20th APMC 1997 Problem 3 The 97 numbers 49/1, 49/2, 49/3, ... , 49/97 are written on a blackboard. We repeatedly pick two numbers a, b on the board and replace them by 2ab - a - b + 1 until only one number remains. What are the possible values of the final number?
We have (2a-1)(2b-1) = 2(2ab-a-b+1) - 1, so ∏ (2a-1) does not change. Its initial value is (97/1)(96/2)... (1/97) = 1. Hence its final value is also 1. So the final number satisfies 2a-1 = 1, so it must be 1. The original question had 48/1, ... , 48/97, but that allows an easier solution. Note that the original numbers include 1/2. But a, 1/2 → 1/2. So whatever order we pick the numbers 1/2 ends up gobbling all the others to leave 1/2. 20th APMC 1997 © John Scholes jscholes@kalva.demon.co.uk 28 Nov 2003 Last corrected/updated 28 Nov 03