Question

一个三位数的百位数字比个位数字大 $2$ 。将这个三位数的数字倒序排列，再用原三位数减去倒序后的数。结果的个位数字是多少？

The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

$0$

$2$

$4$

$6$

$8$

答案：E

知识点：位值数字

难度评级：1260

解答：

设个位数字为 $u$ ，十位数字为 $t$ 。则百位数字为 $u+2$ 。原数为倒序数为两数之差为因此结果的个位数字是 $8$ 。 $100(u+2) +10t+u$ $=101u+200+10t$ $100u +10t+(u+2)$ $=101u+2+10t.$ $(101u+10t+200)-$ $(101u+10t+2)=198.$

所以正确答案是 E。

Let the units digit be $u,$ and let the tens digit be $t.$ This makes the hundreds digit be $u+2.$ This makes the number equal to $100(u+2) +10t+u$ $=101u+200+10t$ and the reversed number is $100u +10t+(u+2)$ $=101u+2+10t.$ This makes the difference equal to $(101u+10t+200)-$ $(101u+10t+2)=198.$ This makes the units digit $8.$

Therefore, the units digit is E .

2010 AMC 8 第 22 题

答案：E

难度评级：1260

解答：

其他年份的第 22 题