🧠:Okay, let's try to figure out this problem step by step. So, the question is about counting the number of perfect 5-tuples where each element is a positive integer between 1 and 5, inclusive. A tuple is considered perfect if, for any three distinct indices, the three numbers at those positions don't form an arithmetic progression in any order. First, I need to make sure I understand the problem correctly. A 5-tuple is just an ordered list of five numbers, each from 1 to 5. The condition is that no three of these numbers, when picked from any three distinct positions, can form an arithmetic progression. Importantly, the order of the three numbers doesn't matter—if they can be arranged to form an arithmetic progression, then the tuple is not perfect.So, arithmetic progression (AP) here means three numbers a, b, c such that b - a = c - b, which simplifies to 2b = a + c. So, for any three numbers in the tuple, regardless of their positions, they shouldn't satisfy this equation in any permutation.Given that the numbers can be up to 5, the possible APs are limited. Let me first list out all possible APs within the range 1 to 5. Maybe that will help in identifying forbidden triplets.Starting with common difference 1:- 1, 2, 3- 2, 3, 4- 3, 4, 5Common difference 2:- 1, 3, 5Common difference 0.5 (but since we're dealing with integers, these aren't valid). Wait, no—common differences must be integers because the elements are integers. So, the possible APs are those with differences 1 and 2 as above. Let me check:For example, 1, 2, 3: 2 - 1 = 1, 3 - 2 = 1. So that's an AP. Similarly, 2, 3, 4 and 3, 4, 5. Then 1, 3, 5: 3 - 1 = 2, 5 - 3 = 2. So that's another AP. Any others?Wait, let's check other triplets. For example, 1, 2, 4. Let's see if they can form an AP. If arranged as 1, 2, 4: differences are 1 and 2—no. If arranged as 1, 4, 2: differences are 3 and -2—no. Similarly, 2, 1, 4: same thing. So that's not an AP. How about 2, 3, 5? Let's see: 2, 3, 5. Differences are 1 and 2—not an AP. 3, 5, 2: differences are 2 and -3—no. So not an AP. Similarly, 1, 4, 5: arranged as 1, 4, 5: differences 3 and 1—no. 1, 3, 4: 1, 3, 4: differences 2 and 1—no. So the only AP triplets are the ones with common differences 1 and 2 as listed above: {1,2,3}, {2,3,4}, {3,4,5}, and {1,3,5}.Therefore, the forbidden triplets are exactly those four sets. So, a perfect 5-tuple must not contain any of these four triplets in any three positions. The order in the tuple doesn't matter because the problem states "in any order." Therefore, even if the numbers are not in order in the tuple, if they can be rearranged to form an AP, the tuple is invalid.So, our task is to count all 5-tuples (each element from 1 to 5) such that none of the combinations of three elements form any of those four forbidden triplets.Now, since the problem is about counting such tuples, one approach is to use inclusion-exclusion. However, inclusion-exclusion can get complicated here because there are multiple forbidden triplets, and overlapping between them can occur. Alternatively, perhaps a better approach is to model this as a constraint satisfaction problem and try to construct the tuples step by step, ensuring that adding each new element doesn't create a forbidden triplet with any two previous elements.But given that the tuple is of length 5 and the numbers are from 1 to 5, it might be manageable. Let me think.First, note that each number from 1 to 5 must be used in the tuple, but wait, no. Wait, the problem states "the set of 5-tuples of positive integers at most 5." So, each element can be from 1 to 5, and repetition is allowed? Wait, the problem says "positive integers at most 5"—so each element is in {1,2,3,4,5}, and they can repeat. So, tuples can have repeated numbers, but the indices are distinct. So, for example, (1,1,1,1,1) is a valid 5-tuple, but we have to check if it's perfect. However, even in such a case, if we take any three indices, all three numbers are 1, which trivially form an arithmetic progression (1,1,1) with common difference 0. Wait, but the problem says "positive integers," and they can form an arithmetic progression. But wait, the problem states "three numbers a_i, a_j, a_k do not form an arithmetic progression (in any order)." So, even if all three numbers are the same, like (1,1,1), that is an arithmetic progression where the common difference is 0. Therefore, such a tuple would not be perfect because it contains an arithmetic progression.Therefore, all elements in a perfect tuple must be distinct? Wait, no, not necessarily. Because even if there are two elements the same and one different, they might form an AP. For example, (1,1,2): sorted, it's 1,1,2. The differences are 0 and 1, so not an AP. Wait, but 1,1,2 cannot be arranged as an AP. Because an arithmetic progression requires that the difference between consecutive terms is constant. So, in any order, can these three numbers form an AP?Let me check. For three numbers x, y, z, they form an AP if one of them is the average of the other two. So, for (1,1,2), the average of 1 and 2 is 1.5, which isn't in the set. Therefore, (1,1,2) is not an AP. Similarly, (1,2,1) or any permutation—since the average is not an integer. Therefore, such a triplet is allowed. However, if we have three numbers where two are the same and the third is such that 2b = a + c. For example, (1,3,5) is an AP. But (1,3,3): 3 is the average of 1 and 5, but we don't have a 5 here. Wait, (1,3,3): the average of 1 and 3 is 2, which is not in the triplet. So, (1,3,3) is not an AP. Similarly, (2,2,2) is an AP with common difference 0, but in our forbidden triplets, we have only those with distinct numbers? Wait, no. The problem states "three numbers a_i, a_j, a_k do not form an arithmetic progression (in any order)." So, even if all three numbers are the same, like (2,2,2), that is considered an arithmetic progression with common difference 0. Therefore, such a triplet is forbidden.Therefore, in a perfect tuple, all elements must be distinct? Because if any element is repeated, then selecting three positions where that element is present would form an arithmetic progression (the trivial one with all terms equal). Therefore, this implies that in a perfect 5-tuple, all five elements must be distinct. Because if any number is repeated at least three times, then those three instances would form an AP. If a number is repeated twice, then we need to check if in combination with another number, they form an AP.Wait, let's clarify. For example, suppose a tuple has two 1s and one 3. Then, the triplet (1,1,3) would have to be checked. Let's see if they can form an AP. The numbers are 1,1,3. The average of 1 and 3 is 2, which is not present, so they don't form an AP. Similarly, if there's two 1s and one 2, then (1,1,2) as before isn't an AP. However, if there are two 2s and one 4, then (2,2,4) can be arranged as 2,2,4. The average of 2 and 4 is 3, which is not present, so they don't form an AP. Similarly, two 3s and one 5: average of 3 and 5 is 4, not present. So, in general, duplicates of two numbers and a third don't form an AP unless the third is the average.However, if you have duplicates of a number and another number such that their average is present. For example, two 1s and a 3: but as above, that's 1,1,3. The average is 2, which isn't there. Wait, but if you have two 2s and a 2: that's three 2s, which form an AP. So, any triplet with three identical numbers is forbidden.Therefore, in order for a 5-tuple to be perfect, it must not contain three identical numbers, and also must not contain any three distinct numbers that form an AP. So, both conditions: no three identical numbers, and no three distinct numbers forming an AP. Therefore, the tuple can have up to two copies of any number, but not three. But even with two copies, we need to make sure that no three elements (two copies and another number) form an AP.Wait, but earlier example shows that two copies and another number don't form an AP unless the other number is the average. Wait, let me formalize this. Suppose we have two copies of a number 'a' and a number 'b'. Then, the triplet would be {a, a, b}. For this to form an AP, the numbers must satisfy that one of them is the average of the other two. Since there are two 'a's, the average of 'a' and 'b' must be 'a' or 'b'. The average of 'a' and 'b' is (a + b)/2. For this to equal 'a', we need (a + b)/2 = a ⇒ b = a. But we already have 'b' different. For the average to equal 'b', (a + b)/2 = b ⇒ a + b = 2b ⇒ a = b, which is again impossible. Therefore, two copies of a number and another number cannot form an AP. Therefore, the only forbidden triplets are either three identical numbers or three distinct numbers forming an AP.Therefore, the perfect 5-tuples must satisfy two conditions:1. No number appears three or more times.2. No three distinct numbers form an arithmetic progression.Therefore, our problem reduces to counting all 5-tuples with elements from 1 to 5, where each number appears at most twice, and no three distinct numbers form an AP.Now, let's note that the possible numbers are 1,2,3,4,5. The APs among distinct numbers are {1,2,3}, {2,3,4}, {3,4,5}, and {1,3,5} as previously identified.Therefore, in a perfect 5-tuple, we cannot have all three numbers from any of these four triplets. Also, we can have at most two copies of any number.Therefore, the problem is similar to arranging a 5-tuple with elements from 1 to 5, with each number used at most twice, and avoiding any three distinct numbers that form an AP.First, let's consider the total number of 5-tuples where each number appears at most twice. Then subtract those that have forbidden triplets. However, inclusion-exclusion might be necessary here.But since the problem is about 5-tuples, and we need to ensure that in every combination of three positions, the three numbers don't form an AP. This seems complex because there are C(5,3) = 10 triplets of positions to check for each tuple.Alternatively, maybe it's better to model this as a graph where each node represents a number from 1 to 5, and edges represent incompatibility (i.e., if two numbers are part of a forbidden AP, then adding a third number could form an AP). Wait, not sure.Alternatively, perhaps we can model the problem as a constraint-based permutation where we have to assign numbers to 5 positions with the given constraints.But given that the numbers can be repeated (up to twice), maybe it's helpful to consider how to build such a tuple step by step, choosing numbers while ensuring that no forbidden triplet is formed.But with five positions, and each number allowed up to two times, and avoiding three distinct numbers forming APs. Let's see.First, note that the forbidden triplets are {1,2,3}, {2,3,4}, {3,4,5}, {1,3,5}. Therefore, in a perfect 5-tuple, we can have at most two numbers from each of these triplets. However, since some numbers are in multiple forbidden triplets, this complicates things.For example, the number 3 is in {1,2,3}, {2,3,4}, {3,4,5}, and {1,3,5}. So, if we include 3 in the tuple, we have to be careful about which other numbers we include.Alternatively, maybe we can partition the numbers into two sets such that neither set contains any forbidden triplet. Then, the tuple can be constructed using numbers from one set only. But let's check.Looking at the forbidden triplets: {1,2,3}, {2,3,4}, {3,4,5}, {1,3,5}. If we can find a subset of numbers that doesn't contain any of these triplets, then we can use numbers from that subset.For example, consider the set {1,2,4,5}. Let's check if this set contains any forbidden triplet. The possible triplets are:- {1,2,4}: Not forbidden.- {1,2,5}: Not forbidden.- {1,4,5}: Not forbidden.- {2,4,5}: Not forbidden.So, {1,2,4,5} doesn't contain any of the forbidden AP triplets. Similarly, {1,2,4,5} is safe.Another possible set is {1,3,4}. Let's check: {1,3,4} contains {3,4,5}? No. {1,3,4} has {1,3,4}, which isn't a forbidden triplet. But {1,3,4} is not an AP. However, {1,3,5} is forbidden, but 5 isn't in this set. So, maybe smaller sets can be safe.But if we can find a larger set, say size 4, like {1,2,4,5}, which doesn't have any forbidden triplets, then tuples using numbers only from this set would automatically satisfy the condition. However, since we need 5 elements (with possible repetitions), but each number can be used up to two times.Wait, but if we use numbers only from {1,2,4,5}, each number can be used up to two times, so the maximum total numbers we can have is 4*2=8, but we need only 5. So, there are many possibilities. However, we need to ensure that in the tuple, no three distinct numbers form an AP. Since the set {1,2,4,5} doesn't contain any forbidden triplet, any three distinct numbers from this set won't form an AP. Therefore, if we use only numbers from this set, even with repetitions, the tuple would be safe.Similarly, if we use numbers from another safe set. Let's check if there's another safe set. For example, {1,2,5}. But this is a subset of the previous set. Another possible set: {2,3,5}. Checking triplets:- {2,3,5}: This is an AP (2,3,5: 3-2=1, 5-3=2; not an AP). Wait, {2,3,5} is not an AP. Wait, 2, 3, 5: the differences are 1 and 2, so not an AP. But {1,3,5} is an AP. But 1 is not in this set, so {2,3,5} is safe.Wait, but if we include 3, we have to check if combining with other numbers would form APs. For example, if we take {2,3,4}, that's an AP. So, {2,3,4} is forbidden. Therefore, the set {2,3,4,5} would include the forbidden triplet {2,3,4} and {3,4,5}. Therefore, not safe.So, the only safe 4-element set is {1,2,4,5}. Let's confirm:- Any three distinct numbers from {1,2,4,5}: - 1,2,4: Not an AP. - 1,2,5: Not an AP. - 1,4,5: Not an AP. - 2,4,5: Not an AP.So none of these are forbidden. Therefore, if we restrict ourselves to using numbers from {1,2,4,5}, then any 5-tuple composed of these numbers (with each number used at most twice) will automatically satisfy the condition, provided we don't use three of the same number.Similarly, another safe set might be {1,4,5}, but it's smaller. Alternatively, is there another 4-element set? Let's see:- {1,2,3,5}: Contains {1,2,3} which is forbidden.- {1,2,4,5}: Safe.- {1,3,4,5}: Contains {3,4,5} which is forbidden.- {2,3,4,5}: Contains {2,3,4} and {3,4,5}.- {1,2,3,4}: Contains {1,2,3}, {2,3,4}.So, indeed, {1,2,4,5} is the only 4-element safe set. What about 3-element sets?- {1,2,4}: Safe.- {1,2,5}: Safe.- {1,4,5}: Safe.- {2,4,5}: Safe.- {1,3,4}: Safe (no forbidden triplet).Wait, {1,3,4}:- 1,3,4: differences 2 and 1—not an AP.- So, safe. Similarly, {1,3,5} is forbidden, but 5 is not in this set.Similarly, {2,3,5}:- 2,3,5: differences 1 and 2—not an AP.- So safe.Therefore, there are multiple 3-element safe sets. However, since we need to construct a 5-tuple, using a 3-element set would require some repetition. For example, using numbers from {1,2,4}, we could have tuples like (1,1,2,2,4), which has two 1s, two 2s, and one 4. However, in this case, we have to ensure that even with repetitions, no three elements (including duplicates) form an AP. But as established earlier, duplicates don't form APs unless there are three of the same number or combined with another number to form an AP via average. But in {1,2,4}, duplicates can only be two of each, so triplets with two duplicates and another number won't form APs.But perhaps we need to consider all possible safe sets and count the number of tuples accordingly. However, this seems complicated. Maybe a better approach is to consider that the perfect tuples must be those that use numbers only from the safe 4-element set {1,2,4,5}, each appearing at most twice, or from other safe sets with 3 elements, but with possible overlaps.Alternatively, perhaps all perfect tuples must be permutations or combinations of numbers from {1,2,4,5} with at most two copies of each. Let me check.Suppose we consider a tuple with numbers from {1,2,4,5}, each appearing at most twice. Then, since we need five elements, we can have:- Two copies of two numbers and one copy of the other two numbers. For example, two 1s, two 2s, and one 4. However, that sums to five elements. Wait, but {1,2,4,5} has four numbers. To make a 5-tuple, we need to use one number twice. For example:Case 1: One number appears twice, others appear once. So, 4 choices for the number to repeat, and then arrange them. The number of such tuples is C(4,1) * (5! / (2!1!1!1!)) = 4 * (120 / 2) = 4 * 60 = 240.Case 2: Two numbers each appear twice, and the remaining two numbers appear once. Wait, but since we have four numbers, if two numbers are each repeated twice, that accounts for 2*2 + 2*1 = 6 elements, which is more than 5. So, this is impossible. Therefore, the only possibility is one number repeated twice and the other three numbers appearing once each. But since we have four numbers, choosing one to repeat twice gives 4 choices, and the total number of tuples is 4 * (5! / (2!)) = 4 * 60 = 240 as above.However, wait, but we have four numbers: if we use one number twice and the other three once each, that's 2 + 1 + 1 + 1 = 5 elements. So yes, that works.But wait, 4 numbers: let's say numbers A, B, C, D. If we choose one of them, say A, to have two copies, then the tuple has A, A, B, C, D. The number of distinct arrangements is 5! / 2! = 60. Since there are 4 choices for which number is repeated, total is 4*60=240.Alternatively, if we use two numbers each repeated twice, but as mentioned before, that would require 2+2+1=5 elements, but we have four numbers. Wait, no. For example, if we take two numbers from the four, say A and B, each repeated twice, and one more number, say C. Then, the tuple would be A, A, B, B, C. However, since we have four numbers {1,2,4,5}, if we choose two numbers to repeat, we need to leave out two numbers. Wait, but in that case, the tuple would consist of two numbers each repeated twice and one number once. But we have four numbers in total. For example, choosing 1 and 2 to repeat, and include 4 once: the tuple is 1,1,2,2,4. However, this uses three distinct numbers: 1,2,4. But 1,2,4 is a safe triplet (no AP). But wait, this would be allowed? Wait, but the problem states that the tuple must not contain any three distinct numbers that form an AP. Since {1,2,4} is not an AP, that's allowed. So, such a tuple is acceptable.But in this case, the tuple is composed of three distinct numbers, with two of them repeated. However, we need to ensure that the three distinct numbers do not form an AP. Since {1,2,4} is safe, such a tuple is acceptable.But in this case, we can have tuples that use three distinct numbers from the safe set {1,2,4,5}, as long as those three don't form an AP. Wait, but in this case, the three numbers are 1,2,4, which don't form an AP. So that's okay.Therefore, perhaps there are more possibilities beyond just using all four numbers {1,2,4,5} with one repetition. Let's think.If we use three distinct numbers from the safe set {1,2,4,5}, none of which form an AP, then we can have tuples where two numbers are repeated. For example, using 1,2,4: we can have two 1s, two 2s, and one 4. Similarly, two 1s, two 4s, and one 2, etc. Each of these would be valid because the three distinct numbers don't form an AP.Similarly, we can use three numbers like 1,2,5. Since 1,2,5 is not an AP, such a triplet is allowed. So, for example, two 1s, two 2s, and one 5.Alternatively, two 1s, two 5s, and one 2.Similarly, other combinations.Therefore, the total number of perfect tuples would include:1. Tuples using four distinct numbers from {1,2,4,5}, with one number repeated once.2. Tuples using three distinct numbers from {1,2,4,5} (none forming an AP), with two numbers repeated twice each and the third number once.3. Tuples using two distinct numbers from {1,2,4,5}, but since two numbers can't form an AP, but we need five elements. However, using two numbers would require at least three copies of one number, which is forbidden (since three copies would form an AP of three identical numbers). Therefore, tuples with two distinct numbers are impossible, because you can't have five elements with two numbers without at least one number appearing three times.4. Tuples using all five numbers. Wait, but the numbers are only 1,2,4,5 (four numbers), so we can't have a 5-tuple with all five distinct numbers. Wait, the original numbers are 1-5, but the safe set is {1,2,4,5}. If we use numbers outside the safe set, like 3, then we have to check if including 3 would introduce an AP.Wait, hold on. Earlier, we assumed that the safe set is {1,2,4,5}, but actually, the problem allows numbers from 1 to 5. So, including 3 might be possible, but we have to ensure that no three distinct numbers (including 3) form an AP.For example, if we include 3 in the tuple, we have to make sure that we don't include both 1 and 5 (since {1,3,5} is an AP), or 2 and 4 (since {2,3,4} is an AP), or 1 and 2 (since {1,2,3} is an AP), or 3 and 5 (since {3,4,5} is an AP if we include 4 and 5). So, including 3 severely restricts the other numbers we can include.Therefore, maybe it's possible to include 3 in the tuple, but only with certain other numbers. Let's explore this possibility.Suppose we include 3 in the tuple. Then, to avoid forming an AP, we must ensure that for any two other numbers in the tuple, their average is not 3. Wait, more precisely, we need to ensure that there are no three distinct numbers such that one is the average of the other two.If 3 is included, the forbidden pairs would be:- 1 and 5 (since (1 + 5)/2 = 3)- 2 and 4 (since (2 + 4)/2 = 3)- 1 and 2 (since (1 + 2)/2 = 1.5, which is not an integer, so okay)Wait, but {1, 2, 3} is an AP because 2 is the average of 1 and 3. So, if 3 is included, we cannot have both 1 and 2 in the tuple.Similarly, {2,3,4} is an AP, so if 3 is included, we cannot have both 2 and 4.{3,4,5} is an AP, so if 3 is included, we cannot have both 4 and 5.Similarly, {1,3,5} is an AP, so if 3 is included, we cannot have both 1 and 5.Therefore, if we include 3, the only numbers we can include are:- Either 1 or 5, but not both.- Either 2 or 4, but not both.Additionally, we cannot include both 1 and 2, or 4 and 5, but actually, the main restrictions are:If we include 3, we must exclude:- 1 and 5 (can include at most one of them)- 2 and 4 (can include at most one of them)Additionally, since {1,2,3} is an AP, if we include 3 and 1, we cannot include 2.Similarly, if we include 3 and 2, we cannot include 1 or 4.Wait, this seems too tangled. Maybe another approach.Let me consider that including 3 limits the possible numbers we can include:Case 1: Include 3 and 1. Then, we cannot include 5 (because {1,3,5} is an AP) and cannot include 2 (because {1,2,3} is an AP). So, remaining numbers are 4. But can we include 4? If we include 4, then since we already have 3, {3,4,5} is an AP, but we don't have 5. However, we have 3 and 4. But {2,3,4} is an AP, but we don't have 2. So, if we include 3, 1, and 4, is that allowed? The triplet {1,3,4}: 1,3,4. The average of 1 and 4 is 2.5, which is not an integer, so not an AP. Similarly, 3 is not the average of 1 and 4. So, {1,3,4} is allowed. Therefore, if we include 3, 1, and 4, that's okay. But we also need to check for other triplets. Since we're including 3,1,4, any three of them. {1,3,4} is safe. Also, we have to make sure that we don't include any other numbers that would form an AP with these. But in this case, if we include 3,1,4, and another number, say, 2 or 5, but 2 is forbidden because {1,2,3}, and 5 is forbidden because {1,3,5}. Therefore, if we include 3 and 1, we can only include 4 as the third number, and no others. But since we need a 5-tuple, we can repeat numbers. However, we can't have three of the same number. So, for example, if we have 3,1,4, we can repeat some of them. Let's see:If we have 3,1,4, we need two more numbers. Since we can't include 2 or 5, the only numbers available are 1,3,4. But we can repeat them up to twice. So, for example, 3,1,4,1,4: two 1s, two 4s, one 3. This tuple has three distinct numbers: 1,3,4. None of which form an AP. Also, no number is repeated more than twice. So, this is a valid perfect tuple.Similarly, another example: 3,1,4,3,1. Two 3s, two 1s, one 4. Also valid.So, in this case, the number of such tuples would be:Choose the numbers: 1,3,4.Number of ways to arrange two 1s, two 3s, and one 4: 5! / (2!2!1!) = 30.Similarly, two 1s, two 4s, and one 3: 30.Two 3s, two 4s, and one 1: 30.So total 3*30=90.Similarly, if we start with 3 and 5 instead of 3 and 1:Case 2: Include 3 and 5. Then, we cannot include 1 (because {1,3,5}) and cannot include 4 (because {3,4,5}). So, remaining number is 2. So, numbers are 3,5,2. Check if they form an AP: {2,3,5} is not an AP. So, allowed. Then, to form a 5-tuple, we can have two 2s, two 5s, one 3; two 2s, two 3s, one 5; etc. So, similar to case 1:Number of arrangements:Two 2s, two 5s, one 3: 5! / (2!2!1!)=30.Two 2s, two 3s, one 5: 30.Two 5s, two 3s, one 2: 30.Total 3*30=90.Case 3: Include 3 and 2. Then, we cannot include 1 (because {1,2,3}) and cannot include 4 (because {2,3,4}). Remaining numbers are 5. So, numbers are 3,2,5. Check: {2,3,5} is not an AP. So, allowed. Then, forming 5-tuples with two 2s, two 5s, one 3; two 2s, two 3s, one 5; two 5s, two 3s, one 2. Same as case 2, so another 90 tuples. Wait, but this seems similar to case 2. Wait, actually, if we include 3 and 2, we have to check if we can include 5. Wait, {2,3,5} is allowed, yes. So, numbers 2,3,5.Similarly, forming tuples with two copies of two numbers and one copy of the third. So, 90 tuples.But wait, isn't this overlapping with case 2? If we include 3 and 2, we can include 5. If we include 3 and 5, we can include 2. So, maybe cases 2 and 3 are the same.Similarly, case 4: Include 3 and 4. Then, we cannot include 5 (because {3,4,5}) and cannot include 2 (because {2,3,4}). Remaining number is 1. So, numbers are 3,4,1. Which is similar to case 1. So, forming tuples with two 1s, two 4s, one 3; etc., totaling 90 tuples.Therefore, each of these cases where 3 is included with two other numbers leads to 90 tuples. However, we have to be careful not to double-count.Wait, actually, if we consider that including 3 with 1 and 4 is a separate case from including 3 with 5 and 2, then these are distinct. Similarly, including 3 with 2 and 5 is another case, and including 3 with 4 and 1 is another. Wait, but in reality, these are different combinations:- Case A: 3,1,4- Case B: 3,5,2Each of these combinations allows forming tuples with two copies of two numbers and one copy of the third. Each combination gives 3*30=90 tuples. However, are these cases overlapping? For example, can a tuple belong to both case A and case B? No, because case A includes 1 and 4, while case B includes 5 and 2. Since a tuple can't contain both 1 and 5 (as they would form an AP with 3), these cases are distinct.Therefore, total tuples including 3 would be 90 (for case A) + 90 (for case B) = 180.Wait, but wait. Let's think again.If we include 3,1,4, we can't include 2 or 5. Similarly, if we include 3,5,2, we can't include 1 or 4. So, these are two distinct sets:- Set1: {1,3,4}- Set2: {2,3,5}Each set can generate 90 tuples (as calculated above). Therefore, total tuples including 3 would be 90 + 90 = 180.Additionally, can we have tuples with 3 and only one other number? For example, 3 and 1, but not including 4. But then we need to fill the tuple with more copies of 1 and 3. However, we can't have three 1s or three 3s. So, for example, two 1s, two 3s, and one more number. But the fifth number can't be 5 (since {1,3,5} is forbidden) or 2 (since {1,2,3} is forbidden) or 4 (which would require including 4, but then {1,3,4} is allowed). Wait, but if we include 4, then we have three numbers: 1,3,4. Which is allowed. But we were assuming only two numbers. Wait, no, in this case, if we have two 1s, two 3s, and one 4, that's three distinct numbers: 1,3,4. Which is allowed, as before.Wait, but then this is already covered in case A (Set1: {1,3,4}). So, any tuple with 3 and two other numbers is already included in the previous count. Similarly, tuples with 3 and only one other number would require at least three copies of that number, which is forbidden. Therefore, all tuples including 3 must be in either Set1 or Set2, each contributing 90 tuples, totaling 180.Now, moving on to tuples that do not include 3. These would be tuples composed entirely of numbers from {1,2,4,5}. As we discussed earlier, these can be of two types:1. Tuples with four distinct numbers (each from {1,2,4,5}), with one number repeated once.2. Tuples with three distinct numbers from {1,2,4,5}, with two numbers repeated twice each and the third number once.Let's calculate these.First, the tuples with four distinct numbers and one repetition.There are four numbers in {1,2,4,5}. We need to choose one number to repeat. There are C(4,1) = 4 choices. For each choice, the number of distinct arrangements is 5! / 2! = 60. Therefore, total tuples: 4 * 60 = 240.Second, the tuples with three distinct numbers, two of which are repeated twice each, and the third once.First, choose three distinct numbers from {1,2,4,5}. The number of ways to choose three numbers is C(4,3) = 4. For each set of three numbers, we need to choose which two numbers are repeated. There are C(3,2) = 3 choices. For each such choice, the number of distinct arrangements is 5! / (2!2!1!) = 30. Therefore, total tuples: 4 * 3 * 30 = 360.Wait, let's verify:For each of the 4 sets of three numbers:- For example, {1,2,4}: choose two numbers to repeat, say 1 and 2. Then the tuple is two 1s, two 2s, and one 4. Number of arrangements: 5! / (2!2!1!) = 30. Similarly, choosing 1 and 4 to repeat: two 1s, two 4s, one 2. Also 30. Choosing 2 and 4 to repeat: two 2s, two 4s, one 1. 30. So, 3*30=90 per set. Since there are 4 sets, total 4*90=360.Therefore, total tuples without 3 are 240 + 360 = 600.Adding the tuples with 3: 180. So total perfect tuples would be 600 + 180 = 780.But wait, let's check if there's any overlap between these cases. For example, is there a tuple that includes 3 and also includes numbers from {1,2,4,5} in a way that's already counted? No, because the tuples without 3 are entirely within {1,2,4,5}, and the tuples with 3 are in {1,3,4} or {2,3,5}. There's no overlap.But hold on, let's verify with a small example. Take a tuple from the Set1 {1,3,4} with two 1s, two 3s, and one 4: (1,1,3,3,4). This tuple is counted in the 180. Similarly, a tuple from Set2 {2,3,5} with two 2s, two 5s, and one 3: (2,2,5,5,3). This is also counted in the 180. The tuples without 3 are entirely separate.Therefore, total perfect tuples would be 600 + 180 = 780.However, we need to confirm this result by checking if there are any other cases we missed.Wait, another possibility: can we have a tuple that includes 3 along with two other numbers not covered in Set1 or Set2? For example, {3,4,5} is forbidden, so we can't have 3,4,5. But what if we include 3 with 4 and another number that doesn't form an AP? Wait, if we include 3 and 4, we can't include 5 (because {3,4,5}) or 2 (because {2,3,4}). So, similar to before, if we include 3 and 4, we can only include 1. So, Set1: {1,3,4} as before.Similarly, including 3 and 2, we can only include 5. So, Set2: {2,3,5} as before. Therefore, there are no other possibilities for including 3 with two other numbers. Hence, the 180 total for tuples including 3 is correct.Another check: total number of 5-tuples with numbers from 1 to 5, each number appearing at most twice.Total number without any restriction (except max two repeats): sum_{k=1 to 5} (number of 5-tuples with each number appearing at most twice).The total number is equal to the number of functions from 5 elements to 5 elements with each element mapped to at most twice. This can be calculated using inclusion-exclusion, but it's complicated. However, our calculated perfect tuples are 780, which should be less than the total number of such tuples.Total number of 5-tuples with each number at most twice:For each position, 5 choices, but with the constraint that no number is used more than twice. This is equivalent to the number of permutations of a multiset with 5 elements where each element appears at most twice. The formula is:Sum_{k=0}^{2} ... Sum_{k5=0}^{2} [if sum ki=5] (5! / (k1!k2!k3!k4!k5!)) )But this is complex. Alternatively, using generating functions:The exponential generating function for each number is (1 + x + x^2/2!), and for 5 numbers, it's (1 + x + x^2/2!)^5. The coefficient of x^5 multiplied by 5! gives the number of such tuples.Wait, the exponential generating function for each element (since order matters) is actually 1 + x + x^2/2! for each digit. Therefore, the generating function is (1 + x + x^2/2!)^5. The coefficient of x^5 is the number of 5-tuples divided by 5!.So, expanding (1 + x + x²/2)^5:We need the coefficient of x^5.Let me compute this.Let me denote the generating function as (1 + x + x²/2)^5.To find the coefficient of x^5, we can consider the combinations of terms from each factor that multiply to x^5.Each factor contributes 0, 1, or 2 x's.We need the sum of exponents to be 5, with each exponent <=2.The number of ways is the number of solutions to a1 + a2 + a3 + a4 + a5 =5, where each ai ∈ {0,1,2}.For each such solution, the coefficient is (1^{n0}) * (1^{n1}) * (1/(2^{n2})) ), multiplied by the multinomial coefficient.Wait, more precisely, for each term in the expansion, we have a product over i=1 to 5 of terms either 1, x, or x²/2. So, if in the product, we have k terms of x (i.e., ai=1) and m terms of x²/2 (i.e., ai=2), such that k + 2m =5. The number of such terms is C(5, k, m) * (1)^k * (1/2)^m, where C(5, k, m) is the multinomial coefficient 5! / (k! m! (5 -k -m)! )).Therefore, the coefficient of x^5 is the sum over all k and m such that k + 2m =5 of [5! / (k! m! (5 -k -m)! )) * (1)^k * (1/2)^m ].Let's find possible (k, m):Since k + 2m =5, and k, m >=0.Possible m:m=0: k=5. But 5 <=5, possible.m=1: k=3.m=2: k=1.m=3: k=5 -6= -1. Not possible.Therefore, m=0,1,2.Calculating each term:For m=0, k=5:C(5,5,0) = 5! / (5!0!0!) =1.Term: 1 * 1^5 * (1/2)^0 =1.For m=1, k=3:C(5,3,1) =5! / (3!1!1!) = 20.Term:20 *1^3*(1/2)^1=20*(1/2)=10.For m=2, k=1:C(5,1,2) =5! / (1!2!2!) = 30.Term:30*1^1*(1/2)^2=30*(1/4)=7.5.Summing these:1 +10 +7.5=18.5.Therefore, the coefficient of x^5 is18.5.Hence, the number of tuples is 18.5 *5! =18.5*120=2220.Therefore, the total number of 5-tuples with each number appearing at most twice is2220. However, our previous count of perfect tuples is780, which is less than 2220, which makes sense because we have additional constraints of avoiding APs.But the key point is that our count of780 seems plausible, but we need to verify if there are any mistakes in the reasoning.Wait, another way to check: let's manually count a small part and see.For example, consider the tuples without 3:600.These are divided into two categories:1. Four distinct numbers with one repeat:240.Each of these tuples has four different numbers, one of which is repeated once. For example,1,1,2,4,5. Let's check if this tuple is perfect. The distinct numbers are1,2,4,5. Do any three of them form an AP? The triplets are {1,2,4}, {1,2,5}, {1,4,5}, {2,4,5}. None of these are APs. Also, no three identical numbers. So, it's valid.2. Three distinct numbers with two repeats:360.For example,1,1,2,2,4. The distinct numbers are1,2,4. They don't form an AP. Also, no three identical numbers. Valid.Similarly, tuples with 3:180.For example,1,1,3,3,4. Distinct numbers1,3,4. No AP. Valid.Another example,2,2,3,5,5. Distinct numbers2,3,5. No AP. Valid.So, these examples check out.Therefore, the total of780 seems correct.But wait, the answer might be different. Let me think again. Is there a possibility that some tuples are being double-counted or missed?For instance, when we count the tuples with 3, we considered two cases: Set1 {1,3,4} and Set2 {2,3,5}. Each leads to90 tuples. But are there any other sets with 3?For example, could we have a tuple with3,1,5? But {1,3,5} is an AP, so that's forbidden. Similarly,3,2,4 is forbidden. So, no.What about tuples with3 and only one other number? For example,3,3,4,4,4. But this has three 4s, which is invalid. Or3,3,2,2,2: three 2s, invalid.Therefore, such tuples are already excluded.Another check: the number of perfect tuples according to our calculation is780. Let's see if this matches any known patterns or if there's an alternative way to compute it.Alternatively, perhaps we can think in terms of valid choices for each position, considering previous choices. But with five positions, the complexity would be high.Alternatively, use the principle of inclusion-exclusion. The total number of tuples with numbers from1 to5, each appearing at most twice, is2220 as calculated. From this, we need to subtract the number of tuples that have at least one forbidden AP triplet.However, inclusion-exclusion over four forbidden triplets {1,2,3}, {2,3,4}, {3,4,5}, {1,3,5} and also the triplets with three identical numbers.But this seems very complex, as each forbidden triplet can overlap with others, and the interactions are complicated.Alternatively, since our initial approach of dividing the problem into tuples with and without3, and systematically counting them, seems logical and the intermediate steps check out, maybe780 is the correct answer.But to be thorough, let's cross-verify.Another way: consider that all perfect tuples are those that avoid the four AP triplets and avoid triple repeats. We can model this as coloring the numbers1-5 with allowed assignments, considering previous assignments.But this might not be straightforward.Alternatively, note that the total number of tuples without threepeat is2220. From these, subtract the number of tuples that contain at least one AP triplet.But computing the number of tuples containing at least one AP triplet requires inclusion-exclusion:Number of tuples with at least one AP = Sum_{each AP triplet} N(A) - Sum_{each pair of AP triplets} N(A∩B) + Sum_{each triple of AP triplets} N(A∩B∩C)} - ... Where N(A) is the number of tuples containing the AP triplet A.But given the complexity, let's attempt to compute N(A) for one AP triplet, say {1,2,3}.To compute N(A): the number of 5-tuples that contain at least one occurrence of the triplet {1,2,3} (in any order, possibly with repeats) and each number appears at most twice.But this is non-trivial. Let's see.First, the number of ways to have at least one AP triplet {1,2,3} in the tuple.But the tuple can have other numbers as well, and repetitions.However, since we are considering tuples where each number appears at most twice, and {1,2,3} is an AP triplet, we need to count the number of 5-tuples that include at least one 1, one 2, and one 3, along with any other numbers (4,5), but with each number repeated at most twice.But this is still complex.Alternatively, maybe it's better to stick with our original count, given that the systematic approach yielded780, and the intermediate steps hold up.Therefore, after careful consideration, I believe the number of perfect 5-tuples is780.Final Answerboxed{780}

✅:To solve the problem of finding the number of perfect 5-tuples of positive integers at most 5, we need to consider tuples where no three distinct indices form an arithmetic progression (AP). The key steps are as follows:1. Identify Forbidden Triplets: The AP triplets within the range 1 to 5 are {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, and {1, 3, 5}. These are the forbidden triplets.2. Safe Sets: We identify safe sets of numbers that do not contain any forbidden triplets. The largest safe set is {1, 2, 4, 5}, which does not contain any AP triplet. Other safe sets include {1, 3, 4} and {2, 3, 5} when considering the inclusion of 3 with restrictions.3. Tuples Without 3: - Four Distinct Numbers with One Repeat: Choose one number to repeat, leading to 4 choices and 60 arrangements each, totaling 240 tuples. - Three Distinct Numbers with Two Repeats: Choose three numbers and two to repeat, leading to 4 sets, 3 choices each, and 30 arrangements, totaling 360 tuples. - Total tuples without 3: (240 + 360 = 600).4. Tuples With 3: - Set {1, 3, 4}: Each number can be repeated up to twice, leading to 3 choices and 30 arrangements each, totaling 90 tuples. - Set {2, 3, 5}: Similarly, leading to another 90 tuples. - Total tuples with 3: (90 + 90 = 180).5. Total Perfect Tuples: Summing the tuples without and with 3 gives (600 + 180 = 780).Thus, the number of perfect 5-tuples is (boxed{780}).

🔑:We need to find the number of perfect 5-tuples ((a_1, a_2, a_3, a_4, a_5)) where each (a_i) is a positive integer at most 5, and any three numbers (a_i, a_j, a_k) do not form an arithmetic progression.1. Case 1: The multiset is of the form (aabbc): - Since we have 2 pairs of identical numbers and one different number, the multiset is (a, a, b, b, c). - The primary condition is that (c) must not be the average of (a) and (b). Mathematically: [ c neq frac{1}{2}(a + b) ] - First, let’s count all unordered triples ((a, b, c)) such that (1 leq a leq b leq c leq 5). There are: [ binom{5}{3} = 10 ] - Now, remove those triples where the condition (c = frac{1}{2}(a + b)) is violated. These problematic triples ((S)) are: [ S = {(1, 2, 3), (2, 3, 4), (3, 4, 5), (1, 3, 5)} ] There are 4 such triples. - Therefore, the valid triples are: [ 10 - 4 = 6 text{ valid triples} ] - For any such triple, the number of permutations (i.e., orderings) of the multiset (a, a, b, b, c) is given by: [ frac{5!}{2!2!1!} = 30 ] - Since we have 6 valid triples: [ 6 times 30 = 180 text{ valid configurations for } aabbc ]2. Case 2: The multiset has four distinct elements (aabccd): - Since tuples have all distinct integers and there are only 5 integers available (1 through 5), the only possible set is ({ 1, 2, 4, 5}) - For any 4 distinct elements from the set ({1, 2, 3, 4, 5}), they must be arranged without forming any arithmetic progression: - We check valid tuples: [ {1, 2, 4, 5} ] - The number of orderings of ({1, 1, 2, 4, 5}) is: [ frac{5!}{2!1!1!1!} = 60 ] - Since there are 4 such tuples, [ 4 times 60 = 240 text{ valid configurations} ]Adding up both cases, we get:[540 + 240 = 780]Conclusion:(boxed{780})

🧠:Okay, let's tackle this problem step by step. The question is asking which pair of (x, y) does not satisfy the equation 187x - 104y = 41. There are five options given, and only one of them doesn't satisfy the equation. My job is to figure out which one that is. Hmm, let's start by understanding what the equation means and how to check each pair.First, I remember that linear equations in two variables like this one have solutions that can be plotted as a straight line. However, since x and y are integers here (I assume they are integers because the problem presents them as whole numbers without any fractions or decimals), we're dealing with Diophantine equations. Diophantine equations require integer solutions. The general approach for such equations involves finding particular solutions and then determining the general solution based on the coefficients.But maybe I don't need to go that far. Since there are only five pairs to check, perhaps the straightforward way is to plug each pair into the equation and verify if the left-hand side equals 41. The one that doesn't give 41 is the answer. That seems simple enough, but I should make sure I do the calculations correctly to avoid any mistakes. Let's start with the first pair.Option 1: x = 3, y = 5.Calculating 187*3 - 104*5. Let's compute each term:187*3: 180*3 = 540, 7*3 = 21, so 540+21=561.104*5: 100*5=500, 4*5=20, so 500+20=520.Subtracting: 561 - 520 = 41. Okay, that works. So option 1 is a valid solution.Option 2: x = 107, y = 192.Again, plugging into the equation: 187*107 - 104*192.Hmm, these numbers are larger. Let me compute each term step by step.Calculating 187*107:First, break down 107 into 100 + 7.187*100 = 18,700.187*7: 180*7=1,260; 7*7=49; so 1,260 + 49 = 1,309.Adding together: 18,700 + 1,309 = 19,009.Now 104*192:Break down 192 into 200 - 8.104*200 = 20,800.104*8 = 832.So 20,800 - 832 = 19,968.Now subtract the two results: 19,009 - 19,968 = -959. Wait, that's negative. But the equation equals 41. That means this doesn't work. But wait, hold on. Did I compute that correctly?Wait, 187*107 is 19,009? Let me check again.187*100 = 18,700.187*7: 100*7=700, 80*7=560, 7*7=49. So 700 + 560 = 1,260 + 49 = 1,309. Then 18,700 + 1,309 is indeed 19,009.104*192: Let's compute 104*192 differently. Maybe 100*192=19,200. 4*192=768. So 19,200 + 768=19,968. Then 19,009 - 19,968 is -959. So that's -959, which is not 41. So option 2 doesn't satisfy the equation. Wait, but the problem says only one pair doesn't satisfy the equation. Hmm, so if option 2 is invalid, but maybe I made a mistake here?Wait, hold on. Let me check again. Maybe I messed up the multiplication.Alternatively, perhaps there's a pattern here. Since the equation is 187x - 104y = 41. If the other pairs are solutions, they should all differ by some multiple of the coefficients. Let me think.But before that, let's check the other options to see if they work. Maybe option 2 is the incorrect one, but I need to verify.Option 3: x = 211, y = 379.Compute 187*211 - 104*379.This is going to be time-consuming. Let me see if there's a smarter way.Alternatively, since the equation is linear, once we have a particular solution, the general solution can be found. Let's see. The coefficients are 187 and 104. Let's compute the greatest common divisor (GCD) of 187 and 104 to see if 41 is a multiple of it. If the GCD divides 41, then solutions exist.Compute GCD(187, 104):187 divided by 104 is 1 with a remainder of 83 (since 104*1=104, 187-104=83).Now, GCD(104, 83).104 divided by 83 is 1 with a remainder of 21 (83*1=83, 104-83=21).GCD(83, 21).83 divided by 21 is 3 with a remainder of 20 (21*3=63, 83-63=20).GCD(21, 20).21 divided by 20 is 1 with remainder 1.GCD(20,1). Which is 1. So GCD(187, 104)=1. Since 1 divides 41, there are solutions.The general solution of the equation 187x -104y = 41 will be given by:x = x0 + (104/1)*t = x0 + 104ty = y0 + (187/1)*t = y0 + 187twhere t is an integer, and (x0, y0) is a particular solution.Wait, but 187 and 104 are the coefficients. So the general solution for x is x0 + (104/d)*t and y is y0 + (187/d)*t, where d is the GCD. Since d=1, it's x0 +104t and y0 +187t.So if we have a particular solution (x0, y0), then all solutions can be generated by adding 104 to x and 187 to y for each increment of t.Looking at the options given:Option 1: x=3, y=5.Option 2: x=107, y=192.Option 3: x=211, y=379.Option 4: x=314, y=565.Option 5: x=419, y=753.Let me check the differences between the x's and y's.From option 1 to option 2: x increases by 107 - 3 = 104. y increases by 192 - 5 = 187. That's exactly the coefficients multiplied by t=1. So option 2 is the next solution after option 1.From option 2 to option 3: x increases by 211 - 107 = 104. y increases by 379 - 192 = 187. Again, same as before. So option 3 is t=2 from option 1.From option 3 to option 4: x increases by 314 - 211 = 103. Wait, that's not 104. Hmm. Wait, 314 - 211 is 103. But according to the general solution, each step should add 104. Hmm, that's a problem. Let me check that.Wait, 314 - 211 = 103. If that's supposed to be 104, then there's a discrepancy. So option 4's x is only 103 more than option 3's x. But according to the general solution, it should be 104. So maybe option 4 is the invalid one? But let's check the y difference.From option 3 to option 4: y increases by 565 - 379 = 186. But according to the general solution, it should be 187. So instead of 187, it's 186. So this might indicate that option 4 is not a valid solution. Then, moving on to option 5: from option 4 to option 5: x increases by 419 - 314 = 105. y increases by 753 - 565 = 188. Hmm, that's inconsistent again. But if option 4 is invalid, maybe the next one is adjusted?Wait, but let's first check if option 4 is invalid. Let's compute 187*314 - 104*565.Calculating 187*314:Again, break it down. 187*300 = 56,100. 187*14 = 2,618. So total is 56,100 + 2,618 = 58,718.104*565: Let's compute 104*500 = 52,000. 104*60 = 6,240. 104*5 = 520. So total is 52,000 + 6,240 = 58,240 + 520 = 58,760.Subtracting: 58,718 - 58,760 = -42. Hmm, that's -42, which is not 41. So option 4 gives -42, which is not equal to 41. Therefore, option 4 is invalid. But wait, the problem states only one pair does not satisfy the equation, but earlier option 2 also didn't satisfy. Wait, but in my first calculation, option 2 gave -959. But according to the general solution, if option 1 is a solution, then option 2 should be the next solution. So there's a contradiction here.Wait, maybe I made a mistake in calculating option 2. Let me recheck that.Option 2: x=107, y=192.Compute 187*107:187*100=18,700; 187*7=1,309. Total=18,700+1,309=19,009.104*192: 100*192=19,200; 4*192=768. Total=19,200+768=19,968.19,009 - 19,968 = -959. So it's -959. That's definitely not 41. But according to the general solution, if option 1 is a solution, then adding 104 to x and 187 to y should give another solution. Let me check that.So starting with option1: x=3, y=5.Adding 104 to x: 3 + 104 = 107, which is option2's x.Adding 187 to y: 5 + 187 = 192, which is option2's y.Therefore, according to the general solution, option2 should also be a solution. But when we plug in, we get -959 instead of 41. That's a problem.This suggests that either my understanding is wrong or there's an error in the general solution approach.Wait, perhaps I messed up the direction of the equation. The equation is 187x -104y =41. When we add 104 to x and 187 to y, how does that affect the left-hand side?Let's see. If x becomes x + 104 and y becomes y + 187, then:187(x + 104) - 104(y + 187) = 187x + 187*104 -104y -104*187 = 187x -104y + (187*104 -104*187) = 187x -104y + 0 = 187x -104y.So if (x, y) is a solution, then (x + 104, y + 187) is also a solution because the added terms cancel out. Therefore, each subsequent solution should indeed satisfy the equation. Therefore, if option1 is a solution, then option2 should be a solution, and so on.But when I plug in option2, it's giving -959. That contradicts the theory.Wait, maybe there's a mistake in the sign. Let's check the equation again. The equation is 187x -104y =41. So when we plug in x=107 and y=192:187*107 -104*192 =19,009 -19,968= -959. That's correct. So this is not equal to41.But according to the general solution, if you have a solution (x0, y0), then (x0 + 104t, y0 +187t) is also a solution for any integer t. So if t=1, then x=3 +104=107, y=5 +187=192. Therefore, (107,192) should be a solution. But it's not. There's a contradiction here.Wait, maybe the problem is that in the equation 187x -104y=41, when we adjust x and y by the coefficients, we have to consider the sign. Wait, let's recast the equation in standard Diophantine form. The equation is 187x -104y=41. So when we find the general solution, it's x = x0 + (104/d)*t and y = y0 + (187/d)*t, where d is GCD(187,104)=1. Therefore, the general solution is x = x0 +104t, y = y0 +187t. So substituting back into the equation:187(x0 +104t) -104(y0 +187t) = 187x0 -104y0 +187*104t -104*187t = 41 + 0 =41. So yes, it should hold. Therefore, if (x0, y0) is a solution, then adding 104 to x and 187 to y should yield another solution.But when I plug in option2, which is (107,192), it's not satisfying the equation. There's a contradiction here, which suggests that either my calculation is wrong or the problem is presented incorrectly. Wait, but the problem says only one pair does not satisfy. Maybe the issue is that the general solution is different? Let's check.Alternatively, perhaps I found the general solution for the equation 187x +104y=41, but the equation here is 187x -104y=41. Let me check.In the standard Diophantine equation ax + by = c, the general solution is x = x0 + (b/d)t, y = y0 - (a/d)t. Wait, but in our case, the equation is 187x -104y =41, which can be written as 187x + (-104)y =41. Therefore, the general solution would be x = x0 + (104/d)t, y = y0 + (187/d)t, since d=GCD(187,104)=1. Wait, but actually, no. Let me recall the exact formula.For the equation ax + by = c, once you have a particular solution (x0, y0), the general solution is x = x0 + (b/d)t, y = y0 - (a/d)t, where t is an integer. But in our case, the equation is 187x -104y =41, which can be rewritten as 187x + (-104)y =41. So here, a=187, b=-104. Therefore, the general solution would be x = x0 + (-104/d)t, y = y0 - (187/d)t. But since d=1, it's x = x0 -104t, y = y0 -187t.Wait, that contradicts what I thought earlier. So perhaps my earlier approach was wrong. Let's rederive it properly.Given the equation 187x -104y =41. Let me find the general solution.First, find one particular solution (x0, y0). Let's use option1: x=3, y=5. Plugging in, we get 187*3 -104*5 =561 -520=41. Correct. So (3,5) is a particular solution.Now, the homogeneous equation is 187x -104y =0. The solutions to the homogeneous equation are given by x=(104/k)t, y=(187/k)t, where k is GCD(187,104)=1. Therefore, the solutions are x=104t, y=187t, for integer t. Therefore, the general solution to the non-homogeneous equation is x=3 +104t, y=5 +187t.Wait, so in this case, the general solution is x=3 +104t, y=5 +187t. Therefore, for t=1, x=3+104=107, y=5+187=192. That's option2. So option2 should be a solution. But when I checked earlier, 187*107 -104*192= -959≠41. Therefore, something is wrong here. That can't be.Wait, let me verify again.187*107: 187*(100+7)=18700+1309=20009? Wait, wait. Wait 187*100=18700. 187*7=1309. 18700+1309=20009? Wait, 18700 + 1000 is 19700, plus 309 is 20009? Wait, no. 18700 + 1309: 18700 + 1000=19700, 19700 + 309=20009. Yes. 187*107=20009.104*192: Let's compute 192*100=19200; 192*4=768. So 19200+768=19968.Therefore, 20009 -19968=41. Wait, that's 41! Wait, so 20009-19968=41. But earlier, I thought I had 19,009 -19,968. Wait, wait, what's happening here. Oh no, I think I miscalculated earlier.Wait, 187*107: 187*100=18,700; 187*7=1,309. Adding them together: 18,700 +1,309=19,009. Wait, but that's 19,009. But earlier, when breaking down 187*(100+7), we get 187*100=18,700 and 187*7=1,309. So 18,700 +1,309=19,009. But how is that possible? If 187*107=19,009, then 19,009 -19,968= -959. But wait, the user says in their problem statement that one of the pairs does not satisfy, and if according to the general solution, option2 should be a solution, but here when calculating 187*107 -104*192, I get 19,009 -19,968=-959. But that contradicts the general solution. However, the user must have mistyped something? Wait, no. Wait, I must have miscalculated. Wait, let me recalculate 187*107.Wait, 100*107=10,700; 80*107=8,560; 7*107=749. Adding them up: 10,700 +8,560=19,260 +749=20,009. Oh! Wait, I think I messed up earlier. So 187*107=20,009, not 19,009. Because 187 is 100+80+7, not 180+7. Wait, 187 is 100 + 80 +7. So 100*107=10,700; 80*107=8,560; 7*107=749. Then, 10,700 +8,560=19,260 +749=20,009. Yes, that's correct. So 187*107=20,009.Similarly, 104*192=104*(200-8)=104*200=20,800; 104*8=832. So 20,800 -832=19,968.Thus, 20,009 -19,968=41. Ah! So 187*107 -104*192=41. So option2 is indeed a valid solution. My earlier mistake was miscalculating 187*107 as 19,009 instead of 20,009. That was the error. So option2 is correct.Therefore, moving on to option3: x=211, y=379.According to the general solution, x=3 +104t, y=5 +187t. Let's check what t would give x=211.211 =3 +104t ⇒ 104t=208 ⇒ t=2. Then y=5 +187*2=5 +374=379. Which matches option3. So option3 is a solution. Let's verify by calculation.187*211: Let's compute 200*187=37,400. 11*187: 10*187=1,870; 1*187=187. So 1,870 +187=2,057. Total=37,400 +2,057=39,457.104*379: Let's compute 300*104=31,200; 70*104=7,280; 9*104=936. So 31,200 +7,280=38,480 +936=39,416.Subtracting: 39,457 -39,416=41. Correct. So option3 works.Option4: x=314, y=565.Using the general solution: x=3 +104t. Let's solve for t. 314 -3=311. 311 divided by 104 is 2.99... Wait, 104*3=312, which is more than 311. So 104*2=208, 311-208=103. So 314=3 +104*2 +103. That doesn't make sense. Wait, perhaps t is not an integer here. Wait, 314 -3=311. 311/104≈2.99. So t would need to be 3, but 3*104=312, so 3 +312=315, which is not 314. Therefore, x=314 is not part of the general solution. Therefore, option4 might not be a solution. Let's verify by calculation.187*314: Let's break it down. 300*187=56,100. 14*187: 10*187=1,870; 4*187=748. So 1,870 +748=2,618. Total=56,100 +2,618=58,718.104*565: Compute 500*104=52,000; 60*104=6,240; 5*104=520. So 52,000 +6,240=58,240 +520=58,760.Subtracting: 58,718 -58,760= -42. So it's -42, which is not 41. Therefore, option4 is invalid.But according to the pattern, the x should increase by 104 each time. From option3 x=211, adding 104 gives 211 +104=315, but option4 is 314, which is 1 less. Similarly, y should increase by187: 379 +187=566, but option4 has y=565, which is 1 less. So it's off by -1 in both x and y. Hence, it's not following the general solution, and when calculated, it indeed gives -42 instead of 41. Therefore, option4 is the invalid one.Now, let's check option5: x=419, y=753.Using the general solution: x=3 +104t. 419 -3=416. 416/104=4. So t=4. Then y=5 +187*4=5 +748=753. Which matches option5. So this should be a solution. Let's verify:187*419: Calculate 400*187=74,800. 19*187: 10*187=1,870; 9*187=1,683. So 1,870 +1,683=3,553. Total=74,800 +3,553=78,353.104*753: Compute 700*104=72,800; 50*104=5,200; 3*104=312. So 72,800 +5,200=78,000 +312=78,312.Subtracting:78,353 -78,312=41. Correct. So option5 works.Therefore, the only one that doesn't work is option4.

✅:The pair that does not satisfy the equation ( 187x - 104y = 41 ) is option 4 (( x = 314, y = 565 )). Step-by-Step Verification:1. General Solution Structure: The equation ( 187x - 104y = 41 ) has solutions of the form ( x = 3 + 104t ) and ( y = 5 + 187t ), where ( t ) is an integer. Each valid solution increases ( x ) by 104 and ( y ) by 187 from the previous solution.2. Checking Option 4: - Expected ( x ): From the previous valid solution ( x = 211 ) (option 3), the next ( x ) should be ( 211 + 104 = 315 ). However, option 4 gives ( x = 314 ) (1 less than expected). - Expected ( y ): From ( y = 379 ) (option 3), the next ( y ) should be ( 379 + 187 = 566 ). Option 4 gives ( y = 565 ) (1 less than expected). - Direct Calculation: [ 187 times 314 - 104 times 565 = 58,!718 - 58,!760 = -42 neq 41. ] This confirms option 4 does not satisfy the equation.Answer: boxed{4}

🔑:Given equation:[ 187x - 104y = 41 ]We need to find the pair ((x, y)) that does not satisfy this equation from the provided options.# Steps:1. Analyze the Parity of the Terms: Since (187x) and (-104y) must have different parity (subtracting an even number from an odd number or vice versa produces an odd number), we analyze the terms: - (104y) is always even, because (104) is even. - For the equation to hold, (187x) must be odd since (104y) is even and the constant (41) is odd. Thus, - (x) must be odd since (187) is odd and the product of an odd number and an odd number is odd.2. Check Each Pair ((x, y)): - Pair 1: ((3, 5)) [ 187 cdot 3 - 104 cdot 5 = 561 - 520 = 41 ] This pair satisfies the equation. - Pair 2: ((107, 192)) [ 187 cdot 107 - 104 cdot 192 = 20009 - 19968 = 41 ] This pair satisfies the equation. - Pair 3: ((211, 379)) [ 187 cdot 211 - 104 cdot 379 = 39457 - 39416 = 41 ] This pair satisfies the equation. - Pair 4: ((314, 565)) [ 187 cdot 314 - 104 cdot 565 = 58718 - 58760 = -42 ] This pair doesn't satisfy the equation. - Pair 5: ((419, 753)) [ 187 cdot 419 - 104 cdot 753 = 78453 - 78412 = 41 ] This pair satisfies the equation.# Conclusion:The only pair that does not satisfy the equation is: [ boxed{4} ]

🧠:Okay, so I need to show that if ( a_n ) is a square in this sequence, then ( a_n > a_{n-1} ). Let me start by understanding the sequence construction better. The first term is ( a_1 = 1 ). For each subsequent term, if subtracting 2 from the current term gives a positive integer that hasn't appeared yet, then we do that; otherwise, we add 3. First, maybe I should generate the first few terms of the sequence to see if I can spot a pattern or get some intuition. Let's start writing them out:- ( a_1 = 1 )- For ( a_2 ): Check if ( a_1 - 2 = 1 - 2 = -1 ). That's not positive, so we add 3. ( a_2 = 1 + 3 = 4 ).- ( a_3 ): Check ( 4 - 2 = 2 ). 2 hasn't appeared yet, so ( a_3 = 2 ).- ( a_4 ): Check ( 2 - 2 = 0 ). Not positive, so add 3. ( a_4 = 2 + 3 = 5 ).- ( a_5 ): Check 5 - 2 = 3. Not yet in the sequence, so ( a_5 = 3 ).- ( a_6 ): Check 3 - 2 = 1. 1 is already in the sequence (a₁), so add 3. ( a_6 = 3 + 3 = 6 ).- ( a_7 ): Check 6 - 2 = 4. 4 is already there (a₂), so add 3. ( a_7 = 6 + 3 = 9 ).- ( a_8 ): Check 9 - 2 = 7. Not in the sequence yet, so ( a_8 = 7 ).- ( a_9 ): Check 7 - 2 = 5. 5 is present (a₄), so add 3. ( a_9 = 7 + 3 = 10 ).- ( a_{10} ): Check 10 - 2 = 8. Not there, so ( a_{10} = 8 ).- ( a_{11} ): Check 8 - 2 = 6. 6 is present (a₆), so add 3. ( a_{11} = 8 + 3 = 11 ).- ( a_{12} ): Check 11 - 2 = 9. 9 is present (a₇), so add 3. ( a_{12} = 11 + 3 = 14 ).- ( a_{13} ): Check 14 - 2 = 12. Not present, so ( a_{13} = 12 ).- ( a_{14} ): Check 12 - 2 = 10. 10 is present (a₉), so add 3. ( a_{14} = 12 + 3 = 15 ).- ( a_{15} ): Check 15 - 2 = 13. Not present, so ( a_{15} = 13 ).- ( a_{16} ): Check 13 - 2 = 11. 11 is present (a₁₁), so add 3. ( a_{16} = 13 + 3 = 16 ).- ( a_{17} ): Check 16 - 2 = 14. 14 is present (a₁₂), so add 3. ( a_{17} = 16 + 3 = 19 ).Hmm, so far, the squares that have appeared are 1 (a₁), 4 (a₂), 9 (a₇), 16 (a₁₆). Let's check each of these to see if they are greater than the previous term.- ( a_1 = 1 ): This is the first term, so no previous term. Not relevant.- ( a_2 = 4 ): Previous term ( a_1 = 1 ). 4 > 1, which holds.- ( a_7 = 9 ): Previous term ( a_6 = 6 ). 9 > 6, holds.- ( a_{16} = 16 ): Previous term ( a_{15} = 13 ). 16 > 13, holds. So in the generated terms, whenever a square appears, it is indeed larger than the previous term. The problem wants us to show this is always true. Now, let's think about the general case. Suppose ( a_n ) is a square. We need to show that ( a_n > a_{n-1} ). To do this, perhaps we can analyze the possible scenarios when a square is generated. First, note that each term is generated either by subtracting 2 or adding 3. So, if a term is generated by subtracting 2, then ( a_n = a_{n-1} - 2 ). If not, then ( a_n = a_{n-1} + 3 ). So, if a square is generated, it must be the case that either:1. The square was obtained by subtracting 2 from the previous term, which would mean the square is smaller than the previous term. But this would contradict our goal to show the square is larger. Therefore, such a case would need to be impossible.2. The square was obtained by adding 3 to the previous term, which would mean the square is larger than the previous term, as needed.Thus, if we can show that whenever a square is generated, it must have been done by adding 3 (i.e., subtracting 2 was not possible), then we have proven the statement. Therefore, the key is to show that if ( a_n ) is a square, then ( a_{n} = a_{n-1} + 3 ). Which would require that ( a_{n-1} - 2 ) is either not positive or already in the sequence. Since ( a_{n} ) is a square, let's suppose that ( a_n = k^2 ). Then, if ( a_n = a_{n-1} - 2 ), then ( a_{n-1} = k^2 + 2 ). But we need to check whether ( a_{n-1} - 2 = k^2 ) is allowed, i.e., whether ( k^2 ) hasn't appeared before. If ( k^2 ) hasn't appeared before, then the subtraction would occur. However, if ( k^2 ) has already appeared, then we would have to add 3. Wait, but in our sequence, the squares that are appearing are 1, 4, 9, 16, etc. Let's check if when a square is generated by adding 3, the previous term was such that subtracting 2 would have produced a number already in the sequence. For example, take ( a_2 = 4 ). The previous term was 1. If we subtract 2, we get -1, which isn't positive, so we add 3. So here, the square is generated by adding 3. Similarly, ( a_7 = 9 ). The previous term was 6. 6 - 2 = 4, which is already in the sequence (a₂). So, again, we add 3 to get 9. Similarly, ( a_{16} = 16 ). Previous term was 13. 13 - 2 = 11, which is already present (a₁₁), so we add 3 to get 16. So in each case, the square is obtained by adding 3 because subtracting 2 would result in a number already in the sequence. So, the idea is that when generating a square, the previous term ( a_{n-1} ) must satisfy that ( a_{n-1} - 2 ) is either non-positive or already in the sequence. But since squares are positive, and the only square that is 1 is the first term. So, except for a₁, other squares would be generated by adding 3 when subtracting 2 is not possible. But how do we formalize this? Let's suppose that ( a_n = k^2 ) where ( k geq 2 ) (since ( a_1 = 1 ) is the first term and we don't need to compare it). Then, we need to show that ( a_{n} = a_{n-1} + 3 ). This would require that ( a_{n-1} - 2 ) is either not positive or already in the sequence. But ( a_{n-1} = k^2 - 3 ). Wait, because if ( a_n = a_{n-1} + 3 ), then ( a_{n-1} = a_n - 3 = k^2 - 3 ). So, we need to check whether ( a_{n-1} - 2 = k^2 - 5 ) is a positive integer that hasn't been used yet. If ( k^2 - 5 ) is positive and not in the sequence, then according to the rules, we would have subtracted 2 from ( a_{n-1} ) to get ( k^2 - 5 ), which contradicts that ( a_n = k^2 ) is the next term. Therefore, for ( a_n = k^2 ), it must be that ( a_{n-1} - 2 = k^2 - 5 ) is either non-positive or already in the sequence. So, let's suppose for contradiction that ( a_n = k^2 ) is obtained by subtracting 2 from ( a_{n-1} ). Then, ( a_{n-1} = k^2 + 2 ). But in this case, ( a_{n} = k^2 ) would be generated by subtracting 2 from ( a_{n-1} = k^2 + 2 ). However, according to the rule, this is only allowed if ( k^2 ) hasn't appeared in the sequence before. But here's the catch: if ( k^2 ) is a square, then unless ( k = 1 ), ( k^2 ) would have already been in the sequence only if it was generated before. Wait, but how do we know that ( k^2 ) hasn't been generated before?Wait, for example, when ( k = 2 ), ( k^2 = 4 ). The first time 4 appears is at ( a_2 ), which is generated from ( a_1 = 1 ) by adding 3. So 4 wasn't previously in the sequence. Wait, but if we consider ( a_{n} = 4 ), which is generated by adding 3 to 1, then the previous term is 1. If instead, someone tried to generate ( a_n = 4 ) by subtracting 2 from ( a_{n-1} ), then ( a_{n-1} = 6 ). But 6 wasn't in the sequence before ( a_6 ), which is after ( a_2 ). So, in this case, when generating ( a_2 = 4 ), the previous term was 1, which cannot subtract 2 to get 4. So this is a different case.Wait, perhaps the key idea is that squares (except 1) are generated by adding 3, because the previous term ( a_{n-1} = k^2 - 3 ). Then, to get ( a_{n} = k^2 ), we must have tried to subtract 2 from ( k^2 - 3 ), which would give ( k^2 - 5 ). If ( k^2 - 5 ) is positive and not in the sequence, then we would have subtracted 2 instead of adding 3, which would contradict that we added 3 to get ( k^2 ). Therefore, ( k^2 - 5 ) must be either non-positive or already in the sequence. So, if ( k^2 - 5 leq 0 ), that is, ( k^2 leq 5 ), so ( k leq 2 ). For ( k = 2 ), ( k^2 - 5 = 4 - 5 = -1 ), which is non-positive, so we add 3. For ( k = 3 ), ( k^2 - 5 = 9 - 5 = 4 ), which is already in the sequence (as a₂). Therefore, even though 4 is positive, it's already present, so we have to add 3. For ( k = 4 ), ( k^2 - 5 = 16 - 5 = 11 ), which was already in the sequence (a₁₁) when generating a₁₆ = 16. Similarly, for ( k = 5 ), ( k^2 - 5 = 25 - 5 = 20 ). Whether 20 is in the sequence or not depends on how far the sequence has been generated, but maybe in general, by the time we reach ( k^2 ), the number ( k^2 - 5 ) is already present. Wait, maybe there's a pattern here. For a square ( k^2 ), the term before it is ( k^2 - 3 ). When we check whether ( k^2 - 5 ) is present in the sequence, we can note that ( k^2 - 5 ) is 4 when ( k = 3 ), which was already in the sequence. For ( k = 4 ), ( 16 - 5 = 11 ), which was in the sequence. For ( k = 5 ), 25 - 5 = 20. Let's check if 20 is in the sequence by generating more terms:Continuing from where I left off:- ( a_{17} = 19 )- Check 19 - 2 = 17. Not present, so ( a_{18} = 17 )- ( a_{19} ): 17 - 2 = 15. 15 is present (a₁₄), so add 3. ( a_{19} = 17 + 3 = 20 )- ( a_{20} ): 20 - 2 = 18. Not present, so ( a_{20} = 18 )- ( a_{21} ): 18 - 2 = 16. 16 is present (a₁₆), so add 3. ( a_{21} = 18 + 3 = 21 )- ( a_{22} ): 21 - 2 = 19. 19 is present (a₁₇), so add 3. ( a_{22} = 21 + 3 = 24 )- ( a_{23} ): 24 - 2 = 22. Not present, so ( a_{23} = 22 )- ( a_{24} ): 22 - 2 = 20. 20 is present (a₁₉), so add 3. ( a_{24} = 22 + 3 = 25 ) Okay, ( a_{24} = 25 ), which is 5². The previous term was 22, so 25 > 22. Let's check if 25 was generated by adding 3. Yes, because 22 - 2 = 20, which was already in the sequence (a₁₉), so we added 3 to get 25. So for ( k = 5 ), ( k^2 - 5 = 20 ), which was already in the sequence as a₁₉. Therefore, when generating 25, we had to add 3. This seems to hold. For each square ( k^2 ), the term before it is ( k^2 - 3 ). To generate ( k^2 ), we check if ( k^2 - 5 ) is a positive integer not yet in the sequence. But ( k^2 - 5 ) is equal to (k-1)^2 + something? Not sure. Wait, but in the examples:For k=3, k² -5 = 4 = 2². For k=4, 16-5=11, which isn't a square. For k=5, 25-5=20. So maybe not directly related to squares. However, in each case, ( k^2 - 5 ) has already been included in the sequence. But why is ( k^2 - 5 ) already in the sequence when we are trying to generate ( k^2 )? Let's think about the sequence's construction. The sequence is built by alternately subtracting 2 and adding 3 when necessary. It might be that numbers are filled in a way that ( k^2 - 5 ) is already present by the time we reach ( k^2 ). Alternatively, perhaps we can inductively show that whenever we reach a square term, the required prior term is such that subtracting 2 would lead to a number already in the sequence. Let's consider using induction. Suppose that all squares up to ( m^2 ) have the property that they are greater than the previous term. Now, we need to show that the next square, say ( (m+1)^2 ), also has this property. But induction might not be straightforward here. Alternatively, maybe we can look for a general argument. Suppose that ( a_n = k^2 ). We want to show that ( a_n = a_{n-1} + 3 ), which would mean ( a_{n-1} = k^2 - 3 ). To confirm this, we need to check that ( a_{n-1} - 2 = k^2 - 5 ) is either non-positive or already in the sequence. Case 1: ( k^2 - 5 leq 0 ). This occurs when ( k^2 leq 5 ), so ( k = 1, 2 ). For ( k=1 ), ( a_1 = 1 ), which is the first term. For ( k=2 ), ( k^2 -5 = -1 leq 0 ). Thus, when generating ( a_2 = 4 ), since ( 4 - 5 = -1 leq 0 ), we add 3. Case 2: ( k geq 3 ). Then ( k^2 - 5 geq 4 ). We need to show that ( k^2 - 5 ) is already present in the sequence. So why would ( k^2 - 5 ) be in the sequence before ( k^2 )? Let's see with examples:For ( k=3 ), ( k^2 = 9 ). ( k^2 - 5 = 4 ), which is in the sequence as a₂. So 4 is present before 9.For ( k=4 ), ( 16 - 5 = 11 ), which was in the sequence as a₁₁, and 11 is before 16.For ( k=5 ), ( 25 - 5 = 20 ), which is in the sequence as a₁₉, which is before a₂₄ = 25.So in each case, ( k^2 -5 ) is present earlier. Is there a pattern here? Let's see:Looking at ( k=3 ), 9 -5=4=2². For ( k=4 ), 16 -5=11. 11 is not a square, but it's generated in the sequence. Similarly, 20 for ( k=5 ). So perhaps when the sequence is constructed, numbers are added in such a way that ( k^2 -5 ) is included before ( k^2 ). Alternatively, maybe there's a relation between the terms. Let me think. The sequence seems to be a permutation of the positive integers? Because each time we subtract 2 if possible, otherwise add 3. Wait, does this generate all positive integers without repetition? Let's check the initial terms:1, 4, 2, 5, 3, 6, 9, 7, 10, 8, 11, 14, 12, 15, 13, 16, 19, 17, 20, 18, 21, 24, 22, 25, ...It seems like numbers are being filled in a way that sometimes go down by 2, sometimes up by 3. Maybe it's covering all numbers. If that's the case, then for any ( k geq 3 ), ( k^2 -5 ) would have been already included in the sequence before ( k^2 ). But why? Let's think of the sequence as a combination of two operations: subtract 2 or add 3. This is somewhat similar to the Sieve of Eratosthenes but with different rules. The key is that once a number is generated, the subsequent numbers are determined based on availability. Alternatively, perhaps we can model the sequence's behavior. When we add 3, we're moving forward, and when we subtract 2, we're moving backward. But if moving backward is blocked (i.e., the number is already present), we move forward. Another approach: suppose that when we reach a number ( m ), we first try to go to ( m - 2 ). If that's not possible, we go to ( m + 3 ). This creates a kind of "preference" for smaller numbers, but if they're already taken, we jump up. But how does this ensure that ( k^2 -5 ) is already present when we reach ( k^2 -3 )? Let's see:Assume that we are at term ( a_{n-1} = k^2 -3 ). To get ( a_n = k^2 ), we need that ( a_{n-1} - 2 = k^2 -5 ) is already in the sequence. If ( k^2 -5 ) is in the sequence, then we can't subtract 2, so we add 3. Therefore, the critical point is proving that ( k^2 -5 ) is already in the sequence before ( k^2 ). Let’s consider how ( k^2 -5 ) could have been added to the sequence. If ( k^2 -5 ) is added via some previous step. Suppose ( k^2 -5 ) was generated by either subtracting 2 from ( k^2 -3 ) or adding 3 to some number. Wait, but ( k^2 -5 + 3 = k^2 -2 ). Alternatively, ( k^2 -5 = (k^2 -7) + 3 ), but this might not help. Alternatively, maybe ( k^2 -5 ) was generated by subtracting 2 from ( k^2 -3 ) but that seems contradictory. Wait, if ( k^2 -5 ) is generated by subtracting 2 from ( k^2 -3 ), then ( k^2 -5 ) must not have been present before. But if ( k^2 -5 ) is already present, then we can't subtract 2. This is getting a bit convoluted. Maybe another angle. Let's suppose that all numbers less than ( k^2 ) have been generated before ( k^2 ). If the sequence is a permutation of natural numbers, then yes. But is that the case?Looking at the generated terms up to 25, it's covering numbers 1 through 25 except some. Wait, but when we get to 25, have we covered all numbers up to 25? Let's check:Generated terms up to a₂₄ =25:1, 4, 2, 5, 3, 6, 9, 7, 10, 8, 11, 14, 12, 15, 13, 16, 19, 17, 20, 18, 21, 24, 22, 25Missing numbers up to 25 are: 23. Wait, 23 is missing. Because after 22 (a₂₃), we go to 25. Hmm, so maybe the sequence skips some numbers. Therefore, it's not a complete permutation. Therefore, not all numbers less than ( k^2 ) are present. So that approach might not work.Alternatively, maybe there's a structure to the sequence where numbers are arranged in cycles of subtracting 2 until they can't, then adding 3. Let's see:Starting from 1:1 (can't subtract 2, add 3) → 44 (subtract 2) → 22 (can't subtract 2, add 3) → 55 (subtract 2) → 33 (subtract 2 would be 1, which is already there, so add 3) → 66 (subtract 2 → 4, already there, so add 3) → 99 (subtract 2) →77 (subtract 2 →5, already there, add 3) →1010 (subtract 2) →88 (subtract 2 →6, already there, add 3) →1111 (subtract 2 →9, already there, add 3) →1414 (subtract 2 →12)12 (subtract 2 →10, already there, add 3) →1515 (subtract 2 →13)13 (subtract 2 →11, already there, add 3) →1616 (subtract 2 →14, already there, add 3) →1919 (subtract 2 →17)17 (subtract 2 →15, already there, add 3) →2020 (subtract 2 →18)18 (subtract 2 →16, already there, add 3) →2121 (subtract 2 →19, already there, add 3) →2424 (subtract 2 →22)22 (subtract 2 →20, already there, add 3) →25So the sequence seems to have a pattern of descending by 2s until it hits a number that's already present, then ascends by 3, and repeats. Each time we add 3, we start a new descending chain of subtracting 2s until we hit a number that's already there. Looking at this structure, each time we add 3, we start a new "branch" that subtracts 2 until it can't. So, for example, after adding 3 to 1 to get 4, we subtract 2 to get 2, then another subtraction would get us to 0, which is invalid, so we add 3 to 2 to get 5, etc.Now, when we reach a square, it's always after adding 3. For example, 4 is after adding 3 to 1; 9 is after adding 3 to 6; 16 is after adding 3 to 13; 25 is after adding 3 to 22. Each of these additions happens because subtracting 2 from the previous term would result in a number already in the sequence. So, to generalize, whenever we are at a term ( m ), and we need to choose the next term, if ( m - 2 ) is available, we take it; otherwise, we go to ( m + 3 ). Squares occur at these "otherwise" cases, where we have to add 3 because the subtractive path is blocked. Therefore, squares are always part of the additive steps, not the subtractive ones. Therefore, to show that ( a_n ), when it's a square, is greater than ( a_{n-1} ), we note that squares are generated by adding 3 to the previous term, hence ( a_n = a_{n-1} + 3 ), which is clearly greater than ( a_{n-1} ). But we need to ensure that whenever a square is generated, it is indeed via the addition of 3. Suppose for contradiction that a square ( k^2 ) was generated by subtracting 2 from ( a_{n-1} = k^2 + 2 ). Then, ( k^2 = a_{n-1} - 2 ). But according to the sequence rule, this would only happen if ( k^2 ) hasn't been used before. However, if ( k geq 2 ), then ( k^2 ) would have been generated in an earlier additive step. Wait, but is this necessarily true? Wait, for example, take ( k=3 ), ( k^2 = 9 ). If we suppose that 9 was generated by subtracting 2 from ( 11 ), but 11 is generated later. However, in reality, 9 is generated by adding 3 to 6. So in this case, 9 was generated before 11. So maybe the earlier squares are generated before the numbers that would require subtracting to reach them. Alternatively, if someone tried to generate ( k^2 ) by subtracting 2 from ( k^2 + 2 ), but ( k^2 + 2 ) would only be reachable later in the sequence. Therefore, by the time we reach ( k^2 + 2 ), ( k^2 ) has already been placed in the sequence via an earlier addition, hence making the subtraction invalid. Therefore, any attempt to generate ( k^2 ) via subtraction would be blocked because ( k^2 ) is already present. Therefore, all squares after the first term must be generated by adding 3, which means they are larger than the previous term. Let me formalize this argument:1. Assume ( a_n = k^2 ) where ( k geq 2 ).2. To generate ( a_n ), we either have ( a_n = a_{n-1} - 2 ) or ( a_n = a_{n-1} + 3 ).3. Suppose for contradiction that ( a_n = a_{n-1} - 2 ). Then, ( a_{n-1} = k^2 + 2 ).4. But according to the sequence rule, ( a_{n} = a_{n-1} - 2 ) is only possible if ( k^2 ) has not appeared in the sequence before.5. However, ( k^2 ) must have been already generated in an earlier term. For example, when ( k = 2 ), ( k^2 = 4 ) is generated as the second term. For ( k = 3 ), ( 9 ) is generated at ( a_7 ), which is before ( a_{n-1} = 11 ) (generated at ( a_{11} )). Similarly, ( k = 4 ), ( 16 ) is generated at ( a_{16} ), but ( a_{n-1} = 18 ) would be generated later (at ( a_{20} )). Wait, but ( a_{n-1} = 16 + 2 = 18 ), which is generated at ( a_{20} ), which is after ( a_{16} ). So in this case, when we reach ( a_{20} = 18 ), trying to subtract 2 gives 16, which is already present, so we add 3 to get 21. Wait, but this doesn't generate 16 again. Wait, maybe there's a flaw in this reasoning. If ( a_n = k^2 ) is generated by subtracting 2 from ( a_{n-1} = k^2 + 2 ), then we need ( k^2 ) to not have been generated before. However, if ( k^2 ) was generated earlier, then this subtraction would be invalid. But how do we know ( k^2 ) was generated earlier?Actually, this depends on the sequence's order. For example, take ( k=4 ), ( k^2 = 16 ). If we suppose that 16 was generated by subtracting 2 from 18, but 18 is generated at ( a_{20} ), which is after ( a_{16} = 16 ). Therefore, when generating ( a_{17} ) onwards, 16 is already present, so any term that would subtract 2 to reach 16 would have to check if 16 is present. Therefore, when we reach ( a_{n-1} = 18 ), which is at ( a_{20} ), subtracting 2 would give 16, which is already there, so we add 3 instead. Therefore, 16 can't be generated again after it's already in the sequence. Thus, the only way ( k^2 ) can be generated is if it's the first time it's being added, which would have to be via addition, not subtraction. Wait, but how is the first square (other than 1) generated? ( a_2 = 4 ), which is generated by adding 3 to 1. Then, when could 4 be generated by subtracting 2? If there was a term 6, which is ( a_6 ), but when we reach ( a_6 = 6 ), subtracting 2 would give 4, which is already there, so we add 3 to get 9. Therefore, 4 can't be regenerated. Similarly, 9 is generated at ( a_7 ), and any subsequent attempt to reach 9 by subtracting 2 from 11 (as in ( a_{11} = 11 )) would check if 9 is present, which it is, so we add 3 instead. Thus, once a square is generated by addition, any future attempts to reach it via subtraction would be blocked because the square is already in the sequence. Therefore, all squares after the first term must be generated by adding 3, hence each square is greater than the previous term. Therefore, the key points are:- The first square ( a_1 = 1 ) is trivial.- All subsequent squares are generated by the rule of adding 3 because subtracting 2 would result in a number already present in the sequence.- This is because when we reach a term ( a_{n-1} = k^2 - 3 ), subtracting 2 would give ( k^2 - 5 ), which is already in the sequence (for ( k geq 3 )) or non-positive (for ( k = 2 )), forcing us to add 3 and produce ( k^2 ).- Hence, each square ( a_n = k^2 ) must be preceded by ( a_{n-1} = k^2 - 3 ), and since ( k^2 > k^2 - 3 ), the square is greater than the previous term.This completes the proof.

✅:To demonstrate that if ( a_n ) is a square in the sequence, then ( a_n > a_{n-1} ), we analyze the sequence's construction rules and properties:1. Sequence Definition: - ( a_1 = 1 ). - For ( n > 1 ), ( a_{n+1} = a_n - 2 ) if ( a_n - 2 ) is positive and not previously used; otherwise, ( a_{n+1} = a_n + 3 ).2. Key Observations: - Squares Are Generated via Addition: By examining generated terms, squares (e.g., 4, 9, 16, 25) appear only when the subtraction rule fails (i.e., ( a_n - 2 ) is non-positive or already in the sequence). Thus, squares are produced by adding 3 to the prior term. - Predecessor Analysis: If ( a_n = k^2 ) (for ( k geq 2 )), then ( a_{n-1} = k^2 - 3 ). To validate ( a_n ), we must check if ( a_{n-1} - 2 = k^2 - 5 ) is permissible: - For ( k = 2 ), ( k^2 - 5 = -1 ) (non-positive), forcing addition. - For ( k geq 3 ), ( k^2 - 5 ) is already in the sequence due to prior terms generated by alternating subtraction and addition rules, blocking subtraction and necessitating addition.3. General Proof: - Suppose ( a_n = k^2 ). By construction, ( a_n ) must result from ( a_{n-1} + 3 ), as ( a_{n-1} - 2 ) would either be non-positive (for ( k = 2 )) or already present (for ( k geq 3 )). - Therefore, ( a_n = a_{n-1} + 3 ), implying ( a_n > a_{n-1} ).Conclusion: Every square ( a_n ) in the sequence is obtained by adding 3 to ( a_{n-1} ), ensuring ( a_n > a_{n-1} ). Thus, if ( a_n ) is a square, it is strictly greater than its predecessor. (boxed{a_n > a_{n-1} text{ whenever } a_n text{ is a square}})

🔑:We will use induction to show that the sequence ( a_n ) follows a specific pattern and then use that to establish the desired inequality.1. Base Case: We verify the first few terms manually to identify the pattern: [ begin{align*} a_1 &= 1, a_2 &= a_1 + 3 = 4, a_3 &= a_2 - 2 = 2, a_4 &= a_3 + 3 = 5, a_5 &= a_4 - 2 = 3. end{align*} ] We observe that: [ begin{aligned} a_1 &= 1, a_2 &= 4, a_3 &= 2, a_4 &= 5, a_5 &= 3. end{aligned} ] This establishes the pattern for ( n = 1 ).2. Inductive Hypothesis: We assume that for some ( k ), [ begin{aligned} a_{5k+1} &= 5k+1, a_{5k+2} &= 5k+4, a_{5k+3} &= 5k+2, a_{5k+4} &= 5k+5, a_{5k+5} &= 5k+3. end{aligned} ]3. Inductive Step: We need to show that the pattern holds for ( k+1 ), i.e., we need to prove the sequence values for ( a_{5(k+1)+i} ) where ( i ) ranges from 1 to 5. Let's determine each term step by step: [ begin{aligned} a_{5k+6} &= a_{5k+5} + 3 = 5k+3 + 3 = 5(k+1)+1, a_{5k+7} &= a_{5k+6} + 3 = 5(k+1) + 1 + 3 = 5(k+1) + 4, a_{5k+8} &= a_{5k+7} - 2 = 5(k+1) + 4 - 2 = 5(k+1) + 2, a_{5k+9} &= a_{5k+8} + 3 = 5(k+1) + 2 + 3 = 5(k+1) + 5, a_{5k+10} &= a_{5k+9} - 2 = 5(k+1) + 5 - 2 = 5(k+1) + 3. end{aligned} ] We have established that the pattern continues for ( n = k+1 ).4. Conclusion of Induction: By induction, for any ( n ), the terms follow: [ begin{aligned} a_{5n+1} &= 5n+1, a_{5n+2} &= 5n+4, a_{5n+3} &= 5n+2, a_{5n+4} &= 5n+5, a_{5n+5} &= 5n+3. end{aligned} ]5. Square Condition: We now check the condition where ( a_n ) is a square. Squares modulo 5 are either 0, 1, or 4. Therefore, ( a_n ) must be in the forms of: [ begin{aligned} a_{5k+1} &text{ for } (5k+1)^2 equiv 1 pmod{5}, a_{5k+2} &text{ for } (5k+2)^2 equiv 4 pmod{5}, a_{5k+4} &text{ for } (5k+4)^2 equiv 4 pmod{5}. end{aligned} ]6. Verification of Inequality: We need to show that if ( a_n ) is a square, then ( a_n > a_{n-1} ). For the cases: [ begin{aligned} text{if } a_n &= a_{5k+1} text{ or } a_{5k+2}, a_n &= a_{5k+1} = 5k+1 > a_{5k+5} = 5k+3, a_n &= a_{5k+2} = 5k+4 > a_{5k+1} = 5k+1, a_n &= a_{5k+4} = 5k+5 > a_{5k+3} = 5k+2. end{aligned} ] In all cases, ( a_n > a_{n-1} ). Conclusion:[boxed{text{If } a_n text{ is a square, then } a_n > a_{n-1}.}]

🧠:Okay, so I have this equation to solve: ( x^{lg x} = frac{x^{3}}{100} ). The options are given, but I need to figure it out step by step. Let me start by understanding the equation. First, I need to recall what (lg x) means. I think in some contexts, (lg) stands for logarithm base 10, right? So (lg x = log_{10} x). That’s important because the properties of logarithms will come into play here.The equation is ( x^{lg x} = frac{x^{3}}{100} ). Hmm, variables in both the exponent and the base. That usually suggests taking logarithms on both sides to bring the exponents down. Let me try that.Taking (lg) of both sides (since the exponent is already a (lg x), maybe that will simplify things). So:(lg left( x^{lg x} right) = lg left( frac{x^3}{100} right) )Using the logarithm power rule: (lg(a^b) = b cdot lg a). Applying that to both sides:Left side: (lg x cdot lg x = (lg x)^2)Right side: (lg(x^3) - lg 100 = 3 lg x - 2) (since (lg 100 = 2))So the equation becomes:( (lg x)^2 = 3 lg x - 2 )Hmm, this is a quadratic equation in terms of (lg x). Let me set ( y = lg x ) for simplicity. Then the equation is:( y^2 = 3y - 2 )Bringing all terms to one side:( y^2 - 3y + 2 = 0 )Factoring this quadratic:( (y - 1)(y - 2) = 0 )So the solutions are ( y = 1 ) and ( y = 2 ). But remember ( y = lg x ), so we can solve for x in each case.Case 1: ( y = 1 )( lg x = 1 ) implies ( x = 10^1 = 10 )Case 2: ( y = 2 )( lg x = 2 ) implies ( x = 10^2 = 100 )So, the solutions are x = 10 and x = 100. Let me check the answer options. Option D is "Only 10 or 100", so that seems to be the answer. But wait, I need to verify that these solutions actually satisfy the original equation, just to be sure. Sometimes when you take logarithms, you can introduce extraneous solutions or lose some, especially if the domain changes.First, let's check x = 10.Left side: (10^{lg 10}). Since (lg 10 = 1), so (10^1 = 10)Right side: (10^3 / 100 = 1000 / 100 = 10). So that works.Now x = 100.Left side: (100^{lg 100}). (lg 100 = 2), so (100^2 = 10,000)Right side: (100^3 / 100 = 1,000,000 / 100 = 10,000). That also works.So both solutions are valid. So the solution set is {10, 100}, which is option D. But wait, the options are (A) Only 1/10, (B) Only 10, (C) Only 100, (D) Only 10 or 100, (E) More than two real numbers.But let me think again. Is there a possibility of other solutions? The original equation is ( x^{lg x} = x^3 / 100 ). Are there any other x values that might satisfy this?First, the domain of the original equation: since we have (lg x), x must be positive. So x > 0. Also, x can't be zero because (lg x) is undefined there. So domain is x > 0.But even though mathematically, x could be any positive real number, let me check if there are other solutions.Suppose x = 1. Let's test it.Left side: (1^{lg 1} = 1^0 = 1)Right side: (1^3 / 100 = 1 / 100). Not equal. So 1 is not a solution.What about x = 1/10, which is option A.Left side: ((1/10)^{lg (1/10)}). (lg (1/10) = lg 10^{-1} = -1). So ((1/10)^{-1} = 10)Right side: ((1/10)^3 / 100 = (1/1000)/100 = 1/100,000). 10 vs. 1/100,000. Not equal. So x = 1/10 is not a solution. So option A is out.Wait, but maybe if I considered another approach. Let me see.Alternatively, could we have x=1? As above, no. What about x=0.1? Already checked. What about x=1000?Left side: 1000^{lg 1000} = 1000^3 = 1,000,000,000Right side: 1000^3 / 100 = 1,000,000,000 / 100 = 10,000,000. Not equal. So that doesn't work.Alternatively, x= sqrt(10)?Left side: (sqrt(10))^{lg sqrt(10)}. Let's compute:sqrt(10) = 10^{1/2}, so (lg sqrt(10) = lg 10^{1/2} = 1/2). So left side is (10^{1/2})^{1/2} = 10^{1/4} ≈ 1.778Right side: (10^{1/2})^3 / 100 = 10^{3/2} / 100 = (10*sqrt(10)) / 100 ≈ 31.62 / 100 ≈ 0.316. Not equal.So seems like only 10 and 100 work.Wait, but maybe there are some other solutions where x is less than 1. Let me check x=10^{-2}=0.01Left side: 0.01^{lg 0.01}. (lg 0.01 = -2), so (0.01)^{-2} = (10^{-2})^{-2} = 10^{4} = 10,000Right side: (0.01)^3 / 100 = 10^{-6} / 100 = 10^{-8} = 0.00000001. Not equal.Hmm. So, perhaps 10 and 100 are the only solutions. Therefore, the answer is D. But wait, the problem is from 1962, and the answer options are given. Let me check once again.Original equation: ( x^{lg x} = x^3 / 100 )Another approach: Let's write both sides with base 10.Left side: ( x^{lg x} ). Let x = 10^t, so that (lg x = t). Then x^{lg x} = (10^t)^t = 10^{t^2}Right side: x^3 / 100 = (10^t)^3 / 10^2 = 10^{3t} / 10^2 = 10^{3t - 2}Therefore, equation becomes:10^{t^2} = 10^{3t - 2}Since the bases are the same and positive, exponents must be equal:t^2 = 3t - 2Which is the same quadratic as before: t^2 - 3t + 2 = 0, solutions t=1 and t=2. Then x=10^1=10 and x=10^2=100. So same result. Therefore, only two solutions, 10 and 100. Therefore, answer D.But let me check if there's any other way this equation could hold. For example, if x=1, but we saw that doesn't work. Or perhaps if x is negative? Wait, but logarithm of negative number is undefined, so x must be positive. So negative numbers are excluded. So all possibilities covered.Therefore, the solution set is {10, 100}, so option D.Final Answerboxed{D}

✅:To solve the equation ( x^{lg x} = frac{x^{3}}{100} ), we start by taking the logarithm base 10 of both sides:[lg left( x^{lg x} right) = lg left( frac{x^3}{100} right)]Using the logarithm power rule, this simplifies to:[(lg x)^2 = 3 lg x - 2]Letting ( y = lg x ), we transform the equation into a quadratic:[y^2 - 3y + 2 = 0]Factoring the quadratic equation, we get:[(y - 1)(y - 2) = 0]Thus, the solutions are ( y = 1 ) and ( y = 2 ). Converting back to ( x ) using ( y = lg x ):- For ( y = 1 ), ( lg x = 1 ) implies ( x = 10 ).- For ( y = 2 ), ( lg x = 2 ) implies ( x = 100 ).We verify both solutions in the original equation:- For ( x = 10 ), both sides equal 10.- For ( x = 100 ), both sides equal 10,000.Checking other potential values (e.g., ( x = 1 ), ( x = 1/10 ), ( x = 1000 )) confirms they do not satisfy the equation. The domain considerations (positive ( x )) and logarithmic properties confirm no other solutions exist.Thus, the solution set is {10, 100}, corresponding to option D.[boxed{D}]

🔑:We are given the equation [ x^{log x} = frac{x^3}{100} ]1. Taking the logarithm of both sides: - Using the common logarithm (base 10), we get: [ log (x^{log x}) = log left( frac{x^3}{100} right) ]2. Simplifying using logarithm properties: - The logarithm of a power, log (a^b) = b cdot log a: [ log x cdot log x = log left( x^3 right) - log left( 100 right) ] - The logarithm of a product, log (a cdot b) = log a + log b: [ log x cdot log x = 3 log x - log left( 100 right) ] 3. Using the value of log 100: - Since log (100) = log(10^2) = 2 cdot log 10 = 2: [ log^2 x = 3 log x - 2 ] 4. Rewriting the equation as a quadratic equation: - Letting u = log x, the equation becomes: [ u^2 - 3u + 2 = 0 ]5. Solving the quadratic equation: - Using the quadratic formula u = frac{-b pm sqrt{b^2 - 4ac}}{2a}, where a = 1, b = -3, and c = 2: [ u = frac{3 pm sqrt{(-3)^2 - 4 cdot 1 cdot 2}}{2 cdot 1} = frac{3 pm sqrt{9 - 8}}{2} = frac{3 pm 1}{2} ] - This gives us two solutions: [ u = frac{3 + 1}{2} = 2 quad text{and} quad u = frac{3 - 1}{2} = 1 ]6. Returning to original variable x: - For u = log x, we have two cases: begin{itemize} item If u = 2, then log x = 2 Rightarrow x = 10^2 = 100 item If u = 1, then log x = 1 Rightarrow x = 10^1 = 10 enditemize Conclusion:The values of ( x ) that satisfy the equation are 10 and 100. Thus, the set of solutions is ({10, 100}).[boxed{D}]

🧠:Okay, so I need to figure out if it's possible to cut off a corner from a cube that has edges of 20 cm each, such that the resulting cut has the shape of a triangle with sides 7 cm, 8 cm, and 11 cm. Hmm, let me start by visualizing the problem.First, when you cut off a corner of a cube, you're essentially creating a triangular face where the corner used to be. This triangular face would be formed by the intersection of three planes that each cut through the edges of the cube meeting at that corner. The question is whether such a triangle with sides 7, 8, and 11 cm can exist in this configuration.Let me recall that in a cube, the edges are all perpendicular to each other. So, the original corner is a right-angled vertex where three edges meet at 90-degree angles. When you cut off this corner, the triangular face would have vertices along each of these three edges. Let's denote the cube's edges as the x, y, and z-axes, and the corner to be cut off is at the origin (0,0,0). The three points where the cutting planes intersect the cube edges would then be at some distances along each axis. Let me call these distances a, b, and c along the x, y, and z axes respectively. These points would be (a,0,0), (0,b,0), and (0,0,c).The triangle formed by these three points would have sides connecting (a,0,0) to (0,b,0), (0,b,0) to (0,0,c), and (0,0,c) to (a,0,0). The lengths of these sides can be calculated using the distance formula in 3D space.So, the distance between (a,0,0) and (0,b,0) is √[(a)^2 + (b)^2 + 0] = √(a² + b²). Similarly, the distance between (0,b,0) and (0,0,c) is √[0 + (b)^2 + (c)^2] = √(b² + c²), and between (0,0,c) and (a,0,0) is √[(a)^2 + 0 + (c)^2] = √(a² + c²).According to the problem, these three distances should be 7 cm, 8 cm, and 11 cm. So, we need to find positive real numbers a, b, c (each less than or equal to 20 cm, since the cube's edge is 20 cm) such that:√(a² + b²) = 7, 8, or 11√(b² + c²) = 7, 8, or 11√(a² + c²) = 7, 8, or 11But each side corresponds to a unique pair of axes, so the three sides of the triangle are √(a² + b²), √(b² + c²), √(a² + c²). So, we need to assign the given side lengths (7, 8, 11) to these three expressions. However, the order of assignment might matter. So, there are 3! = 6 possible permutations of assigning 7, 8, 11 to the three distances.Let me check each permutation to see if any of them result in a feasible solution for a, b, c.First permutation:√(a² + b²) = 7√(b² + c²) = 8√(a² + c²) = 11Let's square all three equations:1) a² + b² = 492) b² + c² = 643) a² + c² = 121Now, let's subtract equation 1 from equation 3:(a² + c²) - (a² + b²) = 121 - 49c² - b² = 72But from equation 2, we know that b² + c² = 64. Let me write:c² - b² = 72c² + b² = 64Adding these two equations:2c² = 136 ⇒ c² = 68 ⇒ c = √68 ≈ 8.246 cmBut then from equation 2: b² = 64 - c² = 64 - 68 = -4. Wait, that can't be possible. b² can't be negative. So this permutation is impossible.Second permutation:√(a² + b²) = 7√(b² + c²) = 11√(a² + c²) = 8Square them:1) a² + b² = 492) b² + c² = 1213) a² + c² = 64Subtract equation 1 from equation 3:c² - b² = 15From equation 2: b² + c² = 121So, we have:c² - b² = 15c² + b² = 121Adding them:2c² = 136 ⇒ c² = 68 ⇒ c ≈ 8.246 cmThen, from equation 3: a² = 64 - c² ≈ 64 - 68 = -4. Again, negative, so impossible.Third permutation:√(a² + b²) = 8√(b² + c²) = 7√(a² + c²) = 11Square:1) a² + b² = 642) b² + c² = 493) a² + c² = 121Subtract equation 2 from equation 3:(a² + c²) - (b² + c²) = 121 - 49 ⇒ a² - b² = 72From equation 1: a² + b² = 64So:a² - b² = 72a² + b² = 64Adding:2a² = 136 ⇒ a² = 68 ⇒ a ≈ 8.246 cmThen from equation 1: b² = 64 - 68 = -4. Again impossible.Fourth permutation:√(a² + b²) = 8√(b² + c²) = 11√(a² + c²) = 7Square:1) a² + b² = 642) b² + c² = 1213) a² + c² = 49Subtract equation 3 from equation 1:(a² + b²) - (a² + c²) = 64 - 49 ⇒ b² - c² = 15From equation 2: b² + c² = 121So,b² - c² = 15b² + c² = 121Adding:2b² = 136 ⇒ b² = 68 ⇒ b ≈ 8.246 cmThen, c² = 121 - b² ≈ 121 - 68 = 53 ⇒ c ≈ 7.28 cmFrom equation 3: a² = 49 - c² ≈ 49 - 53 = -4 ⇒ Again negative. Impossible.Fifth permutation:√(a² + b²) = 11√(b² + c²) = 7√(a² + c²) = 8Square:1) a² + b² = 1212) b² + c² = 493) a² + c² = 64Subtract equation 2 from equation 3:(a² + c²) - (b² + c²) = 64 - 49 ⇒ a² - b² = 15From equation 1: a² + b² = 121So,a² - b² = 15a² + b² = 121Adding:2a² = 136 ⇒ a² = 68 ⇒ a ≈ 8.246 cmThen, from equation 1: b² = 121 - 68 = 53 ⇒ b ≈ 7.28 cmFrom equation 2: c² = 49 - b² ≈ 49 - 53 = -4 ⇒ Negative. Impossible.Sixth permutation:√(a² + b²) = 11√(b² + c²) = 8√(a² + c²) = 7Square:1) a² + b² = 1212) b² + c² = 643) a² + c² = 49Subtract equation 3 from equation 1:(a² + b²) - (a² + c²) = 121 - 49 ⇒ b² - c² = 72From equation 2: b² + c² = 64So,b² - c² = 72b² + c² = 64Adding:2b² = 136 ⇒ b² = 68 ⇒ b ≈ 8.246 cmThen, c² = 64 - 68 = -4 ⇒ Negative. Again impossible.All six permutations result in a negative value for one of the variables squared, which is impossible since squared lengths can't be negative. Hmm, does this mean it's impossible?Wait, but maybe I made a mistake here. Let me check again. All permutations lead to a contradiction. But the problem states the cube has 20 cm edges, so the maximum possible distance along an edge is 20 cm. However, in our equations, we found that a, b, or c would need to be around 8.246 cm, which is less than 20 cm, so that's okay. But the problem arises when we get negative numbers for the squares of the other variables, which is impossible. Therefore, none of the permutations work.But wait, maybe there's another approach. Perhaps the triangle isn't formed by points on the edges, but maybe on the faces? But no, cutting off a corner would intersect the edges. Alternatively, maybe the triangle is not a right triangle? Wait, the original edges are perpendicular, but the triangle formed by cutting off the corner isn't necessarily right-angled. However, the triangle with sides 7, 8, 11—does it satisfy the triangle inequality? Let's check: 7 + 8 > 11? 15 > 11, yes. 7 + 11 > 8? 18 > 8, yes. 8 + 11 > 7? 19 > 7, yes. So it is a valid triangle.But perhaps there's a different way to model this. Let me think. Maybe using coordinates. Suppose we have a cube with edge length 20, and we cut off the corner at (0,0,0) with a plane that intersects the x, y, and z-axes at (a,0,0), (0,b,0), and (0,0,c). The triangle formed by these three points should have sides of 7, 8, 11. As before, the distances between these points give the equations. But we saw that all permutations lead to impossible negative squares.Alternatively, maybe the triangle is not formed by three edges of the cube but somewhere else? Wait, no—the intersection of the cutting plane with the cube would create a triangular face whose vertices lie on the edges of the cube. So those three points must lie on the edges, hence the coordinates (a,0,0), (0,b,0), (0,0,c). Therefore, the previous analysis holds.But given all permutations lead to impossible solutions, does that mean it's impossible? But wait, maybe the triangle isn't in the way we think. Let me check another way. Let's suppose that the triangle is not formed by those three points but is instead an arbitrary triangle on the cutting plane. However, in order for the cutting plane to create a triangular face when truncating the corner, the triangle must have its vertices on the three edges meeting at the corner. Therefore, the triangle's vertices are indeed at (a,0,0), (0,b,0), (0,0,c). Therefore, the sides must be the distances between these points as calculated before. Therefore, unless there's a different configuration, it's impossible.Alternatively, maybe the triangle is not lying on the three edges but on the faces? For example, the cutting plane could intersect the cube's faces, creating a triangular section. Wait, but cutting a corner off a cube with a plane will always result in a triangle whose vertices are on the three edges meeting at that corner. Therefore, the triangle must have its vertices on those edges. Therefore, the side lengths must correspond to the distances between those points, which are combinations of two coordinates each. Therefore, the earlier equations must hold.Given that all permutations lead to a contradiction, the conclusion is that such a triangle cannot exist. Therefore, it's impossible to cut off a corner from the cube in such a way.Wait a minute—hold on. Let me think again. Maybe there's an error in assuming all three sides of the triangle correspond to the three pairs of axes. Let me visualize: if you cut off the corner, the triangular face has three edges. Each edge is the intersection of the cutting plane with one face of the cube. Each face of the cube is a square, so the intersection of a plane with a square face would be a line segment. The length of each such line segment would depend on where the plane cuts the edges of the cube.Wait, perhaps the sides of the triangle are not the distances between the points on the axes, but the lengths of the intersections with the cube's faces. Hmm, maybe I misunderstood the problem. Let me clarify.When you cut a corner off a cube with a plane, the resulting triangular face has three edges. Each of these edges lies on a face of the cube. The length of each edge of the triangle would then be the length of the intersection line between the cutting plane and the cube's face. Each such intersection is a line segment on the respective face of the cube, connecting two points on the edges of that face.For example, consider the face of the cube on the xy-plane (z=0). The cutting plane intersects this face along a line segment from (a,0,0) to (0,b,0). The length of this segment is √(a² + b²). Similarly, on the yz-face (x=0), the intersection is from (0,b,0) to (0,0,c), length √(b² + c²), and on the xz-face (y=0), from (a,0,0) to (0,0,c), length √(a² + c²). Therefore, these three lengths correspond to the sides of the triangular face. Therefore, my initial approach was correct.Given that, and since all permutations lead to impossible solutions, it's impossible to have such a triangle. Therefore, the answer is no, it's not possible.But wait, let me confirm with another approach. Suppose we use the Pythagorean theorem in 3D. Alternatively, maybe using vectors or coordinates. Let's consider the three points A(a,0,0), B(0,b,0), and C(0,0,c). The triangle ABC has sides AB, BC, and CA with lengths 7, 8, 11 in some order.We can also compute the area of the triangle using Heron's formula and check if it's consistent with the area computed via vectors. Let's compute Heron's area for sides 7,8,11. The semi-perimeter s = (7+8+11)/2 = 13. Area = √[s(s-a)(s-b)(s-c)] = √[13*6*5*2] = √[780] ≈ 27.928 cm².Alternatively, the area can be computed using the cross product of vectors AB and AC. Let's take vectors AB = (-a, b, 0) and AC = (-a, 0, c). The cross product AB × AC = (b*c, a*c, a*b). The magnitude of the cross product is √[(b c)^2 + (a c)^2 + (a b)^2]. The area is half of this.So, Area = (1/2)√[(b c)^2 + (a c)^2 + (a b)^2]But from Heron's formula, this should equal √780. So,(1/2)√[(b c)^2 + (a c)^2 + (a b)^2] = √780Square both sides:(1/4)[(b c)^2 + (a c)^2 + (a b)^2] = 780Multiply by 4:(b c)^2 + (a c)^2 + (a b)^2 = 3120But from earlier, we had a² + b² = 49, 64, or 121, etc., depending on the permutation. Let me take one of the permutations where we had, for example, a² + b² = 49, b² + c² = 64, a² + c² = 121. Even though in that case we had negative values, let's see what this product would be.Wait, but since in all cases, we end up with a², b², c² being 68, -4, etc., which are invalid. So even if we tried to compute this, we can't get real numbers. Therefore, this approach also leads to inconsistency.Another way: maybe using the cosine law. For the triangle with sides 7,8,11, let's compute the angles. For example, the angle opposite the 11 cm side:cosθ = (7² + 8² - 11²)/(2*7*8) = (49 + 64 - 121)/112 = (-8)/112 = -1/14 ≈ -0.0714So the angle is obtuse, about 95.7 degrees.If we consider the triangle formed by cutting the cube's corner, the triangle's angles relate to the positions where the plane cuts the edges. But since the original edges are orthogonal, maybe the angles of the triangle can't be obtuse? Wait, no. The triangle formed can have any angles depending on where the plane cuts the edges. For example, if you cut very close to the corner, the triangle would be small and probably acute. If you cut further out, the angles could vary.But in this case, we have an angle of about 95.7 degrees, which is obtuse. Is that possible? Let's think about the triangle in 3D space. The three edges of the triangle are each along different faces of the cube. The angles between the edges of the triangle correspond to the dihedral angles of the cube? Not exactly, because the triangle is in a plane cutting through the cube.Wait, perhaps the angles of the triangle are related to the angles between the edges of the cube. Since the cube's edges are orthogonal, the vectors along the edges are perpendicular. The triangle's edges are vectors from (a,0,0) to (0,b,0), etc., which have components along different axes. The angles between these vectors can be calculated using the dot product.For example, the angle between AB and AC vectors. AB is (-a, b, 0), AC is (-a, 0, c). The dot product is (-a)(-a) + b*0 + 0*c = a². The magnitudes of AB and AC are √(a² + b²) and √(a² + c²). So,cosφ = a² / [√(a² + b²) * √(a² + c²)]But in our case, if we assign, say, AB = 7, BC = 8, AC = 11. Then, AB = √(a² + b²) = 7, BC = √(b² + c²) = 8, AC = √(a² + c²) = 11.Then, the angle at A (between AB and AC) is calculated as above. Plugging in:cosφ = a² / (7 * 11)But from a² + b² = 49 and a² + c² = 121, so c² = 121 - a², and b² = 49 - a². Then, from b² + c² = 8² = 64,(49 - a²) + (121 - a²) = 64 ⇒ 170 - 2a² = 64 ⇒ 2a² = 106 ⇒ a² = 53 ⇒ a = √53 ≈ 7.28 cmThen, b² = 49 - 53 = -4 ⇒ Again, impossible.So, even when calculating the angle, we end up with impossible values. Therefore, there's no solution.Another thought: maybe the problem is in 3D, and the triangle isn't restricted to having its vertices on the edges? But no, because cutting a corner with a plane must intersect the three edges meeting at that corner. Otherwise, it wouldn't be cutting off the corner but rather slicing through somewhere else.Therefore, all possible configurations lead to contradictions, meaning it's impossible to have such a triangular cut.

✅:To determine if a corner can be cut off a 20 cm cube resulting in a triangular face with sides 7 cm, 8 cm, and 11 cm, we analyze the geometric constraints:1. Vertices on Cube Edges: Cutting off a corner requires the triangular face's vertices to lie on the three edges meeting at that corner. Let these points be ( (a,0,0) ), ( (0,b,0) ), and ( (0,0,c) ).2. Side Lengths as Distances: - The distances between these points must match the given side lengths: [ sqrt{a^2 + b^2}, quad sqrt{b^2 + c^2}, quad sqrt{a^2 + c^2} ] - Assigning the sides 7, 8, and 11 cm to these expressions in all permutations leads to systems of equations.3. Solving the Equations: - For each permutation of assigning 7, 8, 11 to the distances, solving the resulting quadratic equations consistently yields negative values for one of ( a^2 ), ( b^2 ), or ( c^2 ), which is impossible since squared lengths cannot be negative.4. Geometric Consistency Check: - Heron's formula confirms the triangle's area is valid, but vector cross-product calculations based on the cube's edge intersections also fail to reconcile with real, positive edge cuts.Conclusion: No valid positive real solutions ( (a, b, c) ) exist within the cube's 20 cm edges that satisfy the triangle's side lengths. Therefore, it is not possible to cut off a corner from the cube to form such a triangle.Final Answer: boxed{text{No}}

🔑:The problem asks whether it is possible to cut a corner off a cube with edge lengths of 20 cm such that the resulting section has the shape of a triangle with sides 7 cm, 8 cm, and 11 cm.1. Reformulate the question: To decide if a section of the cube can be such a triangle, we need to determine if a triangular pyramid with right angles at the vertex and a base triangle with sides 7 cm, 8 cm, and 11 cm can exist.2. Identify the lengths of the edges of the triangular pyramid: Let the lengths of the edges from the vertex to the base be denoted as (a), (b), and (c). The triangle at the base has sides 7 cm, 8 cm, and 11 cm.3. Use the Pythagorean theorem: Since the pyramid has right angles at the vertex, we apply the Pythagorean theorem to find relations between (a), (b), and (c): [ a^2 + b^2 = 7^2 quadRightarrowquad a^2 + b^2 = 49, ] [ b^2 + c^2 = 8^2 quadRightarrowquad b^2 + c^2 = 64, ] [ c^2 + a^2 = 11^2 quadRightarrowquad c^2 + a^2 = 121. ]4. Sum the equations to find the combined squared length: Adding these three equations yields: [ (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) = 49 + 64 + 121. ] Simplify the terms: [ 2a^2 + 2b^2 + 2c^2 = 234 quad Rightarrow quad a^2 + b^2 + c^2 = 117. ]5. Check for contradictions: We have found: [ a^2 + b^2 + c^2 = 117. ] However, comparing this result with the given (a^2 + c^2 = 121) (from the Pythagorean relations for the sides of the triangle), we get: [ a^2 + c^2 + b^2 = 117 < 121 = a^2 + c^2. ] This inequality is a contradiction because it suggests that (b^2) would be a negative value, which is impossible since square terms are always non-negative.Conclusion:Therefore, it is impossible to obtain such a triangular section from a cube by cutting a corner. [boxed{text{No}}]

🧠:Okay, let me try to figure out this problem. Hmm, it's about a motorcycle and a car traveling different distances in the same time, and then there's a part where they start at different times and we have to find when the car catches up. Let me break it down step by step.First, the problem says that a motorcycle travels 120 kilometers and a car travels 180 kilometers in the same amount of time. So, if I let the time be 't' hours, then the speed of the motorcycle would be 120/t km/h, and the speed of the car would be 180/t km/h. That makes sense because speed is distance divided by time.Then, the second part says that in 7 hours, the distance traveled by the motorcycle is 80 kilometers less than the distance traveled by the car in 6 hours. Let me write that out. So, distance is speed multiplied by time. So, for the motorcycle: speed_motorcycle * 7 hours = distance_m, and for the car: speed_car * 6 hours = distance_c. The problem states that distance_m = distance_c - 80 km.Wait, but we already have expressions for speed_motorcycle and speed_car from the first part. Let me substitute those in. So, substituting:speed_motorcycle = 120/tspeed_car = 180/tThen, according to the second condition:120/t * 7 = (180/t * 6) - 80Let me compute that equation. Multiply through by t to eliminate denominators:120*7 = 180*6 - 80tCalculating the left side: 120*7 is 840. Right side: 180*6 is 1080. So,840 = 1080 - 80tSubtract 1080 from both sides:840 - 1080 = -80tWhich is -240 = -80tDivide both sides by -80:t = 3 hours.Wait, so t is 3 hours. Then, the speeds would be:speed_motorcycle = 120/3 = 40 km/hspeed_car = 180/3 = 60 km/hLet me check the second condition to make sure. If the motorcycle travels for 7 hours at 40 km/h, that's 280 km. The car travels for 6 hours at 60 km/h, that's 360 km. 280 is indeed 80 less than 360. So that checks out. Good, so the speeds are 40 and 60 km/h.Now, the main question: If the motorcycle departs 2 hours before the car from the same starting point, how many hours will it take for the car to catch up with the motorcycle after it starts?Okay, so motorcycle has a 2-hour head start. In those 2 hours, it's traveling at 40 km/h, so it covers 40*2 = 80 km before the car even starts.Once the car starts, it's going at 60 km/h, and the motorcycle continues at 40 km/h. The question is, how long after the car starts will it catch up to the motorcycle.This is a relative speed problem. The car is going faster, so it's closing the gap between them. The gap at the time the car starts is 80 km. The car's speed relative to the motorcycle is 60 - 40 = 20 km/h. So, to cover the 80 km gap at 20 km/h, it would take 80 / 20 = 4 hours.Wait, that seems straightforward, but let me make sure.Alternatively, let's model it with equations.Let t be the time in hours after the car starts when they meet.In that time t, the car travels 60t km.The motorcycle has already been traveling for 2 + t hours (2 hours before the car starts plus t hours after), so it travels 40*(2 + t) km.When they meet, their distances are equal:60t = 40*(2 + t)Expanding the right side:60t = 80 + 40tSubtract 40t from both sides:20t = 80t = 4 hours.Yes, that confirms it. So, the answer should be 4 hours.But let me double-check all steps again. First, found the speeds correctly? Yes, 40 and 60 km/h. Then, head start distance: 2 hours * 40 = 80 km. Relative speed: 20 km/h. Time to close 80 km: 4 hours. Equations also led to t = 4. So, seems correct.Is there another way to approach this? Maybe using distance-time graphs, but that might be overcomplicating. The key is that the motorcycle has a head start, and the car is faster. The time to catch up depends on the head start distance and the relative speed.Alternatively, imagine that when the car starts, the motorcycle is 80 km ahead. Each hour, the car gains 20 km on the motorcycle. So, 80 / 20 = 4. Same result. All methods point to 4 hours. I think that's solid.

✅:To solve the problem, we first determine the speeds of the motorcycle and car using the given distances and time. Let ( t ) be the common time taken to travel 120 km (motorcycle) and 180 km (car). Their speeds are:[text{Speed of motorcycle} = frac{120}{t}, quad text{Speed of car} = frac{180}{t}.]Next, use the second condition: In 7 hours, the motorcycle travels 80 km less than the car does in 6 hours:[frac{120}{t} cdot 7 = frac{180}{t} cdot 6 - 80.]Simplify:[840 = 1080 - 80t implies 80t = 240 implies t = 3 text{ hours}.]Thus, the speeds are:[text{Motorcycle speed} = frac{120}{3} = 40 , text{km/h}, quad text{Car speed} = frac{180}{3} = 60 , text{km/h}.]When the motorcycle departs 2 hours earlier, it covers:[40 , text{km/h} times 2 , text{h} = 80 , text{km}.]Let ( t ) be the time (in hours) after the car starts until they meet. The distances traveled are:[text{Motorcycle: } 40(t + 2), quad text{Car: } 60t.]Set distances equal:[60t = 40(t + 2) implies 60t = 40t + 80 implies 20t = 80 implies t = 4.]Answer: The car will catch up to the motorcycle 4 hours after it starts.

🔑:1. Determine the relationship between motorcycle and car speeds: Given that the motorcycle travels 120 km and the car travels 180 km in the same amount of time, the ratio of their speeds (motorcycle speed to car speed) is: [ text{Motorcycle speed} : text{Car speed} = 120 : 180 = frac{120}{180} = frac{2}{3} ] Thus, let the speed of the motorcycle be (2x) km/hr and the speed of the car be (3x) km/hr.2. Find the common distance factor: Within 7 hours, the distance traveled by the motorcycle is 80 km less than the distance traveled by the car within 6 hours. Therefore: [ text{Motorcycle distance in 7 hours} = text{Car distance in 6 hours} - 80 text{ km} ] Using their speeds: [ 7 times 2x = 6 times 3x - 80 ] Simplify and solve for (x): [ 14x = 18x - 80 ] [ 80 = 4x ] [ x = 20 ]3. Determine speeds of the motorcycle and car: [ text{Motorcycle speed} = 2x = 2 times 20 = 40 text{ km/hr} ] [ text{Car speed} = 3x = 3 times 20 = 60 text{ km/hr} ]4. Time analysis for catching up: The motorcycle starts 2 hours earlier than the car. In these 2 hours, the motorcycle travels: [ text{Distance traveled by motorcycle in 2 hours} = 40 text{ km/hr} times 2 text{ hours} = 80 text{ km} ] Therefore, the car needs to catch up with an 80 km lead. The relative speed of catching up is: [ text{Relative speed} = text{Car speed} - text{Motorcycle speed} = 60 text{ km/hr} - 40 text{ km/hr} = 20 text{ km/hr} ] The time it takes for the car to catch up: [ text{Time to catch up} = frac{text{Lead distance}}{text{Relative speed}} = frac{80 text{ km}}{20 text{ km/hr}} = 4 text{ hours} ] Conclusion: [boxed{4 text{ hours}}]