Question

一个两位数与把它的数字倒序得到的数之差，是这两个数任一者数字和的 $5$ 倍。这个两位数与其倒序数之和是多少？

The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number. What is the sum of the two-digit number and its reverse?

$44$

$55$

$77$

$99$

$110$

答案：D

知识点：位值数字

难度评级：1270

解答：

设较大的数为 $10a+b$ ，则化简得 $2a=7b$ 。 $\begin{aligned} &(10a+b)-(10b+a)=9(a-b)\\ &\quad{}=5(a+b), \end{aligned}$

满足条件的非零数字只有 $a=7$ 、 $b=2$ ，所以这个数是 $72$ ，倒序数是 $27$ 。

两数之和为 $72+27=99$ 。

所以正确答案是 D。

Let the larger number be $10a+b.$ Then $\begin{aligned} &(10a+b)-(10b+a)=9(a-b)\\ &\quad{}=5(a+b), \end{aligned}$ which simplifies to $2a=7b.$

The only nonzero digits satisfying this are $a=7$ and $b=2,$ so the number is $72$ and its reverse is $27.$

Their sum is $72+27=99.$

Thus, the correct answer is D.

2014 AMC 12A 第 6 题

答案：D

难度评级：1270

解答：

其他年份的第 6 题