How should I approach using two 8s and two 3s to make the number 24?

Question

Use two $8$s, two $3$s, and basic arithmetic operators ($ +, -, \times , \div$, parentheses) to make the number $24$.

(You may not join numbers together to form new numbers, like $ 8, 3\rightarrow 83$)

I don't know how to start besides just trying to find the correct answer. Is there a way you can make this equation through small steps or I should just bruteforce it?

Answer 1

While there are some good answers here, it seems like you are asking how to think of the answer. (If so, perhaps the title of this might need to be edited.)

Here's one method of thinking to get to the answer:

1) Is this a trick question?

It appears not - everything seems to be at face value, and there is a mathematics tag not a lateral thinking tag or similar.

2) What do we need to do?

What is the structure of the answer that you need to find? Well, it looks something like $8 + 8 - (3 + 3) = 10$. Except of course, this example equals 10, we need 24. But at least that's what we are going for. Another example is $8 + 8 - (3 \times 3) = 7$, but that doesn't work either. Not to worry just yet, we are just getting a feel of things.

3) Can we simplify the problem down at all?

Well, in this case, we can see that we can generate more potential solutions by changing the operators that we use. In fact, that's what we did above - we changed the $+$ in the brackets to $\times$, which changed the $6$ in the brackets to a $9$, which subtracted an extra $3$ from the result. The $8 + 8 = 16$ didn't change at all. Hmmm... there's something in that which we can use.

4) What components get us closer to the solution?

So the $16$ we had in both the proposals above is like its own starting point - that is, we can swap the two 8s from the original question for a 16, and make the question "Given a 16 and two 3s, make 24". That's not to say that we are going to find a solution to this, but it's one possible statement that will solve the original question. And it comes from us thinking about the number $16$. What other numbers can we make by consuming two of the numbers?

• $1 = 8 \div 8$ with $3,3$ leftover
• $16 = 8 + 8$ with $3,3$ leftover
• $64 = 8 \times 8$
• $0 = 8 - 8$
• $24 = 8 \times 3$ with $8,3$ leftover
• $11 = 8 + 3$
• $5 = 8 - 3$
• $2 \frac{2}{3} = 8 \div 3$
• ...

5) Work from the other end - what do the components of the solution look like?

Consider the solution: $ ? = 24$. What could those components possibly look like? Well, we know that $8 * 3 = 24$ - that's a good start, and can lead us to a potential solution:

$\sqrt{8 * 8 * 3 * 3} = 8 * 3 = 24$

I'm not completely happy with this though - it seems to me that using the square root is a bit of trickery. How else can we make 24 using one of our numbers?

• $8 * 3 = 24$
• $8 / \frac{1}{3} = 24$
• $27 - 3 = 24$
• $21 + 3 = 24$
• $32 - 8 = 24$
• ...

6) Connect the dots.

We now have a list of numbers that can be made with two of our numbers, and a list of numbers that we want to be made with 3 of our numbers. It might take a bit of inspiration, but is there any link we can make between any of them?

From the above, here's the link I've come up with:

$ 3 - 2 \frac{2}{3} = \frac{1}{3}$

That will lead us to a solution by putting it all together:

$8 \div \left(3 - \frac{8}{3} \right) = 24$

Fin

That's the way I think of these things. Hopefully you will get to a point where most of this occurs in your head pretty fast, and not necessarily in that order.

Answer 2

Here is a solution that uses only "elementary" operations (addition, subtraction, multiplication, and division).

$8 \div (3 - (8 \div 3))$ (or alternatively $\frac{8}{3 - \frac{8}{3}}$)
$= 8 \div \frac{1}{3}$
$= 24$

If we allow square roots, a simpler solution is possible.

$\sqrt{8 \times 8 \times 3 \times 3}$
$= 8 \times 3$
$= 24$

In fact, there are many more solutions if you allow more operations.