Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
− | Using the standard multiplication algorithm, <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> |
− | (I hope you didn't seriously multiply it outright... | + | (I hope you didn't seriously multiply it outright...) |
==Solution 2(Pattern)== | ==Solution 2(Pattern)== |
Revision as of 12:11, 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(Pattern)
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, receiving whose digits sum up to
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 |
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