[Legal counting forum game] Counting with four fours

4!*sqrt(4)*sqrt(4)/4 = 24

25 = (4!*4+4)/4

4!+sqrt(4)*4/4 = 26

4!+4-(4/4) = 27

28 = 4!XOR4 + 4 - 4

edit- assuming you treat XOR as a bitwise operator and the numbers as the usual binary

I'm just gonna suggest 4!+4*4/4 for 28

also
4!+4+4/4 = 29

(4^√ 4)*(√ 4)-(√ 4)

Stopped getting alerts from this thread...

30 = 4*4*√4 - √ 4

(4+(4+sqrt(4))!)/4! = 31

(4*4) + (4*4) =32

4!+(4!/4)+4 = 33

4! + 4 + 4 + √4 = 34

I actually had kinda trouble with 35, but eventually I came up with this:
4! + 4 * √4 - 4^0

I think that would actually be 34

EDIT: Whoops, silly me, 0 is a number too
I have to leave for school now so feel free to do a correct 35 or I might edit this when I get home again

Don't worry, it took me some time aswell, I wanted to do it in a more neat way, but this works at least...

..............................4
35 = (lim(x->∞) ((Σ (4n)) + xrt(4) )
..........................n = √4

I feel like I overdid something...

*quiet

Allow me to try again...

4!+(4!+sqrt 4)/sqrt 4

Thanks!
Why is your TI set to Dutch? My father's was too, so I suppose there might be a reason for that.

I actually found two easier ways myself, but this one seems interesting as well.

35 = 4! + (4! - sqrt(4))/sqrt(4)

And even easier

35 = 4! + 44/4

4!+4+4/.4 = 38

wow idk how i didnt see that one lol