2017 AMC 12A 第 12 题

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12.

1010 匹马,名字分别为 Horse 11 Horse 22 \ldots Horse 1010 它们的名字来自它们绕圆形跑道跑一圈所需的分钟数:Horse kk 恰好用 kk 分钟跑一圈。在时刻 00,所有马都在跑道起点。它们沿同一方向开始奔跑,并以各自恒定速度一直绕跑道跑。所有 1010 匹马再次同时到达起点的最小正时间(分钟)为 S>0S\gt0,且 S=2520S=2520。设 T>0T\gt0 是至少 55 匹马再次同时在起点的最小正时间(分钟)。TT 的各位数字之和是多少?

There are 1010 horses, named Horse 1,1, Horse 2,2, ,\ldots, Horse 10.10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse kk runs one lap in exactly kk minutes. At time 00 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time S>0,S\gt0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520.S=2520. Let T>0T\gt0 be the least time, in minutes, such that at least 55 of the horses are again at the starting point. What is the sum of the digits of T?T?

22

33

44

55

66

答案:B
知识点:最小公倍数整除性
难度评级:1630
解答:

Horse kk 在时刻 tt 位于起点,恰好当 ktk\mid t。 因此我们需要最小的 tt, 使它在 1,2,,101,2,\ldots,10 中至少有 55 个因数。

检查较小的数,1212 可被 1,2,3,41,2,3,466 整除,对应恰好 44 匹马;没有更小的 1212 能使 匹马同时到达起点。因此 T=12T=12,它的各位数字之和是 1+2=31+2=3

所以正确答案是 B

Horse kk is at the starting point at time tt precisely when kt.k\mid t. So we want the smallest tt with at least 55 divisors among 1,2,,10.1,2,\ldots,10.

The positive integers below 1212 have at most 44 divisors, while 1212 is divisible by 1,2,3,4,1,2,3,4, and 6.6. Thus T=12,T=12, and the sum of its digits is 1+2=3.1+2=3.

Thus, the correct answer is B.

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