RyanJ Posted October 15, 2005 Author Posted October 15, 2005 Seeing as no-one has posted for a few days I'd say the challange has finished which means the winner was cosine who had 119! Congradulations to you all - this is the highest result I have seen on any forum to date Cheers, Ryan Jones
NPK Posted December 25, 2005 Posted December 25, 2005 (√4 / .4)! + 4 - 4 = 120 (44/4)^ √4 = 121 (√4 / .4)! + 4 - √4 = 122
cosine Posted December 25, 2005 Posted December 25, 2005 (√4 / .4)! + 4 - 4 = 120 (44/4)^ √4 = 121 (√4 / .4)! + 4 - √4 = 122 Wowzers! I was just looking at this thread last night too! (My first SFN thread, sigh...)
RyanJ Posted December 25, 2005 Author Posted December 25, 2005 Hmm, looks like the challange has been re-opened then - fair enough Cheers, Ryan Jones
cosine Posted December 25, 2005 Posted December 25, 2005 Hmm' date=' looks like the challange has been re-opened then - fair enough Cheers, Ryan Jones[/quote'] Yeah... so did we ever decide what functions should be allowed?
RyanJ Posted December 25, 2005 Author Posted December 25, 2005 Yeah... so did we ever decide what functions should be allowed? I belive I did say any function in the original post: 3. You may use any mathematical operations and symbols (not including symbols for other numbers' date=' like Pi, e, etc.) you wish. [/quote'] I think the bold would allow for functions as long as your uing only 4's as the values I see no probles there Cheers, Ryan Jones
NPK Posted December 26, 2005 Posted December 26, 2005 (√√√(√4/.4)^4!) - √4 = 123 or 44/.4' + 4! = 123 (.4' = .4 recurring)
Trond Posted December 26, 2005 Posted December 26, 2005 I saw someone posting this when I made a similar thread on a math-forum: 4/4=1 4+4/4=2 4+4+4/4=3 4+4+4+4/4=4 4+4+4+4+4/4=5 4+4+4+4+4+4/4=6 4+4+4+4+4+4+4/4=7 4+4+4+4+4+4+4+4/4=8 4+4+4+4+4+4+4+4+4/4=9 4+4+4+4+4+4+4+4+4+4/4=10 There are as many fours adding together as the number is in the answer, and it seems to go on forever
the tree Posted December 26, 2005 Posted December 26, 2005 Isn't that a little obvious? [math]\frac{ab}{b}=a[/math]
BigMoosie Posted December 27, 2005 Posted December 27, 2005 A while back I continued with my own list using only basic algebraic expressions (+ - / * ^ ! etc), and I managed to get all but three numbers less than 100. Anyone wanna help get 89, 91, 93 using these rules? An underline indicates repetition: (i.e: .4 = .44444 = 4/9): 1 = 4*4/(4*4) 2 = 4/4+4/4 3 = (4+4+4)/4 4 = (4-4)/4+4 5 = 4^(4-4)+4 6 = (4+4)/4+4 7 = 4+4-4/4 8 = 4+4+4-4 9 = 4/4+4+4 10 = (4*4+4!)/4 11 = (4+4!)/4+4 12 = (4-4/4)*4 13 = (4+4!+4!)/4 14 = 4!/4+4+4 15 = 4*4-4/4 16 = 4*4+4-4 17 = 4*4+4/4 18 = (4*4!-4!)/4 19 = 4!-(4+4/4) 20 = (4/4+4)*4 21 = 4!+4/4-4 22 = 4!-(4+4)/4 23 = 4!-4^(4-4) 24 = 4*4+4+4 25 = 4!+(4/4)^4 26 = 4!+4!/4-4 27 = 4!+4-4/4 28 = (4+4)*4-4 29 = 4/4+4!+4 30 = (4*4!+4!)/4 31 = (4+4!)/4+4! 32 = 4^4/(4+4) 33 = (4-.4)/.4+4! 34 = 4!/4+4+4! 35 = (4.4/.4)+4! 36 = (4+4)*4+4 37 = 4/.[u]4[/u]+4+4! 38 = 44-4!/4 39 = (4*4-.4)/.4 40 = (4^4/4)-4! 41 = (4*4+.4)/.4 42 = 4!+4!-4!/4 43 = 44-4/4 44 = 4*4+4+4! 45 = (4!/4)!/(4*4) 46 = (4!-4)/.4 - 4 47 = 4!+4!-4/4 48 = (4*4-4)*4 49 = 4!+4!+4/4 50 = (4*4+4)/.4 51 = 4!/.4-4/.[u]4[/u] 52 = 44+4+4 53 = 44+4/.[u]4[/u] 54 = (4!/4)^4/4! 55 = (4!-.4)/.4-4 56 = 4!+4!+4+4 57 = 4/.[u]4[/u]+4!+4! 58 = (4^4-4!)/4 59 = 4!/.4-4/4 60 = 4*4*4-4 61 = 4!/.4+4/4 62 = (4!+.4+.4)/.4 63 = (4^4-4)/4 64 = 4^(4-4/4) 65 = 4^4+4/4 66 = (4+4!)/.4-4 67 = (4+4!)/.[u]4[/u]+4 68 = 4*4*4+4 69 = (4+4!-.4)/.4 70 = (4^4+4!)/4 71 = (4!+4.4)/.4 72 = (4-4/4)*4! 73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u] 74 = (4+4!)/.4+4 75 = (4!/4+4!)/.4 76 = (4!-4)*4-4 77 = (4!-.[u]4[/u])/.[u]4[/u]+4! 78 = (4!x.[u]4[/u]+4!)/.[u]4[/u] 79 = ([sup].4[/sup]√4-.4)/.4 80 = (4*4+4)*4 81 = (4/4-4)^4 82 = 4!/.[u]4[/u]+4!+4 83 = (4!-.4)/.4+4! 84 = (4!-4)*4+4 85 = (4/.4+4!)/.4 86 = (4-.4)x4!-.4 87 = 4!x4-4/.[u]4[/u] 88 = 4^4/4+4! [size="4"][color="Red"][b]89[/b][/color][/size] 90 = (4!/4)!/(4+4) [size="4"][color="Red"][b]91[/b][/color][/size] 92 = (4!-4/4)*4 [size="4"][color="Red"][b]93[/b][/color][/size] 94 = (4+4!)/.4 + 4! 95 = 4!*4-4/4 96 = 4!*4+4-4 97 = 4!*4+4/4 98 = (4!+.4)*4+.4 99 = (4!+4!-4)/.[u]4[/u] 100 = 4*4/(.4*.4) Here is a reason why in my opinion using any functions is pointless: [math]0 = \log_{\tfrac{4}{.4}}\frac{4}{4}[/math] [math]1 = \log_{\tfrac{4}{.4}}\frac{4}{.4}[/math] [math]2 = \log_{\tfrac{4}{.4}}\frac{4}{4\%}[/math] [math]3 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%}[/math] [math]4 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%}[/math] [math]5 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%}[/math] [math]6 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%}[/math] [math]7 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%}[/math] [math]8 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%}[/math] [math]9 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%\%}[/math] [math]10 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%\%}[/math]
cosine Posted December 27, 2005 Posted December 27, 2005 A while back I continued with my own list using only basic algebraic expressions (+ - / * ^ ! etc)' date=' and I managed to get all but three numbers less than 100. Anyone wanna help get 89, 91, 93 using these rules? An underline indicates repetition: (i.e: .[u']4[/u] = .44444 = 4/9): 1 = 4*4/(4*4) 2 = 4/4+4/4 3 = (4+4+4)/4 4 = (4-4)/4+4 5 = 4^(4-4)+4 6 = (4+4)/4+4 7 = 4+4-4/4 8 = 4+4+4-4 9 = 4/4+4+4 10 = (4*4+4!)/4 11 = (4+4!)/4+4 12 = (4-4/4)*4 13 = (4+4!+4!)/4 14 = 4!/4+4+4 15 = 4*4-4/4 16 = 4*4+4-4 17 = 4*4+4/4 18 = (4*4!-4!)/4 19 = 4!-(4+4/4) 20 = (4/4+4)*4 21 = 4!+4/4-4 22 = 4!-(4+4)/4 23 = 4!-4^(4-4) 24 = 4*4+4+4 25 = 4!+(4/4)^4 26 = 4!+4!/4-4 27 = 4!+4-4/4 28 = (4+4)*4-4 29 = 4/4+4!+4 30 = (4*4!+4!)/4 31 = (4+4!)/4+4! 32 = 4^4/(4+4) 33 = (4-.4)/.4+4! 34 = 4!/4+4+4! 35 = (4.4/.4)+4! 36 = (4+4)*4+4 37 = 4/.[u]4[/u]+4+4! 38 = 44-4!/4 39 = (4*4-.4)/.4 40 = (4^4/4)-4! 41 = (4*4+.4)/.4 42 = 4!+4!-4!/4 43 = 44-4/4 44 = 4*4+4+4! 45 = (4!/4)!/(4*4) 46 = (4!-4)/.4 - 4 47 = 4!+4!-4/4 48 = (4*4-4)*4 49 = 4!+4!+4/4 50 = (4*4+4)/.4 51 = 4!/.4-4/.[u]4[/u] 52 = 44+4+4 53 = 44+4/.[u]4[/u] 54 = (4!/4)^4/4! 55 = (4!-.4)/.4-4 56 = 4!+4!+4+4 57 = 4/.[u]4[/u]+4!+4! 58 = (4^4-4!)/4 59 = 4!/.4-4/4 60 = 4*4*4-4 61 = 4!/.4+4/4 62 = (4!+.4+.4)/.4 63 = (4^4-4)/4 64 = 4^(4-4/4) 65 = 4^4+4/4 66 = (4+4!)/.4-4 67 = (4+4!)/.[u]4[/u]+4 68 = 4*4*4+4 69 = (4+4!-.4)/.4 70 = (4^4+4!)/4 71 = (4!+4.4)/.4 72 = (4-4/4)*4! 73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u] 74 = (4+4!)/.4+4 75 = (4!/4+4!)/.4 76 = (4!-4)*4-4 77 = (4!-.[u]4[/u])/.[u]4[/u]+4! 78 = (4!x.[u]4[/u]+4!)/.[u]4[/u] 79 = ([sup].4[/sup]√4-.4)/.4 80 = (4*4+4)*4 81 = (4/4-4)^4 82 = 4!/.[u]4[/u]+4!+4 83 = (4!-.4)/.4+4! 84 = (4!-4)*4+4 85 = (4/.4+4!)/.4 86 = (4-.4)x4!-.4 87 = 4!x4-4/.[u]4[/u] 88 = 4^4/4+4! [size="4"][color="Red"][b]89[/b][/color][/size] 90 = (4!/4)!/(4+4) [size="4"][color="Red"][b]91[/b][/color][/size] 92 = (4!-4/4)*4 [size="4"][color="Red"][b]93[/b][/color][/size] 94 = (4+4!)/.4 + 4! 95 = 4!*4-4/4 96 = 4!*4+4-4 97 = 4!*4+4/4 98 = (4!+.4)*4+.4 99 = (4!+4!-4)/.[u]4[/u] 100 = 4*4/(.4*.4) Here is a reason why in my opinion using any functions is pointless: [math]0 = \log_{\tfrac{4}{.4}}\frac{4}{4}[/math] [math]1 = \log_{\tfrac{4}{.4}}\frac{4}{.4}[/math] [math]2 = \log_{\tfrac{4}{.4}}\frac{4}{4\%}[/math] [math]3 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%}[/math] [math]4 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%}[/math] [math]5 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%}[/math] [math]6 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%}[/math] [math]7 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%}[/math] [math]8 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%}[/math] [math]9 = \log_{\tfrac{4}{.4}}\frac{4}{.4\%\%\%\%}[/math] [math]10 = \log_{\tfrac{4}{.4}}\frac{4}{4\%\%\%\%\%}[/math] Good arguement, I think.
RyanJ Posted December 27, 2005 Author Posted December 27, 2005 Good arguement, I think. If you can think of a rule that allows functions and stops them being "abused" then post it Cheers, Ryan Jones
cosine Posted December 27, 2005 Posted December 27, 2005 If you can think of a rule that allows functions and stops them being "abused" then post it Cheers' date=' Ryan Jones[/quote'] Thats the problem, its hard to differentiate one function from another (pun intended). Although, I was thinking, how about this: If a function can be defined by a differential equation, then it should be accepted. And if a function can only be defined by a difference equation, then it shouldn't be accepted. Though it seems a nice enough rule, it allows for what Big moosie's arguement, so...? Edit: I don't like this rule already, cause it would allow: [math]f(x) = N[/math] thus, [math]f(4+4+4+4) = N[/math]
cosine Posted December 29, 2005 Posted December 29, 2005 Thats the problem' date=' its hard to differentiate one function from another (pun intended). Although, I was thinking, how about this: If a function can be defined by a differential equation, then it should be accepted. And if a function can only be defined by a difference equation, then it shouldn't be accepted. Though it seems a nice enough rule, it allows for what Big moosie's arguement, so...? Edit: I don't like this rule already, cause it would allow: [math']f(x) = N[/math] thus, [math]f(4+4+4+4) = N[/math] There's always the 4 or 5 operator rule, where only + - * / and ^ (and parenthesis) are allowed...
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