AMC8 2017

AMC8 2017 · Q12

AMC8 2017 · Q12. It mainly tests GCD & LCM, Remainders & modular arithmetic.

The smallest positive whole number greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

大于1的最小的正整数，除以4、5、6时余数均为1，它位于下列哪一对数的之间？

(A) 2 and 19 2 和 19

(B) 20 and 39 20 和 39

(D) 60 and 79 60 和 79

(E) 80 and 124 80 和 124

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The least common multiple of 4, 5, and 6 is 60. Numbers that leave a remainder of 1 when divided by 4, 5, and 6 are 1 more than a whole number multiple of 60. So the smallest positive number greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 is 61.

Answer (D): 4、5、6 的最小公倍数是 60。被 4、5、6 除都余 1 的数，等于 60 的整数倍再加 1。因此，大于 1 且被 4、5、6 除都余 1 的最小正整数是 61。

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