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

Revision as of 08:06, 5 October 2021

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 $N=\boxed{14}$ .

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 $N=\boxed{14}$

Solution 3 (Stars and Bars)

5 sections ( $x,y,z,w,1$ ) 4 of which need at least 1 object. $\dbinom{N+4-4\cdot1}{4}$ . Test the choices and find that $N=\boxed{\text{(B) }14}$

Solution 4 (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. $\binom{N}{4} = 1001$ , $N=\boxed{\text{(B) }14}$

~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

@@ Line 19: / Line 19: @@
 ==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 <math>N</math> identical balls to <math>5</math> different boxes <math>(x, y, z, w, t)</math> such that boxes <math>(x, y, z, w)</math> have 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.
+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 count the terms with at least <math>1</math> of <math>(a, b, c, d, 1)</math>. There are <math>\binom{N-1}{4}</math> terms.
+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 count the terms with 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.
+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>N=\boxed{\text{(B) }14}</math>

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

Page

Toolbox

Search