Puzzle: You have a common balance with two sides. You can only keep weighs on one side. You have to weigh items from 1 to 1000 weight units with integral weights only.

What is the minimum number of weights required and what will be their weights?

Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.

Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.

Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.

So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.

So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.

Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.

You can check Part 1 of this puzzle How many weights? Puzzle Part 1.

Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.

Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.

We can start from smallest number 1. To measure 1 we definitely need 1.

To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.

Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.

Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.

Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.

You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.

The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.

Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.

Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.

Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.

Source: Classic Puzzles

Solution: Amazing explanation provided by Kasturi.

You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:

Put 1: measure 1, weights in pan 1

To measure further you need to use a new weight.

Put 2: measure 3, weights in pan 1,2

Remove 1: measure 2, weights in pan 2

Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.

To measure further you need to use a new weight.

Put 4: measure 6, weights in pan 2,4

Remove 2: measure 4, weights in pan 4

Put 1: measure 5, weights in pan 1,4

Put 2: measure 7, weights in pan 1,2,4

Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.

Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.

Eventually, we can measure 125 packets in 125 movements.

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