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

Denote this number as $\overline{ab}$ .

Hence, we have $\overline{ab} = a + b^2$ .

This can be written as $10 a + b = a + b^2$ .

Hence, $b \left( b - 1 \right) = 9 a$ . This implies $9 | b \left( b - 1 \right)$ .

Hence, either $9 | b$ or $9 | b - 1$ . Because $b \in \left\{ 0 , 1 , \cdots , 9 \right\}$ , $b = 9$ or 1.

For $b = 9$ , we get $a = 8$ . This is a solution.

For $b = 1$ , we get $a = 0$ . However, recall that $a \neq 0$ . Hence, this is not a solution.

Therefore, the answer is $\boxed{\textbf{(B) }1}$ .

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2021 Fall AMC 10A Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See Also