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

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==Problem==
 
==Problem==
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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 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==
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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>81\longrightarrow \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>\boxed{(E)\text{ }81.}</math>
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(I hope you didn't seriously multiply it outright...)
 
  
 
==Solution 2==
 
==Solution 2==
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Note that  
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We note that
  
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<math>11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.</math>
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<math>1^2 = 1</math>,
  
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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>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==
 
==Solution 3==
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You can see that
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We see that <math>111^2</math> can be written as <math>111(100+10+1)=11100+1110+111=12321</math>.
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<math>111*111</math> can be written as
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 +
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>
  
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<math>111+1110+11100</math>, which is <math>12321</math>.
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<math>=11111111100000000+1111111110000000+\cdots+111111111</math>
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We can apply the same fact into 111,111,111, receiving
 
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<math>111111111+1111111110+11111111100... = 12,345,678,987,654,321</math> whose digits sum up to <math>81\longrightarrow \fbox{E}.</math>
 
  
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==Solution 4==
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<math>=12,345,678,987,654,321</math>
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We can also do something a little bit more clever rather than just adding up the digits though. Realize too that
 
  
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<math>1^2 = 1</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>.
  
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<math>11^2 = 121</math>
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==Solution 4 (Confusing But Fast)==
 +
We note that the only way for the sum of multiple numbers' digits to not equal the sum of the digits after they are all added is for a place (i.e., tens place, ones place, etc.) to be carried over. If that was a little confusing, note that <math>63+22=85,</math> and <math>8+5=13.</math>  <math>6+3+2+2=13,</math> too. The ones don't carry over to the tens and the tens don't carry over to the hundreds. However, <math>63+37=100,</math> and <math>6+3+3+7=19 \neq 1.</math> The ones carry over to the tens and the tens carry over to the hundreds, meaning it loses <math>2(9-1)=18</math> of its digit sum.
  
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<math>111^2 = 12321</math>
 
  
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<math>1,111^2 = 1234321</math>
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If that made sense, note that <math>111111111</math> has <math>9</math> digits. Since the only operation in long multiplication will be <math>1 \cdot 1</math> and the addition tower will be <math>9</math> stacks long, there is no way a <math>1</math> stack can be <math>10</math> long and carry over its value. Thus, it is sufficient to conclude that we will multiply <math>1</math> with <math>1</math> <math>9 \cdot 9= \boxed{\textbf{(E) }81}</math> times, with no <math>1</math> being carried over.
  
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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>
+
~ Wesserwes7254
  
 
==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 19:00, 16 July 2024

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

Solution 4 (Confusing But Fast)

We note that the only way for the sum of multiple numbers' digits to not equal the sum of the digits after they are all added is for a place (i.e., tens place, ones place, etc.) to be carried over. If that was a little confusing, note that $63+22=85,$ and $8+5=13.$ $6+3+2+2=13,$ too. The ones don't carry over to the tens and the tens don't carry over to the hundreds. However, $63+37=100,$ and $6+3+3+7=19 \neq 1.$ The ones carry over to the tens and the tens carry over to the hundreds, meaning it loses $2(9-1)=18$ of its digit sum.


If that made sense, note that $111111111$ has $9$ digits. Since the only operation in long multiplication will be $1 \cdot 1$ and the addition tower will be $9$ stacks long, there is no way a $1$ stack can be $10$ long and carry over its value. Thus, it is sufficient to conclude that we will multiply $1$ with $1$ $9 \cdot 9= \boxed{\textbf{(E) }81}$ times, with no $1$ being carried over.

~ Wesserwes7254

See also

2009 AMC 10A (Problems • Answer Key • Resources)
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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