Question

三个男孩约定按如下方式分一袋弹珠。第一个男孩拿走比总数一半多一颗的弹珠。第二个男孩拿走剩余弹珠的三分之一。第三个男孩发现，留给他的弹珠数是第二个男孩的两倍。原有弹珠数：

Three boys agree to divide a bag of marbles in the following manner. The first boy takes one more than half the marbles. The second takes a third of the number remaining. The third boy finds that he is left with twice as many marbles as the second boy. The original number of marbles:

不属于以下任何一种情况

is none of the following

无法由已知数据确定

cannot be determined from the given data

是 $20$ 或 $26$

is $20$ or $26$

是 $14$ 或 $32$

is $14$ or $32$

是 $8$ 或 $38$

is $8$ or $38$

答案：B

知识点：fractions of a remainder integer constraints underdetermined problem

难度评级：1490

小提示：

设原有弹珠数为 $n$ ，表示第一个男孩拿走后剩余的数量

Let the original number be $n$ and express the remainder after the first boy

大提示：

将后两个男孩所得数量的条件写成方程，看看能否唯一确定 $n$

Translate the last two boys’ share condition into an equation and see whether it determines $n$ uniquely

解答：

第一个男孩拿走 $\frac{n}{2}+1$ ，剩下 $\frac{n}{2}-1$ 。第二个男孩拿走余数的三分之一，第三个男孩得到另外三分之二，自动是第二个男孩所得的两倍。因此分配条件不能唯一确定 $n$ 。它只要求 $\frac{n}{2}-1$ 是 $3$ 的非负倍数，所以 $8,14,20,26,\ldots$ 等许多数值都可行。

因此，原有弹珠数无法确定，正确答案是 B。

The first boy takes $\frac{n}{2}+1,$ leaving $\frac{n}{2}-1.$ The second takes one third of that remainder, and the third receives the other two thirds, automatically twice the second boy’s share. Thus the share condition imposes no unique value of $n.$ It only requires $\frac{n}{2}-1$ to be a nonnegative multiple of $3,$ so many values such as $8,14,20,26,\ldots$ work.

Therefore the original number cannot be determined, and the correct answer is B.

1955 AMC 12 第 35 题

答案：B

难度评级：1490

小提示：

大提示：

解答：

其他年份的第 35 题