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

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==Problem==
 
==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>\text 111,111,111</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
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<!-- 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>
 
<math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81</math>
  
 
==Solution 1==
 
==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\fbox{(E)}.</math>
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Using the standard multiplication algorithm, <math>111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>(E) 81.</math>
(I hope you didn't seriously multiply it outright...)
 
  
 
==Solution 2==
 
==Solution 2==
Note that  
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We note that
  
<math>11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.</math>
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<math>1^2 = 1</math>,
  
We observe a pattern with the squares of number comprised of only <math>1</math>'s and use it to find that <math>111,111,111^2=12,345,678,987,654,321</math> whose digit sum is <math>81\fbox{(E)}.</math>
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<math>11^2 = 121</math>,
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<math>111^2 = 12321</math>,
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and <math>1,111^2 = 1234321</math>.
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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
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<math>1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1</math>
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<math>=(1+2+3\cdots n)+(1+2+3+\cdots n-1)</math>
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<math>=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}</math>
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<math>=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.</math>
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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==
 
==Solution 3==
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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>.
 
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>.
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==Solution 4==
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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==
 
==See also==
 
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:45, 21 July 2021

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 $(E) 81.$

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

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 $9^2=81$.

See also

2009 AMC 10A (ProblemsAnswer KeyResources)
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

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