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\[\sqrt {3 \times 3} \times 3\]

\[3^0 \times 3 \times 3\]

\[\sqrt{3!3!3!}\]

oh damn i tried :)

this probably dosent count as there are four 3's
best i can do \[3 \log_3(3)^3=9\]

\[ \sqrt{3!3!} + 3\]

Excellent. I hadn't thought that one.
Also, feel free to abuse the floor and ceiling function.

\[ 3^{\frac{3!} 3}\]

Excellent once again!

greatt!!!! sin3 works!!

perhaps log 3 too :D

sin3 is something between 0 and 1, so it will. cos3 unfortunately is negative.

hmm.. the plus sign..

Ah great this works
ceil 9^(-cos(3))

\[ \lceil (3*3)^{-\cos 3} \rceil \]

@experimentX \[\sqrt{3!3!} \times 3\]..youarent allowed to use + lol

i mean \[\sqrt{3!3!} + 3\]

you cant use plus

Should've caught that :/

I have at least 4 more solutions no one has posted so far =D

\[ \left\lceil \sqrt[3]{\left( \sqrt{\left(3!3!\right)}\right)!}\right\rceil\]

\[ (3*3)^{\lfloor \sqrt 3\rfloor }\]

i bet sqrt 3 can be replaced with ln
log using ceil ..

Probably, but let's try and get things that look new, and not just replacing one part with another.

I still have 3 more solutions that look different from any posted above.

http://www.wolframalpha.com/input/?i=3%5Eceil%28log%5B10%2C+3%5E3%5D%29

e^(3ln(3))/3

How about \[\large 3^{\lceil\sqrt3\rceil\cdot\lfloor\sqrt3\rfloor}\]

is e allowed??

Let's restrict it so we don't have \(e\), \(\pi\), \(\phi\), or other constants like that for now.

33 (mod 3)

\(33\equiv0\pmod3\), although you have a case in that \(9\equiv0\pmod3\) as well.

I've got to go to bed now. Keep posting solutions, and I'll post the ones I have left tomorrow.

\[\frac{3\times3!}{\Gamma(3)}=9\]

Nice idea