4th CaMO 1972 Problem 10 Find the longest possible geometric progression in {100, 101, 102, ... , 1000}.
128, 192, 288, 432, 648, 972 has six terms and ratio 3/2. We show it is the best possible. The ratio must be rational. Suppose it is h/k. If there are more than 6 terms, then the first term must be divisible by k 6 and the last term must be divisible by h 6 . If k >= 3, then h > k, so h >= 4. But 4 6 = 4096 > 1000, so the last term is not divisible by h 6 . So k < 3. We cannot have k = 1, because then the last term is at least h 6 times the first term, which is impossible. So k = 2, h = 3. So the first term must be a multiple of 64. The smallest such is 128. But then we only get 6 terms. 4th CaMO 1972 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002