Question

Ringo 把弹珠每 $6$ 个装一袋，会剩下 $4$ 个。Paul 同样装袋会剩下 $3$ 个。两人把弹珠合在一起，仍按每袋 $6$ 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?

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：模运算

难度评级：770

小提示：

把两个余数相加，再看除以 $6$ 的余数。

Add the two remainders modulo $6$

大提示：

$4+3$ 除以 $6$ 余 $1$ 。

$4+3$ leaves remainder $1$ when divided by $6$

解答：

Ringo 的弹珠数除以 $6$ 余 $4$ ，所以可写成 $6x+4$ ，其中 $x$ 为整数。

同理，Paul 的弹珠数可写为 $6y+3$ ，其中 $y$ 为整数。

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

所以正确答案是 A。

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

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

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

Thus, the correct answer is A .

2012 AMC 10B 第 4 题

答案：A

难度评级：770

小提示：

大提示：

解答：

其他年份的第 4 题