2009 AMC 10A Problems/Problem 5: Difference between revisions

Revision as of 17:34, 5 February 2018

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 out right... ;) )

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 $81\longrightarrow \fbox{E}.$

2009 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

@@ Line 5: / Line 5: @@
 ==Solution 1==
-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>
+Using the standard multiplication algorithm, <math>\text 111,111,111^2=12,345,678,987,654,321,</math> whose digit sum is <math>81\longrightarrow \fbox{E}.</math>
 (I hope you didn't seriously multiply it out right... ;) )

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