How To Solve This Tricky Algebra Problem (XIV)

Welcome to the fourteenth entry in the tricky algebra problem series. This series celebrates the joy of algebra, where the featured problems range from beginner to advanced levels.

In this entry, we have the following equation as the starter:

x + xy + y = 64

Given the condition that both ‘x’ and ‘y’ are positive integers, your challenge is to figure out the solution for (x + y).

Do you think you can figure it out?

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Spoiler Alert

Beyond this section, I will be explicitly discussing the solution to this problem. If you feel like giving this problem a shot, I encourage you to pause reading at this point and work it out on your own.

Once you are done with your attempt, you may return and continue reading and comparing our respective approaches. All the best!

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Setting Up the Tricky Algebra Problem

When I originally started solving this problem, what did the trick for me was to take advantage of the only condition that we have in the original problem statement:

Both ‘x’ and ‘y’ are positive integers.

Based on this condition, we can derive a bunch of other useful properties that help us solve this problem. For example:

Since (x > 0) and (y > 0), (x*y > 0).

Another interesting fact to note is that the right-hand side of the equation is ‘64’, an even number. For this to be true, both ‘x’ and ‘y’ need to be (necessarily) even numbers. Here’s why:

x + xy + y = 64 [even]

1. [odd] + [odd * odd] + [odd] = [odd] + [odd] + [odd] = [odd] → Cannot Be 64

2. [odd] + [odd *even] + [even] = [odd] + [even] + [even] = [odd] → Cannot Be 64

3. [even] + [even *even] + [even] = [even] + [even] + [even] = [even] → Can Be 64

With these derived observations, we have more than enough information to help us solve our problem.

Solving the Tricky Algebra Problem

Let us start by factoring the left-hand side of the original equation as follows:

x + xy + y = 64

Factoring ‘x’ on the left-hand side:

x(1+y) + y = 64

The geometrically inclined reader may perceive the above equation as follows (pardon the subtle departure from algebra):

Here, we basically have a rectangle of area (x*y) squared units. On top of it, we have a rectangle of area (x*1) squared units. On the right, we have one last rectangle of area (y*1) squared units.

Together, they comprise the equation: x + xy + y = x(x + 1) + y = 64 (squared units).

Now, notice the notch on the right-hand top corner. It can be filled with a small square of area (1*1) squared units. In other words, all we need to do is to add 1 to both sides of the equation to arrive at the following result:

This result takes us just one final step away from our final solution.

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The Solution to the Tricky Algebra Problem

Weknow from the initial condition that ‘x’ and ‘y’ are positive integers. This means that (x + 1) and (y + 1) are respectively necessarily at least equal to 2.

With this at the back of the mind, let us examine the factors for 65:

1. 65 * 1 = 65

2. 1 * 65 = 65

3. 5 * 13 = 65

4. 13 * 5 = 65

Based on our conditions so far, we know that neither (x + 1) nor (y + 1) can be equal to 1. So, we can immediately eliminate the first two options.

The last two options are symmetrical (as are the first two). If we start with one of these, we arrive at the solution to our problem as follows:

If we had started with the assumption that (x + 1) = 13 and (y + 1) = 5, we would have x = 12 and y = 4, which would lead to the same result of (x + y) = 16.

I hope you enjoyed working on this problem. If this sort of thing interests you, keep an eye on this space for more. As always, thanks for reading!

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Further reading that might interest you:

• How To Really Use Mathematical Induction?
• The Three Prisoners Puzzle: How To Really Solve It?
• How To Really Solve The Dartboard Paradox?

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