How To Really Solve This Tricky Math Problem?

I recently came across this tricky math problem and found it interesting to solve. So, I thought I’d write about it. It has also been a while since I wrote about mathematics. So, here we are.

To start, you are given the following equation:

2^x + 4^y + 8^z = 328

where x, y, and z are positive integers.

Given this equation, your challenge is to figure out the possible values for ‘x’, ‘y’, and ‘z’. Do you think you can solve this problem?

If you wish for it, I can provide you with an optional hint.

Optional Hint

This problem becomes needlessly complicated if you employ algebraic manipulation. This is also one of the reasons why I didn’t slot this problem under my algebra series.

I employed discrete mathematics and combinatoric reasoning to solve this problem very quickly. In fact, you wouldn’t even need discrete mathematics; combinatoric reasoning alone suffices to solve this problem. And that is your hint.

Spoiler Alert

Beyond this section, I will be explicitly discussing the solution to this problem. If you wish to solve it on your own, I recommend that you pause reading this essay at this point and give it a try.

After you are done with your attempt, you may continue reading the essay and compare our approaches.

This essay is supported by Generatebg

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Setting Up the Tricky Math Problem

Let us start by labelling the given equation as equation-1 for ease of reference going forward:

We know that 2² = 4 and 2³=8. Therefore, the left-hand side of equation-1 can be expressed as a sum of powers of 2. Let us label this simplified equation as equation-2:

Note that the left-hand side of equation-2 is a sum of three powers of 2. On the other hand, the right-hand side features an even number. Did you know that any even number can be expressed as a sum of powers of 2?

The Power of Discrete Mathematics

The key to solving this problem is just expressing the right-hand side (328) as a sum of three powers of 2. This is where I employed my knowledge from discrete mathematics.

One of the first things that I did when learning discrete mathematics was to memorise powers of 2 from 0 up to 12. If you are interested in doing this as well, here’s the list (I’ll explain how to use it in a bit):

In discrete mathematics, memorising this list is akin to memorising times-tables in conventional mathematics. Note that proper memorisation allows you to recall any power of 2 without having to work it out from the start or end.

For example, I would be able to tell you directly that 2⁹ = 512 instead of working it out by multiplying with 2 from 2³ forwards or by dividing by 2 from 2¹⁰ backwards. Now, back to our problem.

The right-hand side of equation-1 features 328. From memory, I can tell directly that 328 is not a power of 2. So, the question we need to ask becomes:

Which is the largest power of 2 that is smaller than 328?

The answer is 2⁸ = 256. When you subtract 256 from 328, you get 72. Now, apply the same logic to 72. We know straightaway that 72 is not a power of 2. The largest power of 2 that is smaller than 72 is 2⁶ = 64. When you subtract 64 from 72, you get 8.

Finally, let us apply this logic to 8 as well. We know that 2³ = 8. Ideally, all of this happens in the head. But writing this stuff out is fine as well.

Using what we have figured out thus far, we can transform equation-2’s right-hand side (328) into a sum of three powers of 2. Let us label this new transformed equation as equation-3:

With this equation, we are now ready to solve our problem.

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How to Solve this Tricky Math Problem?

Equation-3 features three powers of 2 on the left-hand side and three powers of 2 on the right-hand side. So, simply by matching, we know for sure that one of the exponents has to be equal to 8, another one has to be equal to 6, and the final one equal to 3.

The exponents in question are ‘x’, ‘2y’, and ‘3z’ respectively. Note here that ‘2y’ can only be an even exponent. Therefore, it cannot be equal to 3. This leads us to two cases:

Case 1: (2y = 6), or

Case2: (2y = 8).

If 2y = 6, then ‘3z’ has to be equal to 3 because 8 is not divisible by 3. This means that x will be equal to 8.

If 2y = 8, then ‘3z’ can either be equal to 3 or 6. If 3z = 6, then x = 3. On the other hand, if 3z = 3, then x = 6.

Consequently, we have three possible solutions to this tricky math problem:

There we go. I hope you enjoyed solving this problem as much as I did.

If solving math problems like these is your thing, then watch this space for more such fun problems in the future!

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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! ?

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