1985 AMC 12 第 18 题

Question

六袋弹珠分别装有

18

、

19

、

21

、

23

、

25

和

34

颗弹珠。其中一袋只装有破损弹珠，另外

5

袋没有破损弹珠。简拿走其中三袋，乔治从其余袋中拿走两袋，只留下装有破损弹珠的那一袋。若简得到的弹珠数是乔治的两倍，那么有多少颗破损弹珠？

Six bags of marbles contain

18,

19,

21,

23,

25

and

34

marbles, respectively. One bag contains chipped marbles only. The other

5

bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?

18

19

21

23

25

答案：D

知识点：整除性模运算分拆与有序分拆

难度评级：1790

小提示：

先求六袋弹珠总数，再减去破损弹珠那一袋

Add all six bag sizes, then remove the chipped bag

大提示：

简与乔治的弹珠数之比为

2:1

，所以二人的总数能被

3

整除

Jane’s and George’s totals are in the ratio

2:1,

so their combined total is divisible by

3

解答：

六袋共有

140

颗弹珠。若破损弹珠袋中有

c

颗，则其余袋的总数是乔治所得数量的三倍，所以

140-c

能被

3

整除。选项中只有

c=23

满足这一条件。它确实可以实现：乔治取

18+21=39

，简取

19+25+34=78

。

所以正确答案是 D。

The six bags total

140.

If the chipped bag contains

c

marbles, then the other bags total three times George’s amount, so

140-c

is divisible by

3.

Of the choices, only

c=23

has this property. It is attainable: George can take

18+21=39,

while Jane takes

19+25+34=78.

Thus the correct answer is D.

1985 AMC 12 第 18 题

答案：D

难度评级：1790

小提示：

大提示：

解答：

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