-5
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Make the numbers 1-50 using 2 0 2 0 in the given order. Use all four digits exactly once.

Allowed operations: +, -, x, /, ! (factorial), double factorial, exponentiation, square root, parentheses. Grouping (e.g. "20") is also allowed, as are decimals (e.g. ".2").

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  • 2
    $\begingroup$ Is squaring also a valid operation? What is the scoring mechanism to determine the "best" answer? $\endgroup$ – Avi Jan 3 at 21:32
  • $\begingroup$ Squaring is not a valid operation. $\endgroup$ – Andrej Jakobčič Jan 3 at 21:44
  • $\begingroup$ Can you group the like this: 0! grouped with 2 makes 12? $\endgroup$ – Duck Jan 3 at 22:25
  • 1
    $\begingroup$ @Duck, no grouping like this! $\endgroup$ – Andrej Jakobčič Jan 3 at 22:40
  • $\begingroup$ If we allow other functions we can put: $\left \lfloor 20^{\sqrt{\sqrt{2}}} \right \rfloor - 0!$ or $-2+.0\bar{2}^{-0!}$ or $-\Gamma(2)+.0\bar{2}^{-0!}$. $\endgroup$ – Andrej Jakobčič Jan 9 at 14:37
2
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Anyone is welcome to edit and improve this answer, so don't checkmark this answer.
Bolded numbers mean they need to be solved or are impossible.

1

$(2+0-2)+0!$

2

$2+(0*2)+0$

3

$2+0!+(2*0)$

4

$2+0+2+0$

5

$2+0!+2+0$

6

$2+0!+2+0!$

7

$(2+(0!))*2)+0!$

8

$2*(0!+2+0!)$

9

$(2+0!)*(2+0!)$

10

$(20/2+0)$

11

$(20/2+0!)$

12

$(2+0!)!*2+0$

13

$((2+0!)!*2)+0!$

14

$20-((2+0!)!)$

15

$(2+0!)/.20$

16

$(2+0!)/.2+0!$

17

$20-2-0!$

18

$20-2+0$

19

$20-2+0!$

20

$20+2*0$

21

$2-0!+20$

22

$20+2+0$

23

$20+2+0!$

24

$(2+0+2+0)!$

25

$((2*0!)*2)!+0!$

26

$20+((2+0!)!)$

27

$(2+0!)^{(2+0!)}$

28

$((2+0!)!)!!-20$

29

$((2+0!)!/.2)-0!$

30

$((2+0!)!/.2)+0$

31

$((2+0!)!/.2)+0!$

32

$2^{((0!+2)!-0!)}$

33

$2^{((0!/.2))}+0!$

34

35

$(2+0!)!^2-0!$

36

$(2+0!)!^2+0$

37

$(2+0!)!^2+0!$

38

$2(-0!+20)$

39

$20*2-0!$

40

$20+20$

41

$(20*2)+0!$

42

$2*(0!+20)$

43

44

45

$((2+0!)!)!!-2-0!$

46

$((2+0!)!)!!-2+0$

47

$((2+0!)!)!!-2+0!$

48

$2*(0!+2+0!)!$

49

$((2+0!)!)!!+2-0!$

50

$((2+0!)!)!!+2+0$

| improve this answer | |
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  • $\begingroup$ I have a premonition that 50 is going to be impossible. Good on you for keeping at it, though: +1 $\endgroup$ – Avi Jan 3 at 21:55
  • 1
    $\begingroup$ Not sure - it's pretty tough to get off of 2 and 0 as the only numbers, with ordering restrictions $\endgroup$ – Avi Jan 3 at 22:31
  • 1
    $\begingroup$ @AndrejJakobčič Clever 27, nice! $\endgroup$ – Avi Jan 3 at 22:37
  • 2
    $\begingroup$ If anyone is welcome to improve this answer, it's better to mark it community wiki. $\endgroup$ – Glorfindel Jan 4 at 8:29
  • 1
    $\begingroup$ @Glorfindel I agree, since Avi and I did significant parts of this. $\endgroup$ – Don Thousand Jan 4 at 13:26

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