AMC 8 2006 Problem 23
AMC8 2006 年第 23 题
余数与同余★★★☆☆
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?
一个盒子内装有金币。若把这些金币均分给 6 个人，那么还剩 4 枚金币。若把这些金币均分给 5 个人，那么还剩 3 枚金币。若盒子里装有的金币个数是满足以上条件的最少的金币个数，那么当把这些金币均分给 7 个人，还剩多少枚金币？
- A.0
- B.1
- C.2
- D.3
- E.5
正确答案 · Correct Answer
A
解析 · Solution
Solution 1
The counting numbers that leave a remainder of 4 when divided by 6 are The counting numbers that leave a remainder of 3 when divided by 5 are So 28 is the smallest possible number of coins that meets both conditions. Because 28 is divisible by 7, there are 0 coins left when they are divided among seven people.
If there were two more coins in the box, the number of coins would be divisible by both 6 and 5. The smallest number that is divisible by 6 and 5 is 30, so the smallest possible number of coins in the box is 28 and the remainder when divided by 7 is 0.
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