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A lot of time, there are standalone questions in QA and DILR that focus on the concept of minimum number of weights required or the minimum number of weighings required to find the object with the different weight. In case you get lucky (or unlucky), you might find an entire set based on the logic. In this part of the article series, I will be covering the minimum weights required part.

### Sample questions:

1. Find the minimum number of weights required so that you can distribute all integral weights between 1 – 40 kg using a spring balance.

2. Find the minimum number of weights required to be able to measure all integral weights between 1 – 40 kg using a beam balance.

As you would have noticed, the only difference in these two questions is the fact that we are using different balances in both the cases. The difference between a beam balance and a spring balance is that in a beam balance, we can use both the pans whereas in a spring balance, you have a scale that shows you how much weight has been attached. Also remember that the cases are independent of each other. If you have to measure say 13 kg you would be using 1 kg, 3 kg and 9 kg in the second case. These three denominations can be used to measure 3 kg or 1 kg or 9 kg etc. as well.

Also, we can create a weight by using three things:

1. Get n times the lowest denomination weight and you would be able to measure everything from 1 till n. So, to measure everything from 1 till 31, you can have 31 weights of 1 kg each

2. Only by addition of weights and formation of clusters satisfying each individual condition

3. Both by addition and subtraction weights and formation of resultants that would satisfy each individual condition

Let’s see how to tackle the individual scenarios.

### The funda of the powers of 2

In the first case, where we had a spring balance, if we want to measure say 1 kg, we would need a 1 kg weight as it cannot be split further. To measure 2 kgs, we would require a 2 kg weight at the very least or two 1 kg weights. Similarly for 3 kgs, you would need either a 3 kg weight OR a 1 kg weight and a 2 kg weight OR three 1 kg weights. The point here is that, if you had a 1 kg weight for each and every denomination, to measure everything from 1 kg to 3 kg, you would need 3 weights of 1 kg each. Now, to minimize the number of weights, we have to check if there is any other combination of weights that ensures a sum from 1 kg till 3 kg. It is pretty much obvious that having weights of 1 kg and 2 kg would suffice in this case. Let’s look at a few more situations:

By now, it is clear that we need to consider powers of 2 in this case. So, whenever the case suggests that we use just one thing to measure a weight (be it in terms of spring balances, giving someone some money on a regular basis, etc.), we use powers of 2.

### The funda of the powers of 3

In the second case, we have a beam balance at our disposal. The good thing about a beam balance is that, we can use two pans and so, distribution of weights, and consequently addition and subtraction of weights is possible. So, if you have to make all possible combinations from say 1 to 13, you can use powers of 3 i.e. 1, 3 and 9.

### Which one to use where?

Simple. If you have two different things/people/objects where you can place your objects in, you go for the powers of 3 thing. If you have just one thing to measure a weight, you have to go for power of 2. A big indicator will be in terms of the value of n. If n is in the form of 2^n – 1, more often than not, you have use the powers of 2. If n is in the form of (3^(n+1) – 1)/2 of course, you have to use powers of 3.

### Cases involving irregular numbers

Say you have something different than what is there in the question. You have to measure everything from 1 till 45. How will you do it in that case? Simple. Just take the highest value of the possible sum using the powers of 2 or 3 depending on the case and take the additional value to be one weight. So, in this case, if we are using powers of 2, we get that 1, 2, 4, 8, and 16 would add up to 31. As the next term would take this sum up to 63, we would add only the remaining amount i.e. 45 – 31 = 14 to the existing set of weights. So, the set {1, 2, 4, 8, 16, 14} should enable us to weigh all possible integral values from 1 till 45. You may try out for some random values and check it out for yourself: 41 will be 14 + 1 + 2 + 8 + 16; 38 will be 14 + 8 + 16 and so on. Basically, we are filling 14 first and the rest can easily be fit because we know that anything from 1 till 31 is already possible.

Similarly for powers of 3, we can go ahead and say that {1, 3, 9, 27, 5} will suffice to measure everything from 1 till 45. So, 41 will be 5 + 9 + 27; 43 will be 5 + 27 + 9 + 3 – 1 and so on.

With practice, you will be able to check for cases more quickly and will be able to distinguish between the powers of 2 and 3 scenarios.

Try this one now:

In Pandora, where I come from, there is a currency called Miso. The Misos are made from Unobtanium which is an extremely rare element and so, needs to be used with utmost caution and minimal spending. The denominations are pretty flexible and I can have the denominations that I demand. As a part of my organization’s CSR initiative, I have to give a Miso a day to a charity. The good thing is that, I need not give the exact change and whatever surplus is there, the charity gives me back that very day, if possible. Assuming that the charity had no Misos to start with and there is only the two of us transacting, how many denominations of Misos do I need to order to fulfill the charity purpose without any shortcoming for a month?

I will cover the concept of minimum weighings in a coming article. Till then, all the best and happy prepping!

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