AMC 8 2025 Problem 16
AMC8 2025 年第 16 题
基本计数原理★★★☆☆
Five distinct integers from 1 to 10 are chosen, and five distinct integers from 11 to 20 are chosen. No two numbers differ by exactly 10 . What is the sum of the ten chosen numbers?
从1到10之间选择五个不同的整数，从11到20之间选择五个不同的整数。其中，没有两个数的差值恰好为10。请问选出的这十个数之和是多少？
- A.95
- B.100
- C.105
- D.110
- E.115
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Note that for no two numbers to differ by 10, every number chosen must have a different units digit. To make computations easier, we can choose (1, 2, 3, 4, 5) from the first group and (16, 17, 18, 19, 20) from the second group. Then the sum evaluates to . Another Way To Compute For 1+2+3+4+5+16+17+18+19+20, we can add the first term and the last term, which is 21. If we add the second term and the second-to-last term it is also 21. There are 5 pairs that sum to 21, so the answer is 21×5 which equals . - leafy Similar solution One efficient method is to quickly add (1, 2, 3, 4, 5, 6, 7, 8, 9, 10), which is 55. Then because you took 50 in total away from (16, 17, 18, 19, 20), you add 50. . To solve this problem, I started with the easiest/smallest case possible. In my opinion, that was 1+2+3+4+5+16+17+18+19+20. I then solved this equation quickly using Little Gauss's method, rearranging that into 1+19+2+18+3+17+4+16+5+20. I simplified this into 20+20+20+20+25. Solving this simple equation gives us the answer, which is . edit by NYCnerd Start with 1. If you choose it, you cannot choose 11, so you must now pick 12. If you choose 12, then you can't choose 2, so you must now pick 13. If you choose 13, you cannot choose 3, so you must pick 4. If you choose 4, then you cannot choose 14 so you have to go to 15. Then you can't choose 5 so you have to go to 6. Then you can't choose 16, so you have to go to 17. Then you can't choose 7, so you have to choose 18. Then you can't choose 8, so you have to go to 9. Finally, you cannot choose 19, so you must go to 20. This makes the equation 1+12+3+14+5+16+7+18+9+20 = 105, hence . Sorry if it is a bit confusing its just how I solved it so wanted to share! -N.
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