Assuming the prices are in whole cents, here is a very silly bruteforce: #ap, bp - prices before/after lunch sell_distrib = [(a,b,c) for a in range(10)

Assuming the prices are in whole cents, here is a very silly bruteforce: #ap, bp - prices before/after lunch sell_distrib = [(a,b,c) for a in range(10) for b in range(16) for c in range(26) if a > b > c] for bp in range(0, 3501): for x in sell_distrib: ap = (3500-x[0]*bp) // (10-x[0]) if x[0]*bp + (10-x[0])*ap == 3500 and \ x[1]*bp + (16-x[1])*ap == 3500 and \ x[2]*bp + (26-x[2])*ap == 3500: print("Prices: {0} {1}".format(bp, ap)) print("Before lunch sales: {}, {}, {}".format(x[0], x[1], x[2])) Note: it's Python 3 (the division!) and first line in the conditional eliminates cases where after lunch price wouldn't be a whole number because then it wouldn't add up to 3500. It doesn't consider cases where one farmer sold everything before lunch (supposedly sales didn't go so well). It's two lines of code more if we want to check those cases. Eliminating of 0 price and requirement that after lunch price is lower than before lunch one is covered by noticing that the solution then must have farmer A selling more than farmer B who in turn sell more than farmer C before lunch. There is a unique solution in whole cents which is nice (no spoilers, you can copy-paste the code to Python interpreter to see it)