- 13 Sep '09 12:26I know four fours have been done here (not sure about 5 5s though)

You have to form the numbers from 0-whatever (in ascending order) using simple arithmetic operations. These being:

addition +

subtraction -

multiplication *

division /

factorial !

concatenation (eg: 55 or.55 etc...)

decimals . (ie .5)

recurring decimals (no overbar (or underline) so will let**.5**denote .555555.... = 5/9)

square roots sqrt() (debatable because of implied 1/2 but then factorial has implied 2,3,4 in it yet no-one says anything about that)

exponentiation ^

Also since the first 30 or so are trivial should we start a bit later? - 13 Sep '09 15:40 / 5 edits

Yeah I agree with you about the LaTeX...the boldface for recuring .5... looks crappy!*Originally posted by Kristaps***5!/5 + 5 + 5/5 = 30**

5*5 + 5 + 5/5 = 31

Hmmm...... = 32

5.5 * 5 + 5.5 = 33

5!/5 + sqrt(5)*sqrt(5) + 5 = 34

5(5 + 5/5) + 5 = 35

LaTeX would make this forum much nicer, though it's not that easy to implement probably.

((sqrt(5*5) + 5)/5)^5 = 32

36 = (5!/(5+5))*(sqrt(5/**.5**)

37 = (((5+5)/5)^5)+5

38 = (5!/(5*.5))-5-5

39 = (5!/(5*.5)) -5/**.5**

40 = ((5*5)/.5) -5-5 - 13 Sep '09 20:39

Thanks for that.*Originally posted by Palynka***LaTeX with greasemonkey for those using Firefox.**

Thread 108499

Now back to the topic

42 = meaning of li.... = ...