AMC 8 2014 Problem 8
AMC8 2014 年第 8 题
余数与同余★★☆☆☆
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 . What is the missing digit of this 3 -digit number?
中学数学俱乐部的 11 个成员为了邀请一个客座演讲人在他们的数学俱乐部会议上讲授解题技巧，每个人都支付了同样的美元金额数且这个数是个整数。他们总共支付给客座演讲人，这个三位数中的 A 代表什么数字？
- A.0
- B.1
- C.2
- D.3
- E.4
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
Since all the eleven members paid the same amount, that means that the total must be divisible by 11. We can do some trial-and-error to get A=3, so our answer is
We know that a number is divisible by 11 if the odd digits added together minus the even digits added together (or vice versa) is a multiple of 11. Thus, we have 1+2-A = a multiple of 11. The only multiple that works here is 0, as . Thus, A =
If you know your multiplication tables up to 12*12, you know that the out of the choices provided, the only number with 1A2 that is divisible by 11 is 132 or 12*11. So the answer is.
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