2
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You may use the following:

  • Addition [e.g. 2+2]
  • Subtraction [e.g. 2-2]
  • Multiplication [e.g. 2x2]
  • Division [e.g. 2/2]
  • Decimal point [e.g. 2+.2]
  • Exponents [e.g. 22] {tetration is NOT allowed}
  • Square roots [e.g. sqrt(2)] {infinite roots sqrt(sqrt(sqrt…sqrt(2))))=1 is NOT allowed}
  • Arbitrary roots [e.g. 0.2th root of 2]
  • Factorial [e.g. (2+2)!]
  • Double factorial [e.g. (2+2)!!]
  • Parentheses [e.g. 2/(2+2)]
  • Concatenation [e.g. 22+20] {2[0!]=21 is NOT allowed}
  • Permutations [e.g. (2+2)P2]
  • Combinations [e.g. (2+2)C2]

I’ve got all the numbers from 1-100, except for 54, 67, 68, 69, 79, 82, 83, 84, 86, 87, 93, 97

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3
  • $\begingroup$ Is this an original puzzle? $\endgroup$ – bobble May 13 at 2:59
  • 2
    $\begingroup$ @bobble This looks like a copy of a MSE question How do I express 67, 69, 83, 84, 86, 87, 88, 93 with 2,0,2,2 only?. Other than same operations, we also have details here like mentioning "infinite square roots" that were discussed there. $\endgroup$ – Vepir May 13 at 10:39
  • $\begingroup$ how should permutations and combinations work in forming a number? $\endgroup$ – Marius May 13 at 13:26
3
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All remaining 12 numbers you listed are doable, only if you allow arbitrary "$.$" and "$!$" uses.


Allowing arbitrary decimal point $.\square$, then 3 of remaining 12 numbers are:

$$\begin{array}{}54 &=& \sqrt{\text{}^{.(0!)}\sqrt[]{2}} + 22\\68 &=& (\sqrt{\text{}^{.(0!)}\sqrt[]{2}}+2) \times 2\\82 &=& ((2 \times 2)!! + .2) / .(0!)\end{array}$$

Additionally allowing repeated-decimals $.\dot{\square}$, then 7 of remaining 9 are:

$$\begin{array}{}67 &=& \left(\left(\sqrt{(.2) ^ {-2}}\right)!! - (.\dot{(0!)})\right) / (.\dot{2}) \\79 &=& (.\dot{(0!)}) ^ {-2} - 2\\83 &=& (.\dot{(0!)}) ^ {-2} + 2 \\84 &=& (.\dot{(0!)}) ^ {-2} + \sqrt{2 / (.\dot{2})} \\86 &=& (.\dot{(0!)}) ^ {-2} + \sqrt{(.2) ^ {-2}} \\87 &=& (.\dot{(0!)}) ^ {-2} + \left(\sqrt{2/(.\dot{2})}\right)! \\97 &=& \left(\left(\sqrt{2 / (.\dot{2})}\right)!\right)!! \times 2 + 0!\end{array}$$

Additionally allowing subfactorial $!\square$, then 2 of remaining 2 are:

$$\begin{array}{}69 &=& \left((0! / .2)!! + \sqrt{.\dot{(!2)}}\right) / (.\dot{2}) \\93 &=& (\text{}^{.2}\sqrt{2}-0!)/ \sqrt{.(\dot{!2})}\end{array}$$


I wrote a python program to solve this.

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1
  • $\begingroup$ Btw, if you use the rules as I used them in this answer, then permutations and combinations are not needed for any of the first 100 numbers. $\endgroup$ – Vepir May 13 at 15:09

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