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

(Solution)
(Solution)
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==Solution==
 
==Solution==
 
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>
 
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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==Solution 2==
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Note that <math>11^2=121</math>, <math>111^2=12321</math>. We see a pattern and find that <math>111,111,111^2=12,345,678,987,654,321</math> whose digit sum is <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 18:47, 13 February 2016

Problem

What is the sum of the digits of the square of $111,111,111$?

$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

Solution

Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $81\longrightarrow \fbox{E}.$

Solution 2

Note that $11^2=121$, $111^2=12321$. We see a pattern and find that $111,111,111^2=12,345,678,987,654,321$ whose digit sum is $81\longrightarrow \fbox{E}.$

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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