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

Revision as of 14:59, 9 February 2023

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution 1==
-Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>\fbox{(E) 81}.</math>
+Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>\boxed{(E)\text{ }81.}</math>
-==Solution 2 -- Nonrigorous solution==
+==Solution 2==
 We note that
@@ Line 42: / Line 42: @@
 The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>.
-==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>.
 ==See also==
 {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

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