Can You Really Solve This Tricky Math Puzzle?

I recently came across this tricky math puzzle that stumped me. So, I thought I’d write about it. On a related note, because of my work with puzzles and math problems, I am slowly starting to receive interesting puzzles and challenges from avid readers and enthusiasts. So, you can safely expect more puzzle-related content from me in the future.

Back to our puzzle now — Say that you are a grocer who owns a weighing scale. You are looking to buy individual weights that enable you to balance objects that weigh up to 40 Kilograms (Kg).

At the same time, you are determined to buy the lowest number of discrete weights necessary to achieve this.

To be clear with the requirements, the weights you are looking for measure only integer units of Kilograms. In other words, you are looking for the minimum number of discrete weights with a resolution of 1 Kilogram that collectively enable you to balance objects that weigh up to 40 Kilograms.

What are these discrete weights and how many of them do you need in total?

Spoiler Alert

Beyond this section, I will be explicitly discussing the solution to this puzzle. So, if you intend to solve the puzzle on your own, I recommend that you pause reading this essay at this point.

After you attempt to solve the puzzle, you may come back and resume reading. All the best!

This essay is supported by Generatebg

---

My Initial Attempt at Solving this Tricky Math Puzzle

When I started, it was immediately obvious to me that the sum of all of my discrete weights needs to be 40 Kilograms. Furthermore, I would also need to pile up weight-combinations such that ALL integers from 1 to 40 could be captured.

To start, I’d need weight combinations that allow me to measure all weights from 1 Kilogram up to 10 Kilograms. From thereon, I could just get weight multiples of 10 Kilograms to cover the rest of my requirement.

For example, consider the following list of weights:

a) 1 Kg.

b) 2 Kg.

c) 2 Kg.

d) 5 Kg.

e) 10 Kg.

f) 10 Kg.

g) 10 Kg.

With such a combination, I would be able to balance any object that weighs up to 40 Kilograms. For example, if I needed to balance an object that weighs 11 Kilograms, I just need to use one 10 Kilogram weight and the one 1 Kilogram weight.

Below are a few other examples to drive home this logic:

a) 39 Kg = (10 + 10 + 10 + 5 + 2 + 2) Kg.

b) 25 Kg = (10 + 10 + 5) Kg.

c) 19 Kg = (10 + 5 + 2 + 2) Kg.

If I were to settle for this solution, I would need a total of seven weights to balance any object that weighs up to 40 Kilograms. The question is: Is this the lowest number of discrete weights necessary?

The answer: of course not!

The Missing Piece of the Puzzle

You see, I had somehow missed an important piece of the puzzle in my first attempt. I had treated this as a uni-directional puzzle. In reality, this is a bi-directional puzzle. Thus far, I have only been manipulating the weights on one of the weighing pans. But the truth is that I have the freedom to manipulate the weights on both pans!

If this sounds confusing, stick with me; it will soon become clear. Let us say that you wish to balance objects that weigh up to 4 Kilograms. What is the minimum number of weights we will need to do this?

Using my former approach (manipulating weights only on one of the pans), I would need a 1 Kilogram weight and two 2 Kilogram weights. This way, I could cover all integers from 1 to 4 as follows:

a) 1 Kg = 1 Kg.

b) 2 Kg = 2 Kg.

c) 3 Kg = (2 + 1) Kg.

d) 4 Kg = (2 + 2) Kg.

But if we manipulate weights on both the pans, all of a sudden we don’t need three weights to balance the scale, but just two: a 1 Kilogram weight and a 3 Kilogram weight. Here’s how:

a) 1 Kg = 1 Kg.

b) 2 Kg = (3–1) Kg.

c) 3 Kg = 3 Kg.

d) 4 Kg = (3 + 1) Kg.

The “(3–1) Kg” above refers to the 3 Kilogram weight placed on an empty weighing pan while the 1 Kilogram weight is placed on the pan holding the object weighing 2 Kilograms (to be balanced).

Now that we have made this realisation, we are ready to solve this puzzle.

---

The Solution to the Tricky Math Puzzle

As soon as we start manipulating weights on both weighing pans, the power of negative integers becomes available to us. We have so far managed to cover up to 4 Kilograms using two weights.

The goal now becomes to jump as far out as possible using the next weight. Consider a 9 Kilogram weight as our next weight. Then, we would be able to balance up to 13 Kilograms using just the three weights:

a) 5 Kg = [9 — (3 + 1)] Kg.

b) 6 Kg = (9–3) Kg.

c) 7 Kg = [(9+1) — 3] Kg.

d) 8 Kg = (9–1) Kg.

e) 9 Kg = 9 Kg.

f) 10 Kg = (9 + 1) Kg.

g) 11 Kg = [(9 + 3) — 1] Kg.

h) 12 Kg = (9 + 3) Kg.

i) 13 Kg = (9 + 3 + 1) Kg.

Since we have covered up to double-digit values with our three weights so far, we can safely jump up 40 Kilograms with our next weight. To do this, all we have to do is compute 40–13 = 27. Therefore, our final weight is a 27 Kilogram weight.

The solution to this puzzle is that we need just 4 weights to balance up to 40 Kilograms: 1, 3, 9, and 27 Kilograms respectively.

Closing Comments

It is interesting to note that all of these weights are powers of three. When I looked into this further, I realised that this puzzle can also be treated as a combinatoric problem.

Each of our weights has three possible states: it can be on the left weighing pan, on the right pan, or completely away from the weighing scale. Correspondingly, when we consider 3 weights, we have 3³ = 27 combinatoric possibilities.

Since we divide our problem space equally into positive and negative integers (by manipulating both weighing pans), we also divide our possibilities into 27/2 = 13 possibilities (rounded down).

Similarly, when we consider 4 weights, we have 3⁴ = 81 combinatoric possibilities. By dividing our problem space equally between positive and negative integers (by manipulating both weighing pans), we also divide our possibilities into 81/2 = 40 possibilities (rounded down). So, 4 weights suffice our requirement!

---

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you:

• How To Really Solve The Monkey And The Coconuts Puzzle?
• Can You Really Solve This Tricky Logic Puzzle (II)?
• What Is So Special About 69! ?

If you would like to support me as an author, consider contributing on Patreon.