Question

Middle School Math Club 的十一名成员每人支付相同金额，请一位嘉宾在数学俱乐部会议上讲解解题方法。他们共付给嘉宾 $\$\underline{1}\underline{A}\underline{2}$ 。这个 $3$ 位数中缺失的数字 $A$ 是多少？

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\$\underline{1}\underline{A}\underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?

$0$

$1$

$2$

$3$

$4$

答案：D

知识点：整除性数字

难度评级：960

解答：

因为 $11$ 个人支付相同金额，所以总额能被 $11$ 整除。因此 $\underline{1} \underline{A} \underline{2}$ 必须能被 $11$ 整除。

$11$ 的整除规则是：交错位数字和之差若能被 $11$ 整除，则原数能被 $11$ 整除。

对 $\underline{1} \underline{A} \underline{2}$ ，这个差为 $1 + 2 - A = 3 - A.$ 要让它能被 $11$ 整除，只能有 $A = 3$ 。

所以正确答案是 D。

Note that since $11$ people paid the same amount, then the resulting sum is divisible by $11.$ Therefore, $\underline{1} \underline{A} \underline{2}$ must be divisible by $11.$

Remember the divisibility rule for $11$ : if we take the difference of the sums of alternating digits, and this difference is divisible by $11,$ the whole number is divisible by $11.$

For $\underline{1} \underline{A} \underline{2},$ the aforementioned difference is $1 + 2 - A = 3 - A.$ The only way for this to be divisible by $11$ is if $A = 3.$

Thus, D is the correct answer.

2014 AMC 8 第 8 题

答案：D

难度评级：960

解答：

其他年份的第 8 题