1989 AMC 12 Problem 27

Attempt Problem 27 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

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27.

Let nn be a positive integer. If the equation 2x+2y+z=n2x+2y+z=n has 2828 solutions in positive integers x,x, yy and z,z, then nn must be either
1414 or 1515
1515 or 1616
1616 or 1717
1717 or 1818
1818 or 1919
Answer: D
Concepts:Diophantine Equationsummationcounting pairs
Small hint:
Group solutions according to s=x+ys=x+y
Big hint:
For fixed s,s, count the positive ordered pairs (x,y)(x,y) and require n−2s≥1n-2s\ge1
Solution:
For s=x+y≥2,s=x+y\ge2, there are s−1s-1 positive ordered pairs (x,y),(x,y), and z=n−2sz=n-2s is positive when s≤⌊n−12⌋.s\le\lfloor\frac{n-1}{2}\rfloor. Writing m=⌊n−12⌋,m=\lfloor\frac{n-1}{2}\rfloor, the number of solutions is ∑s=2m(s−1)=m(m−1)2. \sum_{s=2}^{m}(s-1)=\frac{m(m-1)}2. Setting this equal to 2828 gives m=8.m=8. Hence ⌊n−12⌋=8,\lfloor\frac{n-1}{2}\rfloor=8, so n=17n=17 or 18.18.
Thus the correct answer is D.
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