Difference between revisions of "2016 AMC 10A Problems/Problem 20"
m |
m (→Solution 4 (Casework)) |
||
Line 27: | Line 27: | ||
Put 3 divisors into <math>N-1</math> gaps. It corresponds to each of <math>(a, b, c, d)</math> has at least one term. There are <math>\binom{N-1}{3}</math> terms. | Put 3 divisors into <math>N-1</math> gaps. It corresponds to each of <math>(a, b, c, d)</math> has at least one term. There are <math>\binom{N-1}{3}</math> terms. | ||
− | So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}</math> terms. <math>\binom{N}{4} = 1001</math>, <math>N=\boxed{\textbf{(B) }14.}</math> | + | So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}</math> terms. <math>\binom{N}{4} = 1001</math>, and since we have <math>\binom{14}{4} = 1001, N=\boxed{\textbf{(B) }14.}</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 07:44, 5 September 2022
Contents
[hide]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
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. , and since we have
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 | |
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.