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; \sqrt{4}(4!-\sqrt{4})-\sqrt{4} = 42 44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43 4! + 4*4 +4 = 44 44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45 ...and copying YT2095; \sqrt{4}(4!-\sqrt{4})+\sqrt{4} = 46 4!+4!-\frac{4}{4} = 47 ... With \sqrt{4} it's quite easy
jeheron Posted October 8, 2005 Posted October 8, 2005 Quote ... 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. 4!+4!+4-4 = 48
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 jeheron said: 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; !4+!4 + \frac{4}{4} = 49 [ math]!4+!4 + \frac{4}{4} = 49[ /math] !4+!4 + \frac{4}{\sqrt{4}} = 50 [ 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 Xyph said: 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: \ said: 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 \frac{\pi}{\pi} to get every number.
RyanJ Posted October 8, 2005 Author Posted October 8, 2005 Quote 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 Quote 44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43 ... 44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45 Too many 4s' date=' [math']43 = 44-\tfrac{4}{4}[/math] 45 = 44+\tfrac{4}{4} 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... 4!+4!+\sqrt{4}+\sqrt{4}=52
YT2095 Posted October 8, 2005 Posted October 8, 2005 Quote 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 YT2095 said: 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 BigMoosie said: 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 \frac{4!-\sqrt{4}}{.4} - \sqrt{4} = 53 4! + 4! + \sqrt{4} + 4 = 54 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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