# Difference between revisions of "2009 AMC 10A Problems/Problem 5"

## Problem

What is the sum of the digits of the square of $\text 111,111,111$? $\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

## Solution 1

Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $81\fbox{(E)}.$ (I hope you didn't seriously multiply it outright...)

## Solution 2 -- Find And Harness a Pattern

We note that $1^2 = 1$, $11^2 = 121$, $111^2 = 12321$,

and $1,111^2 = 1234321$.

We can clearly see the pattern: If $X$ is $111\cdots111$, with $n$ ones (and for the sake of simplicity, assume that $n<10$), then the sum of the digits of $X^2$ is $1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1$ $=(1+2+3\cdots n)+(1+2+3+\cdots n-1)$ $=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}$ $=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.$

Aha! We know that $111,111,111$ has $9$ digits, so its digit sum is $9^2=\boxed{81(E)}$.

## Solution 3

We see that $111^2$ can be written as $111(100+10+1)=11100+1110+111=12321$.

We can apply this strategy to find $111,111,111^2$, as seen below. $111111111^2=111111111(100000000+10000000\cdots+10+1)$ $=11111111100000000+1111111110000000+\cdots+111111111$ $=12,345,678,987,654,321$

The digit sum is thus $1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}$.

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