You have been given
sacks having indefinite amount of grains. Except
sack every sack has
weighted grains, and the
exceptional sack has
weighted grains. You have been given a weighing machine, in how many minimum number of usage of weighing machine you can identify the sack.
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Arrange all the sacks in one line and number it 1 , 2 , 3 . . . . , n . Now take 1 grain from 1 s t sack, 2 grains from 2 n d sack and so on.
Now if all the sack would have contained 1 0 g r a m s grains then, weight would have been - 1 0 ∗ ( 1 + 2 + 3 + 4 . . . . + n ) = 1 0 ∗ { 2 n ( n + 1 ) }
but actually weight that we will get is - 1 0 ∗ 1 + 1 0 ∗ 2 . . . . + 9 ∗ k . . . . + 1 0 ∗ n = 1 0 ∗ 1 + 1 0 ∗ 2 . . . . + 1 0 ∗ k − k . . . . + 1 0 ∗ n = 1 0 ∗ { 2 n ( n + 1 ) } − k
Now, on subtracting the weight that we calculated earlier and the weight that we got, we see that we are left with k which is the sack number. So only once the weighing machine was used.