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2021 Fall AMC 10A Problems/Problem 8

Revision as of 19:37, 22 November 2021 by Nh14 (talk | contribs) (Created page with "== Solution 1 == Note that the number <math>\overline{xy} = 10x + y.</math> By the problem statement, <cmath>10x + y = x + y^2 \Rightarrow 9x = y^2 - y \Rightarrow 9x = y(y-1)...")
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Solution 1

Note that the number $\overline{xy} = 10x + y.$ By the problem statement, \[10x + y = x + y^2 \Rightarrow 9x = y^2 - y \Rightarrow 9x = y(y-1).\] 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 $1$ cuddly number which is $89.$ Thus, the answer is $\boxed{\textbf{(B).}}$

~NH14

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