This week’s puzzle was given to us by the same (now retired) maths teacher who gave us the Four Fours Puzzle.

The problem is to make a series of equations containing the numbers two, zero, two and zero (2020), in any order, 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 think that 30 is a good target, but go as far as you can.

In this puzzle you can combine numbers to make 20 or 22 (or 200 or 2020) so 20 + 20 = 40 is a valid equation. And you can raise to the power of 2 or 0, so 200 ^ 2 = 40,000 is also valid. You must use all four numbers each time - you can't use three!

We asked how long a sequence of equations you could produce - here is a list to 32, some from us and some from our solvers:

NB We've added some brackets that aren't really necessary (if you are familiar with BODMAS), but they make things clearer.

20 / 20 = 1

(2 / 2) + 0! + 0 = 2

2 + 2 - 0! + 0 = 3

2.0 + 2.0 = 4

2 + 2 + 0! + 0 = 5

2 + 2 + 0! + 0! = 6

(2 + 0!)! + 2 - 0! = 7

(2 + 0!)! + 2 + 0 = 8

(2 + 0!)! + 2 + 0! = 9

(2 / 0.2) + 0 = 10

(2 / 0.2) + 0! = 11

(2 + 0!)! + (2 + 0!)! = 12

(2 + 0!)! * 2 + 0! = 13

((2 + 0!)! + 0!) * 2 = 14

(2 + 0!) / 0.2 = 15

(2 + 0! + 0!) ^ 2 = 16

20 - (2 + 0!) = 17

20 - 2 + 0 = 18

20 - 2 + 0! = 19

22 - (0! + 0!) = 20

22 - (0! + 0) = 21

22 + 0 + 0 = 22

22 + 0! + 0 = 23

22 + 0! +0! = 24

((2 + 0!)! - 0!) ^ 2 = 25

(2 + 2)! + 0! + 0! = 26

(2 + 0!) ^ (2 + 0!) = 27

((2 + 0!)! / .2 (recurring)) + 0! = 28

(((2 + 0!)!) / .2) - 0! = 29

((2 + 0!)!) / 0.2 = 30

((2 + 0!)!) / .2) + 0! = 31

2 ^ ((2 + 0!)! - 0!) = 32