This week’s puzzle was given to us by a maths teacher who used to set it as a revision exercise.

The problem is to make a series of equations containing four fours 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. The teacher said that the one pupil produced equations from one all the way to 32 in one hour. As you’ve got as much time as you want, how far can you extend the sequence.

In this puzzle you can combines fours to make 44 (or 444 or 4,444) so 44 + 44 = 88 is a valid equation. And you can raise to the power of 4, so 4 ^ 4 + 4 +4 = 264 is also valid. You must use four fours each time - you can't use three!

So how long a sequence of equations can you produce?

As usual you can post the answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

Quiz Master ShopWe didn’t realise that the problem had been solved in this way, but it has proved to be a very popular puzzle

MathleteThis can be done up to infinity, as proven in the Numberphile video: https://www.youtube.com/watch?v=Noo4lN-vSvw

Using the notation log(a, b) as log base a of b, and sqrt© as the square root of c, then we have:

0 – log(sqrt(4)/4, log(4, 4))

1 – log(sqrt(4)/4, log(4, sqrt(4)))

2 – log(sqrt(4)/4, log(4, sqrt(sqrt(4))))

3 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(4)))))

4 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(4))))))

5 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(4)))))))

6 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4))))))))

7 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4)))))))))

8 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4))))))))))

9 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4)))))))))))

10 – log(sqrt(4)/4, log(4, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(4))))))))))))

…and so on. You see the pattern.

GeoffSequences are stupid