Difference between revisions of "2016 AMC 10A Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | All the desired terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math> (the <math>1^t</math> part is necessary to make stars and bars work better.) | ||
− | Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be <math>0</math>), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use stars and bars (also known as ball and urn) to see that there are <math>\binom{(N-4)+4}{4}</math> or <math>\binom{N}{4}</math> solutions to this equation. We notice that <math>1001=7\cdot11\cdot13</math>, which leads us to guess that <math>N</math> is around these numbers. This suspicion proves to be correct, as we see that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math> | + | Since <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math> must be at least <math>1</math> (<math>t</math> can be <math>0</math>), let <math>x' = x - 1</math>, <math>y' = y - 1</math>, <math>z' = z - 1</math>, and <math>w' = w - 1</math>, so <math>x' + y' + z' + w' + t = N - 4</math>. Now, we use stars and bars (also known as ball and urn) to see that there are <math>\binom{(N-4)+4}{4}</math> or <math>\binom{N}{4}</math> solutions to this equation. We notice that <math>1001=7\cdot11\cdot13</math>, which leads us to guess that <math>N</math> is around these numbers. This suspicion proves to be correct, as we see that <math>\binom{14}{4} = 1001</math>, giving us our answer of <math>\boxed{\textbf{(B) }14.}</math> |
* An alternative is to instead make the transformation <math>t'=t+1</math>, so <math>x + y + z + w + t' = N + 1</math>, and all variables are positive integers. The solution to this, by Stars and Bars is <math>\binom{(N+1)-1}{4}=\binom{N}{4}</math> and we can proceed as above. | * An alternative is to instead make the transformation <math>t'=t+1</math>, so <math>x + y + z + w + t' = N + 1</math>, and all variables are positive integers. The solution to this, by Stars and Bars is <math>\binom{(N+1)-1}{4}=\binom{N}{4}</math> and we can proceed as above. | ||
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==Solution 2== | ==Solution 2== | ||
− | 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> | + | 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> |
==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> | + | 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>\boxed{\textbf{(B) }14.}</math> |
==Solution 4 (Casework)== | ==Solution 4 (Casework)== | ||
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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> | + | So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}</math> terms. <math>\binom{N}{4} = 1001</math>, <math>\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 07:41, 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. ,
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 |
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