14th VMO 1976 Problem B1 Find all three digit integers abc = n, such that 2n/3 = a! b! c!
432 Solution 2n/3  444, so b, c ≤ 5. Consider first the case a = 4. Then (2n/3)/a! ≤ 2·455/(3·24)  40, so b, c ≤ 3. Hence b = c = 3. But (3/2)3!3!3! = 324 ≠ 333, so there are no solutions with a = 3. Suppose a = 2. Then n = 3b!c!, so 200  255 and 3·3!3!  103 and 3!3! < 67, so one of b, c must be 4. If the other is x, then 2.7 < x! < 4.3, so x! = 3 or 4. Contradiction. Thanks to Suat Namli 14th VMO 1976 © John Scholes jscholes@kalva.demon.co.uk 13 February 2004 Last corrected/updated 13 Feb 04