Happy New Year!

My solution :

$\huge ( 2 \ + \ 0 \ + \ 1 ) \ ! \ = \ 6$

$\color{skyblue}{⇩} \\$ $\huge\color{#D61F06}{(2+0!)! * 1}\\$ $\color{#EC7300}{OR} \\$ $\huge\color{#69047E}{(2+e^0)!*1}\\$

Other possible solutions:

$\lceil \sqrt{20}\rceil+1=6$

$(\lfloor \sqrt{\sqrt{201}}\rfloor)!=6$

$\lceil |\tan(2)| \rceil \times (0!+1)=6$

My solution: 2^{2} + 0^{0} + 1 =6

the instructions allowing us to use ANY function makes the question trivial since you could define a function to force it. maybe kind of cheating.

$(2+0+1)! = \large \boxed{6}$

This is my solution

$2-0-1 = log_{6} (6)$

Here's a solution:

$\Large 2^{\log_2 6} + \sin0 + \ln1$

It is equal as is in the integers mod 3.

( 2+0+1 )! = 3! = 3 × 2 × 1 = 6 proved

2^4 -10=6 (I invert 01 to 10)

{2 + 0 + 1)! = 3! = 1 x 2 x 3 = 6

My solution: 2^(log7/log2) - 0 - 1 = 6

If anything is impossible 2^2+0+2^1 = 6

2! + 0! +1 = 6

Using factorial a can make the 0 useful

My answer is 2 X mod(sqrt(10)) = 6

Here's my solution: 2^3 - (e^0 + 1) = 6

Or 2^2 +(cos(0))+1=6..... 2^2=4..... cos(0)=1...... So 4+1+1=6

The solution is in the answer itself

You could also do: 2^2 + 0^0 + 1 = 6 I'm not that good at formatting but wanted to show my answer. Anything to the power of 0 is 1 so therefore I've written 4 + 1 + 1 which equals 6.

I thought of calculus and went for 2x0x1=d/dx(6)

If that counts

My solution was:

2×0 + 1 = 6^(0)

1^2=6^0 So 1=1 since anything to the power of 0 equals 1. I think my answer is great because I didn't add or reuse any of the numbers.

I just thought that, we could use greater than operator. it could be 2+0+1 <= 6 GREATER THAN is an operater right??

Solution: ( 2 + 0 + 1 ) ! = 6 Add all of them and then take factorial.