Difference between revisions of "2016 AMC 10A Problems/Problem 20"
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By [[Hockey Stick Identity]], the number of terms that have all <math>a,b,c,d</math> raised to a positive power is <math>\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}</math>. We now want to find some <math>N</math> such that <math>\binom{N}{4} = 1001</math>. As mentioned above, after noticing that <math>1001 = 7\cdot11\cdot13</math>, and some trial and error, we find that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math>\boxed{\textbf{(B) }14.}</math> | By [[Hockey Stick Identity]], the number of terms that have all <math>a,b,c,d</math> raised to a positive power is <math>\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}</math>. We now want to find some <math>N</math> such that <math>\binom{N}{4} = 1001</math>. As mentioned above, after noticing that <math>1001 = 7\cdot11\cdot13</math>, and some trial and error, we find that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math>\boxed{\textbf{(B) }14.}</math> | ||
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==Solution 4 (Casework)== | ==Solution 4 (Casework)== |
Revision as of 19:46, 24 October 2022
Contents
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
Note: An alternative is instead of making the transformation, we "give" the variables 1, and then proceed as above.
~ Mathkiddie
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 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.