Obtain numbers from 0 to 9 with the digits of 2022 in that order

Question

The title says (nearly) all. Write the digits $2$ $0$ $2$ $2$ in that order and apply mathematical operators to obtain all values from 0 to 9. What's allowed:

• The four basic operators, $+ - * /$
• Parenthesis to force the order of operations
• Power (like $2^0$)
• Factorial (like $2!$)
• Concatenation (like $202$)
• Square root ($\sqrt {}$)
• Digital point (with or without leading 0, like $2.02$ or $.2$)
• Periodic numbers (like $.\overline{2}$)

I did not need to use truncation.

Answer 1

0: 2*0*22

1: 2*0+(2/2)

2: 2+0+2-2

3: 2+0+(2/2)

4: 2*0+2+2

5: 2^0+2+2

6: 2+0+2+2

7: 2+0!+2+2

8: (2+0)*2*2

9: (2^0+2)^2  

Answer 2

Since

2x0=0, we can add zero to 2-2=0, 2/2=1, sqrt(2x2)=2, sqrt(2/.2rec)=3, 2x2=4, 2/.2rec=9 to get 0,1,2,3,4,9.

Since

2-0!=1 and 2-0=2, we can add 1 or 2 instead of 0, or (-2+0+ etc.) subtract instead of adding, getting 5,6 as 1+4,2+4 and 8,7 as 9-1,9-2.

And we're done.

Answer 3

Alternate solutions. Mostly.

$0 = 2 \times 0 \times 2 \times 2$
$1 = 2 + 0 - 2/2$
$2 = 2 + 0 \times 2/2$
$3 = 2 + 0 + 2 / 2$
$4 = 2 \times 0 + 2 \times 2$
$5 = 20/2/2$
$6 = 2 + 0 + 2 + 2$
$7 = 2/0.\overline{2} - 2$
$8 = 20/2 - 2$
$9 = (20 - 2)/2$

Answer 4

A bit late but:

$2×0+2-2=0$
$2×0+2/2=1$
$2+0+2-2=2$
$2+0+2/2=3$
$2×0+2+2=4$
$2^0+2+2=5$
$2+0+2+2=6$
$-2+0+2/.\overline{2}=7$
$2×(0+2+2)=8$
$2×0+2/.\overline{2}=9$