.Similarly, for non-adjacent swaps? Wait, but in a race, you can only swap positions with the person directly in front of or behind you, right? So, maybe the swaps between Primus and Tertius (1 and 3) can't happen directly unless they pass through each other via intermediate positions. Hmm, this complicates things.Wait, maybe the problem is abstracted such that when they say "change places exactly X times," it refers to the number of times they overtake each other, regardless of their positions. But overtaking would require moving past each other, which would involve intermediate steps.But perhaps the problem is simplifying it, so that any two runners can swap positions any number of times, regardless of their current positions. That might not be realistic, but perhaps that's how the problem is framed.Alternatively, maybe each "change of places" between two runners is counted whenever they swap positions, regardless of how many steps it takes. So, for example, if Primus and Secundus swap positions once, that's one change. If they swap back, that's another. So, the total number of times their order changes is the count given.But given that the problem says "change places exactly 9 times," which might mean that they end up swapping positions 9 times. Each time they swap, they exchange their current positions. So, for example, if Primus and Secundus start as 1 and 2, after one swap, they become 2 and 1. After another swap, back to 1 and 2. So, after an odd number of swaps, their order is reversed; after even, original.But here's the catch: if other swaps happen in between, say, Tertius swaps with Secundus, that could affect the positions of Primus and Tertius indirectly.Wait, this seems complicated. Let me think. Suppose all swaps are independent. But in reality, swapping Secundus and Tertius affects their positions relative to Primus as well. For example, if Secundus and Tertius swap, Tertius moves up to second, which means now Tertius is adjacent to Primus (if Primus is first), and then Primus and Tertius can swap.But the problem doesn't specify the order of the swaps. So, perhaps the answer can be determined purely by considering the parity of swaps between each pair, similar to a permutation parity.Wait, another thought: Each swap between two runners is a transposition. The total number of transpositions will determine the overall permutation. But since transpositions can overlap (i.e., involving different pairs), the total permutation is a combination of these transpositions. However, since we don't know the order of the transpositions, maybe the parity of the number of transpositions for each pair can help determine the final permutation.But permutations can be decomposed into cycles. Let me try to model this.Let's denote the initial order as [1, 2, 3].Each swap between 1 and 2 is a transposition (1 2).Each swap between 2 and 3 is a transposition (2 3).Each swap between 1 and 3 is a transposition (1 3).Now, the total number of transpositions is 9 (1-2 swaps) + 10 (2-3 swaps) + 11 (1-3 swaps) = 30 transpositions.But permutations can be represented as products of transpositions, and the parity of the permutation (even or odd) depends on the number of transpositions: even if even number, odd if odd number. However, different decompositions can have different numbers of transpositions, but the parity is fixed. Here, 30 transpositions is even, so the total permutation is even.But the problem is not about the parity of the permutation, but the final order. Let's try to compute the final permutation by considering the effect of each type of transposition.Wait, if we have 9 transpositions of (1 2), 10 of (2 3), and 11 of (1 3). Each transposition can be considered as a generator, and we can compute the net effect.Alternatively, maybe we can compute the total number of times each runner is swapped.Wait, for example, each (1 2) swap moves 1 and 2. Each (2 3) swap moves 2 and 3. Each (1 3) swap moves 1 and 3.But since we have multiple swaps, the positions of each runner can be affected multiple times.But perhaps the key is that each transposition affects the order. Let's model the permutations step by step.Alternatively, note that each transposition between 1 and 2 can be considered as swapping the first and second positions. Each transposition between 2 and 3 swaps the second and third positions. Each transposition between 1 and 3 swaps the first and third positions.But in permutation terms, the transpositions (1 2), (2 3), and (1 3) generate the symmetric group S_3, which has 6 elements.So, the total permutation is a combination of these transpositions. But without knowing the order, we can't compute the exact permutation. However, perhaps we can consider that the number of transpositions between each pair corresponds to the number of times they swap, and the parity (even or odd) would determine whether their order is preserved or reversed.But in permutation terms, applying a transposition (1 2) an odd number of times would result in swapping 1 and 2, while even would leave them as is. Similarly for the others.But if we consider the swaps as independent, which they are not, because swapping 1 and 2 then swapping 2 and 3 is different than swapping 2 and 3 then swapping 1 and 2.Therefore, this complicates things. However, maybe the problem is designed such that the order of swaps does not affect the final result, only the number of swaps between each pair. That would be a simplification, but perhaps it's the case here.Let me think: If we have 9 swaps between 1 and 2, 10 between 2 and 3, and 11 between 1 and 3. Since each swap is a transposition, the total permutation is the product of all these transpositions. However, since transpositions don't commute, the order matters, but maybe the problem expects us to consider the parity for each pair independently.Wait, perhaps in this problem, when they say "change places exactly X times," it's equivalent to the number of inversions between that pair. So, for example, if 1 and 2 swap 9 times, regardless of the other swaps, the relative order between 1 and 2 is flipped 9 times. Since starting with 1 before 2, after an odd number of flips, 2 is before 1; even number, 1 remains before 2.Similarly, for 2 and 3 swapping 10 times (even) so their order remains 2 before 3. For 1 and 3 swapping 11 times (odd), so their order is flipped from 1 before 3 to 3 before 1.But if we consider that, then:- 1 vs 2: 9 swaps (odd) → 2 before 1- 2 vs 3: 10 swaps (even) → 2 before 3- 1 vs 3: 11 swaps (odd) → 3 before 1But now we have conflicting orders. If 2 is before 1, and 2 is before 3, but 3 is before 1. So, 2 > 1, 2 > 3, and 3 > 1. Is that possible? Let's see.If 2 is first, then 3 is before 1, so 2, 3, 1. But in this case, 3 is before 1, which satisfies 3 > 1. Also, 2 is before 3, which satisfies 2 > 3. Wait, but 2 is before 3 and 3 is before 1, so the order is 2, 3, 1.But does this satisfy all the parity conditions?Wait, let's check:- Between 1 and 2: started as 1 before 2, ended as 2 before 1 → 1 swap (odd), which matches 9 swaps (odd).- Between 2 and 3: started as 2 before 3, ended as 2 before 3 → 0 swaps (even), but they had 10 swaps (even). Wait, but 10 is even, so the relative order should remain the same. Which it does here.- Between 1 and 3: started as 1 before 3, ended as 3 before 1 → 1 swap (odd), which matches 11 swaps (odd). Wait, 11 is odd, but here it's only swapped once. Hmm, conflict.So, if we end up with 231, then between 1 and 3, we have one swap (odd), but the problem says 11 swaps. That doesn't align.Wait, maybe this approach is wrong. Because each swap between 1 and 3 would require moving past 2, unless they can directly swap. But in a race, you can't overtake someone without moving through intermediate positions. So, perhaps swapping 1 and 3 requires moving through 2, which would involve swapping with 2.But the problem states that there are 9 swaps between 1 and 2, 10 between 2 and 3, and 11 between 1 and 3. So, in reality, the 11 swaps between 1 and 3 might be composed of swaps via 2.Wait, this is getting more complicated. Maybe we need to model the number of inversions each pair has undergone.Alternatively, consider that each time two runners swap, it affects their relative positions. The total number of swaps between each pair will determine their final order.But in the problem, it's possible that swaps between different pairs affect each other. For example, a swap between 1 and 2 followed by a swap between 2 and 3 results in 1 moving to position 2, 2 moving to position 3, and 3 moving to position 1. So, effectively, a cyclic permutation. But the count of swaps is given per pair.Alternatively, perhaps the key is that the number of swaps between each pair affects their relative ordering, and we can determine the final order by considering the parity for each pair.But in the example of 231, the relative order between 1 and 2 is reversed (odd), between 2 and 3 remains (even), and between 1 and 3 is reversed (odd). However, the problem states 9 swaps (odd) between 1-2, 10 swaps (even) between 2-3, and 11 swaps (odd) between 1-3. So, in that case, 231 would be the result. But in that case, the total number of swaps for 1-3 is 11, but the actual number of inversions is 1. So, that approach might not work.Alternatively, maybe each swap between 1 and 3 is equivalent to two swaps: first swapping 1 and 2, then swapping 2 and 3, effectively moving 1 to position 3. So, each direct swap between 1 and 3 would require two adjacent swaps, which would increment the counts for both 1-2 and 2-3. But in the problem, the counts for 1-2, 2-3, and 1-3 are given as separate numbers. So, perhaps the 1-3 swaps are considered as separate from the adjacent swaps.But this seems conflicting. If a 1-3 swap is considered a single event, even though in reality it would require passing through position 2, then the counts are independent. But in that case, how can 1 and 3 swap 11 times, while 1 and 2 swap 9 times, and 2 and 3 swap 10 times.Alternatively, the problem might be using a different model where any two runners can swap regardless of their positions, so a 1-3 swap is possible without involving 2. In that case, each swap is counted independently.But if that's the case, then the total number of swaps for each pair can be considered as independent transpositions, and the final permutation can be computed by applying all these transpositions. However, since transpositions don't commute, the order matters, but since we don't know the order, perhaps the problem expects us to assume that the order doesn't matter, and we can compute the parity for each pair.Wait, in permutation group theory, the number of transpositions between each pair would generate a certain permutation. For example, if you have x transpositions of (1 2), y transpositions of (2 3), and z transpositions of (1 3), the total permutation can be computed as follows:Each transposition (1 2) can be represented as swapping 1 and 2.Similarly, (2 3) swaps 2 and 3, and (1 3) swaps 1 and 3.But since permutations don't commute, the order matters. However, if we consider the total number of each transposition, maybe we can compute the net effect.Alternatively, note that in the symmetric group S_3, any permutation can be written as a product of transpositions. The three transpositions (1 2), (2 3), and (1 3) generate S_3. So, the total permutation would be some element of S_3, which has 6 elements.But since we don't know the order of the transpositions, we can't determine the exact permutation. However, perhaps we can use the fact that the number of transpositions for each pair corresponds to their parity, regardless of order.Alternatively, think of each transposition as toggling the state between two elements. For example, each (1 2) swap toggles the order of 1 and 2, each (2 3) swap toggles the order of 2 and 3, and each (1 3) swap toggles the order of 1 and 3.But since these toggles are not independent—because swapping 1 and 2 affects their relationship with 3—the problem is more complex.Alternatively, model the relative orderings as a series of toggles.Let me consider the relative order between each pair:1. Between 1 and 2: Start with 1 < 2. After 9 swaps (odd), the order is 2 < 1.2. Between 2 and 3: Start with 2 < 3. After 10 swaps (even), the order remains 2 < 3.3. Between 1 and 3: Start with 1 < 3. After 11 swaps (odd), the order is 3 < 1.Now, if we have 2 < 1, 2 < 3, and 3 < 1, what's the possible ordering?If 2 < 3 and 3 < 1, then 2 < 3 < 1. So, the order would be 2, 3, 1. Which also satisfies 2 < 1 (since 2 is first and 1 is third). Wait, but in the order 2, 3, 1, we have:- 2 before 3: correct (2 < 3)- 3 before 1: correct (3 < 1)- 2 before 1: correct (2 < 1)So, this ordering 2, 3, 1 satisfies all the pairwise conditions. Therefore, the result would be 231.But let's verify this with the given number of swaps. If we end up with 231, how does that relate to the number of swaps?Wait, the problem states that there are 9 swaps between 1 and 2, 10 between 2 and 3, and 11 between 1 and 3. If we end up with 231, how does each swap count affect that.Starting from 123:- 9 swaps between 1 and 2: Each swap toggles their order. 9 is odd, so 1 and 2 end up reversed: 213.- Then, 10 swaps between 2 and 3: 10 is even, so their order remains the same. If we are at 213, swapping 2 and 3 ten times would leave it as 213 (since even number of swaps).- Then, 11 swaps between 1 and 3: Each swap toggles their order. Starting from 213 (1 is second, 3 is third), swapping 1 and 3 once would result in 231. Swapping again would go back to 213. So, 11 swaps (odd) would result in 231.But this assumes that we first do all 1-2 swaps, then all 2-3 swaps, then all 1-3 swaps. But in reality, the order of swaps could be interleaved. However, the problem doesn't specify the order, so maybe the answer is still 231 regardless of the order.But is this the case? Let's test with a different order.Suppose we start with 123.First, do all 11 swaps between 1 and 3. Each swap toggles their positions. Starting with 1 first and 3 third. After 11 swaps (odd), 3 and 1 have swapped: 321.Then, do 9 swaps between 1 and 2. But in 321, 1 is third and 2 is second. Swapping 1 and 2 would exchange their positions. Each swap toggles their order. Starting with 2 before 1. After 9 swaps (odd), 1 would be before 2: 312.Then, do 10 swaps between 2 and 3. In 312, 3 is first, 1 is second, 2 is third. Swapping 2 and 3 (positions 1 and 3) ten times. Wait, but 2 is third and 3 is first. To swap 2 and 3, you need to swap positions 1 and 3. But if each swap between 2 and 3 is a transposition (2 3), which in the current order would be swapping the second and third positions. Wait, in the current order 312, the second position is 1, third is 2. Swapping 2 and 3 (positions 2 and 3) would swap 1 and 2. So, swapping 2 and 3 in the current order would swap 1 and 2.Wait, this seems like the labels 1, 2, 3 are fixed to the runners, not their positions. So, regardless of their current positions, a swap between Secundus (2) and Tertius (3) would exchange their positions, wherever they are.So, starting from 312 (positions: 3, 1, 2), swapping Secundus (2) and Tertius (3) would swap their positions. So, Tertius is in position 1, Secundus is in position 3. After swapping, Tertius moves to position 3 and Secundus to position 1. So, the new order is 213.Wait, but starting from 312 (which is Tertius, Primus, Secundus), swapping Secundus and Tertius would result in Secundus moving to position 1, Tertius moving to position 3. So, the order becomes 213 (Secundus, Primus, Tertius). Then, swapping them again would revert to 312. So, each swap between 2 and 3 toggles between these two orders.But we have 10 swaps between 2 and 3. Starting from 312, after 10 swaps (even), we would end up back at 312. Then, after doing the 9 swaps between 1 and 2: in 312, swapping 1 and 2 (Primus and Secundus) would swap positions 2 and 3 (since 1 is in position 2, and 2 is in position 3). Wait, this is getting confusing.Alternatively, maybe the labels refer to the runners, not the positions. So, a swap between Primus (1) and Secundus (2) exchanges their positions, wherever they are. Similarly for others.So, starting from 123 (Positions: 1-1, 2-2, 3-3).First, perform 11 swaps between 1 and 3. Each swap exchanges their positions. Starting with 1 in position 1 and 3 in position 3. After one swap: 3 in position 1, 1 in position 3. After two swaps: back to original. So, 11 swaps would result in 1 swap (since 11 is odd). So, 3, 2, 1.Wait, but swapping 1 and 3 in 123 would result in 321. Then, swapping again would result back to 123. So, after 11 swaps, it would be 321 (since 11 is odd).Then, perform 9 swaps between 1 and 2. In 321, 1 is in position 3, 2 is in position 2. Swapping 1 and 2 would exchange positions 2 and 3. So, after one swap: 312. After two swaps: 321 again. 9 swaps (odd) would result in 312.Then, perform 10 swaps between 2 and 3. In 312, 2 is in position 3, 3 is in position 1. Swapping 2 and 3 would exchange positions 1 and 3. So, after one swap: 213. After two swaps: 312. 10 swaps (even) would bring it back to 312. So, the final result is 312.But this contradicts the previous result of 231. Hmm.Wait, so depending on the order of the swaps, we can get different results. This is a problem because the problem doesn't specify the order of the swaps, only the total number between each pair. Therefore, there must be another approach.Perhaps the key is to realize that each swap between two runners affects their relative positions, but the total number of swaps can be considered as the net number of times they pass each other, regardless of the order.In such a case, the parity (odd or even) of the number of swaps between each pair will determine their final order. This is similar to considering the permutation parity for each pair independently.So, for each pair:- Primus and Secundus swapped 9 times (odd): their order is reversed.- Secundus and Tertius swapped 10 times (even): their order remains the same.- Primus and Tertius swapped 11 times (odd): their order is reversed.So, applying these three conditions:1. Primus and Secundus are reversed: so instead of 1 before 2, it's 2 before 1.2. Secundus and Tertius remain the same: 2 before 3.3. Primus and Tertius are reversed: instead of 1 before 3, it's 3 before 1.So, combining these:- 2 before 1 (from 1)- 2 before 3 (from 2)- 3 before 1 (from 3)This gives us the order 2, 3, 1 (since 2 is first, 3 is second because 2 is before 3, and 3 is before 1). So, 231.But earlier, when we considered a different order of swaps, we got 312. Which contradicts this.But perhaps the answer should be 231, and the discrepancy arises because in the previous example, we applied the swaps in a specific order, but the problem might intend for the net effect based on parity regardless of order.Alternatively, maybe the key is that the number of swaps between each pair directly translates to their relative order, and we need to arrange them such that all the parity conditions are satisfied.So, if we have:- 2 before 1 (due to 9 swaps)- 2 before 3 (due to 10 swaps)- 3 before 1 (due to 11 swaps)Then the only ordering that satisfies all three is 2, 3, 1.Because:- 2 is before 3 (satisfies 2 before 3)- 3 is before 1 (satisfies 3 before 1)- 2 is before 1 (satisfies 2 before 1)So, the order 231.But in the other scenario where swaps were applied in a different order, we ended up with 312. But why is that?Ah, perhaps because when we perform swaps between 1 and 3 first, it affects the positions such that subsequent swaps between 1 and 2 are actually different.This suggests that the order of swaps affects the result, which complicates the problem. However, the problem doesn't specify the order, so maybe we are to assume that the net effect is based purely on the parity of swaps between each pair, leading to 231.But to verify, let's think of another way. The total number of position swaps between each pair should affect their relative order. Let's model the initial order as 1, 2, 3.Each swap between 1 and 2 will flip their order. After an odd number, it's reversed; after even, same.Each swap between 2 and 3 similarly.Each swap between 1 and 3 will flip their order.But in reality, swapping 1 and 3 requires them to pass each other, which would involve either moving through position 2 (which would require swaps with 2), but in this problem, perhaps they can swap directly.Assuming direct swaps, the parity determines their order.Thus, if all swaps are independent, then:- 1 and 2 swap 9 times: reversed.- 2 and 3 swap 10 times: same.- 1 and 3 swap 11 times: reversed.Therefore, final order 2, 3, 1.But let's confirm with another example. Suppose we have 1, 2, 3.Swap 1 and 2 once: 2, 1, 3.Swap 1 and 3 once: 2, 3, 1.So, after one swap each between 1-2 and 1-3, you get 2, 3, 1. Which is the same as if you reversed 1-2 and 1-3.Similarly, if you do multiple swaps:If you swap 1-2 nine times, you end up with 2, 1, 3 if 9 is odd.Then swap 2-3 ten times, which is even, so remains 2, 1, 3.Then swap 1-3 eleven times: starting from 2,1,3. Swapping 1 and 3 once gives 2,3,1. Swapping again gives 2,1,3. So, 11 swaps (odd) would result in 2,3,1.Thus, regardless of the order, if you apply all swaps of one pair, then another, then another, you still get 2,3,1.But if you interleave the swaps, does it change?Let's say starting with 1,2,3.Swap 1-2: 2,1,3.Swap 2-3: 2,3,1.Swap 1-3: 3,2,1.Swap 1-2: 3,1,2.Swap 2-3: 3,2,1.Etc. It's complicated, but the key is that each swap toggles the order between two runners. So, the total number of toggles for each pair determines their final order.Therefore, regardless of the order in which the swaps are performed, the parity (odd or even) of the number of swaps between each pair will determine whether their order is reversed or remains the same.Therefore, even if you interleave the swaps, the final result depends only on the parity of the number of swaps between each pair. Therefore, the answer should be 231.Therefore, the final result is recorded as 231.Final AnswerThe result of Sunday's race is recorded as boxed{231}.</think>To determine the final order of the sportsmen after Sunday's race, we need to consider the number of times each pair of sportsmen swapped places and the parity (odd or even) of those swaps. 1. Primus (1) and Secundus (2) swapped 9 times: Since 9 is odd, their order is reversed from the original (1 before 2 to 2 before 1).2. Secundus (2) and Tertius (3) swapped 10 times: Since 10 is even, their order remains the same (2 before 3).3. Primus (1) and Tertius (3) swapped 11 times: Since 11 is odd, their order is reversed from the original (1 before 3 to 3 before 1).Combining these results:- Secundus (2) must be before Primus (1) due to the odd number of swaps between them.- Secundus (2) must be before Tertius (3) due to the even number of swaps between them.- Tertius (3) must be before Primus (1) due to the odd number of swaps between them.The only order that satisfies all these conditions is 2 (Secundus), 3 (Tertius), 1 (Primus).Thus, the result of Sunday's race is recorded as boxed{231}.]

🔑:2. 231 At the start of the race, the runners are in order 123. By the end of the race Primus and Tertius have exchanged places 11 times, so Tertius ends the race ahead of Primus. Also, since Secundus and Tertius have exchanged places 10 times, Secundus ends ahead of Tertius. So the result of the race is 231 .

🧠:[<think>Okay, let me try to figure out this problem. So, we have a brick factory where each worker produces ten bricks per day. But each worker also steals as many bricks per day as there are workers employed. The question is asking for the maximum number of workers that can be hired so that by the end of the day, there are at least 13 more bricks at the factory than at the beginning. Hmm, let me break this down step by step.First, let's denote the number of workers as 'n'. Each worker produces 10 bricks, so the total number of bricks produced per day would be 10 multiplied by the number of workers, which is 10n. That makes sense. Now, each worker also steals bricks equal to the number of workers. So, each worker steals 'n' bricks. Since there are 'n' workers, the total number of bricks stolen per day would be n multiplied by n, which is n². So, the net gain of bricks for the factory each day would be the total produced minus the total stolen. That is, net gain = 10n - n². The problem states that this net gain needs to be at least 13 bricks. So, we need to find the maximum integer value of 'n' such that 10n - n² ≥ 13.Wait, that sounds like a quadratic inequality. Let me write that down:10n - n² ≥ 13I can rearrange this inequality to standard quadratic form. Subtract 13 from both sides:-n² + 10n - 13 ≥ 0Multiplying both sides by -1 (which will reverse the inequality sign):n² - 10n + 13 ≤ 0Now, I need to solve the quadratic inequality n² -10n +13 ≤ 0. To find the values of 'n' that satisfy this, I first find the roots of the equation n² -10n +13 = 0.Using the quadratic formula, n = [10 ± sqrt(100 - 52)] / 2. Since the discriminant is 100 - 4*1*13 = 100 -52 = 48. So sqrt(48) is 4*sqrt(3), which is approximately 6.928.So the roots are (10 + 6.928)/2 ≈ 16.928/2 ≈ 8.464 and (10 - 6.928)/2 ≈ 3.072/2 ≈ 1.536.Since the quadratic coefficient is positive, the parabola opens upwards. Therefore, the inequality n² -10n +13 ≤ 0 is satisfied between the two roots. That is, for n between approximately 1.536 and 8.464.But since the number of workers must be an integer, the possible integer values of 'n' are from 2 up to 8 (since 1.536 rounds up to 2 and 8.464 rounds down to 8). However, we need to check which of these integer values actually satisfy the original inequality 10n -n² ≥13.Let me test n =8 first since we need the maximum number of workers. Plugging n=8 into 10n -n²:10*8 - 8² =80 -64=16. 16 is greater than 13, so that's good.What about n=9? 10*9 -9²=90 -81=9. 9 is less than 13, so that's not acceptable. So 9 is too many. Therefore, 8 is the maximum number of workers? Wait, but let me check n=8 again. Yes, 16 is indeed ≥13.But let's check n=7 just to be thorough. 10*7 -7²=70-49=21. 21 is also greater than 13. Similarly, n=6: 60-36=24. So all the way down to n=2.Wait, but the problem says "the maximum number of workers", so the answer is 8. But wait, let me confirm again. Wait, when n=8, the net gain is 16, which is more than 13. If n=9, it's 9, which is less. So 8 is the highest possible.But wait, when I solved the inequality, the upper bound was approximately 8.464, which would mean that up to 8.464 workers would still satisfy the inequality, but since you can't have a fraction of a worker, you have to take the integer part. So 8 workers would be the maximum integer value where the net gain is still above 13. So the answer is 8.But just to make sure, maybe I made a mistake in the quadratic equation. Let me recheck:Original inequality: 10n -n² ≥13.Let me check n=8: 80 -64=16 ≥13: yes.n=9:90-81=9 <13: no.Hence, 8 is correct. So the maximum number of workers is 8.But wait, let me check the roots again. The roots were at approximately 1.536 and 8.464. Since the quadratic is negative between the roots, but we want n where 10n -n² ≥13. That's equivalent to the original equation. So when n is between 1.536 and 8.464, the quadratic is ≤0. Wait, hold on, let me clarify.Wait, after rearrangement, we had n² -10n +13 ≤0. So the solution set is between the two roots. So n between approximately 1.536 and 8.464. Since the problem is asking for the maximum number of workers (integer), the maximum integer less than or equal to 8.464 is 8. So that's correct.Therefore, the answer is 8.Final AnswerThe maximum number of workers is boxed{8}.</think>Let ( n ) be the number of workers employed at the factory. Each worker produces 10 bricks per day, so the total production is ( 10n ) bricks. Each worker also steals ( n ) bricks per day, so the total number of bricks stolen is ( n^2 ). The net gain of bricks at the factory is given by the production minus the theft:[10n - n^2]We need this net gain to be at least 13 bricks:[10n - n^2 geq 13]Rearranging this inequality into standard quadratic form:[-n^2 + 10n - 13 geq 0]Multiplying by -1 (which reverses the inequality):[n^2 - 10n + 13 leq 0]Solving the quadratic equation ( n^2 - 10n + 13 = 0 ) using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[n = frac{10 pm sqrt{100 - 52}}{2} = frac{10 pm sqrt{48}}{2} = frac{10 pm 4sqrt{3}}{2} = 5 pm 2sqrt{3}]Approximating the roots:[5 + 2sqrt{3} approx 8.464 quad text{and} quad 5 - 2sqrt{3} approx 1.536]The quadratic inequality ( n^2 - 10n + 13 leq 0 ) holds between the roots ( 1.536 ) and ( 8.464 ). Since ( n ) must be an integer, the maximum integer value is 8. Checking ( n = 8 ):[10 times 8 - 8^2 = 80 - 64 = 16 geq 13]Checking ( n = 9 ):[10 times 9 - 9^2 = 90 - 81 = 9 < 13]Thus, the maximum number of workers that can be hired is (boxed{8}).,<think>Okay, let's try to figure out this problem. So, the question is about a brick factory where each worker produces ten bricks per day but also steals as many bricks per day as there are workers employed. We need to find the maximum number of workers that can be hired so that by the end of the day, there are at least 13 more bricks at the factory than at the beginning.Hmm, let's break it down. Let me denote the number of workers as 'n'. Each worker produces 10 bricks, so total production would be 10n bricks per day. But each worker also steals 'n' bricks, right? Wait, the problem says each worker steals as many bricks per day as there are workers. So if there are 'n' workers, each one steals 'n' bricks. Therefore, total stolen bricks would be n workers * n bricks each = n² bricks stolen per day.So, the net gain for the factory would be the total production minus the total stolen bricks. That is, net gain = 10n - n². We need this net gain to be at least 13. So, the inequality is 10n - n² ≥ 13.Wait, but let me check if I'm interpreting the stealing correctly. Each worker steals 'n' bricks, so yes, per worker. So, total stolen is n * n. So, that part seems correct.So the equation is net bricks = 10n - n². We need 10n - n² ≥ 13. Let's rearrange this inequality. Let's move everything to one side:-n² + 10n - 13 ≥ 0 Multiply both sides by -1 (remember to reverse the inequality sign):n² - 10n + 13 ≤ 0So, we need to solve the quadratic inequality n² -10n +13 ≤0.To find the values of n where this inequality holds, first find the roots of the equation n² -10n +13 =0.Using the quadratic formula:n = [10 ± sqrt(100 - 52)] / 2 Since 10^2 = 100, and 4*1*13=52, so discriminant is 100 - 52 = 48.So sqrt(48) = 4*sqrt(3) ≈ 6.928.Therefore, the roots are:n = [10 + 4√3]/2 = 5 + 2√3 ≈5 + 3.464≈8.464 n = [10 -4√3]/2 =5 -2√3≈5 -3.464≈1.536So the quadratic equation n² -10n +13=0 has roots at approximately 1.536 and 8.464.Since the coefficient of n² is positive, the parabola opens upwards. Therefore, the inequality n² -10n +13 ≤0 is satisfied between the roots. So, n must be between approximately 1.536 and 8.464.But since n is the number of workers, it must be an integer. So the possible integer values of n are 2,3,4,5,6,7,8.But the question asks for the maximum number of workers. So the largest integer less than or equal to 8.464 is 8. So n=8 is the maximum number? Wait, but we need to check if at n=8, the net gain is at least 13.Let's compute 10n -n² when n=8:10*8 -8²=80 -64=16. 16 is greater than 13, so that's okay.If n=9, which is beyond 8.464, so the inequality is not satisfied. Let's check n=9: 10*9 -81=90-81=9, which is less than 13, so 9 is too many.So the maximum number is 8. So the answer is 8. Therefore, the maximum number of workers that can be hired is 8, since at 8 workers, net gain is 16 bricks, which is above 13. If they hire 9, the net gain drops to 9, which is below 13. Therefore, 8 is the maximum.Wait, let me double-check with exact values. Since the roots are 5 + 2√3 and 5 - 2√3. 2√3 is approximately 3.464. So 5+3.464=8.464, and 5-3.464=1.536. Therefore, the quadratic is negative between these two roots. So any integer n between 2 and 8 inclusive will satisfy the inequality. So 8 is the maximum. Therefore, the answer is 8.But just to make sure, let's verify for n=8:Each worker produces 10 bricks, so 8 workers make 80 bricks. Each worker steals 8 bricks (since the number of workers is 8), so each worker steals 8, so total stolen is 8*8=64 bricks. Net gain is 80-64=16, which is 16≥13, so yes, that works.For n=9:Total production 90 bricks, each steals 9, total stolen 81. Net gain 90-81=9<13. Not enough. So 9 is too much.Therefore, 8 is the maximum.So the answer should be boxed{8}</think>Let ( n ) be the number of workers. Each worker produces 10 bricks per day and steals ( n ) bricks per day. The total production is ( 10n ) bricks, and the total stolen bricks are ( n^2 ). The net gain in bricks is given by:[10n - n^2]We need this net gain to be at least 13:[10n - n^2 geq 13]Rearranging the inequality:[n^2 - 10n + 13 leq 0]Solving the quadratic equation ( n^2 - 10n + 13 = 0 ) using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[n = frac{10 pm sqrt{100 - 52}}{2} = frac{10 pm sqrt{48}}{2} = frac{10 pm 4sqrt{3}}{2} = 5 pm 2sqrt{3}]The roots are approximately ( 1.536 ) and ( 8.464 ). Since the quadratic opens upwards, the inequality ( n^2 - 10n + 13 leq 0 ) holds between these roots. The integer values of ( n ) within this interval are 2, 3, 4, 5, 6, 7, and 8. The maximum integer value is 8.Verification for ( n = 8 ):- Total production: ( 10 times 8 = 80 ) bricks- Total stolen: ( 8 times 8 = 64 ) bricks- Net gain: ( 80 - 64 = 16 geq 13 )Verification for ( n = 9 ):- Total production: ( 10 times 9 = 90 ) bricks- Total stolen: ( 9 times 9 = 81 ) bricks- Net gain: ( 90 - 81 = 9 < 13 )Thus, the maximum number of workers is (boxed{8}).]

🔑:# Problem 2. (1 point)At a brick factory, each worker produces ten bricks per day and steals as many bricks per day as there are workers employed at the factory. What is the maximum number of workers that can be hired so that by the end of the day, there are at least 13 more bricks at the factory than at the beginning.Answer: 8

🧠:[<think>Okay, so the problem is: 6 people each have a hat. They shuffle the hats and redistribute them. What is the probability that exactly one person gets their own hat back?Hmm. Let's see. Probability problems can be tricky, but let me try to break this down step by step. First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. So I need to find out how many ways exactly one person gets their own hat back, and then divide that by the total number of ways the hats can be redistributed.Total number of redistributions is 6 factorial, right? Because each person can get any hat, so that's 6! = 720. That's straightforward.Now, the tricky part is figuring out the number of favorable outcomes where exactly one person gets their own hat. Let me recall permutations here. When we talk about permutations where no element appears in its original position, those are called derangements. But here, we want exactly one person in the original position and the rest deranged. So perhaps the number we need is the number of derangements for 5 elements, multiplied by the number of ways to choose which one person stays fixed.Wait, let me confirm that. If exactly one person gets their own hat, then we fix that one person, and the remaining five must not get their own hats. So yes, the number should be C(6,1) * D(5), where D(5) is the derangement number for 5 elements.So first, let's compute D(5). The formula for derangements is D(n) = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!). Alternatively, it's also D(n) = floor(n!/e + 0.5) for a quick approximation, but since n is 5 here, let's compute it exactly.Calculating D(5):D(5) = 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]Compute each term:5! = 120Then inside the brackets:1 - 1/1! = 1 - 1 = 0+ 1/2! = 0 + 0.5 = 0.5- 1/6 = 0.5 - 0.166666... = 0.333333...+ 1/24 = 0.333333... + 0.041666... ≈ 0.375- 1/120 ≈ 0.375 - 0.008333... ≈ 0.366666...So multiplying by 5! which is 120:0.366666... * 120 = let's calculate that.0.366666... is 11/30. Because 11 divided by 30 is approximately 0.366666...So 11/30 * 120 = 11 * 4 = 44. So D(5) = 44.Wait, let me check that again. Alternatively, maybe my step-by-step calculation was wrong. Let's do the exact calculation.Compute the sum inside the brackets:1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!Calculate each term:1 = 1-1/1! = -1+1/2! = +0.5-1/6 ≈ -0.166666...+1/24 ≈ +0.041666...-1/120 ≈ -0.008333...Adding these up step by step:1 - 1 = 00 + 0.5 = 0.50.5 - 0.166666... ≈ 0.333333...0.333333... + 0.041666... ≈ 0.3750.375 - 0.008333... ≈ 0.366666...So the total sum is approximately 0.366666..., which is exactly 11/30. Then, D(5) = 5! * 11/30 = 120 * 11 /30 = (120/30)*11 = 4*11=44. Yes, that's correct. So D(5)=44.Therefore, the number of favorable permutations where exactly one person gets their own hat is C(6,1)*D(5) = 6*44=264.Therefore, the probability is 264 / 720. Let's simplify that. Both numerator and denominator are divisible by 24: 264 ÷24=11, 720÷24=30. So 11/30. Wait, 264 divided by 24 is 11? Let me check: 24*10=240, 24*11=264. Yes. So 264/720=11/30≈0.3666...So the probability is 11/30.Wait a second, but let me verify if this is correct. Let me think again.Alternatively, is there another way to compute the number of permutations with exactly one fixed point?Yes, in combinatorics, the number of permutations of n elements with exactly k fixed points is C(n,k)*D(n -k). So here, n=6, k=1, so it's C(6,1)*D(5)=6*44=264, which we already have.Therefore, yes, that's correct. So probability is 264/720=11/30≈0.3667.But just to be thorough, let's consider another approach. Let's compute the probability step by step.First, choose one person to fix: 6 choices. Then, for the remaining 5 people, we need a derangement. So for each of those 5, the probability that none of them get their own hat is D(5)/5! = 44/120=11/30. Wait, but here we are not dealing with probability yet. Wait, the number of derangements is 44, so the number of favorable permutations is 6*44=264. So probability is 264/720=11/30≈0.3667.Alternatively, if we think in terms of probability step by step:Probability that exactly one person gets their own hat. So first, pick a person, say person A. The probability that person A gets their own hat is 1/6. Then, given that person A has their own hat, the probability that none of the remaining 5 people get their own hats is D(5)/5! = 44/120=11/30. So the probability that person A gets their own hat and others don't is (1/6)*(11/30). But since there are 6 different people who could be the one that gets their own hat, we multiply by 6, so 6*(1/6)*(11/30)=11/30. That's the same answer.Yes, that's another way to get there. So whether we compute it combinatorially or step by step using probabilities, we get 11/30.Therefore, the probability is 11/30.But just to make sure, let me check with small numbers. Let's take n=2. What's the probability that exactly one person gets their own hat? Wait, with n=2, there are two permutations: [1,2] and [2,1]. The first one has both fixed points, the second one has none. So there are zero permutations where exactly one person gets their own hat. So the probability is zero. Using the formula: C(2,1)*D(1)=2*0=0. Correct. D(1)=0 because if one person is left, they can't derange. So that works.Another example, n=3. How many permutations have exactly one fixed point. Let's list all permutations:123 – all fixed.132 – fixes 1, swaps 2 and 3. So fixed points: 1. So exactly one.213 – fixes 3, swaps 1 and 2. Fixed points: 3. So exactly one.231 – derangement, fixes none.312 – derangement, fixes none.321 – swaps 1 and 3, fixes 2. So fixed points: 2. Exactly one.So out of 6 permutations, there are 3 with exactly one fixed point. So probability 3/6=1/2. Using the formula: C(3,1)*D(2)=3*1=3. 3/6=1/2. Correct. Since D(2)=1 (derangements of two elements is 1: swapping them). So yes, formula works here. So that gives me more confidence.Therefore, going back to the original problem with 6 people, the formula holds. So the answer is 11/30.Final AnswerThe probability is boxed{dfrac{11}{30}}.</think>To find the probability that exactly one person gets their own hat back when 6 people shuffle and redistribute their hats, we need to consider the number of favorable outcomes and divide it by the total number of possible outcomes.1. Total number of permutations: This is (6! = 720).2. Number of favorable outcomes: We need to count the permutations where exactly one person gets their own hat. This can be calculated by: - Choosing 1 person out of 6 to be fixed: (binom{6}{1} = 6) ways. - Deranging the remaining 5 hats (so none of the remaining 5 people get their own hats). The number of derangements of 5 elements, (D(5)), is calculated as follows: [ D(5) = 5! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!}right) ] Calculating each term: [ 1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120} = frac{11}{30} ] Multiplying by (5!): [ D(5) = 120 times frac{11}{30} = 44 ] - Therefore, the number of favorable permutations is (6 times 44 = 264).3. Probability calculation: The probability is the number of favorable outcomes divided by the total number of permutations: [ frac{264}{720} = frac{11}{30} ]Thus, the probability that exactly one person gets their own hat back is (boxed{dfrac{11}{30}}).,<think>Okay, so I have this probability problem here: there are 6 people, each with their own hat. They shuffle the hats and redistribute them. I need to find the probability that exactly one person gets their own hat back. Hmm, let's see. I remember something about permutations and derangements here. Let me try to work through it step by step.First, probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, in this case, the total number of ways the hats can be redistributed is 6 factorial, which is 6! = 720. That's straightforward because each of the 6 hats can go to any of the 6 people, and we assume they're all shuffled equally likely.Now, the tricky part is finding the number of favorable outcomes where exactly one person gets their own hat back. Let's break that down. If exactly one person gets their own hat, then that person is fixed, and the remaining 5 people must have a derangement of their hats. A derangement is a permutation where no element appears in its original position. So, the number of derangements of n elements is denoted by !n (subfactorial n). The formula for derangements is !n = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!). Wait, but here we need exactly one person fixed and the rest deranged. So, first, we need to choose which one person is fixed. There are 6 choices for that. Then, for the remaining 5 people, we need a derangement. So, the number of favorable outcomes would be 6 * !5. Then, the probability is (6 * !5)/6!.But let me confirm this. So, first, choosing 1 person out of 6 to have their correct hat: C(6,1) = 6. Then, the remaining 5 hats must be deranged. The derangement number for 5 is !5. So total favorable is 6 * !5. The total permutations are 6!.Therefore, probability is (6 * !5)/6! = !5 /5! because 6/6! = 1/5!.But what is !5? Let me compute that. The formula for derangements is !n = n! * [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!]So for n=5:!5 = 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]Calculating each term:5! = 1201 - 1/1! = 1 - 1 = 0Wait, no, let's write all terms:1 = 1-1/1! = -1+1/2! = +1/2-1/3! = -1/6+1/4! = +1/24-1/5! = -1/120So adding these up:1 -1 = 00 + 1/2 = 1/21/2 -1/6 = 1/2 - 1/6 = 1/31/3 + 1/24 = 8/24 + 1/24 = 9/24 = 3/83/8 -1/120 = (45/120 - 1/120) = 44/120 = 11/30So, !5 = 5! * (11/30) = 120 * (11/30) = 4 * 11 = 44Wait, let's check that again. 120 * (11/30) = (120/30)*11 = 4*11=44. Yes.So !5 = 44. Therefore, the number of favorable outcomes is 6*44=264.Total permutations: 720.So the probability is 264/720. Let's simplify that. Divide numerator and denominator by 24: 264/24=11, 720/24=30. So 11/30. Therefore, the probability is 11/30. Let me check that again.Alternatively, since the formula for the number of permutations with exactly k fixed points is C(n, k) * !(n - k). So here, n=6, k=1. So C(6,1)*!5=6*44=264. 264/720=11/30. That seems right.Alternatively, I can recall that the probability of exactly m fixed points in a permutation of n elements is given by (C(n, m) * !(n - m))/n!. So substituting n=6, m=1: (C(6,1)*!5)/6! = (6*44)/720 = 264/720 = 11/30 ≈ 0.3667.Alternatively, I can think of it another way. Suppose we fix one person to have their own hat. The probability that person 1 gets their own hat is 1/6. Then, the remaining 5 hats must be deranged. The probability that the remaining 5 are deranged is !5 /5! = 44/120 = 11/30. But since we have 6 choices for which person is fixed, we multiply by 6: 6*(1/6)*(11/30) = (1)*(11/30)=11/30. Yes, that matches. So that's another way to see it.So the answer should be 11/30. Let me check with n=6, the formula for the number of permutations with exactly m fixed points. The general formula is C(n, m) * !(n - m). So for m=1, it's C(6,1)*!5=6*44=264. Divide by 720: 264/720=11/30. Yep, seems correct.Alternatively, maybe I can compute inclusion-exclusion for exactly one fixed point. Let me try that.The number of permutations with exactly one fixed point can be calculated using inclusion-exclusion. The idea is that we want permutations with at least one fixed point, but subtract those with at least two, add those with at least three, etc., but restricted to exactly one.Alternatively, the formula for exactly m fixed points is sum_{k=0}^{n - m} (-1)^k /k!} * C(n, m) * (n - m)! / n! )Wait, maybe that's getting too complicated. Alternatively, for exactly m fixed points, the number is C(n, m) multiplied by the number of derangements of the remaining n - m elements. So in our case, exactly 1 fixed point is C(6,1)*!5, which is 6*44=264, as before. So 264/720=11/30.Alternatively, maybe I can check the total probability over m=0 to 6 fixed points. The probabilities should sum to 1. Let me recall that the number of derangements for n=6 is !6=265. So probability of 0 fixed points is 265/720. Then, probability of exactly 1 fixed point is 264/720. Then, exactly 2 fixed points would be C(6,2)*!4. Let's compute that: C(6,2)=15, !4=9. So 15*9=135. 135/720=3/16. Then, exactly 3 fixed points: C(6,3)*!3=20*2=40. 40/720=1/18. Exactly 4 fixed points: C(6,4)*!2=15*1=15. 15/720=1/48. Exactly 5 fixed points: C(6,5)*!1=6*0=0. Exactly 6 fixed points: 1. So sum up all these:265 +264+135+40+15+0+1=265+264=529; 529+135=664; 664+40=704; 704+15=719; 719+0=719; 719+1=720. So total is 720, which checks out. So the counts are correct.So, yes, exactly one fixed point is 264/720=11/30≈0.3667. Therefore, the probability is 11/30.Wait, but let me just think through once more to be thorough. If we have 6 people, each with their own hat. The question is, after shuffling, what's the probability exactly one person gets their own hat. So, to compute this, we can fix a particular person, say person A. The probability that person A gets their own hat is 1/6. Then, given that person A has their own hat, we need the probability that none of the other 5 people get their own hats. That is a derangement of 5 elements, which we already calculated as 44. So, the probability for person A is (1/6)*(44/120)= (1/6)*(11/30)=11/180. But since there are 6 different people who could be the one who gets their own hat, we multiply by 6: 6*(11/180)=11/30. That's the same result. So that's another way to think about it.So, that reinforces that 11/30 is correct.Alternatively, if I didn't remember the derangement formula, maybe I can compute the derangement number for 5 manually. Let's try that. For 5 elements, the number of derangements is !5. The formula is 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! -1/5!] as we did earlier. Let's compute that again:5! = 120.Inside the brackets: 1 - 1 + 1/2 - 1/6 + 1/24 - 1/120.Compute step by step:1 - 1 = 00 + 1/2 = 1/21/2 - 1/6 = (3/6 - 1/6) = 2/6 = 1/31/3 + 1/24 = (8/24 + 1/24) = 9/24 = 3/83/8 - 1/120 = convert to common denominator 120: 3/8 = 45/120, so 45/120 - 1/120 = 44/120 = 11/30.Multiply by 5! = 120: 120*(11/30) = 4*11=44. Yep, so !5=44. So that's correct.Therefore, the number of derangements for 5 elements is 44, so the number of favorable permutations is 6*44=264. The total number of permutations is 6!=720. So 264/720 reduces to dividing numerator and denominator by 24: 264 ÷24=11, 720 ÷24=30. So 11/30. So yes, 11/30 is the correct probability.Alternatively, if I think of the probability for exactly k fixed points in a permutation, the formula is (1/k! ) * sum_{i=0}^{n -k} (-1)^i /i! } But perhaps that's more complicated. Wait, actually, the formula for the number of permutations of n elements with exactly m fixed points is given by:C(n, m) * !(n - m)Which is exactly what we used. So for m=1, it's C(6,1)*!5=6*44=264.Therefore, probability is 264/720=11/30.So, after verifying through multiple approaches—using derangement formula, inclusion-exclusion, checking total permutations—the answer consistently comes out to 11/30. Therefore, I can be confident that the probability is 11/30.Final AnswerThe probability is boxed{dfrac{11}{30}}.</think>To find the probability that exactly one person gets their own hat back when 6 people shuffle and redistribute their hats, we start by noting the total number of permutations of the hats, which is (6! = 720).Next, we need to determine the number of favorable permutations where exactly one person gets their own hat. This involves choosing 1 person out of 6 to receive their correct hat, which can be done in (binom{6}{1} = 6) ways. The remaining 5 hats must form a derangement (a permutation where no one gets their own hat). The number of derangements of 5 items, denoted (!5), is calculated as follows:[!5 = 5! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!}right)]Calculating each term step by step:[5! = 120][1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120} = 0 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120}][= frac{1}{2} - frac{1}{6} = frac{1}{3}][frac{1}{3} + frac{1}{24} = frac{8}{24} + frac{1}{24} = frac{9}{24} = frac{3}{8}][frac{3}{8} - frac{1}{120} = frac{45}{120} - frac{1}{120} = frac{44}{120} = frac{11}{30}]Multiplying by (5!):[!5 = 120 times frac{11}{30} = 44]Thus, the number of favorable permutations is (6 times 44 = 264). The probability is then:[frac{264}{720} = frac{11}{30}]Therefore, the probability that exactly one person gets their own hat back is (boxed{dfrac{11}{30}}).,<think>Okay, so I have this probability problem here: 6 people each have a hat. They shuffle their hats and redistribute them. What's the probability that exactly one person gets their own hat back? Hmm, let me think about how to approach this.First, I remember that when dealing with probabilities involving permutations where elements are rearranged such that none are in their original position, those are called derangements. The classic problem is the hat check problem, where the probability that no one gets their own hat back is calculated using derangements. But in this case, the question is slightly different—it's asking for the probability that exactly one person gets their own hat back. So, it's a derangement problem but with exactly one fixed point. Let me recall the formula for derangements.The number of derangements of n objects is given by n! times the sum from k=0 to n of (-1)^k / k!, which approximates n!/e for large n. But here, we need the number of permutations with exactly one fixed point. I think the formula for that is C(n,1) multiplied by the number of derangements of the remaining n-1 elements. Let me verify that.Suppose we have n people. The number of permutations where exactly one person gets their own hat would be choosing 1 person out of n to have the correct hat, and then deranging the remaining n-1 hats so that none of the others get their own hat. So, the formula would be C(n,1) * D_{n-1}, where D_{n-1} is the derangement number for n-1 elements. Then, the probability would be [C(n,1) * D_{n-1}] divided by the total number of permutations, which is n!.Let me check if that makes sense. For example, if n=1, then the probability is 1 (since the only person must get their own hat). Plugging into the formula, C(1,1) * D_0 / 1! = 1 * 1 / 1 = 1, which is correct. D_0 is 1 by convention, right? Because derangement of 0 objects is 1. What about n=2? The possible permutations are (1,2) and (2,1). If exactly one person gets their own hat, there are no such permutations because if one person gets their own hat, the other must also get theirs. So, the probability should be 0. Let's check the formula: C(2,1) * D_1 / 2! = 2 * 0 / 2 = 0. That works. D_1 is 0 because there's no derangement of 1 object. So, the formula seems valid for n=2.Another test case: n=3. Let's compute manually. The permutations of 3 elements:1. (1,2,3) – all fixed points, 3 fixed points.2. (1,3,2) – fixed point 1.3. (2,1,3) – fixed point 3.4. (2,3,1) – no fixed points.5. (3,1,2) – no fixed points.6. (3,2,1) – fixed point 2.So permutations with exactly one fixed point are (1,3,2), (2,1,3), and (3,2,1). Wait, that's 3 permutations. According to the formula: C(3,1) * D_2. D_2 is 1 (derangements of 2 elements: only (2,1)), so 3 * 1 = 3. Total permutations are 6, so probability 3/6 = 1/2. Wait, but when I listed them, there are 3 permutations with exactly one fixed point. Yes, 3/6 is 1/2, but when I check the permutations:Wait, (1,3,2) has fixed point 1.(2,1,3) has fixed point 3.(3,2,1) has fixed point 2.So, three permutations, each with exactly one fixed point. The others: (1,2,3) has three fixed points, (2,3,1) and (3,1,2) are derangements with no fixed points. So the formula gives 3/6=1/2, which matches. So the formula works here. So, the formula seems correct for n=3.Therefore, for n=6, the number of permutations with exactly one fixed point is C(6,1)*D_5. Then the probability is [6 * D_5] / 6! So we need to compute D_5, the number of derangements of 5 elements.The derangement number D_n can be calculated using the formula D_n = n! * [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!]. For n=5:D_5 = 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]Compute that:First, 5! = 120.Then, the series inside the brackets:1 - 1 + 1/2 - 1/6 + 1/24 - 1/120Let me compute each term:1 = 1-1/1! = -1+1/2! = +0.5-1/3! = -0.166666...+1/4! = +0.041666...-1/5! = -0.008333...Adding them up:1 -1 = 00 +0.5 = 0.50.5 -0.166666... ≈ 0.333333...0.333333... +0.041666... ≈ 0.3750.375 -0.008333... ≈ 0.366666...So approximately 0.366666..., which is 11/30. Wait, 11/30 is approximately 0.366666...Check:11/30 = 0.366666...Yes. Let me verify:1 -1 = 00 +1/2 = 1/21/2 -1/6 = (3/6 -1/6) = 2/6 = 1/31/3 +1/24 = (8/24 +1/24) = 9/24 = 3/83/8 -1/120 = (45/120 -1/120) = 44/120 = 11/30. Yes, exactly. So D_5 = 120 * (11/30) = 120*(11)/30 = 4*11 = 44. Therefore, D_5 is 44.Therefore, the number of permutations with exactly one fixed point is 6 * 44 = 264.Total number of permutations is 6! = 720.Therefore, the probability is 264 / 720. Simplify that:Divide numerator and denominator by 24: 264 ÷24=11, 720 ÷24=30. So 11/30 ≈ 0.366666...So the probability is 11/30.Wait, but let me confirm once again. Because sometimes when you fix one element, derangements of the rest, but is there a possibility of overcounting? For example, if two people have fixed points, but we are counting exactly one. Hmm. Wait, but when you choose one person to fix and derange the rest, derangements of the rest mean that none of the others are fixed. So exactly one fixed point in total. So the count is correct.Yes, because when you fix one specific person, say person A, and derange the remaining n-1, then in those derangements, none of the other people get their own hats. So combined with person A having their own hat, the total number of fixed points is exactly 1.Therefore, the formula is correct. So, in our case, n=6, so C(6,1)*D_5 = 6*44=264. Total permutations=720. 264/720 simplifies to 11/30. So the probability is 11/30.Alternatively, another way to compute D_5 is using the recursive formula D_n = (n - 1)*(D_{n-1} + D_{n-2}). Let's check if D_5=44 using that.Known that D_1=0, D_2=1, D_3=2, D_4=9.Wait, D_1=0D_2=(2-1)*(D_1 + D_0)=1*(0 +1)=1D_3=(3-1)*(D_2 + D_1)=2*(1 +0)=2D_4=(4-1)*(D_3 + D_2)=3*(2 +1)=9D_5=(5-1)*(D_4 + D_3)=4*(9 +2)=4*11=44. Yes, that's correct. So D_5=44. Hence, the formula holds.Therefore, the probability is 6*44 / 720 = 264 /720 = 11/30.Alternatively, another approach to compute the probability: for exactly one fixed point, we can consider the probability that a specific person gets their hat and the others don't, then multiply by the number of people.So, the probability that person 1 gets their hat is 1/6. Then, the probability that the remaining 5 people derange their hats is D_5 /5! =44/120=11/30. So the probability for person 1 and derangements of others is (1/6)*(11/30). Then, since there are 6 people, multiply by 6: 6*(1/6)*(11/30)=11/30. Same result. So this confirms it.Therefore, the probability is 11/30. So the answer should be boxed{dfrac{11}{30}}.Final AnswerThe probability that exactly one person gets their own hat back is boxed{dfrac{11}{30}}.</think>To determine the probability that exactly one person gets their own hat back when 6 people shuffle and redistribute their hats, we need to consider the number of permutations with exactly one fixed point (derangement of the remaining 5 hats).1. Number of ways to choose the person who gets their own hat: This is given by ( binom{6}{1} = 6 ).2. Number of derangements for the remaining 5 hats: The derangement number ( D_5 ) can be calculated using the formula for derangements: [ D_n = n! left( 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + cdots + (-1)^n frac{1}{n!} right) ] For ( n = 5 ): [ D_5 = 5! left( 1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!} right) ] Calculating each term: [ 5! = 120 ] [ 1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120} = frac{11}{30} ] Therefore, [ D_5 = 120 times frac{11}{30} = 44 ]3. Total number of permutations with exactly one fixed point: Multiply the number of ways to choose the person by the number of derangements of the remaining 5 hats: [ 6 times 44 = 264 ]4. Total number of permutations of 6 hats: This is ( 6! = 720 ).5. Probability: The probability is the ratio of the number of favorable permutations to the total number of permutations: [ frac{264}{720} = frac{11}{30} ]Thus, the probability that exactly one person gets their own hat back is boxed{dfrac{11}{30}}.,<think>Okay, so I have this probability problem here: 6 people each have a hat. They shuffle the hats and redistribute them. What is the probability that exactly one person gets their own hat back? Hmm, let me think about how to approach this.First off, probability problems often require figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, I need to find the number of ways exactly one person gets their own hat back, and then divide that by the total number of ways the hats can be redistributed. That makes sense.The total number of ways to redistribute the hats is 6 factorial, which is 6! So, 6! = 720. That's straightforward. Now, the tricky part is figuring out the number of favorable outcomes where exactly one person gets their own hat.Wait a second, when they say exactly one person, that means one specific person gets their own hat, and the remaining five people do not get their own hats. But actually, no. It's not a specific person; it's any one person. So, first, I might need to choose which one person gets their own hat, and then make sure the other five don't get theirs. Is that right?Yeah, that seems right. So, the number of favorable outcomes would be the number of ways to choose one person out of six, multiplied by the number of derangements (permutations where no element appears in its original position) for the remaining five hats. So, mathematically, it would be C(6,1) multiplied by D(5), where D(n) is the derangement number for n elements.But wait, I need to recall what the derangement number is. The derangement number D(n) is calculated as n! times the sum from k=0 to n of (-1)^k / k! So, for D(5), it would be 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]. Let me compute that. Alternatively, I remember that D(n) can be approximated as floor(n!/e + 0.5), where e is the base of the natural logarithm. But maybe I should compute it directly to be precise.So, let's compute D(5):D(5) = 5! [1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!]First, compute each term:1 = 11/1! = 1/1 = 11/2! = 1/2 = 0.51/3! = 1/6 ≈ 0.16671/4! = 1/24 ≈ 0.04171/5! = 1/120 ≈ 0.0083Now, plugging into the sum:1 - 1 + 0.5 - 0.1667 + 0.0417 - 0.0083Let's compute step by step:1 - 1 = 00 + 0.5 = 0.50.5 - 0.1667 ≈ 0.33330.3333 + 0.0417 ≈ 0.3750.375 - 0.0083 ≈ 0.3667So, the sum is approximately 0.3667.Multiply that by 5! = 120:0.3667 * 120 ≈ 44.004Hmm, since derangements must be an integer, so D(5) is 44. Let me check another way. Alternatively, D(n) = (n - 1) * (D(n - 1) + D(n - 2)). Let's see if that works.Starting with D(1) = 0 (since one item can't be deranged), D(2) = 1 (swap the two items). Then D(3) = 2*(D(2) + D(1)) = 2*(1 + 0) = 2. D(4) = 3*(D(3) + D(2)) = 3*(2 + 1) = 9. D(5) = 4*(D(4) + D(3)) = 4*(9 + 2) = 4*11 = 44. Yes, that matches. So D(5) is indeed 44. Great.Therefore, the number of favorable outcomes is C(6,1) * D(5) = 6 * 44 = 264.Therefore, the probability is 264 / 720. Let me simplify that. Divide numerator and denominator by 24: 264 ÷ 24 = 11, 720 ÷ 24 = 30. So, 11/30 ≈ 0.3667. Wait, interestingly, that's the same as the sum we computed earlier. But let me confirm.Wait, 264 divided by 720: 264/720 = (264 ÷ 24)/(720 ÷ 24) = 11/30. Yes, 11/30 is approximately 0.3667, which is about 36.67%.But let me think again if this is correct. So, the logic is: choose 1 person out of 6 to have their hat fixed, then derange the remaining 5 hats. So, 6 * 44 = 264. Total permutations: 720. So 264/720 = 11/30. Hmm.But wait, I should make sure that in derangements of the remaining 5 hats, none of them get their own. So, exactly one person gets their own hat, and the rest get someone else's, but not their own. So, that seems correct.But let me verify with a smaller number to see if the logic holds. Let's take n=2. If there are 2 people, what is the probability that exactly one person gets their own hat? Wait, but with 2 people, if one person gets their own hat, the other must also get their own hat. Because there's only two hats. So, in that case, the number of derangements where exactly one person gets their own hat is zero. So, probability zero.But according to the formula: C(2,1)*D(1). But D(1) is 0. So 2*0=0, which matches. So probability 0/2=0. Correct.Another test: n=3. What is the probability of exactly one person getting their own hat. Let's compute via formula: C(3,1)*D(2). D(2)=1. So 3*1=3. Total permutations 6. So probability 3/6=1/2. Let's check manually.For 3 people, let's list all permutations. There are 6 permutations. The identity permutation has all three fixed. Then, derangements (permutations with no fixed points) are two: (2 3 1) and (3 1 2). Then, the permutations with exactly one fixed point: the other two must be swapped. So, for each person fixed, there's one permutation. For example, fix person 1, swap 2 and 3: (1,3,2). Similarly, fix person 2, swap 1 and 3: (3,2,1). Fix person 3, swap 1 and 2: (2,1,3). So that's three permutations. So exactly three permutations with exactly one fixed point. Therefore, 3/6=1/2. Which matches the formula. So, formula works here.Similarly, for n=4. Let me see if the formula gives the correct count. Number of permutations with exactly one fixed point: C(4,1)*D(3). D(3)=2. So 4*2=8. Total permutations 24. Let's check. The derangements D(3)=2. For each fixed person, the other three must be deranged. For each of the 4 people, there are 2 derangements, so 8. Let's see, how many permutations of 4 elements have exactly one fixed point. Yes, I think that's correct. So, 8/24=1/3. So, the formula seems valid.Therefore, returning to the original problem, n=6. Using the formula: 6 * D(5) = 6*44=264. Total permutations 720. Therefore, probability 264/720=11/30≈0.3667.But wait, just to make sure there isn't an alternative approach or a potential pitfall here. For example, sometimes inclusion-exclusion can be tricky. Let me think.Alternatively, the probability that exactly one person gets their own hat can be calculated by considering that for each person, the probability that they get their own hat and nobody else does. Then, since these events are mutually exclusive for different people, we can sum the probabilities.So, for a specific person, say person 1, the probability that person 1 gets their own hat and others do not. The probability that person 1 gets their own hat is 1/6. Then, given that person 1 has their own hat, the remaining 5 hats must be deranged among the remaining 5 people. The probability of that derangement is D(5)/5! = 44/120 = 11/30. So, the probability for person 1 is (1/6)*(11/30) = 11/180. Then, since there are 6 people, we multiply by 6, getting 6*(11/180) = 11/30. Which matches the previous result. So, this alternative approach confirms the answer.Therefore, the probability is 11/30.Wait, so I used two methods: one combinatorial counting, another probabilistic reasoning with conditional probability. Both give the same result, so that's reassuring.Therefore, I think the answer is 11/30. So, boxed, it's boxed{dfrac{11}{30}}.Final AnswerThe probability that exactly one person gets their own hat back is boxed{dfrac{11}{30}}.</think>To find the probability that exactly one person gets their own hat back when 6 people redistribute their hats, we need to consider the number of favorable outcomes divided by the total number of possible outcomes.1. Total number of permutations: This is (6! = 720).2. Number of favorable outcomes: We need to choose 1 person out of 6 to receive their own hat, and then derange the remaining 5 hats. The number of ways to choose 1 person is ( binom{6}{1} = 6 ). The number of derangements of 5 hats (denoted (D(5))) is 44. Therefore, the number of favorable outcomes is (6 times 44 = 264).3. Probability calculation: The probability is the ratio of favorable outcomes to the total number of permutations: [ frac{264}{720} = frac{11}{30} ]We verified this result using combinatorial counting and probabilistic reasoning with conditional probability, both leading to the same conclusion.The probability that exactly one person gets their own hat back is (boxed{dfrac{11}{30}}).]

🔑:1. Fix one person with their own hat: Let's denote the people as ( P_1, P_2, P_3, P_4, P_5, P_6 ) and their corresponding hats as ( H_1, H_2, H_3, H_4, H_5, H_6 ). Assume ( P_1 ) gets ( H_1 ).2. Derangement of the remaining hats: We need to find the number of ways to distribute the remaining 5 hats such that none of the remaining 5 people get their own hat. This is known as the derangement of 5 items, denoted as ( D_5 ).3. Calculate the derangement ( D_5 ): The formula for the derangement of ( n ) items is given by: [ D_n = n! sum_{k=0}^{n} frac{(-1)^k}{k!} ] For ( n = 5 ): [ D_5 = 5! left( frac{(-1)^0}{0!} + frac{(-1)^1}{1!} + frac{(-1)^2}{2!} + frac{(-1)^3}{3!} + frac{(-1)^4}{4!} + frac{(-1)^5}{5!} right) ] [ D_5 = 120 left( 1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120} right) ] [ D_5 = 120 left( 1 - 1 + 0.5 - 0.1667 + 0.0417 - 0.0083 right) ] [ D_5 = 120 left( 0.3667 right) = 44 ]4. Total number of ways to distribute hats: The total number of ways to distribute 6 hats to 6 people is ( 6! ): [ 6! = 720 ]5. Probability calculation: The probability that exactly one person gets their own hat is the number of favorable outcomes divided by the total number of outcomes. Since any one of the 6 people can be the one who gets their own hat, we multiply the derangement by 6: [ text{Probability} = frac{6 cdot D_5}{6!} = frac{6 cdot 44}{720} = frac{264}{720} = frac{11}{30} ]The final answer is ( boxed{ frac{11}{30} } )

🧠:[<think>Okay, let me try to figure out these two problems step by step. First, problem (1): Find the general term formula of the sequence {a_n} where the sum of the first n terms S_n = n² + 2n.Hmm, I remember that the sum of the first n terms S_n is related to the individual terms a_n. Specifically, the nth term a_n can be found by subtracting S_{n-1} from S_n. So, the formula should be a_n = S_n - S_{n-1}. Let me verify that.Yes, because S_n is the sum up to the nth term, and S_{n-1} is the sum up to the (n-1)th term. So their difference gives the nth term. Let me compute that.Given S_n = n² + 2n. Then S_{n-1} would be (n-1)² + 2(n-1). Let's expand that:(n-1)² = n² - 2n + 12(n-1) = 2n - 2So S_{n-1} = n² - 2n + 1 + 2n - 2 = n² -1Therefore, a_n = S_n - S_{n-1} = (n² + 2n) - (n² -1) = n² + 2n - n² +1 = 2n +1Wait, that seems straightforward. Let me check for n=1 to make sure. If n=1, then S_1 = 1² +2*1 = 3. So a_1 should be 3. Using the formula, a_1 = 2*1 +1=3. Correct. For n=2, S_2 = 4 +4=8. Then a_2 = S_2 - S_1 =8 -3=5. The formula gives 2*2 +1=5. Good. For n=3, S_3=9 +6=15, a_3=15-8=7. Formula: 2*3+1=7. Perfect. So the general term is a_n=2n+1. That seems right.Now problem (2): If b_n = 1/(a_n² -1), find the sum of the first n terms T_n.First, since we already found a_n=2n+1, let's compute a_n² -1. That would be (2n+1)² -1. Let's expand (2n+1)²: 4n² +4n +1. Subtract 1: 4n² +4n. So a_n² -1=4n² +4n=4n(n+1). Therefore, b_n=1/(4n(n+1)).So T_n is the sum from k=1 to n of 1/(4k(k+1)). We can factor out the 1/4: T_n= (1/4) * sum_{k=1}^n [1/(k(k+1))].Now, the sum of 1/(k(k+1)) is a telescoping series. I remember that 1/(k(k+1)) can be written as 1/k - 1/(k+1). Let me confirm that:1/k - 1/(k+1) = (k+1 -k)/(k(k+1)) =1/(k(k+1)). Yes, that's correct.Therefore, the sum from k=1 to n of [1/k -1/(k+1)] telescopes: the terms cancel out sequentially. Let's write out the first few terms:For k=1: 1/1 -1/2k=2: 1/2 -1/3k=3: 1/3 -1/4...k=n: 1/n -1/(n+1)When we add all these up, the -1/2 cancels with +1/2, -1/3 cancels with +1/3, and so on, leaving the first term 1/1 and the last term -1/(n+1). So the total sum is 1 -1/(n+1) = (n+1)/ (n+1) -1/(n+1)= n/(n+1).Therefore, the sum sum_{k=1}^n [1/(k(k+1))] = n/(n+1). Then T_n is (1/4) times that, so T_n= (1/4)*(n/(n+1)) = n/(4(n+1)).Wait, let me check with n=1. If n=1, then b_1 =1/(a_1² -1)=1/(9 -1)=1/8. So T_1=1/8. According to the formula, 1/(4*(1+1))=1/(8). Correct.For n=2, b_1=1/8, b_2=1/(25-1)=1/24. So T_2=1/8 +1/24= (3/24 +1/24)=4/24=1/6. The formula gives 2/(4*3)=2/12=1/6. Correct.For n=3, b_3=1/(49 -1)=1/48. T_3=1/8 +1/24 +1/48= (6/48 +2/48 +1/48)=9/48=3/16. The formula: 3/(4*4)=3/16. Correct. So the formula seems to hold.Therefore, T_n= n/(4(n+1)).Final Answer(1) The general term is boxed{2n + 1}.(2) The sum of the first ( n ) terms is boxed{dfrac{n}{4(n + 1)}}.</think>Given that the sum of the first ( n ) terms of the sequence ({a_n}) is ( S_n = n^2 + 2n ):(1) To find the general term formula of the sequence ({a_n}):The ( n )-th term ( a_n ) can be found by subtracting ( S_{n-1} ) from ( S_n ):[a_n = S_n - S_{n-1}]Given ( S_n = n^2 + 2n ), we calculate ( S_{n-1} ):[S_{n-1} = (n-1)^2 + 2(n-1) = n^2 - 2n + 1 + 2n - 2 = n^2 - 1]Thus,[a_n = (n^2 + 2n) - (n^2 - 1) = 2n + 1]The general term formula of the sequence is (boxed{2n + 1}).(2) If ( b_n = frac{1}{a_n^2 - 1} ), find the sum of the first ( n ) terms of the sequence ({b_n}), denoted as ( T_n ):First, compute ( a_n^2 - 1 ):[a_n^2 - 1 = (2n + 1)^2 - 1 = 4n^2 + 4n + 1 - 1 = 4n(n + 1)]Thus,[b_n = frac{1}{4n(n + 1)}]The sum ( T_n ) is:[T_n = sum_{k=1}^n b_k = frac{1}{4} sum_{k=1}^n left( frac{1}{k(k + 1)} right)]Using the telescoping series property:[frac{1}{k(k + 1)} = frac{1}{k} - frac{1}{k + 1}]The sum becomes:[sum_{k=1}^n left( frac{1}{k} - frac{1}{k + 1} right) = 1 - frac{1}{n + 1} = frac{n}{n + 1}]Therefore,[T_n = frac{1}{4} cdot frac{n}{n + 1} = frac{n}{4(n + 1)}]The sum of the first ( n ) terms of the sequence ({b_n}) is (boxed{dfrac{n}{4(n + 1)}}).,<think>Okay, so I need to solve two parts here. Let's start with the first one: finding the general term of the sequence {a_n} where the sum of the first n terms is S_n = n² + 2n. Hmm, I remember that the general term a_n can be found by subtracting S_{n-1} from S_n. Because the sum up to n terms minus the sum up to (n-1) terms should give the nth term. Let me verify that.So, S_n = sum_{k=1}^n a_k, and S_{n-1} = sum_{k=1}^{n-1} a_k. Therefore, a_n = S_n - S_{n-1}. Yeah, that makes sense. Let me apply this.Given S_n = n² + 2n, so S_{n-1} would be (n - 1)² + 2(n - 1). Let me compute that.First, expand (n - 1)²: that's n² - 2n + 1. Then 2(n - 1) is 2n - 2. So adding those together: n² - 2n + 1 + 2n - 2 = n² - 1. So S_{n-1} = (n - 1)² + 2(n - 1) = n² - 1.Therefore, a_n = S_n - S_{n-1} = (n² + 2n) - (n² - 1) = n² + 2n - n² + 1 = 2n + 1. Wait, so a_n = 2n + 1? Let me check if this works for the first few terms.For n = 1, S_1 should be 1² + 2*1 = 3. So a_1 = 3. According to the formula, a_1 = 2*1 + 1 = 3. Correct.For n = 2, S_2 = 2² + 2*2 = 4 + 4 = 8. Then a_2 = S_2 - S_1 = 8 - 3 = 5. Using the formula, a_2 = 2*2 + 1 = 5. Good.For n = 3, S_3 = 9 + 6 = 15. a_3 = 15 - 8 = 7. Formula gives 2*3 + 1 = 7. Perfect.So part (1) seems to be a_n = 2n + 1. Alright, that was straightforward.Now part (2): If b_n = 1 / (a_n² - 1), find the sum of the first n terms T_n.First, let's write down b_n. Since a_n = 2n + 1, then a_n² = (2n + 1)² = 4n² + 4n + 1. Therefore, a_n² - 1 = 4n² + 4n + 1 - 1 = 4n² + 4n = 4n(n + 1). So b_n = 1 / [4n(n + 1)].Thus, T_n = sum_{k=1}^n b_k = sum_{k=1}^n [1 / (4k(k + 1))] = (1/4) sum_{k=1}^n [1 / (k(k + 1))].I need to compute this sum. The sum of 1/(k(k + 1)) is a telescoping series. I remember that 1/(k(k + 1)) can be written as 1/k - 1/(k + 1). Let me check:1/k - 1/(k + 1) = (k + 1 - k) / [k(k + 1)] = 1 / [k(k + 1)]. Yes, that's correct.Therefore, the sum becomes (1/4) sum_{k=1}^n [1/k - 1/(k + 1)]. This telescopes, so most terms cancel out. Let's write it out:sum_{k=1}^n [1/k - 1/(k + 1)] = [1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... + [1/n - 1/(n + 1)]. All the intermediate terms cancel, leaving 1 - 1/(n + 1).Therefore, sum_{k=1}^n [1/(k(k + 1))] = 1 - 1/(n + 1) = (n + 1 - 1)/(n + 1) = n/(n + 1).Hence, T_n = (1/4) * [n/(n + 1)] = n / [4(n + 1)].Wait, is that correct? Let me verify with a small n.Take n = 1: T_1 = b_1 = 1 / (3² - 1) = 1/(9 - 1) = 1/8. According to the formula, T_1 = 1 / [4(1 + 1)] = 1/8. Correct.n = 2: T_2 = 1/8 + 1/(5² -1) = 1/8 + 1/24 = (3 + 1)/24 = 4/24 = 1/6. Formula: 2 / [4(2 + 1)] = 2 / 12 = 1/6. Correct.n = 3: T_3 = 1/8 + 1/24 + 1/(7² -1) = 1/8 + 1/24 + 1/48. Let's convert to 48 denominator: 6/48 + 2/48 + 1/48 = 9/48 = 3/16. Formula: 3 / [4(3 +1)] = 3 / 16. Correct. So the formula works.Therefore, T_n = n / [4(n + 1)].But let me just recheck the steps once more. So starting from b_n = 1/(a_n² -1) = 1/[4n(n +1)], then sum from 1 to n. Then factoring out 1/4, sum [1/(n(n +1))], which telescopes to 1 - 1/(n +1). Therefore, sum is (1 - 1/(n +1)) = n/(n +1). Multiply by 1/4: n/[4(n +1)]. Yes, all steps are solid.Therefore, part (2) answer is T_n = n / [4(n +1)].Final Answer(1) The general term is boxed{2n + 1}.(2) The sum of the first ( n ) terms of ({b_n}) is boxed{dfrac{n}{4(n + 1)}}.</think>Given that the sum of the first ( n ) terms of the sequence ({a_n}) is ( S_n = n^2 + 2n ):(1) To find the general term formula of the sequence ({a_n}):We use the relationship ( a_n = S_n - S_{n-1} ). First, compute ( S_{n-1} ):[S_{n-1} = (n-1)^2 + 2(n-1) = n^2 - 2n + 1 + 2n - 2 = n^2 - 1]Then, subtract ( S_{n-1} ) from ( S_n ):[a_n = S_n - S_{n-1} = (n^2 + 2n) - (n^2 - 1) = 2n + 1]Thus, the general term formula of the sequence ({a_n}) is (boxed{2n + 1}).(2) If ( b_n = frac{1}{a_n^2 - 1} ), find the sum of the first ( n ) terms of the sequence ({b_n}), denoted as ( T_n ):First, express ( b_n ) using the general term ( a_n = 2n + 1 ):[a_n^2 - 1 = (2n + 1)^2 - 1 = 4n^2 + 4n + 1 - 1 = 4n(n + 1)][b_n = frac{1}{4n(n + 1)}]Next, we find the sum ( T_n ):[T_n = sum_{k=1}^n b_k = sum_{k=1}^n frac{1}{4k(k + 1)} = frac{1}{4} sum_{k=1}^n left( frac{1}{k} - frac{1}{k + 1} right)]This sum telescopes:[sum_{k=1}^n left( frac{1}{k} - frac{1}{k + 1} right) = 1 - frac{1}{n + 1}]Thus,[T_n = frac{1}{4} left( 1 - frac{1}{n + 1} right) = frac{1}{4} cdot frac{n}{n + 1} = frac{n}{4(n + 1)}]Therefore, the sum of the first ( n ) terms of the sequence ({b_n}) is (boxed{dfrac{n}{4(n + 1)}}).]

🔑:(1) When n=1, a_1=3.When ngeqslant 2, a_n=S_n-S_{n-1}=n^{2}+2n-left[(n-1)^{2}+2(n-1)right]=2n+1,This holds true when n=1,Thus, a_n=2n+1.(2) From (1), we know that b_n=frac{1}{a_n^{2}-1}=frac{1}{(2n+1)^{2}-1}=frac{1}{4n(n+1)}=frac{1}{4}left(frac{1}{n}-frac{1}{n+1}right),Thus, the sum of the first n terms of the sequence {b_n} is T_n=frac{1}{4}left(1-frac{1}{2}+frac{1}{2}-frac{1}{3}+frac{1}{3}-frac{1}{4}+ldots+frac{1}{n}-frac{1}{n+1}right)=frac{1}{4}left(1-frac{1}{n+1}right)=boxed{frac{n}{4n+4}}.

🧠:[<think>Okay, let's see. The problem is about finding the actual depth of Lake Superior at its lowest point. Hmm. The question says that the water surface is at an elevation of 180 meters above sea level, and the lowest point of the lake floor is 220 meters below sea level. So I need to figure out how deep the lake is there.First, let me visualize this. If the surface of the lake is 180 meters above sea level, that's like the top of the water being pretty high up. Then the bottom of the lake, the floor, is 220 meters below sea level. Wait, so sea level is like the reference point here. So from the surface down to the floor, how much is that?Let me think. If the surface is 180 m above sea level, and the floor is 220 m below sea level, then the total depth would be the distance from 180 m above to 220 m below. Since sea level is in the middle, maybe I need to add those two numbers together?Wait, yeah. Because if you start at 180 m above sea level and go down to sea level, that's 180 meters. Then from sea level down to 220 m below, that's another 220 meters. So total depth would be 180 + 220. Let me check that.So, surface elevation: +180 m. Lowest point: -220 m. The difference between these two is 180 - (-220) = 180 + 220 = 400 meters. Wait, but that seems like a lot. But maybe that's right. Wait, the answer options include 400 m as option D. Hmm.But wait, maybe I should double-check. If the surface is above sea level, and the floor is below, then the depth is the total distance from the surface to the floor. So you have to go from +180 to -220. The difference in elevation is 180 - (-220) = 400 m. So that would be the depth. So 400 meters. So answer D.But let me make sure. Suppose sea level is 0. The surface is at +180, and the floor is at -220. The vertical distance between them is 180 - (-220) = 400 m. Yes, that's correct. Because when you subtract the two elevations, you get the difference. Since one is positive and one is negative, subtracting them is like adding their absolute values.Alternatively, imagine a vertical number line. The surface is at 180, the floor is at -220. The distance between them is how far apart they are. So from 180 down to 0 is 180 meters, then from 0 down to -220 is another 220 meters. So total depth is 180 + 220 = 400. That makes sense. So I think 400 meters is the answer, which is option D.But just to make sure there's no trick in the question. It says "actual depth of the lake at this point". So maybe "actual depth" is different? Hmm. Wait, depth is from the surface to the floor, regardless of sea level. So even if the surface is above sea level, the depth is still the total distance from surface to floor. So yes, adding both the 180 above sea level and the 220 below sea level would be correct.Wait, another way to think: If the entire lake was above sea level, then the depth would just be the difference between the surface elevation and the floor elevation. But since the floor is below sea level, you have to account for both the part above and below sea level.So in this case, the depth is 180 (from surface to sea level) plus 220 (from sea level to the lowest point), which totals 400 meters. Yeah, that seems right.Therefore, the answer should be D) 400 m.Final Answerboxed{D}</think>The water surface of Lake Superior is at an elevation of (180 mathrm{~m}) above sea level, and the lowest point of the lake floor is (220 mathrm{~m}) below sea level. To find the actual depth of the lake at this point, we need to calculate the total vertical distance from the surface to the floor.1. The elevation of the water surface is (180 mathrm{~m}) above sea level.2. The elevation of the lowest point of the lake floor is (220 mathrm{~m}) below sea level.3. The total depth is the sum of the distance from the surface to sea level and from sea level to the lowest point of the lake floor.[text{Depth} = 180 mathrm{~m} + 220 mathrm{~m} = 400 mathrm{~m}]Thus, the actual depth of the lake at the lowest point is (boxed{D}).,<think>Okay, so I need to figure out the actual depth of Lake Superior at its lowest point. Let me read the problem again carefully. It says the water surface of Lake Superior is at an elevation of 180 meters above sea level. The lowest point of the lake floor is 220 meters below sea level. The question is asking for the actual depth of the lake at that point. The options are A through E, with answers ranging from 40 meters to 500 meters. Hmm, okay.First, let me visualize this. Imagine sea level is like a reference point, right? So if the surface of the lake is 180 meters above sea level, that means the water starts at +180 meters. Then, the bottom of the lake at its deepest point is 220 meters below sea level, which would be -220 meters relative to sea level. So to find the depth of the lake, I need to calculate the vertical distance from the surface down to the lowest point of the floor.Wait, so if the surface is at 180 meters above sea level and the floor is at 220 meters below sea level, then the total depth would be the distance between these two points. Since one is above and the other is below sea level, I can't just subtract them directly, can I? Let me think. If both were above sea level, you'd subtract the lower elevation from the higher one. But here, the floor is below sea level, so maybe I need to add the two distances together?Let me break it down. The elevation of the surface is 180 m above sea level. The elevation of the floor is -220 m (since it's below sea level). The depth would be the difference between these two elevations. So how do you calculate the difference between a positive and a negative number? In mathematics, the difference between two numbers is the absolute value of their subtraction. So if I take the elevation of the surface (180 m) and subtract the elevation of the floor (-220 m), that would be 180 - (-220) = 180 + 220 = 400 m. Wait, that seems like a lot. But is that right?Alternatively, since the surface is 180 m above sea level, and the bottom is 220 m below sea level, then from the surface to sea level is 180 m, and from sea level down to the bottom is another 220 m. So total depth would be 180 + 220 = 400 m. That gives me the same answer. So maybe 400 meters is the depth. But let me check the answer options. Option D is 400 m, so that's a possible choice.But wait, let me make sure I didn't make a mistake here. If the surface is at 180 m above sea level, and the floor is 220 m below sea level, then the total depth is the sum of those two distances because they're on opposite sides of sea level. So 180 m down to sea level, and then another 220 m down to the lowest point. So total depth is indeed 180 + 220 = 400 m. That seems to make sense.Alternatively, if both elevations were above sea level, you would subtract. For example, if the surface was at 300 m and the floor was at 100 m, the depth would be 200 m. But since one is above and the other is below, adding them gives the total vertical distance.Another way to think about it: The elevation of the lake floor relative to the lake surface. The surface is at +180 m, the floor is at -220 m. So the difference is 180 - (-220) = 180 + 220 = 400 m. So that's another way to calculate it. So the depth is 400 meters. So the answer should be D.Wait, but let me check the options again. The options are A) 40 m, B) 300 m, C) 380 m, D) 400 m, E) 500 m. So D is 400 m, which matches my calculation. Therefore, the correct answer is D.But hold on, just to be absolutely sure. Let's consider sea level as zero. The lake surface is at +180 m. The lake floor is at -220 m. The distance between these two points is 180 - (-220) = 400 m. Yes, that's correct. Because when you subtract the two elevations, you get the vertical distance between them. So if you have two points on a vertical line, their elevations are 180 and -220. The difference is 180 - (-220) = 400. So yes, that's how depth is calculated. So 400 meters is the actual depth. I think that's solid. So the answer must be D) 400 m.Final Answerboxed{D}</think>The water surface of Lake Superior is at an elevation of (180 mathrm{~m}) above sea level, and the lowest point of the lake floor is (220 mathrm{~m}) below sea level. To find the actual depth of the lake at this point, we need to calculate the vertical distance between these two elevations.1. The elevation of the lake surface is (180 mathrm{~m}) above sea level.2. The elevation of the lake floor is (-220 mathrm{~m}) (below sea level).The depth of the lake is the difference between these two elevations. Since one is above sea level and the other is below, we add the absolute values of these elevations:[180 mathrm{~m} + 220 mathrm{~m} = 400 mathrm{~m}]Thus, the actual depth of the lake at the lowest point is (boxed{D}).]

🔑:The water surface has an elevation of +180 mathrm{~m}, and the lowest point of the lake floor has an elevation of -220 mathrm{~m}. Therefore, the actual depth of the lake at this point is 180-(-220)=400 mathrm{~m}.Answer: (D)

🧠:[<think>Okay, let's see. So, Nikolai Petrovich missed his riverboat by 5 minutes, and he hired a motorboat to catch up. The motorboat catches up with the riverboat, drops him off, and then immediately returns. We need to find out how much time passes from when the motorboat departs until it returns. First, let me note down the given information:- The riverboat's speed relative to the water is 3 times the speed of the river current. Let's denote the river current's speed as ( v ). So, the riverboat's speed relative to the water is ( 3v ).- The motorboat's speed is 5 times the river current's speed, so that's ( 5v ).- Nikolai was late by 5 minutes, which is ( Delta t = 5 ) minutes. I should probably convert that to hours since speed is in terms of ( v ) per hour, but maybe we can handle the units later.So, let's visualize this. The riverboat is going downstream. When Nikolai is late, the riverboat has already left, going downstream at its speed. The motorboat departs later to catch up. Since both the riverboat and motorboat are going downstream, but the motorboat is faster, it will eventually catch up. Then after dropping off Nikolai, the motorboat turns around and goes back upstream to its original starting point. We need the total time from when the motorboat departs until it returns.Hmm. Let's define some variables. Let's denote:- ( v ) = speed of the river current.- Speed of riverboat relative to water = ( 3v ). Therefore, its speed relative to the ground (or the shore) is ( 3v + v = 4v ) downstream, since the current is aiding it.Wait, hold on. If the riverboat's speed is 3v relative to the water, then relative to the ground, its speed is ( 3v + v = 4v ). Similarly, the motorboat's speed is 5v relative to the water. So when it's going downstream, its speed relative to the ground is ( 5v + v = 6v ), and when it's going upstream, its speed relative to the ground is ( 5v - v = 4v ).Right, that makes sense. So, when moving downstream, both boats have their speed relative to water plus the current, and upstream is minus the current.Now, the motorboat departs 5 minutes after the riverboat. Let's convert that 5 minutes into hours because the speeds are in terms of ( v ) per hour. 5 minutes is ( frac{5}{60} = frac{1}{12} ) hours.So, in that ( frac{1}{12} ) hours, the riverboat has already traveled a distance of ( 4v times frac{1}{12} = frac{4v}{12} = frac{v}{3} ).So, when the motorboat starts, the riverboat is ( frac{v}{3} ) distance ahead downstream.Now, the motorboat needs to catch up to the riverboat. Let's denote the time it takes for the motorboat to catch up as ( t_1 ) hours. During this time, the motorboat is going downstream at ( 6v ), and the riverboat is going downstream at ( 4v ). The distance the motorboat travels is ( 6v times t_1 ), and the distance the riverboat travels is ( 4v times t_1 ). But the motorboat has to make up the initial ( frac{v}{3} ) distance. So:Distance covered by motorboat = Distance covered by riverboat + initial gapSo:( 6v times t_1 = 4v times t_1 + frac{v}{3} )Subtract ( 4v times t_1 ) from both sides:( 2v times t_1 = frac{v}{3} )Divide both sides by ( v ):( 2 t_1 = frac{1}{3} )So, ( t_1 = frac{1}{6} ) hours, which is 10 minutes.Okay, so it takes the motorboat 10 minutes to catch up to the riverboat. Then, after dropping off Nikolai, the motorboat immediately turns around and goes back upstream. Now, we need to find the time it takes for the motorboat to return upstream to its original starting point. Let's denote this time as ( t_2 ).But wait, when the motorboat turns around, it's at the point where it caught up with the riverboat. So, how far is that point from the original starting point?At the time of catching up, the motorboat has been traveling for ( t_1 = frac{1}{6} ) hours downstream. So, the distance from the starting point is ( 6v times frac{1}{6} = v ). So, the motorboat is ( v ) distance downstream from the starting point when it catches up.Therefore, to return upstream to the starting point, it needs to cover that ( v ) distance, going against the current. The motorboat's speed upstream is ( 5v - v = 4v ).So, time taken to return is distance divided by speed: ( frac{v}{4v} = frac{1}{4} ) hours, which is 15 minutes.Therefore, total time from departure to return is ( t_1 + t_2 = frac{1}{6} + frac{1}{4} = frac{2}{12} + frac{3}{12} = frac{5}{12} ) hours. Converting that to minutes: ( frac{5}{12} times 60 = 25 ) minutes.Wait, so the total time is 25 minutes? Let me check again.First, motorboat departs 5 minutes after the riverboat. Then, takes 10 minutes to catch up, then 15 minutes to return. So total time from motorboat's departure is 10 + 15 = 25 minutes. That seems right.But let me verify the calculations step by step to ensure no mistakes.1. River current speed = ( v ).2. Riverboat speed relative to water = ( 3v ), so downstream speed = ( 3v + v = 4v ).3. Motorboat speed relative to water = ( 5v ), so downstream speed = ( 6v ), upstream speed = ( 4v ).4. Nikolai is 5 minutes late. In those 5 minutes (which is ( frac{1}{12} ) hours), the riverboat travels ( 4v * frac{1}{12} = frac{v}{3} ).5. Let ( t_1 ) be the time taken by the motorboat to catch up. In that time, motorboat travels ( 6v * t_1 ), riverboat travels ( 4v * t_1 ). The motorboat has to cover the initial gap ( frac{v}{3} ), so:( 6v t_1 = 4v t_1 + frac{v}{3} )Subtract ( 4v t_1 ):( 2v t_1 = frac{v}{3} )Divide by ( 2v ):( t_1 = frac{1}{6} ) hours = 10 minutes.6. During ( t_1 ), motorboat travels ( 6v * frac{1}{6} = v ). So the catch-up point is ( v ) distance downstream.7. Returning upstream, motorboat's speed is ( 5v - v = 4v ). Distance to cover is ( v ). So time taken ( t_2 = frac{v}{4v} = frac{1}{4} ) hours = 15 minutes.8. Total time: 10 + 15 = 25 minutes.That seems consistent. So, the answer is 25 minutes. Therefore, the time that passed from the departure of the motorboat until its return is 25 minutes.Final AnswerThe total time that passed is boxed{25} minutes.</think>Given that Nikolai Petrovich was late by 5 minutes for the departure of his riverboat, we need to determine the total time from the departure of the motorboat until its return. 1. Convert the delay time to hours: [ Delta t = 5 text{ minutes} = frac{5}{60} text{ hours} = frac{1}{12} text{ hours} ]2. Calculate the distance the riverboat traveled in 5 minutes: - Speed of the riverboat relative to the ground (downstream): (4v) - Distance: [ 4v times frac{1}{12} = frac{v}{3} ]3. Determine the time for the motorboat to catch up: - Speed of the motorboat downstream: (6v) - Speed of the riverboat downstream: (4v) - Relative speed: (6v - 4v = 2v) - Time to catch up ((t_1)): [ t_1 = frac{frac{v}{3}}{2v} = frac{1}{6} text{ hours} = 10 text{ minutes} ]4. Calculate the distance from the starting point when caught up: - Distance traveled by the motorboat: [ 6v times frac{1}{6} = v ]5. Determine the time for the motorboat to return upstream: - Speed of the motorboat upstream: (4v) - Time to return ((t_2)): [ t_2 = frac{v}{4v} = frac{1}{4} text{ hours} = 15 text{ minutes} ]6. Total time from departure to return: [ t_1 + t_2 = 10 text{ minutes} + 15 text{ minutes} = 25 text{ minutes} ]The total time that passed from the departure of the motorboat until its return is boxed{25} minutes.,<think>Okay, so Nikolai Petrovich missed his riverboat by 5 minutes, and now he needs to catch up with it using a motorboat. The problem is asking how much time passes from when the motorboat departs until it returns after catching up and disembarking him. The riverboat's speed relative to the water is 3 times the current's speed, and the motorboat's speed is 5 times the current's speed. Hmm, let me break this down step by step.First, let's assign some variables to make this clearer. Let's say the speed of the river current is ( v ) (in some unit, like km/h). Then, the riverboat's speed relative to the water is ( 3v ). Since the riverboat is moving downstream, its actual speed relative to the ground (or the riverbank) would be its own speed plus the current's speed. So, riverboat's ground speed is ( 3v + v = 4v ).Similarly, the motorboat's speed relative to the water is ( 5v ). But when it's going downstream (chasing the riverboat), its ground speed would be ( 5v + v = 6v ). When it's returning upstream after catching up, its ground speed would be ( 5v - v = 4v ). Okay, got that.Now, Nikolai was 5 minutes late. Let's convert that to hours since speed is probably in km/h. 5 minutes is ( frac{5}{60} = frac{1}{12} ) hours. So, by the time the motorboat starts, the riverboat has already been traveling for ( frac{1}{12} ) hours downstream. Therefore, the distance between Nikolai and the riverboat when the motorboat departs is the riverboat's speed multiplied by the time it's been traveling: ( 4v times frac{1}{12} = frac{4v}{12} = frac{v}{3} ). So the distance to catch up is ( frac{v}{3} ).Now, the motorboat needs to cover this distance while both are moving. But wait, the riverboat is still moving downstream at ( 4v ), and the motorboat is going downstream at ( 6v ). So their relative speed is ( 6v - 4v = 2v ). Therefore, the time it takes for the motorboat to catch up is the distance divided by the relative speed: ( frac{v/3}{2v} = frac{1}{6} ) hours. Converting that back to minutes, that's 10 minutes. So, 10 minutes to catch up.Once they catch up, the motorboat disembarks Nikolai and immediately turns around to go back. Now, the motorboat is going upstream, so its ground speed is ( 5v - v = 4v ) upstream. But how much time will it take to return?Wait, while the motorboat is returning, the starting point (where Nikolai was originally) is stationary relative to the ground, right? But the motorboat is now at the point where they caught up with the riverboat. Let me think. The distance the motorboat needs to cover to return is the same as the distance it traveled to catch up, which was ( frac{v}{3} ). Wait, no. Wait, when the motorboat went downstream for 10 minutes (1/6 hours), it traveled ( 6v times frac{1}{6} = v ) kilometers. Meanwhile, the riverboat in those 10 minutes traveled an additional ( 4v times frac{1}{6} = frac{2v}{3} ). Wait, but initially, the riverboat had a 5-minute head start, so the total distance between the starting point and the meeting point is ( 4v times frac{1}{12} + 4v times frac{1}{6} )?Wait, maybe I confused something here. Let's step back.The initial distance between Nikolai and the riverboat when the motorboat starts is ( frac{v}{3} ). Then, during the time ( t ) that the motorboat is chasing, the riverboat moves another ( 4v times t ), and the motorboat moves ( 6v times t ). The motorboat needs to cover the initial distance plus the riverboat's additional distance. So, ( 6v times t = frac{v}{3} + 4v times t ). Solving for ( t ):( 6vt = frac{v}{3} + 4vt )Subtract ( 4vt ) from both sides:( 2vt = frac{v}{3} )Divide both sides by ( v ):( 2t = frac{1}{3} )So, ( t = frac{1}{6} ) hours, which is 10 minutes. That matches the previous result. So, the meeting happens 10 minutes after the motorboat departs. Then, the motorboat needs to go back.But how far is the meeting point from the original starting point? The motorboat traveled ( 6v times frac{1}{6} = v ). So the distance is ( v ). Alternatively, the riverboat had a head start of ( frac{v}{3} ) and then traveled ( 4v times frac{1}{6} = frac{2v}{3} ), so total distance is ( frac{v}{3} + frac{2v}{3} = v ). So yes, the meeting point is ( v ) distance downstream from the starting point.Now, the motorboat needs to go back upstream to the starting point. Its speed upstream is ( 5v - v = 4v ). The distance to cover is ( v ). So, time taken to return is ( frac{v}{4v} = frac{1}{4} ) hours, which is 15 minutes. Therefore, total time from departure to return is 10 minutes (downstream) + 15 minutes (upstream) = 25 minutes. So the answer should be 25 minutes.But wait, let me verify once again. Maybe there's a different approach.Alternatively, let's consider the time taken for the motorboat to go downstream and upstream in terms of the current's speed. Let ( v ) be the speed of the current.Let’s denote:- Speed of riverboat relative to water: ( 3v )- Speed of riverboat relative to ground: ( 3v + v = 4v )- Speed of motorboat downstream: ( 5v + v = 6v )- Speed of motorboat upstream: ( 5v - v = 4v )Time delay of Nikolai: 5 minutes = ( frac{5}{60} = frac{1}{12} ) hours.In that time, the riverboat has moved a distance of ( 4v times frac{1}{12} = frac{v}{3} ).Let ( t ) be the time taken for the motorboat to catch up. In time ( t ), the motorboat moves ( 6v times t ), and the riverboat moves ( 4v times t ). The motorboat has to cover the initial ( frac{v}{3} ) plus the distance the riverboat moves in time ( t ):( 6v t = frac{v}{3} + 4v t )Solving:( 6v t - 4v t = frac{v}{3} )( 2v t = frac{v}{3} )Cancel ( v ):( 2t = frac{1}{3} )( t = frac{1}{6} ) hours = 10 minutes.So, that's 10 minutes to catch up. Then, the motorboat turns around and goes back upstream. The distance back is the same as the distance it went downstream, which is ( 6v times frac{1}{6} = v ).Time to go back upstream: ( frac{v}{4v} = frac{1}{4} ) hours = 15 minutes.Total time: 10 + 15 = 25 minutes.Alternatively, maybe we can model the positions as functions of time.Let’s set the departure time of the motorboat as ( t = 0 ). The riverboat left at ( t = -frac{1}{12} ) hours (i.e., 5 minutes earlier). The position of the riverboat at time ( t ) is ( 4v times (t + frac{1}{12}) ).The position of the motorboat going downstream is ( 6v times t ). Setting them equal to find when they meet:( 6v t = 4v (t + frac{1}{12}) )Divide both sides by ( v ):( 6t = 4t + frac{4}{12} )( 6t - 4t = frac{1}{3} )( 2t = frac{1}{3} )( t = frac{1}{6} ) hours, same as before.Then, the motorboat turns back. The position where they meet is ( 6v times frac{1}{6} = v ). So the distance back is ( v ). The speed upstream is ( 4v ), so time is ( frac{v}{4v} = frac{1}{4} ) hours, which is 15 minutes.Total time: ( frac{1}{6} + frac{1}{4} = frac{2}{12} + frac{3}{12} = frac{5}{12} ) hours. Converting to minutes: ( frac{5}{12} times 60 = 25 ) minutes. Yep, same result.Therefore, the answer should be 25 minutes. Wait, but the problem says "how much time passed from the departure of the motorboat until its return". So that's 25 minutes. So boxed answer is boxed{25}.Final Answerboxed{25}</think>Nikolai Petrovich was 5 minutes late for the departure of his riverboat, which had a speed of (3v) relative to the water (where (v) is the speed of the river current). The motorboat's speed is (5v) relative to the water. 1. Convert the delay time to hours: [ Delta t = 5 text{ minutes} = frac{1}{12} text{ hours} ]2. Calculate the distance the riverboat traveled in 5 minutes: [ text{Distance} = 4v times frac{1}{12} = frac{v}{3} ] (The riverboat's ground speed is (3v + v = 4v).)3. Determine the relative speed between the motorboat and the riverboat: - Motorboat's downstream speed: (5v + v = 6v) - Riverboat's speed: (4v) - Relative speed: (6v - 4v = 2v)4. Calculate the time to catch up: [ t = frac{frac{v}{3}}{2v} = frac{1}{6} text{ hours} = 10 text{ minutes} ]5. Calculate the distance to the meeting point: [ text{Distance} = 6v times frac{1}{6} = v ]6. Calculate the return time upstream: - Motorboat's upstream speed: (5v - v = 4v) [ text{Time} = frac{v}{4v} = frac{1}{4} text{ hours} = 15 text{ minutes} ]7. Total time from departure to return: [ 10 text{ minutes} + 15 text{ minutes} = 25 text{ minutes} ]Thus, the time passed from the departure of the motorboat until its return is (boxed{25}).]

🔑:1. s=left(3 v_{mathrm{T}}+v_{mathrm{T}}right) cdotleft(t_{1}+Delta tright)=left(5 v_{mathrm{T}}+v_{mathrm{T}}right) cdot t_{1} rightarrow 4 Delta t=2 t_{1} rightarrow t_{1}=2 Delta t=10 min t_{2}=frac{s}{5 v_{mathrm{T}}-v_{mathrm{T}}}=frac{6 v_{mathrm{T}} cdot t_{1}}{4 v_{mathrm{T}}}=frac{3}{2} t_{1}=3 Delta t=15 min rightarrow T=t_{1}+t_{2}=5 Delta t=25 min