AMC 8 2025 Problem 6
AMC8 2025 年第 6 题
余数与同余★☆☆☆☆
Sekou writes the numbers 15, 16, 17, 18, 19. After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 4. Which number did he erase?
Sekou写下数字15、16、17、18、19。他擦去其中一个数字后，剩下的四个数字之和是4的倍数。请问他擦去了哪个数字？
- A.15
- B.16
- C.17
- D.18
- E.19
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
The sum of all five numbers is 85. Since 85 is 1 more than a multiple of 4, the number being subtracted must be 1 more than a multiple of 4. Thus, the answer is. - Creativity 💡 Notice the 10 's place as you erase one number; you get 40, which is a multiple of 4, so you can ignore the 10 's place in the numbers provided altogether because it will always be a multiple of 40. This means that the only relevant numbers are the 1 's digit ones. If you add the numbers now, you get 5+6+7+8+9=35. The closest multiple of 4 underneath 35 which can be achieved as a result of subtracting either 5, 6, 7, 8, or 9 is 28, which is achieved by subtracting 7, therefore the answer is.
The sum of the residues of these numbers modulo 4 is . Hence, the number being subtracted must be congruent to 1 modulo 4. The only such answer is.
The sum of all the numbers here is equivalent to . Subtracting 17 gives us , which is clearly a multiple of 4. Therefore, the answer is.
Since the sum of is odd, we immediately exclude B and D. We further note that if A is true, then E is true. Hence, the answer is C.
We can use the fact that . Notice that 15, 16, 17, 18, 19 dividing by 4 have remainders 3, 0, 1, 2, 3. Their sum is 9. It is easy to see that 9-1=8 is divisible by 4 and so is the correct answer. edit by themathkidwoohooooo, just fixed the LaTeX for the answer
If we use brute force and add 15+16+17+18+19, we get 85. When we subtract every number from 85, we get 70, 69, 68, 67, 66, we see that 68 is divisible by 4, so is the correct answer.
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