2009 AMC 10A Problems/Problem 5

Problem

What is the sum of the digits of the square of $\text 111111111$ ?

$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

Solution 1

Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\boxed{(E)\text{ }81.}$

Solution 2

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

Solution 4 (Confusing But Fast)

We note that the only way for the sum of multiple numbers' digits to not equal the sum of the digits after they are all added is for a place (i.e., tens place, ones place, etc.) to be carried over. If that was a little confusing, note that $63+22=85,$ and $8+5=13.$ $6+3+2+2=13,$ too. The ones don't carry over to the tens and the tens don't carry over to the hundreds. However, $63+37=100,$ and $6+3+3+7=19 \neq 1.$ The ones carry over to the tens and the tens carry over to the hundreds, meaning it loses $2(9-1)=18$ of its digit sum.

digits. Since the only operation in long multiplication will be  and the addition tower will be  stacks long, there is no way a  stack can be  long and carry over its value. Thus, it is sufficient to conclude that we will multiply  with   times, with no  being carried over.

~ Wesserwes7254

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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