Problems and Solutions

mathsyperson · 2006-02-25 09:17:57

Jai Ganesh · 2006-02-25 17:31:21

Jai Ganesh · 2006-02-25 17:38:44

Problem # k + 107

The price of a water-melon is 50 cents, an apple is 10 cents and a plum is 1 cent. Five dollars were used to buy 100 items of different kinds of fruit. How many pieces of each type were bought?

Jai Ganesh · 2006-02-26 00:13:53

Problem # k + 108

What is the sum of the first 50 terms common to the series 15,19,23 ... and 14,19,24 ... ?

mathsyperson · 2006-02-26 06:48:06

Possibly a bit unconventional...

But nothing unconventional here.

MathsIsFun · 2006-02-26 08:38:42

(19 pages, wowee! Maybe you could start a new topic?)

Ricky · 2006-02-26 15:56:27

I was actually wondering that for quite a while, MathIsFun. Wouldn't it be more organized if Ganesh did one question per topic? I mean, we got an entire section of the forum for it, why not?

Jai Ganesh · 2006-02-26 16:49:34

To mathsyperson :- Funnily, your solution to Problem # k + 107 is correct, although unconventional, as you put it. Please read the Problem # k + 108 again before posting your solution.

To MathsIsFun :- Good suggestion, worth considering.

To Ricky :- One question per topic is fine. But what do we do with the problems already posted? Put them all in an 'Assorted' or 'Miscellaneous' topic? Good suggestion, worth considering.

krassi_holmz · 2006-02-26 22:18:34

Do you mean this is the end of "Problems and solutions"?

Jai Ganesh · 2006-02-26 22:26:23

I hope not. Some problems would be posted here in the future too. There are some unanswered problems for which solutions would have to be posted. Hence, this is not the end of 'Problems and solutions'. This topic shall remain the precursor of other topics.

Jai Ganesh · 2006-02-28 16:28:18

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).

mathsyperson · 2006-03-01 03:34:28

It's fairly simple apart from when a ends in 5.

When mod(10) a = 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9, mod(10)(a[sup]4[/sup] + 4) = 4, 5, 0, 5, 0, 9, 0, 5, 0 and 5 respectively.

Numbers that end in 4, 5 or 0 are never prime (apart from 5) so that proves it for all values of a except for ##5. But proving it for that is quite difficult.

Jai Ganesh · 2006-03-01 03:40:37

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

Jai Ganesh · 2006-03-02 18:44:03

Problem # k + 110

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

Daisy · 2006-07-12 01:19:14

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%.

Patrick · 2006-07-12 02:34:31

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

krassi_holmz · 2006-07-22 22:25:17

#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.

JaneFairfax · 2009-01-04 11:03:07

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 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 (2009-01-04 11:48:44)

bobbym · 2010-10-21 13:00:42

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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