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)
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All AMC 10 Problems and Solutions

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@@ Line 1: / Line 1: @@
 ==Problem==
-<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>111,111,111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
+<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>What is the sum of the digits of the square of <math>\text 111111111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
 <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math>
-==Solution==
+==Solution 1==
-Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>81\longrightarrow \fbox{E}.</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==
+We note that
+<math>1^2 = 1</math>,
+<math>11^2 = 121</math>,
+<math>111^2 = 12321</math>,
+and <math>1,111^2 = 1234321</math>.
+We can clearly see the pattern: If <math>X</math> is <math>111\cdots111</math>, with <math>n</math> ones (and for the sake of simplicity, assume that <math>n<10</math>), then the sum of the digits of <math>X^2</math> is
+<math>1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1</math>
+<math>=(1+2+3\cdots n)+(1+2+3+\cdots n-1)</math>
+<math>=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}</math>
+<math>=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.</math>
+Aha! We know that <math>111,111,111</math> has <math>9</math> digits, so its digit sum is <math>9^2=\boxed{81(E)}</math>.
+==Solution 3==
+We see that <math>111^2</math> can be written as <math>111(100+10+1)=11100+1110+111=12321</math>.
+We can apply this strategy to find <math>111,111,111^2</math>, as seen below.
+<math>111111111^2=111111111(100000000+10000000\cdots+10+1)</math>
+<math>=11111111100000000+1111111110000000+\cdots+111111111</math>
+<math>=12,345,678,987,654,321</math>
+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>.
 ==See also==
 {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

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