2016 AMC 10A Problems/Problem 20

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 to see that there are $\binom{N}{4}$ solutions to this equation. We notice that $\binom{14}{4} = 1001$ , partly because 1001=7 * 11 * 13 which gives us some suspicions, so $N = \boxed{14}$ .

Another solution

The number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$ . For $N=\boxed{14}$ , $\binom{N}{4}=1001$ .

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.

Page

Toolbox

Search

2016 AMC 10A Problems/Problem 20

Solution

Another solution

See Also