AMC8 2010

AMC8 2010 · Q22

AMC8 2010 · Q22. It mainly tests Fractions, Digit properties (sum of digits, divisibility tests).

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?

一个三位数的百位数字比个位数字大 2。把这个三位数的各位数字倒过来组成的新数减去原数，结果的个位数字是多少？

(A) 0 0

(B) 2 2

(D) 6 6

(E) 8 8

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let the hundreds, tens, and units digits of the original three-digit number be $a$, $b$, and $c$, respectively. We are given that $a=c+2$. The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$. The hundreds, tens, and units digits of the reversed three-digit number are $c$, $b$, and $a$, respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$. Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$. Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$

设原三位数的百位、十位、个位数字分别为 $a$、$b$、$c$。已知 $a=c+2$。原数为 $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$。反转后的数字为 $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$。两数相减得 $(101c+10b+200) - (101c+10b+2) = 198$。因此，结果的个位数字是 $\boxed{\textbf{(E)}\ 8}$。

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