2010 AMC 10A Problems/Problem 25

Revision as of 23:06, 30 December 2011 by Juluqu (talk | contribs) (Included problem)


Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$, then his sequence contains $5$ numbers:

\[\begin{array}{ccccc} {}&{}&{}&{}&55\\ 55&-&7^2&=&6\\ 6&-&2^2&=&2\\ 2&-&1^2&=&1\\ 1&-&1^2&=&0\\ \end{array}\]

Let $N$ be the smallest number for which Jim’s sequence has $8$ numbers. What is the units digit of $N$?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 3 \qquad \mathrm{(C)}\ 5 \qquad \mathrm{(D)}\ 7 \qquad \mathrm{(E)}\ 9$


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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