Only using: ! ; ( ; ) ; / and digits: 4 ; 5
Can you make an equations to get result = 2
ALSO you need to do it in fewer characters than me.
Only using: ! ; ( ; ) ; / and digits: 4 ; 5
well i didn't use 5 at least
4!/((4)((4!)(4!)/((4)(4)(4))))
so can my 4 year old.
55!5!44!/4!45!
pls. make it pc compatible.
That's interesting, in that it relies on 4 being 2 squared. It suggest that if you only got primes like 5 to work with, you can do almost nothing.
no, actually, after reevaluating your proposal, I see you produce two 3's in
(4!)(4!) = 4^3 * 3^2
and you have no way of getting ride of them
4/4+4/4=2
check mate
No +'s
4/((.5)(4))
how did you get the dot for .5
its on my keyboard
No .'s
>Only using: ! ; ( ; ) ; / and digits: 4 ; 5
5!/5 is just 4! and five being prime, there is no other thing we could with to divide by 5.
Clearly 2=(4!/4)/3 and so it comes down to producing 3 from 4.
We're pretty fucked when producing primes such as 7 by repreaded factorial, but with multiplication of factorials we end up with powers of 3 we can't get rid of.
I'd like to see a formal proof that we can't produce 3 from 4 and *, / and !.
Jesus
(((332/2/2)!/(332/2/2))/((3/((3/3)/3/3/3))!)/2 =3
sorry that was for another website
I'm off by 1
[math]((5)(5))!/((4!)!)[/math]
an equation is not a thing. stop increasing entropy for no reason
((5)(5))!/((4!)!) = 25
so you are off by 23
5!=2
4!/4/(5!!/5)
yeah but 23 is 1 digit away from being only 2.
Or in terms of 5's: (5!!)!/(5!!/5)