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Difference between revisions of "2016 AMC 10A Problems/Problem 20"

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==Solution==
 
==Solution==
 
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 balls 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>N=\boxed{14}</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>N=\boxed{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.

Revision as of 01:35, 9 January 2020

Problem

For some particular value of $N$, when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$, each to some positive power. What is $N$?

$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution

All the desired terms are in the form $a^xb^yc^zd^w1^t$, where $x + y + z + w + t = N$ (the $1^t$ part is necessary to make stars and bars work better.) Since $x$, $y$, $z$, and $w$ must be at least $1$ ($t$ can be $0$), let $x' = x - 1$, $y' = y - 1$, $z' = z - 1$, and $w' = w - 1$, so $x' + y' + z' + w' + t = N - 4$. Now, we use stars and bars (also known as ball and urn) to see that there are $\binom{(N-4)+4}{4}$ or $\binom{N}{4}$ solutions to this equation. We notice that $1001=7\cdot11\cdot13$, which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\binom{14}{4} = 1001$, giving us our answer of $N=\boxed{14}$.

  • An alternative is to instead make the transformation $t'=t+1$, so $x + y + z + w + t' = N + 1$, and all variables are positive integers. The solution to this, by Stars and Bars is $\binom{(N+1)-1}{4}=\binom{N}{4}$ and we can proceed as above.

Solution 2

By Hockey Stick Identity, the number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$. We now want to find some $N$ such that $\binom{N}{4} = 1001$. As mentioned above, after noticing that $1001 = 7\cdot11\cdot13$, and some trial and error, we find that $\binom{14}{4} = 1001$, giving us our answer of $N=\boxed{14}$

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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