2012 AMC 8 第 15 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

15.

大于 22 且除以 33445566 时余数都为 22 的最小数,位于哪两个数之间?

The smallest number greater than 22 that leaves a remainder of 22 when divided by 3,3, 4,4, 5,5, or 66 lies between what numbers?

40405050

4040 and 5050

51515555

5151 and 5555

56566060

5656 and 6060

61616565

6161 and 6565

66669999

6666 and 9999

答案:D
知识点:最小公倍数模运算
难度评级:1240
小提示:

从未知数中减去 22

Subtract 22 from the unknown number.

大提示:

结果必须是 33445566 的公倍数。

The result must be a common multiple of 33, 44, 55, and 66.

视频讲解:
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文字解答:

设这个数为 xx。因为它除以 33445566 时余数都是 22,所以 x2x-233445566 的公倍数。这表示 x2x-2lcm(3,4,5,6)\operatorname{lcm}(3,4,5,6) 的倍数,而这个最小公倍数是 6060。因此 x2x-2 必须是 6060 的倍数。下一个满足条件的数出现在 x2=60    x=62x-2 = 60 \implies x = 62 时。

所以正确答案是 D

Let the number be x.x. Since it leaves a remainder of 22 when divided by 33, 44, 55, and 6,6, we know x2x-2 is a multiple of 33, 44, 55, and 6.6. This means x2x-2 is a multiple of lcm(3,4,5,6)\operatorname{lcm}(3,4,5,6), which is 60.60. Therefore, x2x-2 must be a multiple of 60.60. The next number such that this occurs is when x2=60    x=62.x-2 = 60 \implies x = 62 .

Thus, the answer is D .

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