AMC 8 2004 Problem 19

AMC8 2004 年第 19 题

余数与同余★★★☆☆

A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5, and 6 . The smallest such number lies between which two numbers?

一个大于 2 的正整数分别除以 3，4，5 和 6，所得余数均为 2。那么这个正整数的最小可能值在哪两个数之间?

A. $40\ \text{and}49$
B. $60 \text{ and } 79$
C. $100\ \text{and}129$
D. $210\ \text{and}249$
E. $320\ \text{and}369$

正确答案 · Correct Answer

解析 · Solution

Solution 1

The smallest number divisible by 3,4,5, and 6, or their least common multiple, can be found to be 60. When 2 is added to a multiple of number, its remainder when divided by that number is 2. The number we are looking for is therefore 62, and between

60\ \text{and79}

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