2012 AMC 10B 第 4 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

4.

Ringo 把弹珠每六个装一袋,会剩下四个。Paul 同样装袋会剩下三个。两人把弹珠合在一起,仍按每袋六个尽可能装满,最后会剩下多少个?

When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?

11

22

33

44

55

答案:A
知识点:模运算
难度评级:770
解答:

Ringo 的弹珠数除以 6644,所以可写成 6x+46x+4,其中 xx 为整数。

同理,Paul 的弹珠数可写为 6y+36y+3,其中 yy 为整数。

因此总数为 这表示合在一起后除以 6611(6x+4)+(6y+3)(6x+4)+(6y+3) =6(x+y+1)+1= 6(x+y+1)+1

所以正确答案是 A

As we know that when Ringo's marbles are divided by 6,6, we have a remainder of 4,4, we conclude that he has 6x+46x+4 marbles for some x.x.

Using the same logic, we can also conclude that Paul has 6y+36y+3 marbles for some y.y.

Therefore, the total number of marbles is (6x+4)+(6y+3)(6x+4)+(6y+3) =6(x+y+1)+1= 6(x+y+1)+1 which, when divided by 6,6, only leaves 11 left over.

Thus, the correct answer is A .

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