2010 AMC 8 第 22 题

先试着解答 2010 AMC 8 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2010 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

一个三位数的百位数字比个位数字大 22。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 22 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?
00
22
44
66
88
答案:E
知识点:位值数字
小提示:
倒序数相减时,十位数字会抵消
The tens digit cancels when the reversed number is subtracted.
大提示:
用个位数字表示原数,并让百位数字比个位数字大二
Represent the original number with hundreds digit two more than the units digit.
解答:
设个位数字为 uu,十位数字为 tt。则百位数字为 u+2u+2。原数为 100(u+2)+10t+u100(u+2) +10t+u =101u+200+10t,=101u+200+10t\text{,}倒序数为 100u+10t+(u+2)100u +10t+(u+2) =101u+2+10t。=101u+2+10t\text{。}两数之差为 (101u+10t+200)−(101u+10t+200)- (101u+10t+2)=198。(101u+10t+2)=198\text{。}因此结果的个位数字是 88。
所以正确答案是 E。
Let the units digit be u,u, and let the tens digit be t.t. This makes the hundreds digit be u+2.u+2. This makes the number equal to 100(u+2)+10t+u100(u+2) +10t+u=101u+200+10t=101u+200+10t and the reversed number is 100u+10t+(u+2)100u +10t+(u+2)=101u+2+10t.=101u+2+10t. This makes the difference equal to (101u+10t+200)−(101u+10t+200)-(101u+10t+2)=198.(101u+10t+2)=198. This makes the units digit 8.8.
Therefore, the answer is E .
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