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

Revision as of 01:56, 25 December 2017

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 $81\longrightarrow \fbox{E}.$

Note that:

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

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

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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 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>
 ==Solution 2==
-Note that:
-<math>11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321</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>
-==Solution 3==
 Note that: