Difference between revisions of "2009 AMC 10A Problems/Problem 5"
(→Solution) |
Henry12345 (talk | contribs) (→Solution) |
||
Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
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== | ||
+ | Note that <math>11^2=121</math>, <math>111^2=12321</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> | ||
==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 19:47, 13 February 2016
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.