[math]\frac{4!}{.4}-4+\sqrt{4}=58[/math]

[math]

\frac{4!}{.4}-\frac{4}{4}=59

[/math]

[math]

\frac{\frac{4!}{4}}{.4}*4=60

[/math]

Quick question, do we have to use exactly 4 4s, or just 4 4s or less? Cuz if it is the latter, it is so much easier, simply with:

[math]

\frac{4!}{.4}

[/math]

you have to use exactly 4 4s. also, you have to wait 3 posts between each of your replies. : )

Oh. I didn't read all the other posts from page 2 to 3. I saw some ppl posting consecutive answers, didn't know there was a rule. Lol srry.

[math]\frac{4!}{.4} + \frac{4}{4} = 61[/math]

 

[math]\frac{4!}{.4} + \frac{4}{\sqrt{4}} = 62[/math]

Oh. I didn't read all the other posts from page 2 to 3. I saw some ppl posting consecutive answers, didn't know there was a rule. Lol srry.

 

 

no worries... it wasnt a rule at first. dave introduced it when he saw all those consecutive posts your talking about.

((4^4)-4)/4 = 63

[Math]64 = 4 ^ {(4 - \tfrac{4}{4})}[/Math]

(4^4+4)/4=65

edit: opps, double post

guess I'll go on

[math]\frac{4!}{.4}+4+\sqrt{4}=66[/math]

edit: opps, wrong number...

second attempt :

 

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

4*20-4+4+4=68

Green, you can't use 20.

4!+4!+4!-4=68

[math]\frac{4!+\sqrt(4)}{.4}+4 = 69[/math]

 

[math]\frac{4^4+4!}{4} = 70[/math]

 

[math]\frac{4!+4.4}{.4} = 71[/math]

 

[math]44+4!+4 = 72[/math]

 

73 seems a little harder...

[math]

4!+4+tan^{-1}(\frac{4}{4})=73

[/math]

Do you count the arctan's -1 as invalid???? Hopefully not.

[math]

4!+4+arctan(\frac{4}{4})=73

[/math]

I think arctan is fine, doesnt seem to be any normal solution for it.

 

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

75 = (4!/4+4!)/.4

76 = (4!-4)*4-4

Do you count the arctan's -1 as invalid???? Hopefully not.

 

 

Nope, any mathematical function may be used so thats fine

 

Cheers,

 

Ryan Jones

Not any function that would be too easy, for instance rounding functions:

 

77 = arcCos(sin(4)) - 4 - 4 - ceil(sin(4))

 

I consider that cheating dont you?

 

Also, if I could use any function I wanted I could use:

 

f(x) = 77;

77 = f(4+4+4+4)

 

And when I use sine (or any other trig function) I am really using several different mathematic symbols for example:

 

[math]sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/math]

 

Your rule of "any mathematical function" needs a more accurate description.

edited : wrong again!! god my maths is shocking. I think I need more sleep

While Trig functions are still okay at the moment, let's do 77:

[math]

\frac{sin^{-1}(\frac{\sqrt{4}}{4})}{.4}+\sqrt{4}=77

[/math]

While Trig functions are still okay at the moment' date=' let's do 77:

[math']

\frac{sin^{-1}(\frac{\sqrt{4}}{4})}{.4}+\sqrt{4}=77

[/math]

 

 

very good - I just couldn't get there:

 

[math]4!\sqrt{4}+sin^{-1}(\frac{\sqrt{4}}{4})= 78[/math]

 

Edit : I'm sure there is a lot easier way then this. May as well add 79 too -

 

[math]\frac{sin^{-1}(\frac{\sqrt{4}}{4})}{.4}+4=79[/math]

[math]4!4-4*4=80[/math]

edit:posted RIGHT AFTER some1 else (Ducky) for 80.