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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

Note that $11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.$

We observe a pattern with the squares of number comprised of only $1$'s and use it to find that $111,111,111^2=12,345,678,987,654,321$ whose digit sum is $81\fbox{(E)}.$

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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