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