# Make the numbers 1-50 using the digits 2, 0, 2, 0

Try to make all the numbers from 1-50 using 2, 0, 2, 0 exactly once.

Rules:

1. Use only the operations +, -, *, /, !, exponents and square root
2. No rounding eg. 20 / (2+0!) = 6
3. No adding extra numbers eg. 2 + 10 + 2 + 0 = 14
4. Concatenation is allowed only in combining 2 and 0 to become 20
5. Try to get all in order (I bet you can’t)
6. Parentheses are also allowed
7. .2 is allowed
• Please don't significantly change a question after answers have been given (and even accepted.) – Bass Mar 16 at 9:08
• I’m just expanding the question – Matheinstein Mar 16 at 9:26
• Please don't. The Stack Exchange site format is "question-answer", and it doesn't really work for any other kind of interaction. For example "question-answer-modified question-new answer" will pretty much always change some good, honest, hard-working answers into "doesn't answer the question (anymore) and should be deleted" answers, which isn't fair for the answerers. If you think the expanded question is worth posting on its own, go ahead and post a new question instead. – Bass Mar 16 at 11:09

All 1-50 solved.

Solved but extending meanings of operators: $$28, 34, 38, 43, 44, 45, 46, 47, 49, 50$$

$$\frac{2}{0!}-2+0! = 1$$

$$2+\frac{0}{2}+0 = 2$$

$$2+\frac{0}{2}+0! = 3$$

$$2+0+2+0 = 4$$

$$2+0+2+0! = 5$$

$$2+0!+2+0! = 6$$

$$2^{0!+2} - 0! = 7$$

$$2^{0!+2} - 0 = 8$$

$$\frac{20}{2}-0! = 9$$

$$\frac{20}{2}-0 = 10$$

$$\frac{20}{2}+0! = 11$$

$$2 \cdot (0!+2)! + 0 = 12$$

$$2 \cdot (0!+2)! + 0! = 13$$

$$20 - (2+0!)! = 14$$

$$\frac{2+0!}{.2} + 0 = 15$$

$$2^{0!+2+0!} = 16$$

$$20 - (2+0!) = 17$$

$$20 - 2 + 0 = 18$$

$$20 - 2^0 = 19$$

$$20 - (2 \cdot 0) = 20$$

$$20 + 2 - 0! = 21$$

$$20 + 2 - 0 = 22$$

$$20 + 2 + 0! = 23$$

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

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

$$20 + (2+0!)! = 26$$

$$(2+0!)^{2+0!} = 27$$

$$\text{square}(2) \cdot (0!+(2+0!)!) = 28$$ (using square)
$$\left \lfloor{\sqrt{20}}\right \rfloor! + \left \lfloor{\sqrt{20}}\right \rfloor = 28$$ (using floor)
$$((2+0!)!+2-0!)!!! = 28$$ (using Multifactorial, added by @Vepir)

$$\frac{(2+0!)!}{.2} - 0! = 29$$

$$\frac{(2+0!)!}{.2} + 0 = 30$$

$$\frac{(2+0!)!}{.2} + 0! = 31$$

$$2^{(0!+2)!-0!} = 32$$

$$\sqrt[.2]{0+2} + 0! = 33$$

$$\sqrt[.2]{2} + 0! + 0! = 34$$ (not in order)
$$2\cdot(-0!+((2+0!)!)!!!) = 34$$ (using unary minus and Multifactorial, added by @Vepir in comments)

$$(2+0!)!^2-0! = 35$$

$$\left(2+0!\right)!\cdot \left(2+0!\right)! = 36$$

$$((2+0!)!)^2+0! = 37$$

$$(20-0!) \cdot 2 = 38$$ (not in order, added by @UnidentifiedX)
$$20 + ((2+0!)!)!!! = 38$$ (using Multifactorial)
$$2\cdot (-0!+20)$$ (using unary minus, added by @Vepir in comments)

$$20\cdot 2-0! = 39$$

$$20\cdot 2-0 = 40$$

$$20\cdot 2+0! = 41$$

$$2 \cdot (0!+20) = 42$$ (added by @UnidentifiedX in comments)

$$!(2+0!+2)-0!=43$$ (using derangements, added by @Vepir in comments)

$$!(2+0!+2)+0=44$$ (using derangements, added by @Vepir in comments)

$$!(2+0!+2)+0!=45$$ (using derangements, added by @Vepir in comments)

$$((2+0!)!)!! - 2 + 0 = 46$$ (using Multifactorial)

$$((2+0!)!)!! - !(2 + 0) = 47$$ (using Multifactorial and derangements)

$$2\cdot(0!+2+0!)! = 48$$

$$((2+0!)!+0!)^2 = 49$$ (not in order, added by @JMP)
$$((2+0!)!)!! + !(2 + 0) = 49$$ (using Multifactorial and derangements)

$$\frac{0!+0!}{.2^2} = 50$$ (not in order, added by @JMP)
$$((2+0!)!)!! + 2 + 0 = 50$$ (using Multifactorial)

• $49=((2+0!)!+0!)^2, 50=\frac{0!+0!}{.2^2}$, and, if you allow binomial, $28=\binom{-(2+0!)-0!}{2}$ – JMP Mar 16 at 11:32
• You can use "$!n$" in the context of Derangement's, where !1=0,!2=1,!3=2,!4=9,!5=44,!6=265, $$!(2+0!+2)-0!=43\\ !(2+0!+2)+0=44\\ !(2+0!+2)-0!=45\\$$ – Vepir Mar 26 at 12:37
• For $28$, we can exploit another definition of "$!$". (see my comment on the other answer) – Vepir Mar 26 at 12:52
• @eyl327 Btw, $38=2\cdot (-0!+20)$ without multifactorial, and $34=2\cdot(-0!+((2+0!)!)!!!)$ in order using multifactorial. – Vepir Mar 26 at 14:06
• @eyl327 I cracked it... $42=2*(0!+20)$ – UnidentifiedX Apr 4 at 13:10

Partial answer (all except 28, updated after .2 was allowed)

$$1=(2+0)/(2+0)$$
$$2=2+0\cdot2\cdot0$$
$$3=2+0\cdot2+0!$$
$$4=(2+0)\cdot(2+0)$$
$$5=(2+0)\cdot2+0!$$
$$6=(2+0!)!+2\cdot0$$
$$7=(2+0!)!+2-0!$$
$$8=(2+0!)!+2+0$$
$$9=(2+0!)!+2+0!$$
$$10=20/2+0$$
$$11=20/2+0!$$
$$12=(2+0!)!\cdot2+0$$
$$13=(2+0!)!\cdot2+0!$$
$$14=20-(2+0!)!$$
$$15=(2+0!)/.2+0$$
$$16=2^{0!+2+0!}$$
$$17=20-2-0!$$
$$18=20-2+0$$
$$19=20-2^0$$
$$20=20+2\cdot0$$
$$21=20+2^0$$
$$22=20+2+0$$
$$23=20+2+0!$$
$$24=(2+0+2)!+0$$
$$25=(2+0+2)!+0!$$
$$26=20+(2+0!)!$$
$$27=(2+0!)^{2+0!}$$
$$28=???$$
$$29=((2+0!)!)/.2-0!$$
$$30=((2+0!)!)/.2+0$$

• Your solution for 24 is incorrect... remove the last factorial? – shanylong Mar 16 at 7:44
• Thanks, I forgot to remove it after copypasting. – trolley813 Mar 16 at 7:45
• If you allow "stacking" factorial into Multifactorials, for example $7!!!=7\cdot4\cdot1=28$, then: $$28 = ((2+0!)!+2-0!)!!!$$ – Vepir Mar 26 at 12:42