YT2095 Posted October 8, 2005 Share Posted October 8, 2005 undoubtedly (I suck at maths) that`s why I showed my working out Link to comment Share on other sites More sharing options...
PhDP Posted October 8, 2005 Share 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 Link to comment Share on other sites More sharing options...
YT2095 Posted October 8, 2005 Share Posted October 8, 2005 4!+4! =48 notice how I only do the easy ones! Link to comment Share on other sites More sharing options...
jeheron Posted October 8, 2005 Share Posted October 8, 2005 ... With [math]\sqrt{4}[/math'] it's quite easy With root 4 its way to easy. Link to comment Share on other sites More sharing options...
PhDP Posted October 8, 2005 Share 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] Link to comment Share on other sites More sharing options...
YT2095 Posted October 8, 2005 Share 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 Link to comment Share on other sites More sharing options...
jeheron Posted October 8, 2005 Share 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? Link to comment Share on other sites More sharing options...
YT2095 Posted October 8, 2005 Share 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! Link to comment Share on other sites More sharing options...
PhDP Posted October 8, 2005 Share 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] Link to comment Share on other sites More sharing options...
Xyph Posted October 8, 2005 Share Posted October 8, 2005 [math]\frac{4!-\sqrt{4}}{.4} - 4 = 51[/math] Link to comment Share on other sites More sharing options...
drochaid Posted October 8, 2005 Share 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 Link to comment Share on other sites More sharing options...
Xyph Posted October 8, 2005 Share Posted October 8, 2005 Numbers apart from 4 aren't allowed. Link to comment Share on other sites More sharing options...
RyanJ Posted October 8, 2005 Author Share Posted October 8, 2005 Numbers apart from 4 aren't allowed. The mathematical constants are, the 48 there is not Cheers, Ryan Jones Link to comment Share on other sites More sharing options...
Xyph Posted October 8, 2005 Share 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. Link to comment Share on other sites More sharing options...
RyanJ Posted October 8, 2005 Author Share 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 Link to comment Share on other sites More sharing options...
BigMoosie Posted October 8, 2005 Share 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). Link to comment Share on other sites More sharing options...
Ducky Havok Posted October 8, 2005 Share Posted October 8, 2005 I think we're on 52... [math]4!+4!+\sqrt{4}+\sqrt{4}=52[/math] Link to comment Share on other sites More sharing options...
YT2095 Posted October 8, 2005 Share 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? Link to comment Share on other sites More sharing options...
RyanJ Posted October 8, 2005 Author Share 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 Link to comment Share on other sites More sharing options...
BigMoosie Posted October 8, 2005 Share Posted October 8, 2005 I refute it thus! Your new rule sux. Link to comment Share on other sites More sharing options...
RyanJ Posted October 8, 2005 Author Share 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 Link to comment Share on other sites More sharing options...
Xyph Posted October 8, 2005 Share 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. Link to comment Share on other sites More sharing options...
Ducky Havok Posted October 9, 2005 Share Posted October 9, 2005 [math]\frac{4!-(4-\sqrt{4})}{.4}=55[/math] Link to comment Share on other sites More sharing options...
The Thing Posted October 9, 2005 Share Posted October 9, 2005 Ha I get the easy number: [math] 4!+4!+4+4=56 [/math] Link to comment Share on other sites More sharing options...
The Thing Posted October 9, 2005 Share Posted October 9, 2005 And the next: [math] \frac{4!-\sqrt{4}}{.4}+\sqrt{4}=57 [/math] Link to comment Share on other sites More sharing options...
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