33rd Eötvös 1929 Problem 1 Coins denomination 1, 2, 10, 20 and 50 are available. How many ways are there of paying 100?
For 2n there are n+1 ways using just the 1 and 2 coins (0, 1, 2, ... or n 2s). Similarly, there are m+1 ways of paying 20m or 20m+10 with just the 10 and 20 coins. Hence using just the 10, 20, 50 coins we get (for example there are 4 ways of paying 70 without using a 50 and 2 ways with, total 6): 0 10 20 30 40 50 60 70 80 90 100 0 1 2 2 3 4 5 6 7 8 10 Hence there are 1·51 + 1·46 + 2·41 + 2·36 + 3·31 + 4·26 + 5·21 + 6·16 + 7·11 + 8·6 + 10·1 = 784 ways in total. 33rd Eötvös 1929 © John Scholes jscholes@kalva.demon.co.uk 1 Nov 2003 Last corrected/updated 1 Nov 03