Problem:
Integers a,b, and c satisfy ab+c=100,bc+a=87, and ca+b=60. What is ab+bc+ca ?
Answer Choices:
A. 212
B. 247
C. 258
D. 276
E. 284
Solution:
Notice that the difference between 100 and 87 is 13 , a prime number. This fact will help to simplify the problem. Subtract the second equation from the first to get
13=(ab+c)−(bc+a)=ab−bc−a+c=b(a−c)−(a−c)=(b−1)(a−c)
Thus b−1=±1 or b−1=±13.
- If b−1=−1, then b=0, implying c=100,a=87, and ca=60, which is impossible.
- If b−1=1, then b=2 and a−c=13, implying ca=58=2⋅29, which cannot be true if a−c=13.
- If b−1=13, then b=14 and a−c=1, implying ca=46=2⋅23, which cannot be true if a−c=1.
- If b−1=−13, then b=−12 and a−c=−1, implying ca=72, which is satisfied when a=−9 and c=−8. In fact, a=−9,b=−12, and c=−8 satisfies all three equations.
The requested value is ab+bc+ca=(−9)(−12)+(−12)(−8)+(−8)(−9)=108+96+72=(D)276.
Note: There are also four noninteger solutions. When written in the form (a,b,c), these solutions are approximately (0.594,0.869,99.484),(1.715,57.455,1.484),(7.477,12.525,6.349), and (86.214, 1.152, 0.683).
The problems on this page are the property of the MAA's American Mathematics Competitions