Math Puzzle
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The largest number I can do is: 4194304 (2^22). But, you want ANY positive number.... and only 3 twos.. not 4 twos.. or 5 twos, just 3? I strongly doubt it. Then again, I have been known to be wrong, from time to time. ;P I prefer to wear gloves when using it, but that's merely a matter of personal hygiene [Roger Wright on VB] Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning. [Rich Cook]
Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
An even larger number exists :-D:-D -Dominik
_outp(0x64, 0xAD);and__asm mov al, 0xAD __asm out 0x64, aldo the same... but what do they do?? ;) -
Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
Umm, how many operators may I use? Because if I have an infinite number of operators, I can make an infinite long number... Btw, the number I thought of before was ((2 << 22)!) Where << is binary left shift. Now if I have an infinite number of operators, why not make (((((((((((((2 << 22)!)!)!)!)!)!)!)!)!)!)!)!) ... :confused: -Dominik
_outp(0x64, 0xAD);and__asm mov al, 0xAD __asm out 0x64, aldo the same... but what do they do?? ;) -
Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Umm, how many operators may I use? Because if I have an infinite number of operators, I can make an infinite long number... Btw, the number I thought of before was ((2 << 22)!) Where << is binary left shift. Now if I have an infinite number of operators, why not make (((((((((((((2 << 22)!)!)!)!)!)!)!)!)!)!)!)!) ... :confused: -Dominik
_outp(0x64, 0xAD);and__asm mov al, 0xAD __asm out 0x64, aldo the same... but what do they do?? ;)Aah yes, but the goal of the problem is to specify any number using three 2's and any mathematical operators, not necessarily the largest number. Say the number is n, then n = 2log2(n), but I'm not sure if this is what is wanted. Any number can be represented by either 2n or 2n+2/2, but again I'm not sure if this is what is wanted. I dunno. I'm working on it :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
Well then, try (2^22)! .... that ought to be quite big. Btw. the question was how to write any number, not the biggest... Part of the non-smoking generation since 12/5-2003 22.35.
-- Opinions expressed do not neccecarily reflect those of my -- employer; I do think for myself. Resisting temptation is -- easier when you think you'll maybe get another chance -- later on. -
Well then, try (2^22)! .... that ought to be quite big. Btw. the question was how to write any number, not the biggest... Part of the non-smoking generation since 12/5-2003 22.35.
-- Opinions expressed do not neccecarily reflect those of my -- employer; I do think for myself. Resisting temptation is -- easier when you think you'll maybe get another chance -- later on.Ludvig A. Norin wrote: Btw. the question was how to write any number, not the biggest... Yes I know. I was just surprised he couldn't get anything bigger :). See my reply to Dominik above :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
You have a degree in mathematics, don't you? :) -- I am on fire. Do you need a light?
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Well done :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Genius!:) Brad Jennings "You're mom is nice. Mind if I go out with her?" - Jörgen Sigvardsson
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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You have a degree in mathematics, don't you? :) -- I am on fire. Do you need a light?
No I don't :) I'm a electrical engineer, but used to be fond of these kind of quizzes when I was younger. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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No I don't :) I'm a electrical engineer, but used to be fond of these kind of quizzes when I was younger. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Have you read Gödel Escher Bach - The Eternal Golden Braid? Your solution reminded me a lot of the authors (Hofstadter) representation of the natural numbers. He defined it by an axiom 0 (zero) and an operation S (successor). Your log/sqrt solution for simulating S gave me a flashback.. :) -- I am on fire. Do you need a light?
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
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my answer -~-~...-~2 is simpler and requires only one 2.
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
Brilliant! Well done :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
Regarding, my answer, -~-~-~ ... -~2, it is can be done with log n operations by appropriate using replacing some of the inner operations with the sqr or factorial functions. Thanks, Wes