Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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We can apply this strategy to find <math>111,111,111^2</math>, as seen below. | We can apply this strategy to find <math>111,111,111^2</math>, as seen below. | ||
− | + | <math>111111111^2\=111111111(100000000+10000000\cdots+10+1)\=11111111100000000+1111111110000000+\cdots+111111111\= | |
− | 111111111^2 | + | 12,345,678,987,654,321</math> The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>. |
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==Solution 4== | ==Solution 4== |
Revision as of 11:22, 30 December 2018
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 outright...)
Solution 2
Note that
We observe a pattern with the squares of number comprised of only 's and use it to find that whose digit sum is
Solution 3
We see that can be written as .
We can apply this strategy to find , as seen below. $111111111^2\=111111111(100000000+10000000\cdots+10+1)\=11111111100000000+1111111110000000+\cdots+111111111\=
12,345,678,987,654,321$ (Error compiling LaTeX. ! Please use \mathaccent for accents in math mode.) The digit sum is thus .
Solution 4
We can also do something a little bit more clever rather than just adding up the digits though. Realize too that
We clearly see the pattern, the number of digits determines how high the number goes, as with , it has digits so it goes up to then decreases back down. If we start adding up the digits, we see that the first one is , the second is , the third one is , and the fourth one is . We instantly see a pattern and find that these are all square numbers. If the number you square has digits, you do to see what the added digits of that particular square will be. In this case, we are dealing with which has digits so equals
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.