2017 AMC 8 第 12 题

Question

在大于

1

的正整数中，除以

4

、

5

和

6

时余数都是

1

的最小的那个数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

小提示：

除以

4

、

5

和

6

都余

1

的数，比这些数的最小公倍数的倍数多一

Numbers with remainder

1

after division by

4,

5,

and

6

are one more than multiples of their least common multiple.

大提示：

求

4

、

5

和

6

的最小公倍数，再加

1

Find the least common multiple of

4,

5,

and

6

, then add

1

.

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.