2009 AMC 10A Problems/Problem 5

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, $\text 111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $81\longrightarrow \fbox{E}.$ (I hope you didn't seriously multiply it outright... ;) )

Solution 2(Pattern)

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 $81\longrightarrow \fbox{E}.$

Solution 3

You can see that $111*111$ can be written as

$111+1110+11100$ , which is $12321$ . We can apply the same fact into 111,111,111, receiving $111111111+1111111110+11111111100... = 12,345,678,987,654,321$ whose digits sum up to $81\longrightarrow \fbox{E}.$

2009 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2009 AMC 10A Problems/Problem 5

Contents

Problem

Solution 1

Solution 2(Pattern)

Solution 3

See also