Difference between revisions of "2016 AMC 10A Problems/Problem 20"
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==Solution 3 (Stars and Bars)== | ==Solution 3 (Stars and Bars)== | ||
5 sections (<math>x,y,z,w,1</math>) 4 of which need at least 1 object. <math>\dbinom{N+4-4\cdot1}{4}</math>. Test the choices and find that <math>N=\boxed{\text{(B) }14}</math> | 5 sections (<math>x,y,z,w,1</math>) 4 of which need at least 1 object. <math>\dbinom{N+4-4\cdot1}{4}</math>. Test the choices and find that <math>N=\boxed{\text{(B) }14}</math> | ||
+ | |||
+ | ==Solution 4 (Casework)== | ||
+ | |||
+ | The terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math>. The problem becomes distributing N identical balls to 5 different boxes <math>(x, y, z, w, t)</math>. There are <math>2</math> cases of how we count terms with <math>(a, b, c, d)</math>. | ||
+ | |||
+ | Case <math>1</math>: | ||
+ | Count the terms that have at least <math>1</math> of <math>(a, b, c, d, 1)</math>. There are <math>\binom{N-1}{4}</math> terms. | ||
+ | |||
+ | Case <math>2</math>: | ||
+ | Count the terms that have at least <math>1</math> of <math>(a, b, c, d)</math> but not <math>1</math>. There are <math>\binom{N-1}{3}</math> terms. | ||
+ | |||
+ | So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}=1001</math> terms. Similar to solutions <math>1, 2, 3</math>, we find <math>N=\boxed{\text{(B) }14}</math> | ||
+ | |||
+ | ~isabelchen | ||
==Video Solution== | ==Video Solution== |
Revision as of 08:40, 5 October 2021
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
.
- 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 N identical balls to 5 different boxes
. There are
cases of how we count terms with
.
Case :
Count the terms that have at least
of
. There are
terms.
Case :
Count the terms that have at least
of
but not
. There are
terms.
So, there are terms. Similar to solutions
, we find
~isabelchen
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.