Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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Note that any number when taken <math>\mod{9}</math> yields the digit sum of that number. So, the problem has simplified to finding <math>111,111,111^2 \pmod{9}</math>. We note that <math>111,111,111 \mod{9}</math> is <math>9</math>, so <math>9^2=81</math>. | Note that any number when taken <math>\mod{9}</math> yields the digit sum of that number. So, the problem has simplified to finding <math>111,111,111^2 \pmod{9}</math>. We note that <math>111,111,111 \mod{9}</math> is <math>9</math>, so <math>9^2=81</math>. | ||
Revision as of 11:56, 13 July 2020
Contents
[hide]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 -- Nonrigorous solution
We note that
,
,
,
and .
We can clearly see the pattern: If is , with ones (and for the sake of simplicity, assume that ), then the sum of the digits of is
Aha! We know that has digits, so its digit sum is .
Solution 3
We see that can be written as .
We can apply this strategy to find , as seen below.
The digit sum is thus .
Solution 4
Note that any number when taken yields the digit sum of that number. So, the problem has simplified to finding . We note that is , so .
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.