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