Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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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> | ||
==Solution 2== | ==Solution 2== | ||
| − | Note that: | + | Note that: |
| − | <math>11^2 = 121 | + | |
| − | 111^2 = 12321 | + | <math>11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321</math> |
| − | 1111^2 = 1234321</math> | + | |
We see a pattern and find that <math>111,111,111^2=12,345,678,987,654,321</math> whose digit sum is <math>81\longrightarrow \fbox{E}.</math> | We see a pattern and find that <math>111,111,111^2=12,345,678,987,654,321</math> whose digit sum is <math>81\longrightarrow \fbox{E}.</math> | ||
Revision as of 20:31, 7 February 2017
Contents
Problem
What is the sum of the digits of the square of 
?
Solution
Using the standard multiplication algorithm, 
whose digit sum is 
Solution 2
Note that:
We see a pattern and find that 
whose digit sum is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
