| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to use the Pigeonhole Principle, which states that if there are more items than containers, then at least one container must contain more than one item. First, let's consider the possible remainders when a positive integer is divided by 10. There are 10 possible remainders: 0, 1, 2, ..., 8, 9. We can think of these remainders as 10 "pigeonholes". Now, let's consider the set of 20 positive integers. We want to find pairs of numbers whose difference is a multiple of 10. A difference is a multiple of 10 if and only if the two numbers have the same remainder when divided by 10. So, we want to find pairs of numbers that have the same remainder. If there are at least two numbers in the set with the same remainder, then their difference is a multiple of 10. If there are no pairs of numbers with the same remainder, then we have at least one number in each of the 10 pigeonholes. But since we have 20 numbers, we must have at least two numbers in at least one of the pigeonholes by the Pigeonhole Principle. Therefore, there must be at least one pair of numbers in the set whose difference is a multiple of 10. But we can do better than that. If we have two numbers with the same remainder, then there are at least two pairs of numbers whose difference is a multiple of 10: the pair consisting of those two numbers, and the pair consisting of those two numbers and any other number in the set with a different remainder. So, let's assume that there is no pair of numbers with the same remainder. Then we have at least one number in each of the 10 pigeonholes. But since we have 20 numbers, we must have at least two numbers in at least one of the pigeonholes by the Pigeonhole Principle. Let's say that the two numbers with the same remainder are in pigeonhole i. Then we have at least two pairs of numbers whose difference is a multiple of 10: the pair consisting of those two numbers, and the pair consisting of those two numbers and any other number in the set with a remainder of j, where j is not equal to i. This gives us at least two pairs of numbers whose difference is a multiple of 10. But we can do even better than that. If we have two numbers with the same remainder in pigeonhole i, and two numbers with the same remainder in pigeonhole j, then we have at least four pairs of numbers whose difference is a multiple of 10: the two pairs consisting of the numbers in pigeonholes i and j, and the two pairs consisting of the numbers in pigeonholes i and j and any other number in the set with a remainder of k, where k is not equal to i or j. This gives us at least four pairs of numbers whose difference is a multiple of 10. We can continue this argument, and we will find that if we have at least two numbers in each of the 10 pigeonholes, then we have at least 10 pairs of numbers whose difference is a multiple of 10. This is because we can choose two numbers from each pigeonhole, and the differences between those numbers will all be multiples of 10. So, the answer to the problem is that in a set of 20 positive integers, there must be at least 10 pairs of numbers whose difference is a multiple of 10. |
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| deepseekr1 | To determine the minimum number of pairs of numbers in a set of 20 positive integers where the difference is a multiple of 10, we use the pigeonhole principle and consider the remainders when these numbers are divided by 10. 1. **Possible Remainders**: When any integer is divided by 10, the possible remainders are 0 through 9, giving 10 possible remainders (pigeonholes). 2. **Distribution of Numbers**: With 20 numbers (pigeons) and 10 remainders, by the pigeonhole principle, each remainder must be assigned at least \(\left\lfloor \frac{20}{10} \right\rfloor = 2\) numbers. This is the most even distribution. 3. **Pairs Calculation**: For each remainder, if there are \(n_i\) numbers, the number of pairs is \(\binom{n_i}{2}\). The goal is to minimize the total number of pairs, which occurs when the numbers are distributed as evenly as possible. 4. **Even Distribution**: If each remainder has exactly 2 numbers, the number of pairs for each remainder is \(\binom{2}{2} = 1\). With 10 remainders, the total number of pairs is \(10 \times 1 = 10\). 5. **Verification**: Any other distribution with some remainders having more than 2 numbers would result in more pairs due to the combinatorial increase in pairs (e.g., 3 numbers in a remainder give \(\binom{3}{2} = 3\) pairs), thus increasing the total number of pairs beyond 10. Therefore, the minimum number of pairs of numbers with a difference that is a multiple of 10 is \(\boxed{10}\). |
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| gpto1 | To solve this problem, we need to determine the minimum number of pairs among 20 positive integers where the difference between the numbers in each pair is a multiple of 10. This occurs precisely when two numbers are congruent modulo 10, meaning they leave the same remainder when divided by 10. **Step 1: Understanding Modulo 10 Residues** Every positive integer, when divided by 10, leaves a remainder between 0 and 9. This remainder is called the residue modulo 10. There are 10 possible residues: 0, 1, 2, ..., 9. **Step 2: Grouping Numbers by Their Residue** We can group the 20 positive integers based on their residues modulo 10. Let \( S_i \) represent the number of integers with residue \( i \) (where \( i = 0, 1, ..., 9 \)). Therefore: \[ S_0 + S_1 + S_2 + S_3 + S_4 + S_5 + S_6 + S_7 + S_8 + S_9 = 20 \] **Step 3: Calculating Pairs with Differences Divisible by 10** Two numbers will have a difference divisible by 10 if they belong to the same residue group. The number of such pairs in residue group \( i \) is given by: \[ \text{Number of pairs in group } i = \binom{S_i}{2} = \frac{S_i(S_i - 1)}{2} \] **Step 4: Minimizing the Total Number of Such Pairs** Our goal is to minimize the total number of these pairs across all residue groups: \[ T = \sum_{i=0}^{9} \binom{S_i}{2} = \sum_{i=0}^{9} \frac{S_i(S_i - 1)}{2} \] Since the function \( f(S) = \frac{S(S - 1)}{2} \) increases rapidly with \( S \), to minimize \( T \), we should keep the sizes of the residue groups as small as possible. **Step 5: Distributing the Numbers Evenly** To achieve the minimum \( T \), we distribute the 20 numbers evenly among the 10 residue classes: \[ S_i = 2 \text{ for each } i = 0, 1, ..., 9 \] **Step 6: Calculating the Minimum Total Pairs** With \( S_i = 2 \) for all \( i \): \[ \binom{S_i}{2} = \frac{2(2 - 1)}{2} = 1 \] Therefore, the total minimum number of pairs is: \[ T = \sum_{i=0}^{9} 1 = 10 \] **Conclusion** The minimum number of pairs with differences divisible by 10 in any set of 20 positive integers is **10**. This occurs when each residue class contains exactly 2 numbers. **Answer:** 10 |
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| GPT-4o | To solve this problem, we can use the pigeonhole principle. The principle states that if \( n \) items are put into \( m \) containers, with \( n > m \), then at least one container must contain more than one item. Let's consider the remainders when the numbers are divided by 10. The possible remainders when a number is divided by 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. There are 10 different possible remainders. In this problem, we have 20 positive integers. When we divide each integer by 10, we will place each remainder into one of 10 "containers" (each container representing one of the possible remainders: 0 through 9). Using the pigeonhole principle, since we have 20 numbers (pigeons) and 10 possible remainders (pigeonholes), at least one remainder must be repeated at least \(\lceil \frac{20}{10} \rceil = 2\) times. This means at least 20 - 10 = 10 remainders must be repeated. To find the minimum number of pairs with differences that are multiples of 10, we consider that each repeated remainder forms pairs with every other number having the same remainder. If a particular remainder \( r \) occurs \( k \) times, the number of pairs of numbers having the same remainder is given by the combination formula \( \binom{k}{2} = \frac{k(k-1)}{2} \). To minimize the number of pairs, we distribute the numbers as evenly as possible among the remainders. This gives us two remainders with 2 numbers each and the rest with 1 number each. For each of the remainders with 2 numbers, we have: \[ \binom{2}{2} = 1 \] Since there are 2 such remainders: \[ 2 \times 1 = 2 \] Thus, the minimum number of pairs of numbers whose difference is a multiple of 10 is \( \boxed{10} \). |
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| llama3.1-70B-Instruct | To find the minimum number of pairs of numbers with a difference that is a multiple of 10, we can use the Pigeonhole Principle. Let's consider the remainders of the 20 positive integers when divided by 10. There are 10 possible remainders: 0, 1, 2, ..., 9. Since there are 20 integers, at least two of them must have the same remainder when divided by 10 (by the Pigeonhole Principle). Now, let's consider the differences between these two integers. Since they have the same remainder when divided by 10, their difference must be a multiple of 10. Therefore, we can conclude that there is at least one pair of numbers with a difference that is a multiple of 10. However, we can do better than this. Let's consider the 10 possible remainders again. Since there are 20 integers, at least two of them must have the same remainder when divided by 10, and at least two of them must have a remainder that is 5 more or 5 less than the first remainder (again by the Pigeonhole Principle). For example, if two integers have a remainder of 3 when divided by 10, then at least two integers must have a remainder of 8 (which is 5 more than 3) or 13 (which is 5 less than 3, but equivalent to 3 modulo 10). Now, consider the differences between these two pairs of integers. The difference between the two integers with remainder 3 is a multiple of 10, and the difference between the two integers with remainder 8 (or 13) is also a multiple of 10. Therefore, we can conclude that there are at least two pairs of numbers with a difference that is a multiple of 10. In fact, we can show that there are at least 10 pairs of numbers with a difference that is a multiple of 10. To see this, consider the 10 possible remainders again. For each remainder, there are at least two integers with that remainder (by the Pigeonhole Principle). Therefore, there are at least 10 pairs of integers with the same remainder, and each of these pairs has a difference that is a multiple of 10. Therefore, the minimum number of pairs of numbers with a difference that is a multiple of 10 is 10. |
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| llama3.1-405B-Instruct | To find the minimum number of pairs of numbers in a set of 20 positive integers that have a difference that is a multiple of 10, we can use the following approach: 1. Divide the numbers into 10 groups based on their remainders when divided by 10. For example, group 1 would contain numbers that leave a remainder of 1 when divided by 10, group 2 would contain numbers that leave a remainder of 2 when divided by 10, and so on. 2. Since there are 20 numbers in total, each group can have at most 2 numbers. If a group has more than 2 numbers, then there would be at least one pair of numbers in that group that have a difference that is a multiple of 10. 3. Since there are 10 groups, the maximum number of pairs of numbers that have a difference that is a multiple of 10 is 10 x 2 = 20. Therefore, the minimum number of pairs of numbers in a set of 20 positive integers that have a difference that is a multiple of 10 is 20. |
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| DeepSeek-Coder-V2 | To solve the problem of finding the minimum number of pairs of numbers in a set of 20 positive integers that have a difference that is a multiple of 10, we can use the Pigeonhole Principle. ### Step-by-Step Solution: 1. **Understanding the Problem:** - We have a set of 20 positive integers. - We need to find the minimum number of pairs of these integers such that the difference between the pairs is a multiple of 10. 2. **Key Insight:** - The difference between two numbers is a multiple of 10 if and only if they leave the same remainder when divided by 10. - This means we can categorize the numbers based on their remainders when divided by 10 (i.e., 0, 1, 2, ..., 9). 3. **Applying the Pigeonhole Principle:** - There are 10 possible remainders (0 through 9). - We have 20 numbers to distribute among these 10 categories. 4. **Minimum Pairs Calculation:** - By the Pigeonhole Principle, if we have more items (numbers) than containers (remainders), at least one container must contain more than one item. - Specifically, with 20 numbers and 10 remainders, at least one remainder category will contain at least \(\left\lceil \frac{20}{10} \right\rceil = 2\) numbers. 5. **Conclusion:** - Since at least one remainder category will contain at least 2 numbers, there must be at least one pair of numbers in the set that have the same remainder when divided by 10. - Therefore, the difference between these two numbers will be a multiple of 10. Thus, the minimum number of pairs of numbers in the set that have a difference that is a multiple of 10 is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | To solve this problem, we need to use the Pigeonhole Principle, which states that if there are more items than containers, then at least one container must contain more than one item. First, let's consider the possible remainders when a positive integer is divided by 10. There are 10 possible remainders: 0, 1, 2, ..., 8, 9. We can think of these remainders as 10 "pigeonholes". Now, let's consider the set of 20 positive integers. We want to find pairs of numbers whose difference is a multiple of 10. A difference is a multiple of 10 if and only if the two numbers have the same remainder when divided by 10. So, we want to find pairs of numbers that have the same remainder. If there are at least two numbers in the set with the same remainder, then their difference is a multiple of 10. If there are no pairs of numbers with the same remainder, then we have at least one number in each of the 10 pigeonholes. But since we have 20 numbers, we must have at least two numbers in at least one of the pigeonholes by the Pigeonhole Principle. Therefore, there must be at least one pair of numbers in the set whose difference is a multiple of 10. But we can do better than that. If we have two numbers with the same remainder, then there are at least two pairs of numbers whose difference is a multiple of 10: the pair consisting of those two numbers, and the pair consisting of those two numbers and any other number in the set with a different remainder. So, let's assume that there is no pair of numbers with the same remainder. Then we have at least one number in each of the 10 pigeonholes. But since we have 20 numbers, we must have at least two numbers in at least one of the pigeonholes by the Pigeonhole Principle. Let's say that the two numbers with the same remainder are in pigeonhole i. Then we have at least two pairs of numbers whose difference is a multiple of 10: the pair consisting of those two numbers, and the pair consisting of those two numbers and any other number in the set with a remainder of j, where j is not equal to i. This gives us at least two pairs of numbers whose difference is a multiple of 10. But we can do even better than that. If we have two numbers with the same remainder in pigeonhole i, and two numbers with the same remainder in pigeonhole j, then we have at least four pairs of numbers whose difference is a multiple of 10: the two pairs consisting of the numbers in pigeonholes i and j, and the two pairs consisting of the numbers in pigeonholes i and j and any other number in the set with a remainder of k, where k is not equal to i or j. This gives us at least four pairs of numbers whose difference is a multiple of 10. We can continue this argument, and we will find that if we have at least two numbers in each of the 10 pigeonholes, then we have at least 10 pairs of numbers whose difference is a multiple of 10. This is because we can choose two numbers from each pigeonhole, and the differences between those numbers will all be multiples of 10. So, the answer to the problem is that in a set of 20 positive integers, there must be at least 10 pairs of numbers whose difference is a multiple of 10. |
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| baichuan4 | To find the minimum number of pairs of numbers with a difference that is a multiple of 10, we can consider the most extreme distribution of numbers where the smallest possible number of such pairs exist. Let's start by listing out the first 20 positive integers: 1, 2, 3, ..., 20. Now, let's arrange these numbers in such a way that the difference between consecutive numbers is not a multiple of 10. We can do this by placing them in increasing order, but skipping every 10th number. This way, the difference between any two adjacent numbers will not be a multiple of 10. Our list would then look like this: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22 Note that we have skipped 10 and 20 to avoid having differences that are multiples of 10. Now, let's count the number of pairs with a difference that is a multiple of 10: - 1 and 11 (difference = 10) - 11 and 21 (difference = 10) - 21 and 31 (not in our set, but if we continued the pattern, this would be another pair with a difference of 10) So, in this arrangement, we have 2 pairs with a difference that is a multiple of 10. However, we need to account for the fact that we have 20 numbers, and in our arrangement, we only have 19 (since we skipped 10 and 20). If we add back either 10 or 20, it will create at least one more pair with a difference of 10. For example, if we add 10 back to our list: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21 Now, we have the following additional pairs with a difference of 10: - 10 and 20 (not in our set, but if 20 was included, this would be a pair) - 10 and 20 (not in our set, but if 20 was included, this would be a pair) - 11 and 21 (existing pair) So, even if we add back 10 or 20, we still only have 2 pairs with a difference of 10. Therefore, in a set of 20 positive integers, the minimum number of pairs of numbers with a difference that is a multiple of 10 is 2. |
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| doubao | Consider the remainders when each number is divided by 10. There are only 10 possible remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. By the Pigeonhole Principle, if there are more than 10 numbers, then at least two of them must have the same remainder. Let's say two numbers have the same remainder, say a and b. Then their difference a - b is a multiple of 10. So, in a set of 20 positive integers, at least 10 pairs of numbers have a difference that is a multiple of 10. |
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| minimax | To solve this problem, we can use the Pigeonhole Principle, which states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Let's consider the set of 20 positive integers. We want to find out the minimum number of pairs that have a difference that is a multiple of 10. First, we can divide the set of integers into groups based on their remainder when divided by 10. There are 10 possible remainders: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Each integer in our set will fall into one of these 10 groups. Since we have 20 integers and only 10 groups, by the Pigeonhole Principle, at least one group must contain at least 3 integers (because if each group contained only 2 integers, we would have a maximum of 2*10 = 20 integers, but since we have exactly 20 integers, there must be at least one group with more than 2). Now, let's consider the integers in a group that has at least 3 integers. If we take any two integers from this group, their difference will be a multiple of 10 because they are in the same remainder group when divided by 10. Since we have at least one group with 3 integers, we can form the following pairs: - Pair 1: (Integer 1, Integer 2) - Pair 2: (Integer 1, Integer 3) - Pair 3: (Integer 2, Integer 3) That's a minimum of 3 pairs from just one group. Since we have 20 integers and at least one group with 3 integers, we can't have more than 2 groups with 3 integers each (otherwise, we would have more than 20 integers). Therefore, the minimum number of pairs with a difference that is a multiple of 10 is 3. However, if there are more groups with exactly 2 integers, we can form additional pairs from those groups as well. But since we are looking for the minimum number of such pairs, we don't need to consider those additional pairs. So, the answer to the problem is that in a set of 20 positive integers, there are at least 3 pairs of numbers that have a difference that is a multiple of 10. |
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| kimi | To solve this problem, we can use the Pigeonhole Principle, which states that if more items are put into fewer containers, then at least one container must contain more than one item. Here's the step-by-step solution: 1. **Divide the numbers into categories**: Since we're dealing with positive integers and multiples of 10, we can categorize each number based on its last digit (i.e., the remainder when divided by 10). There are 10 possible categories: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. 2. **Apply the Pigeonhole Principle**: With 20 numbers and only 10 categories, at least two numbers must fall into the same category because 20 divided by 10 is 2 with a remainder of 0. This means that at least one category will have at least 2 numbers. 3. **Calculate the minimum pairs**: If two numbers fall into the same category, their difference is a multiple of 10 (since their last digits are the same). So, for each category that has at least 2 numbers, we have at least 1 pair. 4. **Account for the remainder**: Since we have 20 numbers, after placing them into categories, we might have some categories with more than 2 numbers. For those categories, we can calculate additional pairs. For example, if a category has 3 numbers, we can have 3 pairs (1-2, 1-3, 2-3), and so on. 5. **Calculate the maximum number of pairs**: To find the maximum number of pairs, we can use the formula for combinations, which is \( \binom{n}{2} = \frac{n(n-1)}{2} \), where \( n \) is the number of items. For a category with \( n \) numbers, this formula gives us the number of pairs. 6. **Sum up the pairs**: To find the total number of pairs across all categories, we sum up the pairs from each category. However, to find the minimum number of pairs with differences that are multiples of 10, we only need to consider the first step, which guarantees at least one pair per category with at least two numbers. Since we have 10 categories and at least two numbers per category, we can immediately say that there are at least 10 pairs (one for each category with at least two numbers). Therefore, the minimum number of pairs of numbers with a difference that is a multiple of 10 in a set of 20 positive integers is at least 10. |
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| wenxinyiyan | 为了解决这个问题，我们可以考虑20个正整数中个位数的情况。 首先，我们观察这20个正整数的个位数。个位数只能是0到9这10个数字。由于有20个数，根据抽屉原理，至少有10个数的个位数是重复的。 接下来，我们考虑这10个个位数相同的数。由于它们的个位数相同，它们之间的差必然是10的倍数。因此，我们可以从这10个数中任意选出2个数，它们的差都是10的倍数。 这就意味着，至少存在10选2，即45对数的差是10的倍数。 因此，答案是：在20个正整数中，至少有45对数的差是10的倍数。 |
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| zhipu | To solve this problem, we need to consider the properties of numbers and their differences with respect to divisibility by 10. ### Key Insight: A difference between two numbers is a multiple of 10 if and only if the two numbers have the same remainder when divided by 10. This is because if \( a \equiv b \pmod{10} \), then \( a - b \equiv 0 \pmod{10} \). ### Step-by-Step Solution: 1. **Understanding Remainders:** When any integer is divided by 10, the possible remainders (also known as residues) are \( 0, 1, 2, 3, 4, 5, 6, 7, 8, \) and \( 9 \). There are exactly 10 possible remainders. 2. **Distributing 20 Numbers into Remainder Groups:** Given a set of 20 positive integers, we can distribute these numbers into 10 groups based on their remainders when divided by 10. Let's denote these groups as \( G_0, G_1, G_2, \ldots, G_9 \), where \( G_i \) contains all numbers in the set that have a remainder of \( i \) when divided by 10. 3. **Applying the Pigeonhole Principle:** The Pigeonhole Principle states that if \( n \) items are put into \( m \) containers, with \( n > m \), then at least one container must contain more than one item. In our case, we have 20 numbers (items) and 10 groups (containers). Therefore, by the Pigeonhole Principle, at least one of these groups must contain at least \( \lceil \frac{20}{10} \rceil = 2 \) numbers. 4. **Forming Pairs:** Each group \( G_i \) that contains at least two numbers will have pairs of numbers that differ by a multiple of 10. Specifically, if a group \( G_i \) contains \( k \) numbers, the number of ways to choose 2 numbers out of \( k \) (i.e., the number of pairs) is given by the combination formula \( \binom{k}{2} = \frac{k(k-1)}{2} \). 5. **Minimum Number of Pairs:** Since at least one group must contain at least 2 numbers, the minimum number of pairs of numbers that have a difference which is a multiple of 10 is at least 1. ### Conclusion: Therefore, in a set of 20 positive integers, at least **1 pair** of numbers will have a difference that is a multiple of 10. ### Final Answer: **At least 1 pair** of numbers in the set will have a difference that is a multiple of 10. |
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| gpto1 | To solve this problem, we need to determine the minimum number of pairs among 20 positive integers where the difference between the numbers in each pair is a multiple of 10. This occurs precisely when two numbers are congruent modulo 10, meaning they leave the same remainder when divided by 10. **Step 1: Understanding Modulo 10 Residues** Every positive integer, when divided by 10, leaves a remainder between 0 and 9. This remainder is called the residue modulo 10. There are 10 possible residues: 0, 1, 2, ..., 9. **Step 2: Grouping Numbers by Their Residue** We can group the 20 positive integers based on their residues modulo 10. Let \( S_i \) represent the number of integers with residue \( i \) (where \( i = 0, 1, ..., 9 \)). Therefore: \[ S_0 + S_1 + S_2 + S_3 + S_4 + S_5 + S_6 + S_7 + S_8 + S_9 = 20 \] **Step 3: Calculating Pairs with Differences Divisible by 10** Two numbers will have a difference divisible by 10 if they belong to the same residue group. The number of such pairs in residue group \( i \) is given by: \[ \text{Number of pairs in group } i = \binom{S_i}{2} = \frac{S_i(S_i - 1)}{2} \] **Step 4: Minimizing the Total Number of Such Pairs** Our goal is to minimize the total number of these pairs across all residue groups: \[ T = \sum_{i=0}^{9} \binom{S_i}{2} = \sum_{i=0}^{9} \frac{S_i(S_i - 1)}{2} \] Since the function \( f(S) = \frac{S(S - 1)}{2} \) increases rapidly with \( S \), to minimize \( T \), we should keep the sizes of the residue groups as small as possible. **Step 5: Distributing the Numbers Evenly** To achieve the minimum \( T \), we distribute the 20 numbers evenly among the 10 residue classes: \[ S_i = 2 \text{ for each } i = 0, 1, ..., 9 \] **Step 6: Calculating the Minimum Total Pairs** With \( S_i = 2 \) for all \( i \): \[ \binom{S_i}{2} = \frac{2(2 - 1)}{2} = 1 \] Therefore, the total minimum number of pairs is: \[ T = \sum_{i=0}^{9} 1 = 10 \] **Conclusion** The minimum number of pairs with differences divisible by 10 in any set of 20 positive integers is **10**. This occurs when each residue class contains exactly 2 numbers. **Answer:** 10 |
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| deepseekv3 | To determine the minimum number of pairs in a set of 20 positive integers where the difference between the numbers is a multiple of 10, we can use the **Pigeonhole Principle**. ### Step-by-Step Solution: 1. **Understanding the Problem**: - We need to find pairs of numbers \( (a, b) \) such that \( a - b \) is a multiple of 10. - This is equivalent to saying that \( a \) and \( b \) are congruent modulo 10, i.e., \( a \equiv b \pmod{10} \). 2. **Using the Pigeonhole Principle**: - The possible remainders when a number is divided by 10 are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. These are the **pigeonholes**. - We have 20 numbers (the **pigeons**) to distribute into these 10 pigeonholes. 3. **Calculating the Minimum Number of Pairs**: - By the Pigeonhole Principle, at least one pigeonhole must contain at least \( \lceil \frac{20}{10} \rceil = 2 \) numbers. - More generally, the minimum number of pairs is determined by the number of numbers that fall into the same pigeonhole. - Let \( k \) be the number of numbers in the most populated pigeonhole. Then, \( k \) must satisfy \( k \geq \lceil \frac{20}{10} \rceil = 2 \). - The number of pairs in a pigeonhole with \( k \) numbers is \( \binom{k}{2} \). - To minimize the total number of pairs, we should distribute the numbers as evenly as possible across the pigeonholes. 4. **Optimal Distribution**: - If we distribute the 20 numbers as evenly as possible across the 10 pigeonholes, each pigeonhole will contain either 2 or 3 numbers. - Specifically, 10 pigeonholes with 2 numbers each would give \( 10 \times \binom{2}{2} = 10 \) pairs. - If one pigeonhole has 3 numbers, it would contribute \( \binom{3}{2} = 3 \) pairs, while the others contribute 1 pair each, totaling \( 3 + 9 \times 1 = 12 \) pairs. - The minimum number of pairs occurs when the numbers are as evenly distributed as possible, which is 10 pairs. 5. **Conclusion**: - Therefore, the minimum number of pairs in a set of 20 positive integers where the difference between the numbers is a multiple of 10 is **10**. ### Final Answer: **At least 10 pairs of numbers in the set have a difference that is a multiple of 10.** |
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