37th BMO 2001 Problem 1 A has a marbles and B has b < a marbles. Starting with A each gives the other enough marbles to double the number he has. After 2n such transfers A has b marbles. Find a/b in terms of n.
After 2 transfers A has 2(a - b) = 4a - 2N marbles, where N = a + b. Suppose A has a n marbles after 2n transfers. We have just shown that a n+1 = 4a n - 2N. Put c n = a n - 2N/3, then c n+1 = 4c n . We have c 0 = a - 2N/3, so c n = 4 n a - 4 n 2N/3. Hence a n = 4 n a - 4 n 2N/3 + 2N/3 = (4 n + 2)a/3 - 2(4 n - 1)b/3. We require a n = b and hence a/b = (2.4 n + 1)/(4 n + 2). Note that we have not so far proved that there is a solution for n. We have just showed that if there is a solution then a/b has that ratio. It might be that one person has a negative number of marbles at some point. Fortunately, we do not have to worry whether the question requires us to show that n is possible, because it is easy to show that it is. The key observation is that if one person goes negative, then they remain negative. But after 2n moves neither A nor B is negative, so they cannot have been negative at any intermediate stage. 37th BMO 2001 © John Scholes jscholes@kalva.demon.co.uk 18 Jul 2002