AMC8 2006

AMC8 2006 · Q23

AMC8 2006 · Q23. It mainly tests GCD & LCM, Remainders & modular arithmetic.

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

一个盒子中有金币。如果金币平均分给六个人，分余4枚。如果平均分给五个人，分余3枚。盒子中满足这两个条件的最小金币数是多少？平均分给七个人时，余多少枚？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) The counting numbers that leave a remainder of 4 when divided by 6 are 4, 10, 16, 22, 28, 34, .... The counting numbers that leave a remainder of 3 when divided by 5 are 3, 8, 13, 18, 23, 28, 33, .... So 28 is the smallest possible number of coins that meets both conditions. Because $4\times 7=28$, there are no coins left when they are divided among seven people.

（A）把一个数除以 6 余 4 的正整数有 4，10，16，22，28，34，……；把一个数除以 5 余 3 的正整数有 3，8，13，18，23，28，33，……。因此，满足两个条件的最小硬币数是 28。因为 $4\times 7=28$，把这些硬币分给 7 个人时不会剩下硬币。

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