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Old 2006-06-05, 16:24   #1
mfgoode
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Question 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
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Old 2006-06-05, 18:43   #2
drew
 
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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
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Old 2006-06-05, 19:22   #3
Uncwilly
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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
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Old 2006-06-05, 20:07   #4
drew
 
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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
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Old 2006-06-05, 22:00   #5
retina
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Does not 0 also give a correct answer? No? 0=2*(0/2)^2 etc.
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Old 2006-06-05, 22:32   #6
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Quote:
Originally Posted by retina
Does not 0 also give a correct answer?
Methinks that thou hast "zeroed in" on the answer.
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Old 2006-06-06, 07:33   #7
mfgoode
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Thumbs up 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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