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 $x>1^2$ , which is $x=2$ .

We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ for the row after that. However, at the fourth row, we see that solving $x-1^2=3$ yields $x=4$ , in which case it would be incorrect since $1^2=1$ is not the greatest perfect square less than or equal to $x$ . So we make it a $2^2$ and solve $x-2^2=3$ . We continue on using this same method where we increase the perfect square until $x$ can be made bigger than it. When we repeat this until we have $8$ rows, we get:

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

Hence the solution is the last digit of $7223$ , which is $\boxed{\textbf{(B)}\ 3}$ .

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2010 AMC 10A Problems/Problem 25