1997 AMC 12 Problem 28

Attempt Problem 28 of the 1997 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 1997 AMC 12 solutions, or check the answer key.

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28.

How many ordered triples of integers (a,b,c)(a,b,c) satisfy ∣a+b∣+c=19,ab+∣c∣=97? \begin{aligned} |a+b|+c&=19,\\ ab+|c|&=97? \end{aligned}
00
44
66
1010
1212
Answer: E
Concepts:Diophantine equationsabsolute valuesfactorization
Small hint:
Set s=∣a+b∣s=|a+b|, so c=19−sc=19-s, and split according to the sign of cc
Big hint:
When c<0c\lt0, rewrite the resulting equations as products equal to 117117
Solution:
If c≥0,c\ge0, substitution and factoring lead to no pair consistent with the required sign of a+b.a+b. Thus c<0,c\lt0, so s>19s\gt19 and ab+s=116.ab+s=116. If a+b=s,a+b=s, then (a+1)(b+1)=117,(a+1)(b+1)=117, producing the unordered pairs {0,116},\{0,116\}, {2,38},\{2,38\}, {8,12}.\{8,12\}. If a+b=−s,a+b=-s, then (a−1)(b−1)=117,(a-1)(b-1)=117, producing {−116,0},\{-116,0\}, {−38,−2},\{-38,-2\}, {−12,−8}.\{-12,-8\}. Each unordered pair has two orders, giving 6⋅2=126\cdot2=12 triples. The correct answer is E.
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