Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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==Solution 1== | ==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>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... ;) ) | ||
==Solution 2== | ==Solution 2== |
Revision as of 19:23, 25 December 2017
Problem
What is the sum of the digits of the square of ?
Solution 1
Using the standard multiplication algorithm, whose digit sum is (I hope you didn't seriously multiply it out right... ;) )
Solution 2
Note that:
We see a pattern and find that whose digit sum is
Solution 3
You can see that can be written as
, which is . We can apply the same fact into 111,111,111, receiving whose digits sum up to
See also
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 |
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