2005 AMC 10A Problem 16

Try one number first to see what happens. Take $47$: the digit sum is $4 + 7 = 11$, and $47 - 11 = 36$. Now take $42$: $42 - 6 = 36$ again. The units digit seems not to matter. Here is why.

1. Write the number with digits. A two-digit number with tens digit $a$ and units digit $b$ equals $10a + b$. Its digit sum is $a + b$.
2. Subtract.
$$(10a + b) - (a + b) = 9a.$$
The $b$ cancels, so the result depends only on the tens digit.
3. List the possible results. For $a = 1, 2, \ldots, 9$: $9, 18, 27, 36, 45, 54, 63, 72, 81$.
4. Which ends in $6$? Only $36$, which comes from $a = 4$.
5. Count. The tens digit must be $4$, and the units digit can be any of $0$ to $9$. That gives $40, 41, \ldots, 49$: $10$ numbers.

The answer is $\boxed{\textbf{(D)}\ 10}$.

Taking away the digit sum removes the units digit completely and leaves $9$ times the tens digit. That is why only one digit is restricted and the other is free. It is also a view of the rule "a number and its digit sum leave the same remainder when divided by $9$."

Write the number as $10a + b$ with tens digit $a$ ($1 \leq a \leq 9$) and units digit $b$. Subtracting the digit sum:
$$(10a + b) - (a + b) = 9a.$$
The units digit $b$ drops out entirely, so the result depends only on $a$.

The possible results are $9, 18, 27, 36, 45, 54, 63, 72, 81$. Exactly one of these ends in $6$: $36$, which needs $a = 4$.

Every number with tens digit $4$ works, whatever its units digit: $40, 41, \dots, 49$. That is $10$ numbers.

The answer is $\boxed{\textbf{(D)}\ 10}$.