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

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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 $\fbox{(E) 81}.$

(I hope you didn't seriously multiply it out... right?)

Solution 2 -- Nonrigorous solution

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

Note that any number when taken $\mod{9}$ yields the digit sum of that number. So, the problem has simplified to finding $111,111,111^2 \pmod{9}$ . We note that $111,111,111 \mod{9}$ is $9$ , so $9^2=81$ .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 48: / Line 48: @@
 ==Solution 4==
 Note that any number when taken <math>\mod{9}</math> yields the digit sum of that number. So, the problem has simplified to finding <math>111,111,111^2 \pmod{9}</math>. We note that <math>111,111,111 \mod{9}</math> is <math>9</math>, so <math>9^2=81</math>.
-Peer Edit: This does not work in general; all this approach tells us is that the sum of the digits is <math>0 \mod{9}</math>, and all of the choices satisfy this.
 ==See also==
 {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

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