2010 AMC 10A Problems/Problem 25

We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ (and that does not violate the fact that $1^2$ must be the greatest perfect square less than $x$ ), which is $2$ .

We continue with the same way with the next row, but at the fourth row, we see that solving $x-1^2=3$ yields $x=4$ , but in that case we would have to make the $1^2$ a $2^2$ since $4 \geq 2^2$ . So we make it a $2^2$ and solve $x-2^2=3$ . We continue on using this same method where we move the perfect square up by $1$ each time until we get the $x$ to be a value greater than the RHS plus the perfect square number. When we repeat this until we have $8$ rows, we get:

$\begin{align*} 7223\\ 7223-84^2 &= 167\\ 167-12^2 &= 23\\ 23-4^2 &= 7\\ 7-2^2 &= 3\\ 3-1^2 &= 2\\ 2-1^2 &= 1\\ 1-1^2 &= 0\end{align*}$

Hence the solution is the last digit of $7223$ , which is $3$ .

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2010 AMC 10A Problems/Problem 25