Question

[SOLVED] Let's see some creativity!

Without using "+" make the number 9 using only three 3's, and no other digits using any mathematical symbol you want. So "3+3+3=9" and similar expressions are off limits.

Here are a couple of the examples I've found so far: \[\frac{3^3}{3}\]\[\sqrt{3^3\cdot3}\]I know of several more possibilities (not including various possible applications of negation). Which ones can you get?

PS: \(-(-\sqrt{3^3\cdot3})\), \(-(-3-3-3)\) and similar don't count.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@abb50 That has a 0 in it. Let me correct the problem to specify against that.

@Cortegu10 \(\sqrt{3!3!3!}=\sqrt{6^3}=6\sqrt{6}\neq9\)

Answer 5

oh damn i tried :)

Answer 6

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

Answer 7

Too many 3's there :(

Nice try though. I know of at least 3 more expressions using increasingly convoluted nested functions.

Answer 8

Hint (for a couple): Keep thinking with factorials and exponents. And remember, "-" isn't completely ruled out. Just don't abuse it.

Answer 9

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

Answer 10

Excellent. I hadn't thought that one.

Also, feel free to abuse the floor and ceiling function.

Answer 11

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

Answer 12

Excellent once again!

Answer 13

LOL .. not sure if it works
\[ \left \lfloor {3*3 + \sin 3} \right \rfloor \]
\[ \left \lceil {3*3 + \cos 3} \right \rceil \]

Answer 14

greatt!!!! sin3 works!!

Answer 15

perhaps log 3 too :D

Answer 16

If you could do those without the "+" sign, those would be accepted. I'm pretty sure you can get rid of it however.

Answer 17

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

Answer 18

hmm.. the plus sign..

Answer 19

Just to be clear,

(Partial) list of lesser known functions I will accept:
\[\lfloor9.5=9\rfloor\]\[\lceil8.5=9\rceil\]\[^33=3^{3^3}\]
Also, I will accept \(\ln\) for \(\log_e\) and \(\log\) for \(\log_{10}\) as allowable functions.

Answer 20

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

Answer 21

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

Answer 22

This is probably the most convoluted solution I've come up with \[\left\lceil \sqrt[3]{\left(\left(\lfloor\sqrt3\rfloor3\right)!\right)!}\right\rceil\]

Answer 23

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

Answer 24

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

Answer 25

you cant use plus

Answer 26

Should've caught that :/

Answer 27

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

Answer 28

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

Answer 29

\[y=\left(\frac{x^3}{3}\right)\]dy/dx at x=3 lol but it's identical to (3^3)/3. Not sure if it counts.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

lol ... this ceil function is so useful
http://www.wolframalpha.com/input/?i=3%5Eceil%28+sqrt%283*ln+3%29%5D%29

Answer 36

e^(3ln(3))/3

Answer 37

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

Answer 38

lol ... i think we should ban usage of ceil
http://www.wolframalpha.com/input/?i=3*+ceil%28log+%283*3%29%29

Answer 39

Alright. Let's ban the ceiling function for now. What else have we got?

btw, I still have 2 more different solutions

Answer 40

is e allowed??

Answer 41

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

Answer 42

Also, let's stay out of integrals and derivatives for now as well. Maybe I'll do this again with those allowed.

Answer 43

this seem to have interesting result
http://www.wolframalpha.com/input/?i=floor%28ln%28%283*3%29%21%29+-+3%29

Answer 44

33 (mod 3)

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

Nice idea

Answer 49

Here are the other ideas I've had that look different (mostly) from previous answers\[3^{3!}-(3!)!\]\[\lfloor \log(^33))\rfloor-3\]Recall that \(^33=3^{3^3}=3^{27}\)