Difference between revisions of "2009 AMC 10A Problems/Problem 5"
(→Solution 3) |
(→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== | ||
| + | You can see that | ||
| + | <math>111*111</math> can be written as | ||
| + | |||
| + | <math>111+1110+11100</math>, which is <math>12321</math>. | ||
| + | We can apply the same fact into 111,111,111, recieving | ||
| + | <math>111111111+1111111110+11111111100... = 12,345,678,987,654,321</math> whose digits sum up to <math>81\longrightarrow \fbox{E}.</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:21, 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
You can see that
can be written as
, which is 
.
We can apply the same fact into 111,111,111, recieving
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
