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

Revision as of 08:44, 8 February 2016

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

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

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

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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 ==Solution==
-Using the standard multiplication algorithm, <math>111111111^2=12345678987654321,</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>
 ==See also==
 {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}