A number is framed when, by adding it to the number obtained by reversing the order of its digits, the result is a perfect square. For example, 164 and 461 are framed, since 164+461 = $625=25^{2}$. How many framed numbers are there between 10 and 100?
A) 5
B) 6
C) 8
D) 9
E) 10
Solution

## ALTERNATIVE C

Let $n$ be a number between 10 and 100, $a$ its tens digit, and $b$ its units digit; note that $1 \leq a \leq 9$ and $0 \leq b \leq 9$. Then $n=10a+b$ and the number obtained by reversing the digits of $n$ is $10b+a$. Since $n$ is a framed number, we have that $(10a+b) + (10b+a) = 11a + 11b = 11(a+b)$ is a perfect square.

First, we note that if $b=0$, then it is not possible for $11(a+b)$ to be a perfect square, since $11a$ is never a perfect square for $a$ taking values from 1 to 9. Therefore, we have $b \neq 0$. With this, we see that $2 \leq a+b \leq 18$; among these possible values for $a+b$, the only one that makes $11(a+b)$ a perfect square is 11. Thus, $a+b=11$ and the possibilities for $n$ are then 29 and 92, 38 and 83, 47 and 74, and 56 and 65, for a total of 8.

Observation: We can also conclude that $b \neq 0$ by directly verifying that 10, 20, $30, \ldots, 90$ are not framed numbers.