AMC 8 2018 Problem 21

AMC8 2018 年第 21 题

余数与同余★★★☆☆
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
有多少个 3 位正整数,满足:当除以 6 余数为 2,当除以 9 余数为 5,当除以 11 余数为 7?
  1. A.1
  2. B.2
  3. C.3
  4. D.4
  5. E.5

正确答案 · Correct Answer

E

解析 · Solution

Solution 1

Looking at the values, we notice that 11-7=4, 9-5=4 and 6-2=4. This means we are looking for a value that is four less than a multiple of 11, 9, and 6. The least common multiple of these numbers is 11⋅32⋅2=19811\cdot3^{2}\cdot2=198, so the numbers that fulfill this can be written as 198k-4, where kk is a positive integer. This value is only a three-digit integer when kk is 1, 2, 3, 4 or 5, which gives 194, 392, 590, 788, and 986 respectively. Thus, we have 5 values, so our answer is 5.

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