- 30 Oct '11 19:47The farmer uses the 40 Kg stone on a balance scale to weigh out 40 kg of feed hay bales. His buddy down the lane asks to borrow the balance stone for a while. He says sure. The buddy comes back a few days later, says 'sorry, I broke the stone accidentally into 4 pieces'. He comes back to apologize again a few days later and says sorry again. The farmer says, its ok, the four pieces you broke it into allows me to weigh anything from 1 kg to 40 kg just using the stones and the balance.

What is the weight of each broken piece that allows the farmer to weigh anything in 1 kg increments from 1 to 40 kg just using the pieces and the balance? - 30 Oct '11 19:53

Should be in powers of three. 1, 3, 9 and 27.*Originally posted by sonhouse***The farmer uses the 40 Kg stone on a balance scale to weigh out 40 kg of feed hay bales. His buddy down the lane asks to borrow the balance stone for a while. He says sure. The buddy comes back a few days later, says 'sorry, I broke the stone accidentally into 4 pieces'. He comes back to apologize again a few days later and says sorry again. The farmer says ...[text shortened]... er to weigh anything in 1 kg increments from 1 to 40 kg just using the pieces and the balance?** - 30 Oct '11 22:15 / 1 editA lovely problem. Here's a variant;

a mathematician gets a kick out of the 40lb stone problem, and goes one up. He has a superb balance scale that can differentiate weights all the way to the precision of the weight of a single grain of sand (0.001 grams) but is sturdy enough for a cow, which can weigh a bit over 1,000 kg, the equivalent of over one billion grains of sand. The above method says that it could be possible to have 20 weights that can tell you how many grains' worth any item up to cow size weighs.. but the mathematician is willing to bet that he can do it with less than 20 weights. Can he?