Problem:
How many ordered triples of integers (a,b,c) satisfy
∣a+b∣+c=19 and ab+∣c∣=97?
Answer Choices:
A. 0
B. 4
C. 6
D. 10
E. 12
Solution:
If c≥0, then ab−∣a+b∣=78, so (a−1)(b−1)=79 or (a+1)(b+1)=79. Since 79 is prime, {a,b} is {2,80},{−78,0},{0,78}, or {−80,−2}. Hence ∣a+b∣=78 or ∣a+b∣=82, and, from the first cquation in the problem statement, it follows that c<0, a contradiction.
On the other hand, if c<0, then ab+∣a+b∣=116, so (a+1)(b+1)=117 in the case that a+b>0 and (a−1)(b−1)=117 in the case that a+b<0. Since 117=32⋅13, we distinguish the following cases:
{a,b}={0,116}{a,b}={2,38}{a,b}={8,12}{a,b}={−116,0}{a,b}={−38,−2}{a,b}={−12,−8} yields yields yiclds yields yields yields c=−97;c=−21;c=−1;c=−97;c=−21;c=−1;
Since a and b are interchangeable, each of these cases leads to two solutions, for a total of 12.