Guest Posted February 4, 2009 Report Share Posted February 4, 2009 If is possible to count from 0 to 100 using 4 4's for ever number for example 0 = 4+4-4-4 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 = 7 = 8 = 9 = 10 = . . . . . 100 = Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 (edited) i don't quite understand your question, but you are alound to use any combination of 4's and mathematical operators so perhaps -(sqrt(4))= -2? Nevermind Edited February 6, 2009 by Mr. Vade Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 Sqrt(4) is the same thing as 4^-2, everyone is just writing it in a different form. thats wrong sqrt(4) =2 - sqrt(4) = -2 4^2=16 4^-2 = .0625 however sqrt(4) is the same thing as 4 ^ (1/2) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 thats wrong sqrt(4) =2 - sqrt(4) = -2 4^2=16 4^-2 = .0625 however sqrt(4) is the same thing as 4 ^ (1/2) Sorry, thats what I meant. So aren't we violating the rule by using square roots if its really just a substite for ^(1/2)? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 Sorry, thats what I meant. So aren't we violating the rule by using square roots if its really just a substite for ^(1/2)? Taking that further, isn't 4/4 just a substitute for 1, and isn't 4/4+4/4 just a sustitute for 2? I hope my extreme example shows why we can use sqrt(4) but not 4^(1/2)...we are trying to find expressions that use four 4s and no other numbers....all of them reduce (substitute) to something else (the number we are trying to obtain). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 Taking that further, isn't 4/4 just a substitute for 1, and isn't 4/4+4/4 just a sustitute for 2? I hope my extreme example shows why we can use sqrt(4) but not 4^(1/2)...we are trying to find expressions that use four 4s and no other numbers....all of them reduce (substitute) to something else (the number we are trying to obtain). yes and yes the idea of this lil whatever it is, is to try and find numbers using alternative ways of created them by using four 4's, no more no less 4/4+4/4 is a perfect answer for finding the number two, and although sqrt(4) is and alternative method of 4^(1/2), the idea is to isolate your equation to strictly using only the number 4 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 I'd like to point out that eventually with rounding, sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt( sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(.....sqrt(4)))))))) = 1. So technically, you could obtain 99 by doing 4! * 4 + 4 - sqrt(sqrt(sqrt(....sqrt(4)))))), because that would equal 100 - *approximately* 1 = 99. You could get some of the other troublesome numbers in this manner as well. But I would hope that there's a better solution than that.... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 (edited) (4/.04)-(4/4) That just leaves 73, 77, 87, 93 Edited February 6, 2009 by Hdizzy Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 6, 2009 Report Share Posted February 6, 2009 Nevermind, I realized I used a "0" in my last post. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 Just trying to think outside the box, but maybe I'm too far out. If we can use π, then cos(π)=-1 without having to use any 4s, so we could then take the solutions to 74, 78, 88, 94, and 100 and simply add cos(π) to get 73, 77, 87, 93, and 99. But that does seem a lot like cheating. Similarly, d/dx(x)=1 without using any 4s or any other constants like π for that matter, but still seems like cheating to take a previous solution and subtract 1 without using any additional 4s. Another possibility would be to take the limit as x approaches 4 of x/x, which equals 1 and uses only one 4 instead of the two needed for 4/4 to make 1 and would require coming up with new ways to get to either 1 more than or 1 less than the numbers we have left without using a previous solution. So then we have: 73=4!+4!+4!+limx->4(x/x) 87=44*sqrt(4)-limx->4(x/x) 93=[4!-limx->4(x/x)]*4+limx->4(x/x) 99=(4/.4)^sqrt(4)-limx->4(x/x) I haven't gotten 77 yet. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 OMG!! I remember doing this in 5th grade (last year)!! We got all of them except for a few. I would really like to look back and post them but i don't have my notebook. :[ i think i threw it away. some were like...IMPOSSIBLE!!! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 7, 2009 Report Share Posted February 7, 2009 Just trying to think outside the box, but maybe I'm too far out. If we can use π, then cos(π)=-1 without having to use any 4s, so we could then take the solutions to 74, 78, 88, 94, and 100 and simply add cos(π) to get 73, 77, 87, 93, and 99. But that does seem a lot like cheating. Similarly, d/dx(x)=1 without using any 4s or any other constants like π for that matter, but still seems like cheating to take a previous solution and subtract 1 without using any additional 4s. Another possibility would be to take the limit as x approaches 4 of x/x, which equals 1 and uses only one 4 instead of the two needed for 4/4 to make 1 and would require coming up with new ways to get to either 1 more than or 1 less than the numbers we have left without using a previous solution. So then we have: 73=4!+4!+4!+limx->4(x/x) 87=44*sqrt(4)-limx->4(x/x) 93=[4!-limx->4(x/x)]*4+limx->4(x/x) 99=(4/.4)^sqrt(4)-limx->4(x/x) I haven't gotten 77 yet. I thought about that, too. But now a vinculum to represent a repeating decimal (I'll use an ellipsis since I don't know how to how else to type that). Then: 73 = (4!*sqrt(4) + sqrt(.4...)) / sqrt(.4...) 77 = (4/.4...)^sqrt(4) - 4 87 = (4!/.4... + 4) / sqrt(.4...) 93 = 4!*4 - sqrt(4)/sqrt(.4...) 99 = (4!*sqrt(4) - 4) / .4... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 9, 2009 Report Share Posted February 9, 2009 Apparently we forgot the summation. 99= summ of( x as x goes from 4 to (4!-(4/.4))) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 9, 2009 Report Share Posted February 9, 2009 OMG!! I remember doing this in 5th grade (last year)!! We got all of them except for a few. I would really like to look back and post them but i don't have my notebook. :[ i think i threw it away. some were like...IMPOSSIBLE!!! I see I'm not the only middle-school-er here. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 10, 2009 Report Share Posted February 10, 2009 37 39 41 71 73 77 78 79 83 85 87 93 99 left... starting to get difficult. 51 = (4!-sqrt(4))/.4 - 4 53 = (4!-sqrt(4))/.4 - sqrt(4) 57 = (4!-sqrt(4))/.4 + sqrt(4) 67 = (4!+sqrt(4))/.4 + sqrt(4) 69 = (4!+sqrt(4))/.4 + 4 74 = (4!+4)/.4 + 4 75 = (4!+sqrt(4)+4)/.4 82 = (4!-4)*4 + sqrt(4) 84 = 44*sqrt(4) - 4 86 = 44*sqrt(4) - sqrt(4) 89 = (4!+sqrt(4))/.4 + 4! 53 = 4!+ 4! + ((sqrt(4))/.4) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 10, 2009 Report Share Posted February 10, 2009 (edited) 53 = 4!+ 4! + ((sqrt(4))/.4) Where were you the past few days? ooh I see now. you just started. Anyway, welcome here. Edited February 10, 2009 by phil Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 11, 2009 Report Share Posted February 11, 2009 I see I'm not the only middle-school-er here. Far from it. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 17, 2009 Report Share Posted February 17, 2009 79=(4^4-(44*4))-(4/4) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 17, 2009 Report Share Posted February 17, 2009 79=(4^4-(44*4))-(4/4) Sorry, but you must use the digit '4' exactly four times. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 17, 2009 Report Share Posted February 17, 2009 (edited) My bad. 39 = 44-(sqrt4/0.4) Edited February 17, 2009 by scsw Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 18, 2009 Report Share Posted February 18, 2009 (edited) sorry didnt read the question correctly Edited February 18, 2009 by Doc.OK Quote Link to comment Share on other sites More sharing options...
0 unreality Posted February 19, 2009 Report Share Posted February 19, 2009 so which ones are left? Have we decided for or against limits or repeating decimals? Quote Link to comment Share on other sites More sharing options...
0 unreality Posted February 19, 2009 Report Share Posted February 19, 2009 so which ones are left? Have we decided for or against limits & repeating decimals? Quote Link to comment Share on other sites More sharing options...
Question
Guest
If is possible to count from 0 to 100 using 4 4's for ever number
for example
0 = 4+4-4-4
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 =
7 =
8 =
9 =
10 =
.
.
.
.
.
100 =
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