[Legal counting forum game] Counting with four fours

4!/.4+4-sqrt(4) = 62

I understand now...
(4^4-4)/4

I feel like 64 should be easy...
This isn't an elegant way, but it works. (√4+√4)^4/4

((4^4)+4)/4 = 65

66 = (4^4) / 4 + sqrt(4)

(4!+sqrt(4))/.4+sqrt(4) = 67

4!/.4+(4*sqrt(4)) = 68

(4!+sqrt(4))/.4+4 = 69

70 = 4^4 / 4 + 4 + sqrt(4)

I think those are 5 4's
70 = 4! * 4 - 4! - sqrt(4)

71 =( (4!!)!!!!*4! + 4 )/ 4

72 = 4! * 4 - 4! * 4^0

I don't know about that one. 4!! is 8, 8!!!! is 32. 32*24+4 is 772, and /4 is 193?

Can we use 0 like that?

I'm just gonna propose
71 = (4!+4.4)/.4
72 = (4!+4)/.4+sqrt(4)

Now I'm having trouble getting 73. If I can't get anything with 4!, .4, etc. I'll try nPr and nCr. Someone else can try to beat me to it though.

I think anything should be possible with the rules we've opened up. Worst case we can resort to concatenation?

73 = (4!!)!/(4!!)!!-(4!!)*4

(4!+4)/.4+4 = 74

Yeah, that's kind of the reason why I used it, noone seemed to be able to get it working properly, so I did it more cheety style

Also, my calculator gives ERROR: Overflow on your solution... I now just have to trust you...

This was my original: wrong number...
70 = 4*4! - 4! - Sqrt(4)

HHere is another cheety one: I need to start learning for my Economeny test later today...

((4^2)^2 +44)/4

It's probably doing a factorial of a factorial, instead of a 'double factorial'.
http://mathworld.wolfram.com/DoubleFactorial.html

8! = 8*7*6*5*4*3*2*1
8!! = 8*6*4*2
8!!! = 8*5*2
...and so on

(4!!)!/(4!!)!!-(4!!)*4
=8!/8!!-8*4
=40320/384-32
=73

edit: Also, I learned about it after seeing this, to give credit where it's due

when someone says those words, that's when you know the thread has succeeded

Huh, I didn't know about that. Interesting.

This is hard... What are you guys' strategies? Have you made a document somewhere, or do you always figure things out from scratch?

(4!+sqrt(4)+4)/0.4 = 75
It's fun looking through all the re-tries to find the last actual number...