# 2021 Fall AMC 10A Problems/Problem 8

## Problem

A two-digit positive integer is said to be $\emph{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

## Solution 1

Note that the number $\underline{xy} = 10x + y.$ By the problem statement, $$10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).$$ From this we see that $y(y-1)$ must be divisible by $9.$ This only happens when $y=9.$ Then, $x=8.$ Thus, there is only $\boxed{\textbf{(B) }1}$ cuddly number, which is $89.$

~NH14

## Solution 2

If the tens digit is $a$ and the ones digit is $b$ then the number is $10a+b$ so we have the equation $10a + b = a + b^2$. We can guess and check after narrowing the possible cuddly numbers down to $13,14,24,25,35,36,46,47,57,68,78,89,$ and $99$. (We can narrow it down to these by just thinking about how $a$'s value affects $b$'s value and then check all the possiblities.) Checking all of these we get that there is only $\boxed{\textbf{(B) }1}$ 2-digit cuddly number, and it is $89$. Yay!!!

~Andlind

## Video Solution (HOW TO THINK CREATIVELY!!!)

~Education, the Study of Everything

~IceMatrix

## Video Solution by WhyMath

https://youtu.be/knVmshj9SDs ~savannahsolver

~Charles 3829

~Lucas