# Difference between revisions of "2006 AMC 8 Problems/Problem 11"

## Problem

How many two-digit numbers have digits whose sum is a perfect square?

$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$

## Solution

There is $1$ integer whose digits sum to $1$: $10$.

There are $4$ integers whose digits sum to $4$: $13, 22, 31, \text{and } 40$.

There are $9$ integers whose digits sum to $9$: $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$.

There are $3$ integers whose digits sum to $16$: $79, 88, \text{and } 97$.

Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$.

Thus, the answer is $1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17}$.