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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,614

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,614

Problem # k + 110

Let n be an integer. Can both n + 3 and n2 + 3 be perfect cubes?

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**Daisy****Member**- Registered: 2006-07-12
- Posts: 1

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%.

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

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

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

#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.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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

Since

and we haveNow 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 (2009-01-04 11:48:44)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 94,482

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.

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