A few months ago we shared a puzzle that was given to us by a maths teacher who used to set it as a revision exercise. It involved making equations using four fours to make one, two, three and so on.

He has now given us a second, very similar, and perhaps more challenging problem - do the same with three threes!

So, you have to make a series of equations containing three threes and any number of common mathematical symbols. The first equation must total one, the second equation must total two, and so on . . . as far as you can go. We got to 50 with four fours - how far can you extend this sequence.

In this puzzle you can combines threes to make 33 (or 333) so 33 * 3 = 99 is a valid equation. And you can raise to the power of 3, so 3 ^ 3 * 3 = 81 is also valid. You must use three threes each time - you can't use two!

Below are answers up to 21 - some from us and some from Social Media posts. No one seems to be able to do 22, so if you can do it, let us know. Also, as a bonus, we've done 23.

We have used brackets in some places where they are not needed (if you understand BODMAS) for extra clarity.

$$ 3! / (3 + 3) = 1 $$

$$ 3 - (3 / 3) = 2 $$

$$ 3 + 3 - 3 = 3 $$

$$ 3 + (3 / 3) = 4 $$

$$ 3! - (3 / 3) = 5 $$

$$ 3! + 3 - 3 = 6 $$

$$ 3 / .3 - 3 = 7 $$

$$ 3! + (3! / 3) = 8 $$

$$ 3 + 3 + 3 = 9 $$

$$ (\sqrt{3} * \sqrt{3} / .3 = 10 $$

$$ 33 / 3 = 11 $$

$$ 3 * 3 + 3 = 12 $$

$$ 3 / .3 + 3 = 13 $$

$$ 3! / .3 - 3! = 14 $$

$$ 3! + 3! + 3 = 15 $$

$$ 3 / .3 + 3! = 16 $$

$$ (3! / .3) - 3 = 17 $$

$$ 3! + 3! + 3! = 18 $$

$$ 3 * (3! + .3 [recurring - can't do the dot]) = 19 $$

$$ (3 + 3) / .3 = 20 $$

$$ 3 ^ 3 - 3! = 21 $$

$$ (3! / .3) + 3 = 23 $$