2017 AMC 8 第 17 题

先试着解答 2017 AMC 8 第 17 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2017 AMC 8 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

我有一些金币和一些空宝箱。我试着每个宝箱放九枚金币,但这样会剩下二个宝箱空着。于是我改为每个宝箱放六枚金币,但这样会剩下三枚金币。我有多少枚金币?

Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?

99

2727

4545

6363

8181

答案:C
知识点:方程组
难度评级:1240
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文字解答:

设宝箱数为 nn,金币数为 gg。 则 且 解这个方程组得 n=7n = 7。 因此金币数为 67+3=456 \cdot 7 + 3 = 459(n2)=g9(n - 2) = g 6n+3=g.6n + 3 = g.

所以正确答案是 C

Let nn be the number of treasure chests and gg be the number of gold coins. Then 9(n2)=g9(n - 2) = g and 6n+3=g.6n + 3 = g. Solving this system yields n=7,n = 7, so the number of gold coins is 67+3=45.6 \cdot 7 + 3 = 45.

Thus, C is the correct answer.

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