2009 AMC 10A Problems/Problem 5

Revision as of 02:21, 25 December 2017 by Zebra5ftw (talk | contribs) (Solution)

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 1

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:

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

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 $81\longrightarrow \fbox{E}.$

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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