This is in continuation of 5 weights.

An ancient balance was found along with 6 weights, but only 4 of them were legible and, these were 1 gram, 2 grams, 7 grams and 87 grams. The remaining two weights were not legible and let us denote their respective weights by x grams and y grams.

Objects placed on one side of the balance could be counter-balanced by a weight on the other side of the balance to determine the object's weight. Also, the object and a weight(s) could be placed on one side of the balance and counter balanced to determine its weight, if necessary.

According to an old inscription, the values of x and y are such that:

(i) Using all the 6 weights, the ancients were able to determine the weight of objects weighting 1, 2, 3, ..., 174, 175 grams.

(ii) Each of x and y are integers with 7 < x < y, and:

(iii) x+y is the minimum amongst all possible pairs (x, y) that satisfy (i).

What are the two remaining weights?

As one such object that might be weighed against is 175 grams: 1+2+7+87+(x+y) = 175, the minimum for (x+y) must be 78.

One such pair (x, y) that will satisfy (i), (ii) and (iii) is (21, 57).

Posted by Dej Mar on 2012-06-25 04:08:02