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 $\mathrm{(D)}$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1989 AHSME Problems/Problem 27

Problem

Solution