AMC 8 2017 Problem 12
AMC8 2017 年第 12 题
整除与因数★★★☆☆
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?
大于 1 且除以 4,5 和 6 所得余数均为 1 的最小正整数在下面哪两个数之间？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
Since the remainder is the same for all numbers, then we will only need to find the lowest common multiple of the three given numbers, and add the given remainder.The is 60. Since 60+1=61, that is in the range of
Call the number we want to find . We can say that We can also say that n-1 is divisible by 4,5, and 6. Therefore, n-1=lcm(4,5,6)=60, so n=60+1=61 which is in the range of
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