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

(Solution 4)
(Solution 2 -- Find and Harness a Pattern)
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(I hope you didn't seriously multiply it outright...)
 
(I hope you didn't seriously multiply it outright...)
  
==Solution 2==
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==Solution 2 -- Find And Harness a Pattern==
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>,
 +
 
 +
<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=(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</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==
 
==Solution 3==

Revision as of 11:32, 30 December 2018

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 $81\fbox{(E)}.$ (I hope you didn't seriously multiply it outright...)

Solution 2 -- Find And Harness a Pattern

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

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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