Competition problem — Algebra — Maths Olympiad Prep

Library / /91 of 348

Algebra Difficulty 4.8 AIME Find the answer

Let $a, b, c$ be positive integers such that $\frac{a}{77}+\frac{b}{91}+\frac{c}{143}=1$ . What is the smallest possible value of $a+b+c$ ?

Your answer

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

Solution

We need $13 a+11 b+7 c=1001$ , which implies $13(a+b+c-77)=2 b+6 c$ . Then $2 b+6 c$ must be divisible by both 2 and 13, so it is minimized at 26 (e.g. with $b=10, c=1$ ). This gives $a+b+c=79$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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