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Title: Brick Post by regaleira on Jul 6th, 2005, 1:49pm One brick is one kilogram and half a brick heavy. How heavy is one brick? This is another easy one! (but that´s why im posting it on easy and not on medium or hard) ;) |
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Title: Re: Brick Post by THUDandBLUNDER on Jul 6th, 2005, 2:09pm on 07/06/05 at 13:49:34, regaleira wrote:
:-/ on 07/06/05 at 13:49:34, regaleira wrote:
Got any harder ones? ::) |
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Title: Re: Brick Post by JocK on Jul 6th, 2005, 2:59pm Hmmm.. let's see. I got this far: let's call the weight of a brick in kilograms "b". With the help of my mathteacher we got close to a solution : b = 1 + b/2 which can be re-arranged into: b - 1 - b/2 = 0 Now if you try b =42 (always a good guess) you find: 42 - 1 - 42/2 = 42 - 1 - 21 = 22 =/= 0 Doesn't work... But the following more advanced approach might work: (I will hide this stuff so as not to give too much away !): [hide] First we square both sides: (b - 1 - b/2)^2 = 0 and take the exponent: exp(b - 1 - b/2)^2 = 1 The solution to the problem can then be found by minimising f(b) = (exp(b - 1 - b/2)^2 - 1)^2 I tried b = 42, but it still doesn't work.... [/hide] A nasty problem. Any takers? If not solvable analytically, would a Monte-Carlo simulation help? ??? |
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Title: Re: Brick Post by baddab457 on Jul 6th, 2005, 7:30pm did u think to far or did i not think far enough... brick being b there fore [hide] b=1+b/2 2b=1 b=.5 i really need some sleep now though... [/hide] |
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Title: Re: Brick Post by Icarus on Jul 6th, 2005, 7:31pm Jock, :D Were ever in one of my classes? I recall some approaches like this amongst my students work... baddab457: Trust me - Jock is able to solve a simple equation. (Even if I was not familiar with his other posts, the very things he mentioned in this post would be enough to assure me of his ability to handle the equation here.) Therefore you may infer that he failed to solve this one on purpose. |
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Title: Re: Brick Post by honkyboy on Jul 6th, 2005, 9:04pm Baddabs' answer still doesn't seem complete. b=1/2 doesn't correctly solve the original equation. First I will sove for the 'real' b. (br) br= 1 + b/2 = 1+(1/2)/2=1.4 The weight of the brick (W) now is found if br is subtituted back into the original equation using an unknown multiplier (N) so . . . W=N(1+br/2). substituting W=N(1+1.4N/2) multiply by the squared reciprocal W=20N-0.4N W=19.6N (I haven't figured how to solve for N yet. Icarus, would I get partial credit for this?) |
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Title: Re: Brick Post by paul schmitz on Jul 6th, 2005, 10:24pm on 07/06/05 at 19:30:37, baddab457 wrote:
you guys are way overthinking. the answer is right here... almost. baddab just made a small error. b = 1 + b/2 *2 (multiply every term by 2) 2b = 2 + b -b b = 2 if you think about it, you start with b = 1 + 1/2, then b = 3/2, so you plug it back in and now b = 1 + 3/4, b = 7/8. This is simply 1 + the sumation of [(1/2)^n], from n = 1 to n = infinity, which approaches 2 as n grows. |
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Title: Re: Brick Post by otter on Jul 13th, 2005, 11:52am on 07/06/05 at 14:59:15, JocK wrote:
Perhaps if we restate the problem, the solution will become apparent: 0.999... brick is 0.999... kilograms and half a brick heavy. How heavy is 0.999... brick? ::) |
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Title: Re: Brick Post by xenon_nightmare on Jul 13th, 2005, 12:39pm I believe one brick (now) is ~ 2 kilos ;D |
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Title: Re: Brick Post by raprap on Jul 15th, 2005, 4:13am Is it a regular masonry brick or a paver? They're different dimensions. A masonry brick is 2 1/4x8 1/2 x 4 1/2 while a paver is normally 3x9x5 and that difference would definitely have an effect on the weight of a brick. Rap |
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