2009 AMC 10A Problems/Problem 5

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 $81\longrightarrow \fbox{E}.$

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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2009 AMC 10A Problems/Problem 5

Contents

Problem

Solution

Solution 2

See also