Make numbers 1 - 30 using the digits 2, 0, 1, 8

It's 2018 so let's repeat last year's challenge with new digits.

This is similar to the "Four fours" puzzle, but using the digits 2, 0, 1 and 8.

Rules:

• Use all four digits exactly once
• Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root
• Parentheses and grouping (e.g. "21") are also allowed
• Squaring uses the digit 2 so expressions using multiple twos, like $2^2$ or $1^2 + 8^2$, are not allowed
• Keep the order "2 0 1 8" in at least 25 expressions (and more if you can!)
• The modulus operator is not allowed
• Rounding is not allowed (e.g. 201/8=25)

Good luck and Happy New Year!

Similar question for 2016

• Can we use the unary - operator? (i.e. -2) Jan 2, 2018 at 10:34
• Could we get a special permission to also use the decimal point? Pretty please with perfect solutions and sugar on top? 8-]
– Bass
Jan 2, 2018 at 14:05

Found 29 solutions with the numbers in order. Found two almost acceptable cheats for the remaining one.

$30 = 21 + 0! + 8 = \frac{(2 + 0! + 1)!}{ .8 } = \sqrt{\frac{(2+0+1)!!}{.8}}$ (Cheaty McCheatface)
$29 = 20 + 1 + 8$
$28 = 20 * 1 + 8$
$27 = 20 -1 + 8$
$26 = 2 + \sqrt{\sqrt{(0! + 1)^8}} !$
$25 = \sqrt{\sqrt{((2 + 0!)!-1)^8}}$
$24 = (2+0!+1^8)!$
$23 = 20 + \sqrt{1+8}$
$22 = -2 + \sqrt{\sqrt{(0! + 1)^8}} !$
$21 = 20 + 1^8$
$20 = 20 * 1^8$
$19 = 20 - 1^8$
$18 = 2 * 0 + 18$
$17 = 20 - \sqrt{1+8}$
$16 = 2^{0!+\sqrt{1+8}}$
$15 = -2 - 0! + 18$
$14 = -(2 + 0) * (1 - 8)$
$13 = 20 + 1 -8$
$12 = 20 * 1 - 8$
$11 = 20 - 1 - 8$
$10 = 2^0 + 1 + 8$
$9 = 2 * 0 + 1 + 8$
$8 = 2 * 0 * 1 + 8$
$7 = 2 * 0 - 1 + 8$
$6 = (2 + 0) * \sqrt{1+8}$
$5 = -2 + 0 - 1 + 8$
$4 = 2 * (0! + 1^8)$
$3 = 2 + 0 + 1^8$
$2 = 20 - 18$
$1 = 2 * 0 + 1^8$

• Damn, those nested sqrts are absolutely brilliant. Jan 2, 2018 at 12:19
• @VotBear thanks! (spoiler: I didn't really actually calculate the fourth root of 390625, that's just "five squared" in masquerade)
– Bass
Jan 2, 2018 at 12:52
• @Bass I want to complete your list with 30 in order: "2(0!+1)"+8. We were allowed to group, so I grouped the 2 we got with a 2 I made and got 22. Then add 8 and we are done :) Jan 3, 2018 at 0:44
• @Tweakimp uearggh, that's so nasty. Certainly gets the job done within the rules as they are stated. Probably not within the rules as they were intended, but I agree, technically correct is the best kind of correct :-)
– Bass
Jan 3, 2018 at 6:14
• To create the solution for 30 in order you can use double factorial: ((2 + 0!)!)!! - 18 = 6!! - 18 = 2·4·6 - 18 = 48 - 18 = 30. Full credit goes to the smart mind EmNero from the German math forum. Jan 8, 2018 at 16:49

Finished all 30, with only 4 not in order

$1 = 2 ^{(0 * 1 * 8)}$
$2 = 2 + (0 * 1 * 8)$
$3 = 2 + 0 + (1 ^ 8)$
$4 = -2 - 0! - 1 + 8$
$5 = -2 + 0 - 1 + 8$
$6 = -2 + (0 * 1) + 8$
$7 = (2 * 0) - 1 + 8$
$8 = (2 * 0 * 1) + 8$
$9 = (2 * 0) + 1 + 8$
$10 = 2 + (0 * 1) + 8$
$11 = 2 + 0 + 1 + 8$
$12 = 2 + 0! + 1 + 8$
$13 = (2 + 0!)! - 1 + 8$
$14 = (2 + 0!)! + (1 * 8)$
$15 = (2 + 0!)! + 1 + 8$
$16 = 2 * (0 + (1 * 8))$
$17 = -2 + 0! + 18$
$18 = (2 + 0) * (1 + 8)$
$19 = 2 - 0! + 18$
$20 = 2 + 0 + 18$
$21 = (2 + 0!) * (-1 + 8)$
$22 =$
$23 = 20 + \sqrt{(1 + 8)}$
$24 = (2 + 0 + 1) * 8$
$25 =$
$26 =$
$27 = (2 + 0!) * (1 + 8)$
$28 = 20 + (1 * 8)$
$29 = 20 + 1 + 8$
$30 =$

Without maintaining the 2018 order:

$22 = 21 + 8^0$
$25 = 8 * (2 + 0!) - 1$
$26 = 28 - 1 - 0!$
$30 = 21 + (0! + 8)$

• You have the same calculation for 17 and 19. Jan 2, 2018 at 12:53
• @Thrax He propably intended to switch them like $0! - 2 + 18$, but that would break the rules. Jan 2, 2018 at 23:31
• My bad, my bad. And @ibrahimmahrir i intended it to be -2 + 0! + 18 Jan 3, 2018 at 2:11
• 25: $((2+0!)-1)*8$ Jan 3, 2018 at 3:00
• @OldBunny2800 That will result in (3-1)*8 = 16. Jan 3, 2018 at 4:19

Belated Happy New Year!

Only uses addition, subtraction, multiplication, exponents and factorials! (and brackets):

$$1=2-0!\cdot1^8$$ $$2=20-18$$ $$3=2+0+1^8$$ $$4=-2-0!-1+8$$ $$5=-2+0-1+8$$ $$6=-2+0!-1+8$$ $$7=-2+0+1+8$$ $$8=2-0!-1+8$$ $$9=2+0-1+8$$ $$10=2+0!-1+8$$ $$11=2+0+1+8$$ $$12=2+0!+1+8$$ $$13=20+1-8$$ $$14=(2+0+1)!+8$$ $$15=-2-0!+18$$ $$16=-2+0+18$$ $$17=-2+0!+18$$ $$18=2\cdot0+18$$ $$19=2-0!+18$$ $$20=2+0+18$$ $$21=2+0!+18$$ $$\color{red}{22=21+(0!)^8}$$ $$\color{red}{23=(2+0!)\cdot8-1}$$ $$24=(2+0+1)\cdot8$$ $$\color{red}{25=(2+0!)\cdot8+1}$$ $$\color{red}{26=28-0!-1}$$ $$27=20-1+8$$ $$28=20^1+8$$ $$29=20+1+8$$ $$\color{red}{30=21+0!+8}$$

• Is 6 supposed to be 0!? Jan 3, 2018 at 9:10
• @Lefty What do you mean? Jan 3, 2018 at 10:04
• I'm seeing -2+0-1+8=6 - which is not true. But -2+0!-1+8=6. Maybe it's just a problem with the way the line is displayed...? Jan 3, 2018 at 10:21
• No problem. I also think 8 is wrong...? Jan 3, 2018 at 12:00
• 20=2+0+18, if you want to simplify Jan 3, 2018 at 12:21

Happy 2018! I hope everyone has a wonderful year!

I reached for only short and simple answers. I didn't use square roots.

$1 = 2+0-1^8$
$2 = 20-18$
$3 = 2+0+1^8$
$4 = 2+0!+1^8$
$5 = -2+0-1+8$
$6 = -2+0*1-8$
$7 = 2*0-1+8$
$8 = 2*0*1+8$
$9 = 2*0+1+8$
$10 = 2+0*1+8$
$11 = 20-1-8$
$12 = 2+0!+1+8$
$13 = 20+1-8$
$14 = (2+0!)!*1+8$
$15 = (2+0!)!+1+8$
$16 = (2+0*1)*8$
$17 = -2^0+18$
$18 = 2*0+18$
$19 = 2^0+18$
$20 = 2+0+18$
$21 = 20+1^8$
* $22 = 21+(8*0)!$
* $23 = (2+0!)*8-1$
$24 = (2+0+1)*8$
* $25 = (2+0!)*8+1$
* $26 = -1-0!+28$
$27 = 20-1+8$
$28 = 20*1+8$
$29 = 20+1+8$
* $30 = 28+1+0!$

The starred ones are out of order.

• The expressions for $4$, $5$ and $6$ are also not in order. Jan 3, 2018 at 10:04
• @TheSimpliFire Great catch, thank you! I changed 4 to 2+0!+1^8, but the other two involved simply switching the middle numbers. (Edit, no rep to comment on yours: very nice compression! I like how simple your answers are!)
– Zanz
Jan 3, 2018 at 15:00
• You're welcome. I hope you don't mind my edits :) Jan 3, 2018 at 16:34

$$1=2\cdot0+1^8$$ $$2=2-(0\cdot1)^8$$ $$3=2-0+1^8$$ $$4=2+0!+1^8$$ $$5=-2-0!\cdot1+8$$ $$6=-2-0\cdot1+8$$ $$7=-2-0+1+8$$ $$8=2\cdot0\cdot1+8$$ $$9=2\cdot0+1+8$$ $$10=2+0\cdot1+8$$ $$11=2+0+1+8$$ $$12=2+0!+1+8$$ $$13=(2+0!)!-1+8$$ $$14=(2+0!)!1+8$$ $$15=(2+0!)!+1+8$$ $$16=\sqrt{2^{0\cdot1+8}}$$ $$17=-2^0+18$$ $$18=(2+0!)!\sqrt{1+8}$$ $$19=20-1^8$$ $$20=20\cdot1^8$$ $$21=20+1^8$$ $$22=-2+\sqrt{\sqrt{(0!+1)^8}}!$$ $$23=20+\sqrt{1+8}$$ $$24=(2+0+1^8)!$$ $$25=\sqrt{\sqrt{((2+0!)!-1)^8}}$$ $$26=2+\sqrt{\sqrt{(0!+1)^8}}!$$ $$27=20-1+8$$ $$28=20\cdot1+8$$ $$29=20+1+8$$ $$^*30=28+1+0!$$

• That's definitely the most hard working 18 I've seen here :-)
– Bass
Jan 3, 2018 at 21:37

I tried to solve these on my own and got everything but 25 and 30. I got unique solutions for some of them.

$1 = 2 + 0 - 1^8$
$2 = 20 - 18$
$3 = 2 + 0 + 1^8$
$4 = 2 + 0! + 1^8$
$5 = 2 + 0 + \sqrt{1+8}$
$6 = 2 + 0! + \sqrt{1+8}$
$7 = (2 * 0) - 1 + 8$
$8 = (2 - 0 - 1) * 8$
$9 = 2 + 0 - 1 + 8$
$10 = 2 + (0 * 1) + 8$
$11 = 20 - 1 - 8$
$12 = 20 * 1 - 8$
$13 = 20 + 1 - 8$
$14 = (2 + 0)(-1 + 8)$
$15 = (2! + 0!)! + 1 + 8$
$16 = (2 + 0 * 1) * 8$
$17 = 20 - \sqrt{1+8}$
$18 = (2 - 0!) * 18$
$19 = 20 - 1^8$
$20 = 2 + 0 + 18$
$21 = 20 + 1^8$
$22 = -2 + (0! + \sqrt{1+8})!$
$23 = 20 + \sqrt{1+8}$
$24 = (2! + 0!)! + 18$
$*25 = ((2 + 0!) * 8) + 1$
$26 = 20 + \sqrt{1+8}!$
$27 = 20 - 1 + 8$
$28 = 20 * 1 + 8$
$29 = 20 + 1 + 8$
$*30 = 21 + 0! + 8$

• Well done on the 22 and 26 especially! Very pretty.
– Bass
Jan 3, 2018 at 21:33