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#476 2006-03-02 02:40:37
- ganesh
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Re: Problems and Solutions
mathsyperson, a good attempt! I shall post the proof after a few days (during the weekend, when I am free). 
Character is who you are when no one is looking.
#477 2006-03-03 17:44:03
- ganesh
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Re: Problems and Solutions
Problem # k + 110
Let n be an integer. Can both n + 3 and n2 + 3 be perfect cubes?
Character is who you are when no one is looking.
#478 2006-07-12 23:19:14
- Daisy
- Novice
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Re: Problems and Solutions
ganesh wrote:
Outstanding! You are really supersmart!
Try this one....But don't post your reply immediately.
Let others too try.
(2) A mixture of 40 liters of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture?
I think that you would have to add 5 liters of water to make the solution 20%.
#479 2006-07-13 00:34:31
- Patrick
- Real Member
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Re: Problems and Solutions
Daisy - that is correct. 9liters(the new amount of water) is in fact 1/5 of 45liters(the new total)
#480 2006-07-23 20:25:17
- krassi_holmz
- Real Member
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Re: Problems and Solutions
#k+110
Let
n+3=x^3;
2n+3=y^3
Then
n=x^3-3;
2(x^3-3)+3=y^3
2x^3-3=y^3
The solutions of this diophantine equations are (1,-1) and (4,5)
So we have n=-2 and n=61.
IPBLE: Increasing Performance By Lowering Expectations.
#481 2009-01-05 10:03:07
- JaneFairfax
- Legendary Member
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Re: Problems and Solutions
ganesh wrote:
Problem # k + 109
Prove that every number of the form a4+4 is a composite number (a≠1).
(This problem was posed by the eminent French mathematician Sophie Germain).
The trick is to use complex numbers – or Gaussian integers (complex numbers with integer real and imaginary parts). Thus, factorizing in the ring of Gaussian integers, we have
Since and we have
Now we multiply the factors in a different order!
And it is clear that if , both and are integers greater than 1. Hence is composite if !
Last edited by JaneFairfax (2009-01-05 10:48:44)
#482 2010-10-22 12:00:42
- bobbym
- Administrator
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Re: Problems and Solutions
Hi ganesh;
For k + 42
No method was ever given for this problem. To fill in the gap I provide my solution.
It avoids having to solve a simultaneous set of equations over the integers, which is possible but computer dependent.
If we call the amount of coconuts originally as C0 ( and x for later ) and C1 the operation performed by the first man, with C2 the second etc, We form this group of equations.
It is easy to spot a recurrence form!
We solve this by standard means:
Do not bother to simplity. Just substitute 5 for n. There are 5 guys remember.
You get the fraction:
Set it equal to y, ( I like x and y ). The step is justified because 1024 x - 8404 is obviously a multiple of 3125.
Rearrange to standard form for a linear diophantine equation.
Solve by Brahmagupta's method, continued fraction, GCD reductions...
Whatever you like. You just need 1 solution! I have a small answer found by trial and error of ( x = - 4 , y = - 4 ).
Now if a linear diophantine equation has one solution it has an infinite number of them.
Utilize Bezouts identity, which says if you have one answer (x,y) then you can get another by:
Plug in x = -4, y = - 4, a = 1024, b = -3125
Now it has been solved in terms of a parameter k. Substitute k = -1,-2,-3,-4,-5 ... to get all solutions.
k = -1 yields (3121, 1020) which is the smallest positive solution. So there are 3121 coconuts in the original pile.
It was not necessary to even know of Bezouts identity. From equation A you have the congruence:
Once one answer of x = -4 was found you just have to add 3125 to get x = 3121.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.