Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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+ | Solution 4: | ||
+ | Note that any number when taken (mod 9) yields the digit sum of that number. So, the problem has simplified to finding (111,111,111)^2 (mod 9). We note that 111,111,111 (mod 9) is 9, so 9^2=81. |
Revision as of 22:23, 28 June 2020
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 .
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.
Solution 4: Note that any number when taken (mod 9) yields the digit sum of that number. So, the problem has simplified to finding (111,111,111)^2 (mod 9). We note that 111,111,111 (mod 9) is 9, so 9^2=81.