Maths Olympiad Prep

Track / Stage 4 / 218 of 340 #478 of 1964

Problem 478

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer

6. Problem: How many pairs of positive integers (a,b)(a, b) with b\leq b satisfy 1a+1b=16\frac{1}{a}+\frac{1}{b}=\frac{1}{6} ?

A number or a short expression. Spacing and $ signs are ignored.

Official solution

Solution: 1a+1b=16a+bab=16ab=6a+6bab6a6b=0\frac{1}{a}+\frac{1}{b}=\frac{1}{6} \Rightarrow \frac{a+b}{a b}=\frac{1}{6} \Rightarrow a b=6 a+6 b \Rightarrow a b-6 a-6 b=0. Factoring yields (ab)(b6)36=0(a-b)(b-6)-36=0. Then (a6)(b6)=36(a-6)(b-6)=36. Because aa and bb are positive integers, only the factor pairs of 36 are possible values of a6a-6 and b6b-6. The possible pairs are:
a6=1,b6=36a6=2,b6=18a6=3,b6=12a6=4,b6=9a6=6,b6=6 \begin{array}{l} a-6=1, b-6=36 \\ a-6=2, b-6=18 \\ a-6=3, b-6=12 \\ a-6=4, b-6=9 \\ a-6=6, b-6=6 \end{array}

Because aba \leq b, the symmetric cases, such as a6=12,b6=3a-6=12, b-6=3 are not applicable. Then there are 5 possible pairs.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.