1989 AHSME Problems/Problem 27

Problem

Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has 28 solutions in positive integers $x$ , $y$ , and $z$ , then $n$ must be either

$\mathrm{(A)}\ 14 \text{ or } 15\ \mathrm{(B)}\ 15 \text{ or } 16\ \mathrm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19$

Solution

This is equivalent to finding the powers of $k$ with coefficient $28$ in the expansion of $(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)$ .

But this is $\left(\frac{k^2}{1-k^2}\right)^2\cdot\frac{k}{1-k}\ =\ k^5\cdot\left(\frac{1}{1-k}\right)^3\cdot\left(\frac{1}{1+k}\right)^2\ =\ k^5\cdot(1+k)\cdot\left(\frac{1}{1-k^2}\right)^3$

$=k^5(1+k)(\tbinom{2}{2}+\tbinom{3}{2}k^2+\tbinom{4}{2}k^4+\tbinom{5}{2}k^6+...)$ , the last part having general term $\tbinom{j}{2}k^{2j-4}$ from which it is easy to see that, since $\tbinom{8}{2}=28$ , the last part contains the term $28k^{12}$ and the whole result $28k^{17}+28k^{18}$ . So the answer is $\mathrm{(D)}$ .

Solution 2

Suppose $n$ is even. Thus $z$ is even, so let $z=2t$ . Hence $x+y+t=\frac{n}{2}$ . By stars and bars, this has $\binom{\frac{n}{2}-1}{3-1}$ solutions. This implies that $\frac{n}{2}-1=8\to n=18$ . If $z$ is odd, let $z=2t-1$ . Then $x+y+t=\frac{n+1}{2}$ , in which $n=17$ is a solution. The answer is then $\mathrm{(D)}$ .

~yofro

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1989 AHSME Problems/Problem 27

Contents

Problem

Solution

Solution 2

See also