Math Is Fun Forum / A few questions

Agnishom · 2012-12-03 12:53:37

Well, I will be posting a lot of questions in this thread one by one, after completely clearing my doubts in each of them.

Question1:

There are two urns each containing an arbitrary number of balls. Both are non-empty to begin
with. We are allowed two types of operations:
(a) remove an equal number of balls simultaneously from the urns, and
(b) double the number of balls in any one of them.

Show that after performing these operations finitely many times, both the urns can be made
empty.

scientia · 2012-12-03 14:25:34

Last edited by scientia (2012-12-03 14:31:30)

Agnishom · 2012-12-03 14:38:39

Thank You ---- That was a nice one

Question 2:

If

Prove that (a+b) is a perfect square

bobbym · 2012-12-03 21:05:09

Hi Agnishom;

This problem appears in texts on Diophantine equations and also on the internet but I do not remember where.

All the solutions to

are of the form:

that is proved by expanding using algebra.

Substituting integers into k, m and n will give every solution.

So

If you make k equal to one then

Since they have no factors in common the GCD ( m (m+n), n (m+n), mn) = 1 and so GCD(a,b,c) is 1.

So a + b is a square for every solution. We are done.

Agnishom · 2012-12-03 23:09:29

I understood but how about the proof of that?

By the way, I saw something like that here

bobbym · 2012-12-03 23:19:32

Hi Agnishom;

That is my proof. Not firm enough? Lacking in rigor? Or just out and out silly?

Agnishom · 2012-12-03 23:28:07

No I like your proof. But how do I prove this:

???

bobbym · 2012-12-03 23:42:09

Hi;

Like with a trig identity I am going to make one side look like the other. Just algebra will be required.

Multiply through by k:

Combine the LHS.

After division by ( m + n ) in the numerator and denominator.

So

http://i638.photobucket.com/albums/uu109/susannah_dingley/Aliens.gif

bob bundy · 2012-12-03 23:55:43

eeekkk! bobbym

I think you've been hacked!

Bob

scientia · 2012-12-04 00:11:14

Bobbym is only trying to link to an image but the server on which the image is hosted isn't allowing remote linking without your first visiting the site.

Try this:

bobbym · 2012-12-04 00:14:28

Hi;

I am seeing the image in my post just like yours. Apparently you all are seeing something else. I will upload the image manually in my post.

Scientia;

Thanks for the link. With your permission I am using that.

scientia · 2012-12-04 00:22:05

Agnishom is seeing this image:

That was what I initially saw as well. That's because neither of us had visited the site http://a1star.com/ first. Once I went to the site, I didn't see the remote-linking error image any more.

Some sites which restrict remote linking work like this. The problem is that you can see the correct image because you've visited the site, but others who haven't visited the site see a different image. They have to visit the site in order to see the correct image.

bobbym wrote:
Scientia;

Thanks for the link. With your permission I am using that.

You're welcome.

Last edited by scientia (2012-12-04 00:23:03)

bobbym · 2012-12-04 00:28:24

Hi Scientia;

I did not know any of that. Thanks for telling me. Luckily, they did not put something extremely offensive up there.

I apologize to everyone who got confused.

Agnishom · 2012-12-04 00:28:56

You got me wrong...
I mean how can you say that all solutions of 1/a + 1/b = 1/c can be put into that form?

bob bundy · 2012-12-04 00:31:09

You are easy to forgive. Had me worried there for a moment.

Thought you had gone completely screwy.

Bob

bobbym · 2012-12-04 00:37:35

Hi Bob;

Thought you had gone completely screwy.

Not yet. I honestly was at a bonafide site that I googled for that gif. It seems no matter how smart I think I am and how long I have lived there is still a billion tricks I have never even heard of.

Agnishom · 2012-12-04 00:40:26

How can you say that every solution can be expressed in that form?

bobbym · 2012-12-04 00:46:04

Hi Agnishom;

If this is true and it appears so

then we can say a = k m (m+n) and b = k n (m+n) and c = k m n where k,m, and n are arbitrary integers that do not equal 0. Also, ( m + n ) does not equal 0. Of course this does not apply here because everything is positive.

Agnishom · 2012-12-04 00:51:32

But how do you get that form???

bobbym · 2012-12-04 00:54:09

Hi;

That is a good question. Also, it is a different question. In the book I am looking at he derives it. I just copied it down in my notes long ago because it was a useful relation for solving equations like that.

Generally, when you are doing a proof you use lots of other theorems and axioms. Just knowing they are true is enough, you do not have to derive them.

For instance, if you use the Pythagorean theorem in your proof you do not have to show how it was derived.

I have shown that the above relation is true for the positive integers. That is enough for the original question.

But since you are asking for the derivation, I can copy it down for you. I do not think I understand it well enough to explain it. But you might be right, if the author put it in to the proof there is a good chance it is mandatory. Let me know what you think about that.

Avon · 2012-12-04 12:37:57

I agree with bobbym that you don't need to reprove a result every time you want to use it, but if someone wants to use a result that you have never seen before and doesn't provide a proof of it then it is reasonable to be unhappy with the result, as Agnishom is.

Suppose that

where a, b, c are positive integers such that gcd(a,b,c) = 1.

Let l = gcd(a,b), m = gcd(a,c), n = gcd(b,c)

Since gcd(a,b,c) = 1, it follows that gcd(l,m) = gcd(l,n) = gcd(m,n) = 1 and so there exist positive integers a', b', c' such that

a = lma' b = lnb' c = mnc'

Also gcd(a', b') = gcd(a', c') = gcd(b', c') = gcd(a', n) = gcd(b', m) = gcd(c', l) = 1.

Now if we substitute our expressions for a, b, c into the original equation we obtain

Since gcd(c', la'b') = 1, it follows that c' = 1, and hence ma' +nb' = la'b'.

Also gcd(a', nb') = gcd(b', ma') = 1, so a' = b' = 1 and so l = m + n.

Therefore a = m(m+n), b = n(m+n), c = mn.

Of course we also have that gcd(m,n) = 1, but we don't need to use this to see that a + b = (m + n)^2.

bobbym · 2012-12-04 13:44:51

Hi;

Thanks for posting the derivation but I did not ever think Agnishom's questions were unreasonable. What I did say is that the derivation is not necessary for the proof.

bobbym wrote:
But since you are asking for the derivation, I can copy it down for you. I do not think I understand it well enough to explain it. But you might be right, if the author put it in to the proof there is a good chance it is mandatory. Let me know what you think about that.

I did say I would provide the derivation of this

if he still wanted to see it.

What I am saying is it is an identity for positive integers. This is proved through the algebra in post #8. Where it comes from is not really important. If I had guessed at it and then proved it that would be fine. I could use it as a theorem in Agnishom's question.

Perhaps post#8 is not rigorous enough, but if it is then it holds for all k,m and n that are positive integers because it is an identity. ( Actually it holds for a great deal more but we only need the positive integers. )

To generate all the solutions just means having to iterate k, m and n for 1,2,3,4,5.... sort of like a loop inside a loop inside a loop.
3 nested loops, a structure Agnishom is familiar with I am sure.

Agnishom · 2012-12-04 14:31:41

Thank You, Avon

bobbym wrote:
I did say I would provide the derivation of this, if he still wanted to see it.

Yes I wanted to see it, and Avon's post has helped.
These proofs are quite hard to go through and memorise

Let me post Question 3:
It is probably more simple than the previous ones.
A censusman on duty visited a house which the lady inmates declined to reveal their individual
ages, but said — “we do not mind giving you the sum of the ages of any two ladies you may
choose”. Thereupon the censusman said — “In that case please give me the sum of the ages of
every possible pair of you”. The gave the sums as follows : 30, 33, 41, 58, 66, 69. The censusman
took these figures and happily went away.

What are the individual ages of each of them?

bobbym · 2012-12-04 15:10:41

Hi;

Call there ages a,b,c,d.

There are four girls needed to make 6 pairs.

Avon · 2012-12-04 15:56:35

bobbym, I apologise if I have caused any offence but I think you are missing something.

Here are two statements:

(1) If k,m,n are positive integers and a = km(m+n), b = kn(m+n), c = kmn, then 1/a + 1/b = 1/c.

(2) If a,b,c are positive integers and 1/a + 1/b = 1/c, then there exist positive integers k,m,n such that a = km(m+n), b = kn(m+n), c = kmn.

In post #8 you demonstrate that (1) is true, but in order to solve Agnishom's problem you are using that (2) is true. Hopefully it is clear that (1) does not imply (2).

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#1 2012-12-03 12:53:37

A few questions

#2 2012-12-03 14:25:34

Re: A few questions

#3 2012-12-03 14:38:39

Re: A few questions

#4 2012-12-03 21:05:09

Re: A few questions

#5 2012-12-03 23:09:29

Re: A few questions

#6 2012-12-03 23:19:32

Re: A few questions

#7 2012-12-03 23:28:07

Re: A few questions

#8 2012-12-03 23:42:09

Re: A few questions

#9 2012-12-03 23:55:43

Re: A few questions

#10 2012-12-04 00:11:14

Re: A few questions

#11 2012-12-04 00:14:28

Re: A few questions

#12 2012-12-04 00:22:05

Re: A few questions

bobbym wrote:

#13 2012-12-04 00:28:24

Re: A few questions

#14 2012-12-04 00:28:56

Re: A few questions

#15 2012-12-04 00:31:09

Re: A few questions

#16 2012-12-04 00:37:35

Re: A few questions

#17 2012-12-04 00:40:26

Re: A few questions

#18 2012-12-04 00:46:04

Re: A few questions

#19 2012-12-04 00:51:32

Re: A few questions

#20 2012-12-04 00:54:09

Re: A few questions

#21 2012-12-04 12:37:57

Re: A few questions

#22 2012-12-04 13:44:51

Re: A few questions

bobbym wrote:

#23 2012-12-04 14:31:41

Re: A few questions

bobbym wrote:

#24 2012-12-04 15:10:41

Re: A few questions

#25 2012-12-04 15:56:35

Re: A few questions

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