2016 AMC 10A Problems/Problem 20
Problem
For some particular value of , when is expanded and like terms are combined, the resulting expression contains exactly terms that include all four variables and , each to some positive power. What is ?
Solution 1
All the desired terms are in the form , where (the part is necessary to make stars and bars work better.) Since , , , and must be at least ( can be ), let , , , and , so . Now, we use stars and bars (also known as ball and urn) to see that there are or solutions to this equation. We notice that , which leads us to guess that is around these numbers. This suspicion proves to be correct, as we see that , giving us our answer of .
- An alternative is to instead make the transformation , so , and all variables are positive integers. The solution to this, by Stars and Bars is and we can proceed as above.
Solution 2
By Hockey Stick Identity, the number of terms that have all raised to a positive power is . We now want to find some such that . As mentioned above, after noticing that , and some trial and error, we find that , giving us our answer of
Video Solution
https://www.youtube.com/watch?v=R3eJW3PCYMs
Video Solution 2
https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution.
~IceMatrix
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AMC 10 Problems and Solutions |
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