22nd Australian 2001 Problem A1 Let L(n) denote the smallest positive integer divisible by all of 2, 3, ... , n. Find all prime numbers p, q such that q = p + 2 and L(q) > q L(p). Solution
: p = 3, q = 5 is the only solution. L(q) > q L(p) iff p+1 does not divide q L(p). If p+1 is not a prime power, then it has two relatively prime factors less than p which must divide L(p) and hence p+1 divides L(p). So suppose p+1 = r n . But then r - 1 divides p, which is prime, so either r = 2 or n = 1. We cannot have n = 1, because then we would have three consecutive primes p, p+1, p+2 which is not possible. So p = 2 n - 1, q = 2 n + 1. But if n is odd, then 2 n + 1 = 0 mod 3, and if n is even, then 2 n - 1 = 0 mod 3. So the only possibilities are n = 1 or 2. n = 1 does now work because then p = 1, which is not prime. n = 2 gives p = 3, q = 5. So L(p) = 6, L(q) = 60 > 30 = q L(p). 22nd Aus 2001 (C) John Scholes jscholes@kalva.demon.co.uk 25 July 2002