20060605, 16:24  #1 
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
The King and the jester.
Find me a number" said the king to the jester " half of which is a square". " how simple" said the jester" . "But" continued the king " one third of it must be a cube". The jester looked serious for once. "And finally " remarked the king "One fifth of it must be a fifth power" The jester looked crestfallen. "Solution tomorrow morning" finished the king "Or you must marry my widowed mother in  law". The jester was crushed. But the next morning he had the answer and was thus saved from a fate worse than death. See if you would have to marry the widowed mother  in  law overnight! Mally 
20060605, 18:43  #2  
Jun 2005
17E_{16} Posts 
Quote:
Quote:
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All 'a' through 'i' must be integers: The exponent of 5 is 6*l = 1 (mod 5) l=1 The exponent of 3 is 10*m = 1 (mod 3) m = 1 The exponent of 2 is 15*n = 1 (mod 2) n = 1 So, the number is 2^{15}*3^{10}*5^{6} = 30,233,088,000,000 That's a pretty big number. Last fiddled with by drew on 20060605 at 18:43 

20060605, 19:22  #3  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
10,891 Posts 
Quote:
a[sup](b*c)[/sup] * b[sup](a*c)[/sup] * c[sup](a*b)[/sup] and in our case: a=2, b=3, c=5 Last fiddled with by Uncwilly on 20060605 at 19:24 

20060605, 20:07  #4  
Jun 2005
2×191 Posts 
Quote:
A general solution would require additional coefficients in the exponents to force this condition. Drew Last fiddled with by drew on 20060605 at 20:10 

20060605, 22:00  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{3}×5×167 Posts 
Does not 0 also give a correct answer? No? 0=2*(0/2)^2 etc.

20060605, 22:32  #6  
Jun 2003
The Texas Hill Country
441_{16} Posts 
Quote:


20060606, 07:33  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
The King and the jester
Quote:
Well Drew you have escaped the fate worse than death ! It s tedious going thru your working, so Im giving you the method used for this problem, which, I would say is more precise, but much the same, so you may compare both yours and mine.* As G.H. Hardy said that If the answer obtained by a method is correct then the method should be correct, whatever, unless you have the intuition of Ramanujan and have no method! A number N divisible by 2, 3, and 5 may have the form N = 2^a*3^b*5^c. Then since N/2 is a square 'a' must be odd and b and c even. Similarly a and c must be multiples of 3 and b ==1 mod 3. Also, a and b must be multiples of 5 and c==1 mod 5. The samllest values satisfying these conditions are a=15 , b= 10 , c= 6 and N= 2^15*3^10*5^6 = 30,233,088,000,000. * [Method by Albert H. Beiler] As Wacky said that you have certainly 'zeroed in' on the number. Excellent Drew! keep up the good work! Mally Last fiddled with by mfgoode on 20060606 at 07:36 

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