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#476 2006-03-01 03:40:37

Registered: 2005-06-28
Posts: 13,835

Re: Problems and Solutions

mathsyperson, a good attempt! I shall post the proof after a few days (during the weekend, when I am free). smile

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#477 2006-03-02 18:44:03

Registered: 2005-06-28
Posts: 13,835

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 01:19:14

Registered: 2006-07-12
Posts: 1

Re: Problems and Solutions

ganesh wrote:

Outstanding! You are really supersmart! big_smile
Try this one....But don't post your reply immediately.
Let others too try. big_smile

(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? smile

I think that you would have to add 5 liters of  water to make the solution 20%.


#479 2006-07-12 02:34:31

Real Member
Registered: 2006-02-24
Posts: 1,005

Re: Problems and Solutions

Daisy - that is correct. 9liters(the new amount of water) is in fact 1/5 of 45liters(the new total)

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#480 2006-07-22 22:25:17

Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Problems and Solutions

The solutions of this diophantine equations are (1,-1) and (4,5)
So we have n=-2 and n=61.


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#481 2009-01-04 11:03:07

Registered: 2007-02-23
Posts: 6,868

Re: Problems and Solutions

ganesh wrote:

Problem # k + 109

Prove that every number of the form a[sup]4[/sup]+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


we have

Now we multiply the factors in a different order! big_smile


And it is clear that if

, both
are integers greater than 1. Hence
is composite if
! dizzy

Last edited by JaneFairfax (2009-01-04 11:48:44)

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#482 2010-10-21 13:00:42

From: Bumpkinland
Registered: 2009-04-12
Posts: 85,215

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.


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