# 2005 AMC 10A Problems/Problem 16

## Problem

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$. How many two-digit numbers have this property?

$\textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19$

## Solution 1

Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.

So $(10a+b)-(a+b)=9a$ must have a units digit of $6$

This is only possible if $9a=36$, so $a=4$ is the only way this can be true.

So the numbers that have this property are $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$.

Therefore the answer is $\boxed{\textbf{(D) }10}$

## Solution 2

Let a two-digit number equal $10a+b$, where $a$ and $b$ are the tens and units digits of the number.

From the problem, we have $10a+b-(a+b)=9a$

Now let $9a=10x+y$, where $x$ and $y$ are the tens and units digits of the number. Then it must be that $y=6$ as stated in the problem.

Note that $10a$ ends in $0$, but $9a$ ends in $6$, so $a=4$. We need not to care about $b$, since it cancels out in the calculation.

So the answer is $\boxed{\textbf{(D) }10}$, since there are $10$ numbers that have $a=4$.

~BurpSuite