Problem:
Let p and q be positive integers such that
95<qp<74
and q is as small as possible. What is q−p ?
Answer Choices:
A. 7
B. 11
C. 13
D. 17
E. 19
Solution:
The first inequality is equivalent to 9p>5q, and because both sides are integers, it follows that 9p−5q≥1. Similarly, 4q−7p≥1. Now
631=74−95=(qp−95)+(74−qp)=9q9p−5q+7q4q−7p≥9q1+7q1=63q16
Thus q≥16. Because
168<95<169<74<1610
the fraction 169 lies in the required interval, but 168 and 1610 do not. Therefore when q is as small as possible, q=16 and p=9, and the requested difference is 16−9=7.
Note: A theorem in the study of Farey fractions states that if pa<qb and bp−aq=1, then the rational number with least denominator between pa and qb is p+qa+b.
The problems on this page are the property of the MAA's American Mathematics Competitions