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 2006-06-05, 16:24 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 1000000001002 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
2006-06-05, 18:43   #2
drew

Jun 2005

5768 Posts

Quote:
 Originally Posted by mfgoode Find me a number" said the king to the jester " half of which is a square". " how simple" said the jester"
number = 52a*32b*22c+1
Quote:
 "But" continued the king " one third of it must be a cube". The jester looked serious for once.
number = 53d*33e+1*22f
Quote:
 "And finally " remarked the king "One fifth of it must be a fifth power" The jester looked crestfallen.
number = 55g+1*35h*25i

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 215*310*56 = 30,233,088,000,000

That's a pretty big number.

Last fiddled with by drew on 2006-06-05 at 18:43

2006-06-05, 19:22   #3
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

5×17×131 Posts

Quote:
 Originally Posted by drew So, the number is 215*310*56 = 30,233,088,000,000
In essence you are saying, in general form:
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 2006-06-05 at 19:24

2006-06-05, 20:07   #4
drew

Jun 2005

38210 Posts

Quote:
 Originally Posted by Uncwilly In essence you are saying, in general form: 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
No, that's not a general solution because a * b = 1 (mod c) does not hold for every possible a, b and c. That was coincidentally (or by design) the case for this particular set of a, b and c.

A general solution would require additional coefficients in the exponents to force this condition.

Drew

Last fiddled with by drew on 2006-06-05 at 20:10

 2006-06-05, 22:00 #5 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 11010110010002 Posts Does not 0 also give a correct answer? No? 0=2*(0/2)^2 etc.
2006-06-05, 22:32   #6
Wacky

Jun 2003
The Texas Hill Country

32·112 Posts

Quote:
 Originally Posted by retina Does not 0 also give a correct answer?
Methinks that thou hast "zeroed in" on the answer.

2006-06-06, 07:33   #7
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
The King and the jester

Quote:
 Originally Posted by drew number = 52a*32b*22c+1 number = 53d*33e+1*22f number = 55g+1*35h*25i 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 215*310*56 = 30,233,088,000,000 That's a pretty big number.

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 2006-06-06 at 07:36

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