Hey Kiwi, I agree that the question I gave earlier was of bit high standard. Thought people would try. Anyway?. will come up with easier ones next time. Here is the mathematical proof for your question. It looks ok to me. Let a, b, c, d be the 4 pieces in which the weight is divided. where a < b< c< d. The first equation will be: a + b+ c+d = 40------ (1) (sum of all 4 pieces). The second equation will be: (a +b+ c) = d ? (a +b+ c) -1 (The sum of the first three pieces should be 1 less than the difference between the last piece and the sum of the first three pieces. They can?t be equal other wise we will have one particular weight with two combinations). Simplifying that further, we get, 2(a +b+ c) = d - 1 -------- (2) The third equation will be: (a + b) = c ? (a + b) ? 1 (same reason as above. The sum of the first two pieces should be 1 less than the difference between the third piece and the sum of the first two pieces. Simplifying, 2 (a + b) = c ? 1------- (3) The fourth equation will be: a = b ?a ? 1 (same reason as above) 2a = b-1--------------- (4) There is a nice pattern in these equations. Plug in equation (2) in (1), you get d = 27. Plug in equation (3) in (1), you get c = 9 (By this time, you already know that d = 27). Plug in equation (4) in (1), you get, b = 3 (you already know c, and d) Now find a, which is 1. Kiwi, chiplina chai bhayena ho. Chances are that you might get hurt too. :)