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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
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### Re: Problems and Solutions

#### ganesh wrote:

Outstanding! You are really supersmart!
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)

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## #480 2006-07-23 20:25:17

krassi_holmz
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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)

Q: Who wrote the novels Mrs Dalloway and To the Lighthouse?

## #482 2010-10-22 12:00:42

bobbym

Online

### 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.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.