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

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==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>\fbox{(E) 81}.</math>
(I hope you didn't seriously multiply it outright...)
 
  
==Solution 2==
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(I hope you didn't seriously multiply it out... right?)
Note that
 
  
<math>11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.</math>
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==Solution 2 -- Nonrigorous solution==
 +
We note that
  
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\longrightarrow \fbox{(E)}.</math>
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<math>1^2 = 1</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>.
 +
 
 +
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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We can apply this strategy to find <math>111,111,111^2</math>, as seen below.
 
We can apply this strategy to find <math>111,111,111^2</math>, as seen below.
\begin{align*}
 
111111111^2&=111111111(100000000+10000000\cdots+10+1)\\
 
&=11111111100000000+1111111110000000+\cdots+111111111\\
 
&= 12,345,678,987,654,321\end{align*} 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==
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<math>111111111^2=111111111(100000000+10000000\cdots+10+1)</math>
We can also do something a little bit more clever rather than just adding up the digits though. Realize too that
 
  
<math>1^2 = 1</math>
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<math>=11111111100000000+1111111110000000+\cdots+111111111</math>
  
<math>11^2 = 121</math>
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<math>=12,345,678,987,654,321</math>  
  
<math>111^2 = 12321</math>
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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>.
  
<math>1,111^2 = 1234321</math>
 
  
We clearly see the pattern, the number of digits determines how high the number goes, as with <math>111^2</math>, it has <math>3</math> digits so it goes up to <math>1,2,3</math> then decreases back down. If we start adding up the digits, we see that the first one is <math>1</math>, the second is <math>2 + 1 + 1 = 4</math>, the third one is <math>1 + 2 + 3 + 2 + 1 = 9</math>, and the fourth one is <math>16</math>. We instantly see a pattern and find that these are all square numbers. If the number you square has <math>4</math> digits, you do <math>4^2</math> to see what the added digits of that particular square will be. In this case, we are dealing with <math>111,111,111</math> which has <math>9</math> digits so <math>9^2</math> equals <math>81\longrightarrow \fbox{E}</math>
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==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==
 
==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 14:52, 13 September 2020

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 $9^2=81$.

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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