Create numbers from 1-100 using 1846

Question

Create all numbers $1$ to $100$ using equations made up of $1,4,6,8$.

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Rules:

1. Use all four digits exactly once
2. Allowed operations: $+,\,-,\,\times,\,\div,\,!$ (factorial), exponentiation ($a^b$), square root ($\sqrt{\text{ }}$).
3. Parentheses and grouping (concatenation) (e.g. "$21$" from $2$ and $1$) are also allowed.
4. Exponentiation can only be used in the number order with the numbers provided. (eg. $1^9+6+8$ is allowed, but $1^6+9+8$ isn't).
5. The modulus operator ($a\pmod b$) is not allowed.
6. Rounding is not allowed (e.g. $201/8=25$).
7. Decimal points are allowed.

Answer 1

Assuming there are typos in the examples given in the question.
My answer uses no decimals and no exponentiation.
Only two of the solutions use division.
Only two of the solutions use multiple-digit numbers.

Only $74$ and $77$ need multiple digits.
Only $83$ and $85$ need division.

$ 1 = 4 + 6 - 8 - 1 $
$ 2 = \sqrt{4} \times (6 - 1) - 8 $
$ 3 = 1 + 4 - (8 - 6) $
$ 4 = (4 - 1)! - (8 - 6) $
$ 5 = 8 - 6 + 4 - 1 $
$ 6 = (8 - 6) \times (4 - 1) $
$ 7 = 8 - 6 + 1 + 4 $
$ 8 = (4 - 1)! + 8 - 6 $
$ 9 = 6 - 4 + 8 - 1 $
$ 10 = (4 - 1) \times 6 - 8 $
$ 11 = 6 - 4 + 1 + 8 $
$ 12 = (8 - 6 + 1) \times 4 $
$ 13 = \sqrt{(1 + 8)} + 4 + 6 $
$ 14 = (6 - 4) \times (8 - 1) $
$ 15 = 4 \times 6 - (1 + 8) $
$ 16 = (6 - 1) \times 8 - 4! $
$ 17 = 4 - 1 + 6 + 8 $
$ 18 = (4 - 1) \times 8 - 6 $
$ 19 = 1 + 4 + 6 + 8 $
$ 20 = (1 + 6) \times 4 - 8 $
$ 21 = 4!! - 1 + 6 + 8 $
$ 22 = (8 - 1) \times 4 - 6 $
$ 23 = (8 - 4) \times 6 - 1 $
$ 24 = (6 - 4 + 1) \times 8 $
$ 25 = 4 \times 8 - (1 + 6) $
$ 26 = (4 - 1) \times 6 + 8 $
$ 27 = 4 \times 8 - (6 - 1) $
$ 28 = (6 - 1) \times 4 + 8 $
$ 29 = (1 + 4)!! + 6 + 8 $
$ 30 = (8 - 4 + 1) \times 6 $
$ 31 = 8 - 1 + 4 \times 6 $
$ 32 = (6 - 1) \times 8 - 4!! $
$ 33 = 1 + 8 + 4 \times 6 $
$ 34 = (8 - 1) \times 4 + 6 $
$ 35 = 6!! - (4 + 8) - 1 $
$ 36 = (1 + 6) \times 4 + 8 $
$ 37 = 6 - 1 + 4 \times 8 $
$ 38 = (1 + 4) \times 6 + 8 $
$ 39 = 1 + 6 + 4 \times 8 $
$ 40 = (1 + 6 - \sqrt{4}) \times 8 $
$ 41 = 6 \times 8 - 4!! + 1 $
$ 42 = (1 + 8) \times 4 + 6 $
$ 43 = 6 \times 8 - (1 + 4) $
$ 44 = (6 - 1) \times 8 + 4 $
$ 45 = 6 \times 8 - (4 - 1) $
$ 46 = (1 + 4) \times 8 + 6 $
$ 47 = (6 - 1)!! + 4 \times 8 $
$ 48 = (6 - 4 + 1)! \times 8 $
$ 49 = 6 \times 8 - 1 + \sqrt{4} $
$ 50 = (1 + 8) \times 6 - 4 $
$ 51 = 4 - 1 + 6 \times 8 $
$ 52 = (6 - 1 + 8) \times 4 $
$ 53 = 1 + 4 + 6 \times 8 $
$ 54 = (4 - 1)! + 6 \times 8 $
$ 55 = (6 + 8) \times 4 - 1 $
$ 56 = (\sqrt{4} + 6 - 1) \times 8 $
$ 57 = (6 + 8) \times 4 + 1 $
$ 58 = (1 + 8) \times 6 + 4 $
$ 59 = 4 - 1 + 8 + 6!! $
$ 60 = (1 + 6) \times 8 + 4 $
$ 61 = 1 + 4 + 8 + 6!! $
$ 62 = (1 + 8) \times 6 + 4!! $
$ 63 = (1 + 4)!! + 6 \times 8 $
$ 64 = (6 - 1) \times 8 + 4! $
$ 65 = (\sqrt{4} + 6) \times 8 + 1 $
$ 66 = (4 - 1 + 8) \times 6 $
$ 67 = \sqrt{(8!! - 4! + 1)} + 6!! $
$ 68 = (6 - 1)!! \times 4 + 8 $
$ 69 = 4!! \times 8 + 6 - 1 $
$ 70 = (1 + 4) \times (6 + 8) $
$ 71 = (4 + 8) \times 6 - 1 $
$ 72 = (4 - 1 + 6) \times 8 $
$ 73 = (4 + 8) \times 6 + 1 $
$ 74 = (4 - 1)! + 68 $
$ 75 = (8 - 1)!! - (4! + 6) $
$ 76 = (8 - 1) \times 4 + 6!! $
$ 77 = \sqrt{841} + 6!! $
$ 78 = (1 + 4 + 8) \times 6 $
$ 79 = (4 + 6) \times 8 - 1 $
$ 80 = \sqrt{4} \times (6 - 1) \times 8 $
$ 81 = (4 + 6) \times 8 + 1 $
$ 82 = (1 + 4)!! \times 6 - 8 $
$ 83 = \frac{6!}{8} - (4!! - 1) $
$ 84 = (1 + 6) \times (4 + 8) $
$ 85 = \frac{6!}{8} - (1 + 4) $
$ 86 = (6!! - 1) \times \sqrt{4} - 8 $
$ 87 = \sqrt{4} \times 6!! - (1 + 8) $
$ 88 = (1 + 4 + 6) \times 8 $
$ 89 = \sqrt{4} \times 6!! - (8 - 1) $
$ 90 = (1 + 8) \times (4 + 6) $
$ 91 = (8 - 1)!! - (4!! + 6) $
$ 92 = ((6 - 1)!! + 8) \times 4 $
$ 93 = (1 + 6)!! - (4 + 8) $
$ 94 = (6 \times 8 - 1) \times \sqrt{4} $
$ 95 = \sqrt{4} \times 6 \times 8 - 1 $
$ 96 = ((4 - 1)! + 6) \times 8 $
$ 97 = \sqrt{4} \times 6 \times 8 + 1 $
$ 98 = (1 + 4)!! \times 6 + 8 $
$ 99 = (1 + 6)!! - 8 + \sqrt{4} $
$ 100 = ((\sqrt{(1 + 8)})!)!! + 6!! + 4 $

Answer 2

In case double factorial is not allowed (the provide solution uses it for 27 numbers), here is my suggestion how to replace them based on the given rules:

$ 21 = 4!! - 1 + 6 + 8 = 4! + 6 - 8 - 1 $
$ 29 = (1 + 4)!! + 6 + 8 = 61 - 4 \times 8 $
$ 32 = (6 - 1) \times 8 - 4!! = 4 \times 6 + 8 \times 1 $
$ 35 = 6!! - (4 + 8) - 1 = 6 \times (8- \sqrt{4}) - 1 $
$ 41 = 6 \times 8 - 4!! + 1 = 48 -6 - 1 $
$ 47 = (6 - 1)!! + 4 \times 8 = - (1^4 - 6 \times 8) $
$ 59 = 4 - 1 + 8 + 6!! = 61 - \frac{8}{4} $
$ 61 = 1 + 4 + 8 + 6!! = \frac{6!}{8+4} + 1 $
$ 62 = (1 + 8) \times 6 + 4!! = 14 + 6 \times 8 $
$ 63 = (1 + 4)!! + 6 \times 8 = 8 \times (6 + \sqrt{4}) - 1 $
$ 67 = \sqrt{(8!! - 4! + 1)} + 6!! = - (1^4 - 68) $
$ 68 = (6 - 1)!! \times 4 + 8 = 1^4 \times 68 $
$ 69 = 4!! \times 8 + 6 - 1 = 1^4 + 68 $
$ 75 = (8 - 1)!! - (4! + 6) = (.1)^(-\sqrt{4}) \times \frac{6}{8} $
$ 76 = (8 - 1) \times 4 + 6!! = 14 \times 6 - 8 $
$ 77 = \sqrt{841} + 6!! = 84 -6 - 1 $
$ 82 = (1 + 4)!! \times 6 - 8 = 41 \times (8 - 6) $
$ 83 = \frac{6!}{8} - (4!! - 1) = 81 + 6 - 4 $
$ 86 = (6!! - 1) \times \sqrt{4} - 8 = \frac{6!}{8} - 1 \times 4 $
$ 87 = \sqrt{4} \times 6!! - (1 + 8) = \frac{6!}{8} - 4 + 1 $
$ 89 = \sqrt{4} \times 6!! - (8 - 1) = 41 + 6 \times 8 $
$ 91 = (8 - 1)!! - (4!! + 6) = 81 + 4 + 6 $
$ 92 = ((6 - 1)!! + 8) \times 4 = 14 \times 6 + 8 $
$ 93 = (1 + 6)!! - (4 + 8) = 61 + 4 \times 8 $
$ 98 = (1 + 4)!! \times 6 + 8 = \sqrt{4} \times (6 \times 8 + 1) $
$ 99 = (1 + 6)!! - 8 + \sqrt{4} = 81 + 4! - 6 $
$ 100 = ((\sqrt{(1 + 8)})!)!! + 6!! + 4 = 86 + 14 $