Maths Olympiad Prep

Library / /13 of 13

, 2013

Number theory Difficulty 8.5 Shortlist Prove it Saudi Arabia

For positive integers aa and bb, gcd(a,b)\operatorname{gcd}(a, b) denotes their greatest common divisor and lcm(a,b)\operatorname{lcm}(a, b) their least common multiple. Determine the number of ordered pairs (a,b)(a, b) of positive integers satisfying the equation
ab+63=20lcm(a,b)+12gcd(a,b). a b+63=20 \operatorname{lcm}(a, b)+12 \operatorname{gcd}(a, b).

Solution

Let d=gcd(a,b)d=\operatorname{gcd}(a, b) and a=daa=d a', b=dbb=d b' with a,ba', b' two relatively prime positive integers. The equation becomes
abd2+63=20abd+12d a' b' d^2 + 63 = 20 a' b' d + 12 d
Therefore, dd divides 6363 and we have
abd+63d=20ab+12 a' b' d + \frac{63}{d} = 20 a' b' + 12
If 5<d<205 < d < 20, then abd+63d<20ab+12a' b' d + \frac{63}{d} < 20 a' b' + 12, and this is impossible. Hence, d=1,3,21d = 1, 3, 21, or 6363.

1. If d=1d = 1, the equation is equivalent to 51=19ab51 = 19 a' b', which is impossible since 1919 does not divide 5151.

2. If d=3d = 3, the equation is equivalent to 9=17ab9 = 17 a' b', which is impossible since 1717 does not divide 99.

3. If d=21d = 21, the equation is equivalent to ab=9a' b' = 9. Because a,ba', b' are relatively prime positive integers, either a=1,b=9a' = 1, b' = 9 or a=9,b=1a' = 9, b' = 1. This leads to two solutions (a,b)=(21,189)(a, b) = (21, 189) and (a,b)=(189,21)(a, b) = (189, 21).

4. If d=63d = 63, the equation is equivalent to 43ab=1143 a' b' = 11, which is impossible since 4343 does not divide 1111.

Therefore, the equation has two solutions (a,b)=(21,189)(a, b) = (21, 189) and (a,b)=(189,21)(a, b) = (189, 21).

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: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.