# Make numbers 1 - 30 using the digits 2, 0, 2, 4

Try to make all numbers 1-30 using the digits 2, 0, 2, 4.

Rules:

Use all four digits exactly once.

Allowed operations: $$+, -, \times, ÷, !$$ (factorial), x^y (exponentiation), √ (square root).

Parentheses and concatenation are allowed. (e.g. 20+(2∗4)).

Squaring uses the digit 2.

Keep the order "2, 0, 2, 4" in all expressions.

The modulus operator is not allowed.

Rounding is not allowed.

• Some cosmetic corrections should be applied, as there are 2 twos and no nine, so keep the order "2 0 2 4".
– z100
Commented Jan 2 at 16:44
• Is the − operator supposed to include both subtraction and and arithmetic negation? Commented Jan 5 at 17:44

Complete solution that doesn't use concatenation.

$$1 = 2 + 0! + 2 - 4 \\ 2 = 2 \times 0 - 2 + 4 \\ 3 = 2^0 -2 + 4 \\ 4 = 2 + 0 - 2 + 4 \\ 5 = -2^0 + 2 + 4 \\ 6 = 2 \times 0 + 2 + 4 \\ 7 = 2^0 + 2 + 4 \\ 8 = 2+0+2+4 \\ 9 = 2^0 + 2 \times 4 \\ 10 = 2 + 0 + 2\times 4 \\ 11 = 2 + 0! + 2 \times 4 \\ 12 = 2^{0!+2} + 4 \\ 13 = -2 - 0! + 2^4 \\ 14 = -2 + 0 + 2^4 \\ 15 = -2^0 + 2^4 \\ 16 = 2 \times 0 + 2^4 \\ 17 = 2^0 + 2^4 \\ 18 = 2+0 + 2^4 \\ 19 = 2 + 0! + 2^4 \\ 20 = -2 + 0 - 2 + 4! \\ 21 = -2^0 - 2 + 4! \\ 22 = 2 \times 0 - 2 + 4! \\ 23 = 2^0 - 2 + 4! \\ 24 = 2 + 0 - 2+ 4! \\ 25 = 2 + 0! - 2 + 4! \\ 26 = 2 \times 0 + 2 + 4! \\ 27 = 2^0 + 2 + 4! \\ 28 = 2 + 0 + 2 + 4! \\ 29 = 2 + 0! + 2 +4! \\ 30 = (2 + 0!) \times 2 + 4!$$

• You can avoid leading minus with 5: (2+0)/2 + 4 and 20: (2+0!+2)*4. Commented Jan 3 at 10:08
• @ZizyArcher Thanks. Actually, I was trying to avoid brackets ... Commented Jan 3 at 11:48

I'm making this a community wiki so the answers aren't split among multiple…answers.

1

$$1=2^{0^{2^4}}$$

2

$$2=\frac{2^0}{2}\cdot4=-2^0\cdot2+4$$

3

$$3=2^0-2+4=-(2+0)/2+4$$

4

$$4=2+0-2+4=2+0\cdot 2+\sqrt{4}$$

5

$$5=\frac{(2+0)}{2}+4=-2^0+2+4=2^{0^2}+4$$

6

$$6=2\cdot0+2+4$$

7

$$7=2^0+2+4$$

8

$$8=2+0+2+4=2^{0+2}+4$$

9

$$9=2^0+2\cdot4$$

10

$$10=2+0+2\cdot4$$

11

$$11=2+0!+2\cdot4$$

12

$$12 = (2+0) \times (2+4)$$

13

$$13=(2+0!)^2+4=-2-0!+2^4$$

14

$$14=2(0!+2+4)$$

15

$$15=-2^0 + 2^4$$

16

$$16=2^{0+2}\cdot4$$

17

$$17=2^0+2^4$$

18

$$18=2+0+2^4$$

19

$$19=2+0!+2^4$$

20

$$20=(2+0!+2)\cdot4=-2-0-2+4!$$

21

$$21=-2^0-2+4!$$

22

$$22=2\times0-2+4!$$

23

$$23=2^0-2+4!$$

24

$$24=2+0-2+4!$$

25

$$25=\frac{2+0}2+4!$$

26

$$26=20+2+4$$

27

$$27=2-0!+2+4!$$

28

$$28=2+0+2+4!$$

29

$$29=2+0!+2+4!$$

30

$$30=(2^0+2)!+4!$$

Bonus solutions!

$$31=$$ $$32=(2+0)\cdot2^4$$ $$33=(2+0!)^2+4!$$ $$34=\frac{20}2+4!$$

• The bonus 31 and 33 solutions don't work because the digits aren't in order Commented Jan 4 at 16:11
• Yeah. I'm not sure who wrote those, though. Commented Jan 4 at 16:30

Thanks, this was fun! Here's what I came up with:

$$1 = {(2 + 0 + 2) \over 4}$$ $$2 = {2 \times 0 - 2 + 4}$$ $$3 = {-(2 + 0) \over 2} + 4$$ $$4 = 2 + 0 - 2 + 4$$ $$5 = {(2 + 0) \over 2} + 4$$ $$6 = 2 \times 0 + 2 + 4$$ $$7 = 2^0 + 2 + 4$$ $$8 = 2 + 0 + 2 + 4$$ $$9 = 2^0 + 2 \times 4$$ $$10 = 20 \times 2 / 4$$ $$11 = (20 + 2) / \sqrt{4}$$ $$12 = (2^0 + 2) \times 4$$ Skipped this and came back to it after 19 where I finally realized $$0! = 1$$: $$13 = -2 - 0! + 2^4$$ $$14 = -2 + 0 + 2^4$$ $$15 = -2^0 + 2^4$$ $$16 = 2 \times 0 + 2^4$$ $$17 = 2^0 + 2^4$$ $$18 = 2 + 0 + 2^4$$ $$19 = 2 + 0! + 2^4$$ $$20 = 20 + 2 - \sqrt{4}$$ $$21 = -2^0 - 2 + 4!$$ $$22 = -2 + 0 + 24$$ $$23 = -2^0 + 24$$ $$24 = 2 \times 0 + 24$$ $$25 = 2^0 + 24$$ $$26 = 2 + 0 + 24$$ $$27 = 2 + 0! + 24$$ $$28 = 2 + 0 + 2 + 4!$$ $$29 = 2 + 0! + 2 + 4!$$ $$30 = (2 + 0!) \times 2 + 4!$$

I was struggling for a while at 13 and 19 before I made that discovery I mentioned, that was huge, and made the rest of the numbers a lot easier to figure out.

• I think you meant 2 instead of the first 4 for 1. Commented Jan 4 at 4:20
• Thanks, Pierre, you're correct. I was copying from a piece of paper where I'd worked it all out and didn't double-check. Commented Jan 4 at 13:23

1. (2+0+2)÷4
2. 2+(0×2×4)
3. (2+0)÷2+$$\sqrt{4}$$
4. 2×0×2+4
5. (2+0)÷2+4
6. ($$\sqrt{2}$$+0)×$$\sqrt{2}$$+4
7. $$2^0$$+2+4
8. 2+0+2+4
9. 2+0!+2+4
10. 2+0+2×4
11. 2+$$0!$$+2×4
12. (2+0)(2+4)
13. -2-0!+$$2^4$$
14. -2+0+$$2^4$$
15. -2+0!+$$2^4$$
16. (2×0)+$$2^4$$
17. (2×0)!+$$2^4$$
18. 2+0+$$2^4$$
19. 2!+0!+$$2^4$$
20. -2 + 0 -2 + 4!
21. -2-(0×2)!+4!
22. -2+0×2+4!
23. -2+(0×2)!+4!
24. (2×0×2)+4!
25. (2×0×2)!+4!
26. 2+0+2+4!
27. 2+(0×2)!+4!
28. 2+0!×2+4!
29. 2+0!+2+4!
30. (2+0!)×2+4!

• Oh, nice! I love how you got the square root of 2 in there! +1 for that alone Commented Jan 3 at 15:53
• How do you get $((2+0)*\sqrt{2}))^4 = 20$ ? Commented Jan 3 at 21:19
• @FlorianF My guess is a mistake trying to use $2^4 + \sqrt{2}^4 = 20$ Commented Jan 5 at 10:58
• Yes, ((2+0)*$\sqrt{2}$))$^4$ = 20 is indeed a miscalculation, which I overlooked, while trying to accomplish $2^4$ + $\sqrt(2)^4$ = 20. Thanks for pointing that out. I've changed the 20th number for simpler -2 + 0 -2 + 4!. Commented Jan 5 at 13:13

Great puzzle, thanks for posting!

$$1 = 2^0+2-\sqrt{4}$$
$$2 = 2+0+2-\sqrt{4}$$
$$3=2^0-2+4$$
$$4=2+0-2+4$$
$$5=2^0+2+\sqrt{4}$$
$$6=(2\times0)+2+4$$
$$7=2^0+2+4$$
$$8=2+0+2+4$$
$$9=2^0+2\times4$$
$$10=2+0+2\times4$$
$$11=2+0!+(2\times4)$$
$$12=(2^0/2)\times4!$$
$$13=-2-0!+(2^4)$$
$$14=20-2-4$$
$$15=-(2^0)+2^4$$
$$16=(2\times0)+2^4$$
$$17=2^0+2^4$$
$$18=20+2-4$$
$$19=2+0!+(2^4)$$
$$20=20+2-\sqrt{4}$$
$$21=-(2^0)-2+4!$$
$$22=20-2+4$$
$$23=(2^0)-2+4!$$
$$24=20+2+\sqrt{4}$$
$$25=2^0+24$$
$$26=20+2+4$$
$$27=2^0+2+4!$$
$$28=2+0+2+4!$$
$$29=2+0!+2+4!$$
$$30=(2^0+2)!+4!$$

1: $$(2+0+2)/4=4/4=1$$

2: $$2+(0\times2)/4=2+0=2$$

3: $$2^0-2+4=1-2+4=-1+4=3$$

4: $$2-0-2+4=0+4=4$$

5: $$2^{0\times2}+4=1+4=5$$

6: $$2+(0\times2)+4=2+4=6$$

7: $$2^0+2+4=1+2+4=7$$

8: $$2+0+2+4=4+4=8$$

9: I'll figure something out

10: $$(2^0+2)!+4=3!+4=6+4=10$$

11: I'll figure something out

12: $$(2+0)(2+4)=2(6)=12$$