Difference between revisions of "2009 AMC 10A Problems/Problem 5"
(→Solution 2) |
(→Solution 2) |
||
| Line 12: | Line 12: | ||
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> | ||
| − | |||
| − | |||
==Solution 3== | ==Solution 3== | ||
Revision as of 00:56, 25 December 2017
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
Solution 3
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. 
