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

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<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math>
 
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math>
  
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==Solution==
+
==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>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>\boxed{\textbf{(B) }14.}</math>
  
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* 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.
+
Note: An alternative is instead of making the transformation, we "give" the variables <math>x, y, z, w</math> 1, and then proceed as above.
 +
 
 +
~ Mathkiddie(minor edits by vadava_lx)
  
 
==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>N=\boxed{14}</math>
+
By the [[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>
 +
 
 +
~minor edits by vadava_lx
 +
 
 +
==Solution 3 (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 <math>N</math> identical balls to <math>5</math> different boxes <math>(x, y, z, w, t)</math> such that each of the boxes <math>(x, y, z, w)</math> has at least <math>1</math> ball. The <math>N</math> balls in a row have <math>N-1</math> gaps among them. We are going to put <math>4</math> or <math>3</math> divisors into those <math>N-1</math> gaps. There are <math>2</math> cases of how to put the divisors.
 +
 
 +
Case <math>1</math>:
 +
Put 4 divisors into <math>N-1</math> gaps. It corresponds to each of <math>(a, b, c, d, 1)</math> has at least one term. There are <math>\binom{N-1}{4}</math> terms.
 +
 
 +
Case <math>2</math>:
 +
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>, and since we have <math>\binom{14}{4} = 1001, N=\boxed{\textbf{(B) }14.}</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/yGJwp72qPzk?t=88
 +
 
 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:13, 8 October 2023

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 1

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 $\boxed{\textbf{(B) }14.}$

Note: An alternative is instead of making the transformation, we "give" the variables $x, y, z, w$ 1, and then proceed as above.

~ Mathkiddie(minor edits by vadava_lx)

Solution 2

By the 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 $\boxed{\textbf{(B) }14.}$

~minor edits by vadava_lx

Solution 3 (Casework)

The terms are in the form $a^xb^yc^zd^w1^t$, where $x + y + z + w + t = N$. The problem becomes distributing $N$ identical balls to $5$ different boxes $(x, y, z, w, t)$ such that each of the boxes $(x, y, z, w)$ has at least $1$ ball. The $N$ balls in a row have $N-1$ gaps among them. We are going to put $4$ or $3$ divisors into those $N-1$ gaps. There are $2$ cases of how to put the divisors.

Case $1$: Put 4 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d, 1)$ has at least one term. There are $\binom{N-1}{4}$ terms.

Case $2$: Put 3 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d)$ has at least one term. There are $\binom{N-1}{3}$ terms.

So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}$ terms. $\binom{N}{4} = 1001$, and since we have $\binom{14}{4} = 1001, N=\boxed{\textbf{(B) }14.}$

~isabelchen

Video Solution by OmegaLearn

https://youtu.be/yGJwp72qPzk?t=88

~ pi_is_3.14

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. AMC logo.png

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