# Make 1-100 using 2 0 2 2

You may use the following:

• 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

• Is this an original puzzle? May 13 at 2:59
• @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. May 13 at 10:39
• how should permutations and combinations work in forming a number? May 13 at 13:26

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.

• 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. May 13 at 15:09