YT2095 Posted October 8, 2005 Posted October 8, 2005 undoubtedly (I suck at maths) that`s why I showed my working out
PhDP Posted October 8, 2005 Posted October 8, 2005 Normally, multiplications are made before additions and subtractions. So your equation would result in 24 - 4 - 2 = 18, with only a little correction; [math]\sqrt{4}(4!-\sqrt{4})-\sqrt{4} = 42[/math] [math]44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43[/math] [math]4! + 4*4 +4 = 44[/math] [math]44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45[/math] ...and copying YT2095; [math]\sqrt{4}(4!-\sqrt{4})+\sqrt{4} = 46[/math] [math]4!+4!-\frac{4}{4} = 47[/math] ... With [math]\sqrt{4}[/math] it's quite easy
jeheron Posted October 8, 2005 Posted October 8, 2005 ... With [math]\sqrt{4}[/math'] it's quite easy With root 4 its way to easy.
PhDP Posted October 8, 2005 Posted October 8, 2005 You need to use four 4, even if our definition of "4" is quite... creative. [math]4!+4!+4-4 = 48[/math]
YT2095 Posted October 8, 2005 Posted October 8, 2005 I thought it said in the original post, four 4`s or Less, so I didn`t bother to cancel the remaining ones
jeheron Posted October 8, 2005 Posted October 8, 2005 This may be an inappropriate place to post this question. I will ask anyway. How do you get the small number images to appear when you write an equation?
YT2095 Posted October 8, 2005 Posted October 8, 2005 I`ve no idea, I think it`s something to do with Latex commands. !4+!4 = 48 then 48 + 4/4 =49 same again for 48 then add (sqrrt4) = 50 Grrr... I cannot do 51 without using five 4`s, this one`s got me beat!
PhDP Posted October 8, 2005 Posted October 8, 2005 This may be an inappropriate place to post this question. I will ask anyway. How do you get the small number images to appear when you write an equation? Use the code from; http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf And place your LaTeX code between [ math] and [ /math] without the space after the [ So, rewriting YT2095's solutions for 49 and 50; [math]!4+!4 + \frac{4}{4} = 49[/math] [ math]!4+!4 + \frac{4}{4} = 49[ /math] [math]!4+!4 + \frac{4}{\sqrt{4}} = 50[/math] [ math]!4+!4 + \frac{4}{\sqrt{4}} = 50[ /math]
drochaid Posted October 8, 2005 Posted October 8, 2005 I know 51 has been done, but as this has been relayed and was considered some time ago, we thought it should be added anyway... From YT2095: I can't get out of my shed untill later but I think the answer is 4!+4!=48+int pi =51 and from IRC.... 13:55 <+RICHARDBATTY> he called it in at 12:21 uk time
RyanJ Posted October 8, 2005 Author Posted October 8, 2005 Numbers apart from 4 aren't allowed. The mathematical constants are, the 48 there is not Cheers, Ryan Jones
Xyph Posted October 8, 2005 Posted October 8, 2005 The 48 results from 4!+4!, but in any case: 3. You may use any mathematical operations and symbols (not including symbols for other numbers, like Pi, e, etc.)[/b'] you wish. If we can use numbers like pi and not have them count towards the 4 4s total, then we could just use [math]\frac{\pi}{\pi}[/math] to get every number.
RyanJ Posted October 8, 2005 Author Posted October 8, 2005 The 48 results from 4!+4!, but in any case:If we can use numbers like pi and not have them count towards the 4 4s total, then we could just use [math]\frac{\pi}{\pi}[/math'] to get every number. Good point I thought I put it as you were allowed, after what you said its probably better that its not then Cheers, Ryan Jones
BigMoosie Posted October 8, 2005 Posted October 8, 2005 [math]44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43[/math] ... [math]44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45[/math] Too many 4s' date=' [math']43 = 44-\tfrac{4}{4}[/math] [math]45 = 44+\tfrac{4}{4}[/math] And for anybody complaining that 44 is not valid, you can always replace every instance of it with (4!-4).
Ducky Havok Posted October 8, 2005 Posted October 8, 2005 I think we're on 52... [math]4!+4!+\sqrt{4}+\sqrt{4}=52[/math]
YT2095 Posted October 8, 2005 Posted October 8, 2005 Good point I thought I put it as you were allowed' date=' after what you said its probably better that its not then [/quote'] so is my post (by proxy) #61 allowed then, or not?
RyanJ Posted October 8, 2005 Author Posted October 8, 2005 so is my post (by proxy) #64 allowed then, or not? When it starts to get hard, then you may use mathematical symbols but they will count as a four, that should resolve the problem of overuse of the constants Cheers, Ryan Jones
RyanJ Posted October 8, 2005 Author Posted October 8, 2005 I refute it thus! Your new rule sux. Fine, read my last post again. Last compramise. You can use them in any way but no more than 4 numbers, constants or what ever in total Cheers, Ryan Jones
Xyph Posted October 8, 2005 Posted October 8, 2005 [math]\frac{4!-\sqrt{4}}{.4} - \sqrt{4} = 53[/math] [math]4! + 4! + \sqrt{4} + 4 = 54[/math] I don't think using constants in any form should be allowed... It defeats the purpose of the game. Why not just leave it as 4 4s, other numbers excluded? Constants will make things far too easy.
The Thing Posted October 9, 2005 Posted October 9, 2005 Ha I get the easy number: [math] 4!+4!+4+4=56 [/math]
The Thing Posted October 9, 2005 Posted October 9, 2005 And the next: [math] \frac{4!-\sqrt{4}}{.4}+\sqrt{4}=57 [/math]
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