# 2010 AMC 8 Problems/Problem 22

## Problem

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? $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

## Solution 1

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 $\boxed{\textbf{(E)}\ 8}$

## Solution 2

The result must hold for any three-digit number with hundreds digit being $2$ more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$. Thus, the units digit in the final result is $\boxed{\textbf{(E)}\ 8}$

## Solution 3

Set the units digit of the original number as $x$. Thus, its hundreds digit is $x+2$. After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since $x-(x+2) = -2$, we can do regrouping and "borrow" $1$ from the tens digit and bring it to the units digit as a $10$. Therefore, the units digit will end up as $-2 + 10 = \boxed{\textbf{(E)}\ 8}$

## Video Solution by WhyMath

~savannahsolver

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