The Thing Posted October 9, 2005 Posted October 9, 2005 [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]
Callipygous Posted October 9, 2005 Posted October 9, 2005 you have to use exactly 4 4s. also, you have to wait 3 posts between each of your replies. : )
The Thing Posted October 9, 2005 Posted October 9, 2005 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.
PhDP Posted October 9, 2005 Posted October 9, 2005 [math]\frac{4!}{.4} + \frac{4}{4} = 61[/math] [math]\frac{4!}{.4} + \frac{4}{\sqrt{4}} = 62[/math]
Callipygous Posted October 9, 2005 Posted October 9, 2005 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.
Ducky Havok Posted October 9, 2005 Posted October 9, 2005 edit: opps, double post guess I'll go on [math]\frac{4!}{.4}+4+\sqrt{4}=66[/math]
PhDP Posted October 9, 2005 Posted October 9, 2005 [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...
The Thing Posted October 9, 2005 Posted October 9, 2005 [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]
BigMoosie Posted October 9, 2005 Posted October 9, 2005 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
RyanJ Posted October 9, 2005 Author Posted October 9, 2005 Do you count the arctan's -1 as invalid???? Hopefully not. Nope, any mathematical function may be used so thats fine Cheers, Ryan Jones
BigMoosie Posted October 9, 2005 Posted October 9, 2005 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.
ashennell Posted October 9, 2005 Posted October 9, 2005 edited : wrong again!! god my maths is shocking. I think I need more sleep
The Thing Posted October 9, 2005 Posted October 9, 2005 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]
ashennell Posted October 9, 2005 Posted October 9, 2005 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]
The Thing Posted October 9, 2005 Posted October 9, 2005 edit:posted RIGHT AFTER some1 else (Ducky) for 80.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now