Difference between revisions of "2024 SSMO Team Round Problems/Problem 3"
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==Solution== | ==Solution== | ||
| + | Note that the two-digit perfect square obtained from removing the leading and units digit has to be in the set <math>\{16,25,\dots,81\}.</math> Taking the expected value of <math>N,</math> we have | ||
| + | \begin{align*} | ||
| + | \mathbb{E}(N) &= \mathbb{E}(i\in\{1,2,\dots,9\})\cdot1000+\left(\frac{4^2+5^2+\dots+9^2}{6}\right)\cdot10+\mathbb{E}(i\in\{0,1,2\dots,9\})\\ | ||
| + | &=5\cdot1000+\left(\frac{271}{6}\right)\cdot10+\frac92 = \frac{32737}{6}\implies 32737+6 = \boxed{32733}. | ||
| + | \end{align*} | ||
| + | |||
| + | ~SMO_Team | ||
Latest revision as of 14:36, 10 September 2025
Problem
Consider positive integers \(N\) such that when \(N\)'s units digit and leading nonzero digit are removed, what remains is a two-digit perfect square. The average of all \(N\) can be expressed as 
for relatively prime positive integers 
and 
Find 
Solution
Note that the two-digit perfect square obtained from removing the leading and units digit has to be in the set 
Taking the expected value of 
we have
\begin{align*}
\mathbb{E}(N) &= \mathbb{E}(i\in\{1,2,\dots,9\})\cdot1000+\left(\frac{4^2+5^2+\dots+9^2}{6}\right)\cdot10+\mathbb{E}(i\in\{0,1,2\dots,9\})\\
&=5\cdot1000+\left(\frac{271}{6}\right)\cdot10+\frac92 = \frac{32737}{6}\implies 32737+6 = \boxed{32733}.
\end{align*}
~SMO_Team
