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| Acmlm's Board - I2 Archive - Brain Teasers - Strange Math Riddles |
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JJ64 Boo Level: 45 Posts: 867/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| It's only an hour earlier than most of you. That shouldn't be too bad... Scores 1st. Grey-22 points 2nd. Heian-794-18 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points NSNick, also, you are missing a few 3-digit answers. If you edit the post with the correct answers, I'll give you back the point. Otherwise, I'm making it an open question again. Question 69 (1 point)- If there are 40 isocahedrons with numbers on each side, what is the probability of a certain outcome of numbers? (edited by JJ64 on 07-29-04 10:14 AM) |
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Heian-794 Red Super Koopa Level: 44 Posts: 345/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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| This question is a little hard to understand. I imagine that we've got a particular set of 40 numbers that we want to see on each of the 40 icosahedrons (a. k. a. 20-sided dice). If it matters which numbers are on which icosaehdrons (let's say we want icosahedron #1 to show face "1", #2 to show "2", etc.), then the probability of getting the exact numbers we want is 1 in 2040. If we don't care which numbers are on which icosahedrons, then the probability would be 40! / 2040. But that assumes that we're looking for 20 different numbers, and multiple instances of the same numbers will lower the probability. If our desired outcome were to see 40 of the same number, the odds would be back to 1 in 2040. |
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Grey the Stampede Don't mess with powers you don't understand. And yes. That means donuts. Level: 82  Posts: 343/3770 EXP: 5192909 For next: 16318 Since: 06-17-04 From: Kingston, RI, USA, Earth Since last post: 2 hours Last activity: 1 hour |
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If you multiply the percentage chances (5% is 1/20), another way of putting it would be (4.5*10^-52)%, or .00000000000000000000000000000000000000000000000000045% (is that enough zeros?)  (edited by Grey on 07-29-04 02:54 PM) (edited by Grey on 07-29-04 02:55 PM) |
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NSNick Laidback Admin Level: 85  Posts: 787/3875 EXP: 5895841 For next: 2699 Since: 03-15-04 From: North Side School: OSU Since last post: 9 hours Last activity: 1 hour |
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| Assuming all numbers on each die are different, the chances are 1 out of 2040, or 1 out of 1.099511627776 * 1052. Edit: Never mind, Heian answered it. (edited by NSNick on 07-29-04 04:23 PM) |
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Dylan Devil Trumpets and Angel Trombones ~ Level: 54  Posts: 497/1407 EXP: 1181697 For next: 52173 Since: 06-19-04 From: Ottawa, Canada. Since last post: 1 day Last activity: 6 hours |
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| Technically it's 140/2040 but obviously that doesn't matter. I want a point! *sigh* I need to wake up earlier... (edited by Gb boy on 07-29-04 06:29 PM) (edited by Gb boy on 07-29-04 06:29 PM) |
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JJ64 Boo Level: 45 Posts: 873/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| Heian-794 gets the point... considering it was hard to understand... Scores 1st. Grey-22 points 2nd. Heian-794-19 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points NSNick, if your answer to Question 65 isn't corrected (adding the 3-digit numbers) by 5:00 my time today, it becomes an open question again. Question 70 (1 point)- If 600 dice with all the same number are producted, and the shape of the dice are icodecahedrons, what are the odds of rolling a certain combination of numbers with all 600 dice? |
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Heian-794 Red Super Koopa Level: 44 Posts: 347/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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JJ, don't take this the wrong way, but your question makes no sense. "Icodecahedron" isn't a word (and neither is "producted"   I'll assume you mean "icosahedron" which has 20 faces. When you say "600 dice with all the same number", do you mean that all 600 dice have the same set of numbers on their faces, or do you mean that the same number is on every face of every die? If it's the same set of numbers, and we don't care which die shows which number, the answer will be different depending on how many repetitions there are in the set of numbers we want to see. Specifically, the denominator will always be 20^600, and the numerator will be 1, 600, (600x599), (600x599x598)... taking the Nth term when there are N repetitions of a single number. Maybe. I think. To take a much simpler example, imagine rolling three regular dice. The probability of getting a desired set of three numbers (say 1, 2, and 3) is 3! / 63, because the combinations 123, 132, 213, 231, 312, 321 are all fine. But if our desired set were all the same number (say all fives), 555 would be the only combination possible, and the odds would be 1 in 63. (edited by Heian-794 on 07-30-04 10:32 AM) |
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JJ64 Boo Level: 45 Posts: 875/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| No offense taken, Heian. Personally, I think I should stop posting 2 minutes after I wake up. Also, it's summer, and... Anyway, you get the point! Scores 1st. Grey-22 points 2nd. Heian-794-20 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points Question 71 (1 point)- There is a compound solid. On a rectangular prism, a pyramid rests. Somehow connected at the points, a cone is attached. At the base of a cone is a cylinder. The radius (of the cone and cylinder) and apothem (of the pyramid and rectangular prism) of the whole thing is 5, and each shape has a height of 8. Find the volume of the whole thing. |
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Heian-794 Red Super Koopa Level: 44 Posts: 349/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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| JJ, I covered several possibilities with my answer, which one was what you actually had in mind? As for this one, I'm imagining that the apothem of the pyramid and prism is the distance from the center to the nearest edge point. This would give us: pi*r^2*h = 200pi for the cylinder + (1/3)pi*r^2*h = 200pi/3 for the cone + (1/3)B*W*H = 800/3 for the pyramid + L*W*H = 800 for the prism Grand total: 1904.4247 units And with that I'm off to bed! |
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JJ64 Boo Level: 45 Posts: 883/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| The last one was the one I was thinking of, Heian. Scores 1st. Grey-22 points 2nd. Heian-794-21 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points Originally posted by JJ64 NSNick missed a few triple-digit numbers which I failed to see, so this is the restart of the question. (edited by JJ64 on 07-30-04 10:02 PM) |
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Heian-794 Red Super Koopa Level: 44 Posts: 350/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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| I'm going to go ahead and attempt it, since NSNick hasn't. Forgive me for stealing a point, so to speak. Prime factorization: 1234567890 = 2 * 3 * 3 * 5 * 3607 * 3803 (Those are two big primes, and very convenient for us since there will be fewer factors overall.) 1234567890 1 2 2x3=6 2x5=10 2x3607=7214 2x3803=7606 2x3x3=18 2x3x5=30 2x3x3607=21642 2x3x3803=22818 2x5x3607=36070 2x5x3803=38030 2x3607x3803 = 27434842 2x3x3x5 = 90 2x3x3x3607 = 64926 2x3x3x3803 = 68454 2x3x5x3607 = 108210 2x3x5x3803 = 114090 2x3x3607x3803 = 82304526 2x3x3x5x3607 = 324630 2x3x3x5x3803 =342270 2x3x3x3607x3803 = 246913578 2x3x5x3607x3803 = 411522630 3 3x3 =9 3x5 =15 3x3607 = 10821 3x3803 = 11409 3x3x5 = 45 3x3x3607 = 32463 3x3x3803 = 34227 3x5x3607 = 54105 3x5x3803 = 57045 3x3607x3803 = 41152263 3x3x3607x3803 = 123456789 3x5x3607x3803 = 205761315 5 5x3607 = 18035 5x3803 = 19015 5x3607x3803 = 68587105 3607 3607x3803 = 13717421 3803 Now that was a tedious problem! |
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NSNick Laidback Admin Level: 85 Posts: 803/3875 EXP: 5895841 For next: 2699 Since: 03-15-04 From: North Side School: OSU Since last post: 9 hours Last activity: 1 hour |
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| Yeah, I didn't have the full prime factorization. | |||
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JJ64 Boo Level: 45 Posts: 891/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| At least you realized your mistake, NSNick. Anyway, it's now a tie! Scores 1st. Grey-22 points 1st. Heian-794-22 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points Question 72 (1 point)- Solve the equation: |x^3+2x^2+7x-46|=0 |
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ziratha Ninji Level: 24 Posts: 14/231 EXP: 77808 For next: 317 Since: 06-29-04 Since last post: 4 days Last activity: 17 days |
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| x=aprox 2.51 |
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JJ64 Boo Level: 45 Posts: 900/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| Although your answer is right, try to show at least a little work, ziratha. Scores 1st. Grey-22 points 1st. Heian-794-22 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. Emptyeye-1 point 6th. RavenX-1 point 6th. Slash Dafter-1 point 6th. Xkeeper-1 point 6th. ziratha-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points It's been two days with no answer, so the point value is upgraded! Question 73 (2 points)- Solve the system of equations: |4x|=7y+42 |4xy+x-y|=21 (edited by JJ64 on 08-03-04 12:00 PM) |
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ziratha Ninji Level: 24 Posts: 15/231 EXP: 77808 For next: 317 Since: 06-29-04 Since last post: 4 days Last activity: 17 days |
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| ok, heres the answer, it took a long time to come up with. first failed attempt. |4x|=7y+42 |4xy+x-y|=21 7y=|4x|-42 y=(|4x|-42)/7 y=4|x|/7-6 humm now that gives me. |4x*(|x|4/7-6) +x -(|x|4/7-6)|=21 |x16|x|/7+x-4|x|/7|=21 |16x^2 /7+x-4|x|/7|=21 |16x^2 /7+3x/7|=21 -b+-(b^2-4ac)^(1/2) x^2+3x/16-(9+3/16) note, if x>0 x^2= posative, ifx <0 x^2= negative. y=|x|4/7-6 so lets say x has the following values, +/-(7,3.5,10). |4*7|=7*-2+42 ,1 |4*+/-7*-2++/-7- -2|=21,0 56-14=42 correct answer. (I hope, i almost messed up the quadratic equation.) 7y=|4x|-42 y=|x|4/7-6 |4x|=7y+42 |4xy-y+x|=21 |y(4x-1)+x|=21 |y(7y+42-1)+x|=21 |7yy+41y+7/4y+10.5-21|=0 |7yy+42 3/4y-10.5|=0 (-42 3/4 +-((42+3/4)^2-4*7*-10.5)^0.5)/14 y=.24 x= 10.91 |
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Heian-794 Red Super Koopa Level: 44 Posts: 360/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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Dagnabbit, Ziratha, I was in the middle of solving this!  |
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ziratha Ninji Level: 24 Posts: 16/231 EXP: 77808 For next: 317 Since: 06-29-04 Since last post: 4 days Last activity: 17 days |
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| Lol sorry, if it helps, It took me almost 2 hours, it wouldn't have been so long but i havent worked with absolute values much. so it tooke me some time to see how i could manipulate them. I was able to get a vague idea of the values based on the equations, which helped a lot. my origional estimate was +/-10,2/7 but that wasn't quite right. well, hope another riddle comes soon. good luck to us both... |
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JJ64 Boo Level: 45 Posts: 911/919 EXP: 640956 For next: 19208 Since: 05-22-04 From: Green Bay, WI Since last post: 441 days Last activity: 269 days |
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| The last time you had the right answer, ziratha. Scores 1st. Grey-22 points 1st. Heian-794-22 points 3rd. NSNick-10 points 4th. Kirby all the way-8 points 5th. wolfman2000-5 points 6th. ziratha-3 points 7th. Emptyeye-1 point 7th. RavenX-1 point 7th. Slash Dafter-1 point 7th. Xkeeper-1 point 11th. neotransotaku-0 points 11th. Gb boy-0 points 11th. Teddylot-0 points Question 74 (1 point)- Solve the system of equations (for b and x only): abx+2abx=40a 8b+6=2x |
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Heian-794 Red Super Koopa Level: 44 Posts: 362/896 EXP: 611014 For next: 271 Since: 06-01-04 From: Kyoto, Japan Since last post: 21 days Last activity: 10 days |
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| (1) abx+2abx=40a (2) 8b+6=2x Let's look at (2) first since it'll be simpler (I actually did it the opposite way in the beginning). 8b+6=2x 4b+3=x Now let's substitute X into the other equation: The A's can be eliminated right away, so we get: 3bx=40 and consequently 3b(4b+3) = 40 then 12b2 + 9b - 40 = 0 Using the quadratic formula, we arrive at: b = (-9 +/- sqrt(-1839)) / 24 b = (-3 +/- 613i)/8 and x = [(-3 +/- 613i)/2] + 3 = (3 +/- 613i)/2 |
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