Try to make all numbers 1-30 using the digits 2, 0, 2, 4.
Rules:
Use all four digits exactly once.
Allowed operations: $+, -, \times, ÷, !$ (factorial), x^y (exponentiation), √ (square root).
Parentheses and concatenation are allowed. (e.g. 20+(2∗4)).
Squaring uses the digit 2.
Keep the order "2, 0, 2, 4" in all expressions.
The modulus operator is not allowed.
Rounding is not allowed.
Complete solution that doesn't use concatenation.
$1 = 2 + 0! + 2 - 4 \\ 2 = 2 \times 0 - 2 + 4 \\ 3 = 2^0 -2 + 4 \\ 4 = 2 + 0 - 2 + 4 \\ 5 = -2^0 + 2 + 4 \\ 6 = 2 \times 0 + 2 + 4 \\ 7 = 2^0 + 2 + 4 \\ 8 = 2+0+2+4 \\ 9 = 2^0 + 2 \times 4 \\ 10 = 2 + 0 + 2\times 4 \\ 11 = 2 + 0! + 2 \times 4 \\ 12 = 2^{0!+2} + 4 \\ 13 = -2 - 0! + 2^4 \\ 14 = -2 + 0 + 2^4 \\ 15 = -2^0 + 2^4 \\ 16 = 2 \times 0 + 2^4 \\ 17 = 2^0 + 2^4 \\ 18 = 2+0 + 2^4 \\ 19 = 2 + 0! + 2^4 \\ 20 = -2 + 0 - 2 + 4! \\ 21 = -2^0 - 2 + 4! \\ 22 = 2 \times 0 - 2 + 4! \\ 23 = 2^0 - 2 + 4! \\ 24 = 2 + 0 - 2+ 4! \\ 25 = 2 + 0! - 2 + 4! \\ 26 = 2 \times 0 + 2 + 4! \\ 27 = 2^0 + 2 + 4! \\ 28 = 2 + 0 + 2 + 4! \\ 29 = 2 + 0! + 2 +4! \\ 30 = (2 + 0!) \times 2 + 4!$
I'm making this a community wiki so the answers aren't split among multiple…answers.
1
$$1=2^{0^{2^4}}$$
2
$$2=\frac{2^0}{2}\cdot4=-2^0\cdot2+4$$
3
$$3=2^0-2+4=-(2+0)/2+4$$
4
$$4=2+0-2+4=2+0\cdot 2+\sqrt{4}$$
5
$$5=\frac{(2+0)}{2}+4=-2^0+2+4=2^{0^2}+4$$
6
$$6=2\cdot0+2+4$$
7
$$7=2^0+2+4$$
8
$$8=2+0+2+4=2^{0+2}+4$$
9
$$9=2^0+2\cdot4$$
10
$$10=2+0+2\cdot4$$
11
$$11=2+0!+2\cdot4$$
12
$$12 = (2+0) \times (2+4)$$
13
$$13=(2+0!)^2+4=-2-0!+2^4$$
14
$$14=2(0!+2+4)$$
15
$$15=-2^0 + 2^4$$
16
$$16=2^{0+2}\cdot4$$
17
$$17=2^0+2^4$$
18
$$18=2+0+2^4$$
19
$$19=2+0!+2^4$$
20
$$20=(2+0!+2)\cdot4=-2-0-2+4!$$
21
$$21=-2^0-2+4!$$
22
$$22=2\times0-2+4!$$
23
$$23=2^0-2+4!$$
24
$$24=2+0-2+4!$$
25
$$25=\frac{2+0}2+4!$$
26
$$26=20+2+4$$
27
$$27=2-0!+2+4!$$
28
$$28=2+0+2+4!$$
29
$$29=2+0!+2+4!$$
30
$$30=(2^0+2)!+4!$$
Bonus solutions!
$$31=$$ $$32=(2+0)\cdot2^4$$ $$33=(2+0!)^2+4!$$ $$34=\frac{20}2+4!$$
Thanks, this was fun! Here's what I came up with:
$$1 = {(2 + 0 + 2) \over 4}$$ $$2 = {2 \times 0 - 2 + 4}$$ $$3 = {-(2 + 0) \over 2} + 4$$ $$4 = 2 + 0 - 2 + 4$$ $$5 = {(2 + 0) \over 2} + 4$$ $$6 = 2 \times 0 + 2 + 4$$ $$7 = 2^0 + 2 + 4$$ $$8 = 2 + 0 + 2 + 4$$ $$9 = 2^0 + 2 \times 4$$ $$10 = 20 \times 2 / 4$$ $$11 = (20 + 2) / \sqrt{4}$$ $$12 = (2^0 + 2) \times 4$$ Skipped this and came back to it after 19 where I finally realized $0! = 1$: $$13 = -2 - 0! + 2^4$$ $$14 = -2 + 0 + 2^4$$ $$15 = -2^0 + 2^4$$ $$16 = 2 \times 0 + 2^4$$ $$17 = 2^0 + 2^4$$ $$18 = 2 + 0 + 2^4$$ $$19 = 2 + 0! + 2^4$$ $$20 = 20 + 2 - \sqrt{4}$$ $$21 = -2^0 - 2 + 4!$$ $$22 = -2 + 0 + 24$$ $$23 = -2^0 + 24$$ $$24 = 2 \times 0 + 24$$ $$25 = 2^0 + 24$$ $$26 = 2 + 0 + 24$$ $$27 = 2 + 0! + 24$$ $$28 = 2 + 0 + 2 + 4!$$ $$29 = 2 + 0! + 2 + 4!$$ $$30 = (2 + 0!) \times 2 + 4!$$
I was struggling for a while at 13 and 19 before I made that discovery I mentioned, that was huge, and made the rest of the numbers a lot easier to figure out.
Great puzzle, thanks for posting!
$1 = 2^0+2-\sqrt{4}$
$2 = 2+0+2-\sqrt{4}$
$3=2^0-2+4$
$4=2+0-2+4$
$5=2^0+2+\sqrt{4}$
$6=(2\times0)+2+4$
$7=2^0+2+4$
$8=2+0+2+4$
$9=2^0+2\times4$
$10=2+0+2\times4$
$11=2+0!+(2\times4)$
$12=(2^0/2)\times4!$
$13=-2-0!+(2^4)$
$14=20-2-4$
$15=-(2^0)+2^4$
$16=(2\times0)+2^4$
$17=2^0+2^4$
$18=20+2-4$
$19=2+0!+(2^4)$
$20=20+2-\sqrt{4}$
$21=-(2^0)-2+4!$
$22=20-2+4$
$23=(2^0)-2+4!$
$24=20+2+\sqrt{4}$
$25=2^0+24$
$26=20+2+4$
$27=2^0+2+4!$
$28=2+0+2+4!$
$29=2+0!+2+4!$
$30=(2^0+2)!+4!$
Partial answer:
1: $(2+0+2)/4=4/4=1$
2: $2+(0\times2)/4=2+0=2$
3: $2^0-2+4=1-2+4=-1+4=3$
4: $2-0-2+4=0+4=4$
5: $2^{0\times2}+4=1+4=5$
6: $2+(0\times2)+4=2+4=6$
7: $2^0+2+4=1+2+4=7$
8: $2+0+2+4=4+4=8$
9: I'll figure something out
10: $(2^0+2)!+4=3!+4=6+4=10$
11: I'll figure something out
12: $(2+0)(2+4)=2(6)=12$
−operator supposed to include both subtraction and and arithmetic negation? $\endgroup$