2016 AMC 10A Problems/Problem 20
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 ?
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.
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
Solution 3 (Stars and Bars)
5 sections () 4 of which need at least 1 object. . Test the choices and find that
Solution 4 (Casework)
The terms are in the form , where . The problem becomes distributing identical balls to different boxes such that each of the boxes has at least ball. The balls in a row have gaps among them. We are going to put or divisors into those gaps. There are cases of how to put the divisors.
Case : Put 4 divisors into gaps. It corresponds to each of has at least one term. There are terms.
Case : Put 3 divisors into gaps. It corresponds to each of has at least one term. There are terms.
So, there are terms. ,
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.
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