Question

大于一的最小正整数在除以四、五、六时余数都是一。它位于下列哪一对数之间？

The smallest positive integer 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?

$2$ 和 $19$

$2$ and $19$

$20$ 和 $39$

$20$ and $39$

$40$ 和 $59$

$40$ and $59$

$60$ 和 $79$

$60$ and $79$

$80$ 和 $124$

$80$ and $124$

答案：D

知识点：最小公倍数模运算

难度评级：1020

视频讲解：

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文字解答：

如果一个数除以 $4$ 、 $5$ 和 $6$ 时余数都是 $1$ ，那么它比这些数的最小公倍数多一。最小公倍数是 $60$ ，所以最小的这类正整数是 $60 + 1 = 61$ 。

所以正确答案是 D。

If a number leaves a remainder of $1$ when divided by $4,$ $5,$ and $6,$ then it is one more than the least common multiple of these numbers. The least common multiple is $60,$ so the smallest such positive integer is $60 + 1 = 61.$

Thus, D is the correct answer.

2017 AMC 8 第 12 题

答案：D

难度评级：1020

视频讲解：

文字解答：

其他年份的第 12 题