15th Irish 2002 Problem B2 n = p · q · r · s, where p, q, r, s are distinct primes such that s = p + r, p(p + q + r + s) = r(s - q) and qs = 1 + qr + s. Find n.
2002 Solution s = p + r implies p or r is 2. Since the primes are all distinct, the others must be odd. So s - q is even and (p + q + r + s) is odd. But p(p + q + r + s) = r(s - q) = even. Hence p must be even and hence 2. So s = r + 2. Now qs = 1 + qr + s gives r = 2q - 3, so s = 2q - 1. So finally p(p + q + r + s) = r(s - q) gives 2q 2 - 15q + 7 = 0, or (2q - 1)(q - 7) = 0, so q = 7. Hence the primes are 2, 7, 11, 13 and n = 2002. 15th Irish 2002 © John Scholes jscholes@kalva.demon.co.uk 15 June 2003 Last updated/corrected 15 Jun 03