Math Riddle [modified]
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which is the Greatest 1111! or 11^11! Prove your Answer.. -- modified at 7:33 Tuesday 27th June, 2006
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which is the Greatest 1111! or 11^11! Prove your Answer.. -- modified at 7:33 Tuesday 27th June, 2006
Hmm.. let's see. It's been a while since I dealt with logarithms. Anyway, I'll assume 1111! < 11^11!
1111! < 11^11!
log(1111!) < log(11^11!)
log(1 * 2 * 3 ... * 1111) < log(11 * 11 * 11 ... * 11) // last repetition is 11! times = 39916800
log(1) + log(2) + log(3) ... log(1111) < log(11) + log(11) ... log(11) // 39916800 repetitions
// To make things easy, let's assume that log(1) = log(2) ... = log(1111) - a worst case scenario of sorts
log(1111) * 1111 = 7791 < 39916800 * log(11) = 39916800 * 2.40
// Since log(1) ... log(1110) are all less than log(1111), the lefthand side is less than 7791My assumption seems to be correct. 1111! is smaller than 11^11!
Last modified: den 27 juni 2006 06:49:55 --
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which is the Greatest 1111! or 11^11! Prove your Answer.. -- modified at 7:33 Tuesday 27th June, 2006
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What's the order of ops between ^ and ! ? 1111! < (11^11)! because unfactorialized 1111 < 11^11.
dan neely wrote:
What's the order of ops between ^ and ! ?
It depends on your calculator ;P
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dan neely wrote:
What's the order of ops between ^ and ! ?
It depends on your calculator ;P
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which is the Greatest 1111! or 11^11! Prove your Answer.. -- modified at 7:33 Tuesday 27th June, 2006
Abdul Gafoor wrote:
which is the Greatest 1111! or 11^11!
When I first read this, I assumed that that the question was using binary arithmetic. Translated into decimal, this is 15! versus 3^3! = 3^6. 15! is the product of 15 terms and the first 12 of those terms exceed 3. Thus 15! > 3^12 > 3^6. Thus in binary we have 1111! > 11^11! Jorgen has assumed base 10. For that case, we have 1111! < 11^11! To prove this, observe that the left hand side is the product of 1111 terms and that the largest of these terms has value 1111. Thus LHS < 1111^1111. Next, observe that the RHS can be written as: RHS = (11^3)^(11!/3) = 1331^(11!/3) since 11^3 = 1331 > 1331^10! since 11/3 > 1 > 1331^(10*9*8*7) = 1331^5040 i.e., RHS > 1331^5040 If then follows that LHS < RHS since 1111 < 1331 and 1111 < 5040. John Carson "To argue with a man who has renounced the use and authority of reason is like administering medicine to the dead." Thomas Paine
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Abdul Gafoor wrote:
which is the Greatest 1111! or 11^11!
When I first read this, I assumed that that the question was using binary arithmetic. Translated into decimal, this is 15! versus 3^3! = 3^6. 15! is the product of 15 terms and the first 12 of those terms exceed 3. Thus 15! > 3^12 > 3^6. Thus in binary we have 1111! > 11^11! Jorgen has assumed base 10. For that case, we have 1111! < 11^11! To prove this, observe that the left hand side is the product of 1111 terms and that the largest of these terms has value 1111. Thus LHS < 1111^1111. Next, observe that the RHS can be written as: RHS = (11^3)^(11!/3) = 1331^(11!/3) since 11^3 = 1331 > 1331^10! since 11/3 > 1 > 1331^(10*9*8*7) = 1331^5040 i.e., RHS > 1331^5040 If then follows that LHS < RHS since 1111 < 1331 and 1111 < 5040. John Carson "To argue with a man who has renounced the use and authority of reason is like administering medicine to the dead." Thomas Paine
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which is the Greatest 1111! or 11^11! Prove your Answer.. -- modified at 7:33 Tuesday 27th June, 2006
Some idiot was crazy enough to visit this thread for giving 1.0 to everyone. Oh Fisticuffs, I Need Your Approval For I Am Misguided Without Your Awesome Insight Please Validate My Existence With You're Internet Powers By Pumpkinhead, Age 15 or something
http://weblogs.com.pk/kadnan | kadnan.blogspot.com | AJAX ba
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Some idiot was crazy enough to visit this thread for giving 1.0 to everyone. Oh Fisticuffs, I Need Your Approval For I Am Misguided Without Your Awesome Insight Please Validate My Existence With You're Internet Powers By Pumpkinhead, Age 15 or something
http://weblogs.com.pk/kadnan | kadnan.blogspot.com | AJAX ba
You know those social sciences majors... Tomaz