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 $10+b$ 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 $89$ .

~Andlind

Video Solution by TheBeautyofMath

https://youtu.be/ycRZHCOKTVk?t=391

~IceMatrix

Video Solution by WhyMath

https://youtu.be/knVmshj9SDs ~savannahsolver

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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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