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

m (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>.
 
==Solution 4==
 
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>
 
 
<math>11^2 = 121</math>
 
 
<math>111^2 = 12321</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>
 
  
 
==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}}

Revision as of 11:25, 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

Note that

$11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.$

We observe a pattern with the squares of number comprised of only $1$'s and use it to find that $111,111,111^2=12,345,678,987,654,321$ whose digit sum is $81\fbox{(E)}.$

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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