AMC8 2014

AMC8 2014 · Q21

AMC8 2014 · Q21. It mainly tests Digit properties (sum of digits, divisibility tests).

The $7$-digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$. Which of the following could be the value of $C$?

这两个7位数 $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ 和 $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ 都是3的倍数。以下哪个可能是 $C$ 的值？

(A) 1 1

(B) 2 2

(D) 5 5

(E) 8 8

Answer

Correct choice: (A)

正确答案：（A）

Solution

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$. To be a multiple of $3$, $A + B$ has to be either $2$ or $5$ or $8$... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$. We then add two of the selected values, $5$ to $15$, to get $20$. We then see that C = $1, 4$ or $7, 10$... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$, to get $23$, which shows us that C = $1$ or $4$ or $7$... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$. However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$, but there is a $1$, so $\boxed{\textbf{(A) }1}$ is our answer.

由于这两个数都能被3整除，所以它们各自数字之和必须能被3整除。$7 + 4 + 5 + 2 + 1 = 19$。为了使其成为3的倍数，$A + B$ 必须是 $2$ 或 $5$ 或 $8$ …… 等等。我们将第二个数中的数字相加；$3 + 2 + 6 + 4 = 15$。然后我们把选定的值中的两个数 $5$ 加到 $15$，得到 $20$。我们接着发现，$C$ 可以是 $1, 4$ 或 $7, 10$…… 等等，否则这个数不能被3整除。我们再把 $8$ 加到 $15$，得到 $23$，这说明 $C$ 也可以是 $1$ 或 $4$ 或 $7$ …… 等等。为了是3的倍数，我们选取这两个方程中得到的几个公共数字，可以是 $1, 4$ 和 $7$。然而，在选项中，没有 $7$、$4$ 或更大于 $7$ 的数，但有 $1$，所以答案是 $\boxed{\textbf{(A) }1}$。

Topics

NT.007 Digit properties (sum of digits, divisibility tests)