# How do you make numbers 1 - 100 using the digits "2 0 2 4"? [closed]

I'm trying to complete this "2024 Challenge", but I'm still missing quite a lot of numbers. I need your help to finish this challenge. All answers are appreciated.

Rules:

Use all 4 digits exactly ONCE!

Allowed operations you can use: +, -, x, /, ! (factorial), !! (double factorial), a^b (exponentiation), √ (square root).

Brackets & concatenation are allowed (e.g. 20/(2*4)).

Squaring uses the number 2.

Rounding is NOT allowed.

• I’m voting to close this question because questions from ongoing contests are banned. Mar 27 at 23:46
• Welcome to PSE (Puzzling Stack Exchange)! Is this part of a contest or homework or just an activity for fun or something else? Mar 28 at 2:18
• Deleting the part of the question where you admit it's a contest, doesn't suddenly make the question acceptable - it actually makes it worse, since it seems you're trying to hide something. Mar 29 at 18:17
• You could instead add more details to explain why this doesn't fall under the policy which I linked in my first comment. Mar 29 at 18:26

I got through 30 before I saw that comment about questions for ongoing contests being banned...

1 = (2+2+0)/4
2 = 2+(2×0×4)
3 = 4-(2/2)+0
4 = (2×4)/2+0
5 = 4+(2/2)+0
6 = (2×0)+2+4
7 = (2^0)+2+4
8 = 2+0+2+4
9 = (2^0)+2×4
10 = (2+0)+2×4
11 = 0!+2+2×4
12 = [(2+2)-0!]*4
13 = (4^2)-2-0!
14 = (4^2)-2×0!
15 = (4^2)-2/0!
16 = 2×0+2^4
17 = 2×2×4+0!
18 = (2+0!)×(2+4)
19 = 4!-2-2-0!
20 = (2+0!+2)×4
21 = 4!-(2+2-0!)
22 = 4!-2-(2×0)
23 = 4!-2/2-0
24 = 4!+(2×2×0)
25 = 4!+(2/2)+0
26 = 4!+2+2×0
27 = 4!+2+(2-0!)
28 = 4!+2+2+0
29 = 4!+2+2+0!
30 = 4!+[2×(2+0!)]

Even though I'm missing a lot of numbers, I have numbers 1 - 30.

1 = (2+2+4)^0

2 = [(4*2)*0]+2

3 = [(4+2)/2]+0

4 = (4+2+0)-2

5 = (2/2)+(4+0)

6 = (2*0)+(2+4)

7 = (20/4)+2

8 = 2+0+2+4

9 = 2^0+(2*4)

10 = [(2*4)+2]+0

11 = [(2*4)+2]+0!

12 = (4!!+2^2)+0

13 = [(2+0!)^2]+4

14 = 2(0!+2+4)

15 = (4^2)-(2^0)

16 = 4[(2^2)*0!]

17 = (4^2)+(2^0)

18 = [(4^2)+2]*0!

19 = (40-2)/2

20 = 4!-[(2^2)+0]

21 = 4!-[(2^0)+2]

22 = 4!-2-(2*0)

23 = 4!-(22^0)

24 = 4!+(22*0)

25 = 4!+(22^0)

26 = 24+(2+0)

27 = 4!+[(2^2)-(0!)]

28 = 4!+[(2^2)-0]

29 = 4!+[(2^2)+(0!)]

30 = 5!/4 (2*2+0!=5)