5
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How can you make the numbers 1-100 using only the digits 2, 0, 0, 0? I've already found several, the ones I mostly need are 13, 26, 34-40, and 64-85, but I figure it would be interesting to have a record here for other potentially easier/simpler methods.

Addition, subtraction, multiplication, division, exponentiation, factorial, powers (as long as all powers are from the available numbers), sin, cos, tan (and the inverses of those three), and precedence adjustment (parentheses) are allowed, but not concatenation.

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8
  • 3
    $\begingroup$ What operations do you allow? I assume +,-,*,/, and exponents. What about factorial, absolute value, Knuth arrows, concatenation, square root, other roots, etc? $\endgroup$
    – benzene
    Aug 20 '17 at 20:25
  • 1
    $\begingroup$ (You may want to post the ones you DO have as well, so there is a complete record.) $\endgroup$
    – Rubio
    Aug 20 '17 at 20:43
  • 3
    $\begingroup$ 2/2=1 Therefore, every rational number is trivial. $\endgroup$
    – Carl
    Aug 21 '17 at 1:26
  • 1
    $\begingroup$ In any case, if you have 52, you have 13 and 26. If you read [x] as 'the solution for x', then 26=([52])/2, and 13=([26])/2. $\endgroup$ Aug 21 '17 at 3:45
  • 3
    $\begingroup$ I take "2, 0, 0, 0" to mean you're allowed to use only one 2 and three 0s, so I can't agree with Carl and Jeff that the puzzle is trivial. I agree with Jeff that concatenation of anything other than literal digits is cheating. $\endgroup$
    – Gareth McCaughan
    Aug 21 '17 at 10:45
18
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A more elegant way to get all the numbers from 1 to 100 if we allow logarithms and the use of lg as logarithm in base 10 is this (I know, I know, it looks like cheating a bit but it's beautiful)

$x = \log_{\frac{0!}{2}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{\frac{0!}{0!\%}\,}\,}\,}}_\text{x+1 square roots}}\right)$

This is equivalent

$x = \log_{\frac{0!}{2}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{\frac{1}{\frac{1}{100}}\,}\,}\,}}_\text{x+1 square roots}}\right)$

Moving on

$x = \log_{\frac{1}{2}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{10\,}\,}\,}}_\text{x square roots}}\right)$

$x = \log_{\frac{1}{2}}\left({\lg{10^{\frac{1}{2^x}}}}\right)$

$x = \log_{\frac{1}{2}}\frac{1}{2^x}$

But here is the brute force approach without log.
Operations used: + - $\frac{a}{b}$ $\times$ $\sqrt{a}$ $a^b$ $\lfloor a \rfloor$ $\lceil a\rceil$ $a\%$ (wich is basically division by 100) and trigonometrical functions (sin, cos, ...)

$0 = 0 \times 0 \times 0 \times 2$
$1 = 0 \times 0 + 2^0$
$2 = 0 + 0 + 0 + 2$
$3 = 0 + 0 + 0! + 2$
$4 = 0 + 2^{0!+0!}$
$5 = 0! + 2^{0!+0!}$
$6 = (0! + 0!+0!) \times 2 $
$7 = (0! + 2)! + 0! +0$
$8 = (0! + 2)! + 0! +0!$
$9 = (0! + 0! + 0!)^2$
$10 = \frac{20}{0!+0!}$
$11 = \lceil{\sqrt{(2+0! +0!+0!)!}}\rceil$
$12 = (0! + 0! + 0!)! \times 2$
$13 = \lfloor{\sqrt{200}}\rfloor - 0!$
$14 = \lfloor{\sqrt{200}}\rfloor + 0$
$15 = \lceil{\sqrt{200}}\rceil + 0$
$16 = \lceil{\sqrt{200}}\rceil + 0!$
$17 = \lfloor\frac{(2+0!)!}{0!\%}\rfloor + 0!$
$18 = 20 - 0! - 0!$
$19 = 20 + 0 - 0!$
$20 = 20 + 0 + 0$
$21 = 20 + 0! + 0$
$22 = 20 + 0! + 0!$
$23 = (2 + 0! + 0!)! - 0!$
$24 = (2 + 0! + 0!)! + 0$
$25 = (2 + 0! + 0!)! + 0!$
$26 = \lfloor{\sqrt{((2+0!)!)!}}\rfloor + 0 + 0$
$27 = \lceil{\sqrt{((2+0!)!)!}}\rceil + 0 + 0$
$28 = \lceil{\sqrt{((2+0!)!)!}}\rceil + 0! + 0$
$29 = \lceil{\sqrt{((2+0!)!)!}}\rceil + 0! + 0!$
$30 = \sqrt{\frac{0!}{0!\%}} + 20$
$31 = \lceil\sqrt{\sqrt{\sqrt{\sqrt{((2+0!+0!)!)!}}}}\rceil + 0$
$32 = 2 ^{\lfloor\sqrt{\sqrt{((0!+0!+0!)!)!}}\rfloor}$
$33 = \lfloor \frac{\frac{0!}{0!\%}}{2+0!} \rfloor$
$34 = \lceil \frac{\frac{0!}{0!\%}}{2+0!} \rceil$
$35 = \lfloor-\frac{\sqrt{\frac{0!}{0!\%}}}{\sin((2+0!)!)}\rfloor$
$36 = ((0! + 0! + 0!)! )^2$
$37 = \sqrt{\frac{0!}{0!\%}} + \lceil\sqrt{(2+0!)!}\rceil$
$38 = $
$39 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\frac{0!}{0!\%}}!}}} \times (2+0!)!\rfloor$
$40 = 20 \times (0! + 0!)$
$41 = \lfloor\sqrt{\sqrt{\sqrt{\frac{0!}{0!\%}}!}}\rfloor - 2 + 0$
$42 = \lfloor\sqrt{\sqrt{\sqrt{\frac{0!}{0!\%}}!}}\rfloor - 2 ^ 0$
$43 = \lfloor\sqrt{\sqrt{\sqrt{\frac{0!}{0!\%}}!}}\rfloor + 2 \times 0$
$44 = \lfloor{\sqrt{2000}}\rfloor$
$45 = \lceil{\sqrt{2000}}\rceil$
$46 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\lfloor\sqrt{(2+0!)!}\rfloor!}}}}\rfloor + 0 +0$
$47 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\lfloor\sqrt{(2+0!)!}\rfloor!}}}}\rfloor + 0! +0$
$48 = \frac{0!}{(0! + 0!)\%} - 2$
$49 = \frac{0!}{2 \times 0!\%} -0!$
$50 = \frac{0!}{2 \times 0!\%} +0$
$51 = \frac{0!}{2 \times 0!\%} +0!$
$52 = \frac{0!}{(0! + 0!)\%} + 2$
$53 = $
$54 = \lceil{\sqrt{((0! + 0! + 0!)!)!}}\rceil \times 2$
$55 = \lceil{\sqrt{((0! + 0! + 0!)!)!}} \times 2\rceil$
$56 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\lfloor\sqrt{(2+0!)!}\rfloor!}}}}\rfloor + \sqrt{\frac{0!}{0!\%}}$
$57 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\lceil\sqrt{(2+0!)!}\rceil!}}}}\rfloor + 0 +0$
$58 = \sqrt{\sqrt{\sqrt{\sqrt{\lceil\sqrt{(2+0!)!}\rceil!}}}} + 0! +0$
$59 = \sqrt{\sqrt{\sqrt{\sqrt{\lceil\sqrt{(2+0!)!}\rceil!}}}} + 0! +0!$
$60 = \sqrt{\frac{0!}{0!\%}} \times (2 + 0!)!$
$61 = $
$62 = $
$63 = $
$64 = 2^{(0!+0!+0!)!}$
$65 = \lceil\frac{\frac{0!}{0!\%}}{ctan(((2+0!)!)!)}\rceil$
$66 = $
$67 = \lfloor\sqrt{\sqrt{\sqrt{\sqrt{\lceil\sqrt{(2+0!)!}\rceil!}}}}\rfloor + \sqrt{\frac{0!}{0!\%}}$
$68 = \lceil\sqrt{\sqrt{\sqrt{\sqrt{\lceil\sqrt{(2+0!)!}\rceil!}}}}\rceil + \sqrt{\frac{0!}{0!\%}}$
$69 = \lfloor{\sqrt{(((2+0!)!) + 0!)!}}\rfloor - 0!$
$70 = \lfloor{\sqrt{(((2+0!)!) + 0!)!}}\rfloor + 0$
$71 = \lceil{\sqrt{(((2+0!)!) + 0!)!}}\rceil + 0$
$72 = \lceil{\sqrt{(((2+0!)!) + 0!)!}}\rceil + 0!$
$73 = \frac{0!}{0!\%} - \lceil{\sqrt{((2 + 0!)!)!}}\rceil$
$74 = \frac{0!}{0!\%} - \lfloor{\sqrt{((2 + 0!)!)!}}\rfloor$
$75 = $
$76 = $
$77 = $
$78 = $
$79 = \lfloor\sqrt{\sqrt{\lceil\sqrt{(2+0!+0!+0!)!}\rceil!}}\rfloor$
$80 = \frac{0!}{0!\%} - 20$
$81 = \lceil\sqrt{\sqrt{\sqrt{(\sqrt{\frac{0!}{0!\%}} + 0!)!}}}\rceil ^ 2$
$82 = $
$83 = -\lfloor\cos((2+0!)!)!) * \frac{0!}{0!\%}\rfloor$
$84 = \lfloor\sqrt{(2+0!)!*\sqrt{\frac{0!}{0!\%}}}\rfloor$
$85 = \lceil\sqrt{(2+0!)!*\sqrt{\frac{0!}{0!\%}}}\rceil$
$86 = $
$87 = $
$88 = $
$89 = \lfloor\sin(2) \times \frac{0!}{0!\%}\rfloor - 0!$
$90 = \lfloor\sin(2) \times \frac{0!}{0!\%}\rfloor + 0$
$91 = \lfloor\sin(2) \times \frac{0!}{0!\%}\rfloor + 0!$
$92 = $
$93 = $
$94 = \frac{0!}{0!\%} - (2 + 0!)!$
$95 = \lceil\sqrt{\sqrt{\sqrt{(20 - 0! -0!)!}}}\rceil$
$96 = \lfloor\frac{0!}{0!\%} + ctan((2+0!)!)\rfloor$
$97 = \frac{0!}{0!\%} - 2 - 0!$
$98 = \frac{0!}{0!\%} - 2 + 0$
$99 = \frac{0!}{0!\%} - 2 + 0!$
$100 = \frac{0!}{0!\%} + 2 \times 0$

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2
  • $\begingroup$ Awesome job dude. This is insane. Try arccos for numbers 83-97. Factorial of (2+0!)! was also extremely helpful. $\endgroup$ Aug 22 '17 at 0:48
  • $\begingroup$ You should probably mention the other functions you're using, such as sine. $\endgroup$
    – Draconis
    Aug 22 '17 at 22:48
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Using only the operations +, -, floor, ceiling, factorial (Gamma function extension), and square root:

1 = 2 - 0! + 0 + 0

2 = 2 + 0 + 0 + 0

3 = 2 + 0! + 0 + 0

4 = 2 + 0! + 0! + 0

5 = (2 + 0!)! - 0! + 0

6 = (2 + 0!)! + 0 + 0

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

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

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

10 = floor((sqrt(2) + 0! + 0!)!) + 0

11 = ceil((sqrt(2) + 0! + 0!)!) + 0

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

13 = ceil((sqrt(2)! + 0!)!!!) + 0 + 0

14 = floor((sqrt((2 + 0!)! + 0!) + 0!)!)

15 = ceil((sqrt((2 + 0!)! + 0!) + 0!)!)

16 = floor((sqrt(2 + 0!) + 0! + 0!)!)

17 = ceil((sqrt(2 + 0!) + 0! + 0!)!)

18 = ceil((sqrt(2)!! + 0!)!!!!) + 0 + 0

19 = ceil((sqrt(sqrt((2 + 0!)! + 0!)) + 0!)!!)

20 = floor((sqrt((2 + 0! + 0!)!) - 0!)!)

21 = floor(sqrt((2 + 0!)! + 0!)!!) + 0

22 = ceil((sqrt(2) + 0!)!)! - 0! - 0!

23 = (2 + 0! + 0!)! - 0!

24 = (2 + 0! + 0!)! + 0

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

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

27 = ceil(sqrt((2 + 0!)!!)) + 0 + 0

28 = ceil(sqrt((2 + 0!)!!)) + 0! + 0

29 = ceil(sqrt((2 + 0!)!!)) + 0! + 0!

30 = ceil((sqrt((2 + 0!)!)! + 0!)!) + 0

31 = floor(sqrt((sqrt(2) + 0!)!!!)) - 0! + 0

32 = floor((sqrt(sqrt(2)) + 0! + 0! + 0!)!)

33 = ceil((sqrt(sqrt(2)) + 0! + 0! + 0!)!)

34 = ceil(sqrt((sqrt(2) + 0!)!!!)) + 0! + 0

35 = floor((sqrt(2)! + 0! + 0! + 0!)!)

36 = ceil((sqrt(2)! + 0! + 0! + 0!)!)

37 = ceil(((sqrt(2)!! + 0!)! + 0! + 0!)!)

38 = floor((sqrt(2 + 0!) + 0!)!!) - 0!

39 = floor((sqrt(2 + 0!) + 0!)!!) + 0

40 = ceil((sqrt(2 + 0!) + 0!)!!) + 0

41 = ceil((sqrt(2 + 0!) + 0!)!!) + 0!

42 = floor((sqrt((sqrt(2) + 0!)!) + 0!)!!) + 0

43 = ceil((sqrt((sqrt(2) + 0!)!) + 0!)!!) + 0

44 = floor(((sqrt(sqrt(2)) + 0!)! + 0! + 0!)!)

45 = floor((sqrt(2) + 0! + 0! + 0!)!)

46 = ceil((sqrt(2) + 0! + 0! + 0!)!)

47 = ceil(sqrt(sqrt(((sqrt(sqrt(2)) + 0!)! + 0!)!!))) + 0

48 = floor((sqrt((2 + 0!)!) + 0! + 0!)!)

49 = ceil((sqrt((2 + 0!)!) + 0! + 0!)!)

50 = ceil(sqrt(((sqrt(sqrt(2)) + 0! + 0!)! - 0!)!))

51 = floor((sqrt((sqrt(2) + 0!)!!) + 0! + 0!)!)

52 = floor(sqrt((sqrt(sqrt(2)) + 0! + 0!)!)!!) + 0

53 = floor(sqrt(sqrt((sqrt(2) + 0! + 0!)!!))) + 0

54 = floor(((sqrt(sqrt(2) + 0!) + 0!)! + 0!)!)

55 = ceil(((sqrt(sqrt(2) + 0!) + 0!)! + 0!)!)

56 = floor(((sqrt(2)! + 0!)!! + 0!)!) + 0

57 = ceil(((sqrt(2)! + 0!)!! + 0!)!) + 0

58 = ceil(((sqrt(2)! + 0!)! + 0! + 0!)!)

59 = ceil(((sqrt(sqrt((2 + 0!)!)) + 0!)! + 0!)!)

60 = ceil(sqrt(floor(sqrt((2 + 0!)! + 0!)!!))!) + 0

61 = floor(sqrt(sqrt((2 + 0!)! + 0!)!!)!) + 0

62 = ceil(sqrt(sqrt((2 + 0!)! + 0!)!!)!) + 0

63 = floor((sqrt(sqrt(sqrt(2)) + 0! + 0!) + 0!)!!)

64 = ceil((sqrt(sqrt(sqrt(2)) + 0! + 0!) + 0!)!!)

65 = ceil(sqrt((sqrt((sqrt(2) + 0! + 0!)!)! - 0!)!))

66 = floor((sqrt(ceil((sqrt(2) + 0!)!!)) + 0! + 0!)!)

67 = floor(((sqrt(2 + 0!)! + 0!)! + 0!)!)

68 = ceil(((sqrt(2 + 0!)! + 0!)! + 0!)!)

69 = floor(sqrt(((2 + 0!)! + 0!)!)) - 0!

70 = floor(sqrt(((2 + 0!)! + 0!)!)) + 0

71 = ceil(sqrt(((2 + 0!)! + 0!)!)) + 0

72 = ceil(sqrt(((2 + 0!)! + 0!)!)) + 0!

73 = floor((sqrt(sqrt(sqrt(2) - 0!)) + 0! + 0!)!!)

74 = floor((sqrt(sqrt(2)! + 0! + 0!) + 0!)!!)

75 = ceil((sqrt(sqrt(2)! + 0! + 0!) + 0!)!!)

76 = ceil((((sqrt(2) - 0!)! + 0!)! + 0!)!!)

77 = floor(sqrt(sqrt(sqrt(((2 + 0!)! - 0!)!)!))) + 0

78 = ceil(sqrt(sqrt(sqrt(((2 + 0!)! - 0!)!)!))) + 0

79 = floor(sqrt(sqrt(ceil((sqrt(2) + 0! + 0!)!)!))) + 0

80 = ceil(sqrt(sqrt(ceil((sqrt(2) + 0! + 0!)!)!))) + 0

81 = floor(sqrt((sqrt(2)!! + 0! + 0!)!!)) + 0

82 = ceil(sqrt((sqrt(2)!! + 0! + 0!)!!)) + 0

83 = ceil(((sqrt(2)!! + 0!)!!! + 0!)!) + 0

84 = floor(sqrt(ceil((sqrt(2) + 0!)!)! - 0!)!) - 0!

85 = floor(sqrt((2 + 0! + 0!)! - 0!)!)

86 = ceil(sqrt((2 + 0! + 0!)! - 0!)!)

87 = floor(sqrt(((sqrt(2) + 0!)!! + 0!)!)) + 0

88 = ceil(sqrt(((sqrt(2) + 0!)!! + 0!)!)) + 0

89 = ceil(sqrt(sqrt((2 + 0!)!)!!!)) + 0 + 0

90 = ceil(sqrt(sqrt((2 + 0!)!)!!!)) + 0! + 0

91 = ceil(sqrt(sqrt((2 + 0!)!)!!!)) + 0! + 0!

92 = floor(((sqrt(sqrt(2) - 0!)! + 0!)! + 0!)!!)

93 = ceil(((sqrt(sqrt(2) - 0!)! + 0!)! + 0!)!!)

94 = floor(sqrt((2 + 0!)! + 0! + 0!)!!)

95 = ceil(sqrt((2 + 0!)! + 0! + 0!)!!)

96 = ceil(sqrt(floor((sqrt(2)! + 0! + 0!)!))!!) + 0!

97 = floor(sqrt(sqrt(((sqrt(2) + 0! + 0!)! + 0!)!)))

98 = ceil(sqrt(sqrt(((sqrt(2) + 0! + 0!)! + 0!)!)))

99 = floor(sqrt(((sqrt(2)! + 0! + 0!)! - 0!)!))

100 = floor(sqrt((2 + 0! + 0!)!)!) - 0!

It should be noted that copying one of the above formulas into WolframAlpha might not give the right answer. This is because it interprets '!!' as the "double factorial function": n!! = n(n-2)(n-4)..., where I use n!! to mean (n!)!. Putting spaces between the exclamation points should fix this.

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2
  • $\begingroup$ Just picking one to test: for 100, your $\lfloor \sqrt{(2+0!+0!)!}! \rfloor - 0!$ simplifies to $\lfloor \sqrt{24}! \rfloor - 1$, but $\sqrt{24}$ isn't a whole number, so the usual factorial operator doesn't work. You've also applied factorial to other irrational roots (e.g. see 95). Are you using a different factorial operator? $\endgroup$
    – Lawrence
    Aug 22 '17 at 23:46
  • 1
    $\begingroup$ @Lawrence I'm actually using the Gamma function, the most common extension of factorial to nonintegers. I will edit the answer to specify this. $\endgroup$
    – benzene
    Aug 22 '17 at 23:52
2
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1 to 100 Using Simple Operations - complete

Operators used: +, -, !, $\sqrt{\cdot}$, %, $\lfloor \cdot \rfloor$, $\lceil \cdot \rceil$. I also concatenate 2 to 0 to form 20, and get reciprocals with $a^{-(0!)}$ and use bracketing. Multiplication and division operators are not used in this answer.

Split the given 4 digits into two sets, {0,0} and {2,0}. We can form 1-7 just using either set:

\begin{array}{clll} 1 & = 0! + 0 & = 2 - 0! \\ 2 & = 0! + 0! & = 2 + 0 \\ 3 & = \lfloor √\sqrt{0!\%^{-(0!)}} \rfloor & = 2 + 0! & \\ 4 & = \lceil √\sqrt{0!\%^{-(0!)}} \rceil & = \lfloor \sqrt{20} \rfloor \\ 5 & = \lfloor √(\lceil √\sqrt{0!\%^{-(0!)}} \rceil !) \rfloor & = \lceil \sqrt{20} \rceil \\ 6 & = \lfloor √√√(\sqrt{0!\%^{-(0!)}}!) \rfloor & = (2 + 0!)! \\ 7 & = \lceil √√√(\sqrt{0!\%^{-(0!)}}!) \rceil & = \lfloor \sqrt{2\%^{-(0!)}} \rfloor \\ \end{array}

We now form 8-11 using either definition of 5 and 7 above:

\begin{array}{cccc} 8 = \lfloor √√(7!) \rfloor & 9 = \lceil √√(7!) \rceil & 10 = \lfloor √(5!) \rfloor & 11 = \lceil √(5!) \rceil \end{array}

Now that we can construct all the numbers from 1 to 11 using just {2,0} as well as using just {0,0}, if we have any integer $x$ constructed from either set, we can get all the integers $x + k$, where $-11 \leq k \leq 11$. If $x$ was formed from {2,0}, form $k$ from {0,0}, and vice versa.

Pick any of the above expressions for the numbers in the expansions below.

\begin{array}{c|c|c} x & \text{expansion} & \text{covers} \\ \hline 11 & \lceil √(5!) \rceil & 0 - 22 \\ 24 & 4! & 13 - 35 \\ 44 & \lceil √√(10!) \rceil & 33 - 55 \\ 50 & 2\%^{-1} & 39 - 61 \\ 71 & \lceil √(7!) \rceil & 60 - 82 \\ 80 & \lceil √√(11!) \rceil & 69 - 91 \\ 100 & 0!\%^{-(0!)} & 89 - 111 \end{array}

This produces all the numbers from 0 to 111. If it's required to use all 4 digits in {2,0,0,0}, observe that none of the combinations above use just the 3 zeros. They either use all 4 digits or at least one 0 is unused. If a 0 is unused, multiply that 0 by the sum of all the remaining unused digits, and add the result (also 0) to the original expression.

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5
  • $\begingroup$ If you assume that you can use an arbitrary number of the digits 0 and 2, why don't you simply define 1 as 0! and every following number N to be (N-1)+1? In that case, there is absolutely no need for concatenation or seven nested square roots. $\endgroup$
    – jarnbjo
    Aug 21 '17 at 15:04
  • 2
    $\begingroup$ @jarnbjo That's not my assumption. In my (partial) answer, I assume that there are only 4 literals that can be combined: '2', '0', '0', '0'. However, to avoid writing things like "0!" or "(2 + 1)!", I've allowed the notation to refer to other numbers. So "1" is short-hand for "0!", but if we've used 3 instances of "0!" in the expression, we can't generate any more "1"s to use in that expression. For example, "27" uses only 2 literals: one instance of "2" and one instance of "0", combined to form "3", then "6", then "27" by using factorial and other operators. $\endgroup$
    – Lawrence
    Aug 21 '17 at 15:07
  • $\begingroup$ Note also that in the expressions for 1 to 40, only the original 4 literals are concatenated (e.g. "2" || "0" to form "20"). I've managed to avoid concatenation of generated numerals (e.g. "0!" || "0!" = "11") in the current version. $\endgroup$
    – Lawrence
    Aug 21 '17 at 15:19
  • $\begingroup$ @jarnbjo Just in case you were wondering about the extra '2's: the ones in brackets are actually (0! + 0!). $\endgroup$
    – Lawrence
    Aug 22 '17 at 10:45
  • $\begingroup$ Note that the question has changed since this answer was posted. $\endgroup$
    – Lawrence
    Apr 26 at 6:15

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