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#1 2006-02-03 10:04:26

Enfy
Guest

integer proof

I've been trying to solve this for a while, but without any good result. I could definitely need some help :]

k = (3*9^(p+1)+25^(p+1))/4

p is any number larger or equal to 0. How can I be 100% sure that k is an integer? I've tried different values to test it, and it looks like it is an integer, but I don't know how to actually proof it.

#2 2006-02-03 10:19:01

irspow
Member
Registered: 2005-11-24
Posts: 1,055

Re: integer proof

I hope you get your answer here.  I was looking for a similar definition in an earlier thread, but a clear one never emerged.


I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

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#3 2006-02-03 10:37:42

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: integer proof

Edit: Ah, spotted a large error in my proof, fixing it right now.

Last edited by Ricky (2006-02-03 10:41:21)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2006-02-03 10:55:54

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: integer proof

p is any number? Only integers, right?

I guess it can be  prove that 3*9^(p+1)+25^(p+1) is always a multiple of 4.

You could try an inductive proof:
    1) prove the base case (p=1),
    2) prove that whenever you add 1 to p, the result will still be a multiple of 4 (use p=n p=n+1)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#5 2006-02-03 11:07:23

Enfy
Guest

Re: integer proof

I've tried the induction proof, but I didn't get too far with it (I'm still not too comfortable with it btw). I might try something similar to what Ricky posted before he edited his post, though I was confused with one of the steps he provided.

#6 2006-02-03 11:14:15

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: integer proof

I was skeptical at first, but it seems that induction is the way to go on this one.

Glad you responded, I wasn't sure if you knew what an inductive proof was, so that saved me a lot of typing.

Inductive assumption:

Inductive assumption, 216, and 600 are all divisible by 4.

Last edited by Ricky (2006-02-03 11:24:26)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-02-03 11:16:39

irspow
Member
Registered: 2005-11-24
Posts: 1,055

Re: integer proof

I think Enfy said any number greater than or equal to zero.  I dont think that there is anything to prove otherwise.  If p is an integer, then p + 1 would also be an integer the equation would only produce a series like;

  (27n + 25n)/4 = 54n/4 = 13n, which of course would all be integers.

This is obviously not a proof, because I mistakenly used an arithmetic series example instead of a geometric series, but I think that it can be shown to be similar since the relationship of multiples still holds.  I will try it that way.

Last edited by irspow (2006-02-03 11:30:04)


I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

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#8 2006-02-03 11:21:44

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: integer proof

If p is an integer, then p + 1 would also be an integer the equation would only produce a series like;

  (27n + 25n)/4 = 54n/4 = 13n, which of course would all be integers.

Let n = 1:

So you are saying there is an integer p ≥ 1 such that (3*9^(p+1)+25^(p+1))/4 = 13?

Furthermore, you are saying that 9^(p+1) = 25^(p+1) for all p, since n = n

Last edited by Ricky (2006-02-03 11:22:42)


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2006-02-03 11:55:12

Enfy
Guest

Re: integer proof

Thanks Ricky, I got it now :]

I need to learn those tricks with adding more terms, very clever.

#10 2006-02-03 12:01:51

Enfy
Guest

Re: integer proof

Oh, and I meant to say integer larger or equal to 0 in my post, since this was about integers and nothing else. Sorry if that confused you, irspow.

#11 2006-02-03 12:06:34

irspow
Member
Registered: 2005-11-24
Posts: 1,055

Re: integer proof

I am always very confused, but thanks for the sympathy.


I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

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#12 2006-02-03 12:17:07

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: integer proof

When you have experience with proving statements like this, you know what the poster means even if (s)he doesn't say it.

That's really all it is, experience, nothing else irspow.

Oh, and I meant to say integer larger or equal to 0 in my post, since this was about integers and nothing else.

Was this a question that a teacher/professor gave you?  It should be worded 9^p and 25^p for p ≥ 1, aka, natural numbers.

It's not wrong the way it is, just...weird.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#13 2006-03-28 00:47:35

iai
Member
Registered: 2006-03-28
Posts: 1

Re: integer proof

how do i solve the following?
write down the integer value of n which satisfy the inequality -2<n1

solve these inequalities: 2x - 5 < 10

x/3 > 6

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#14 2016-12-04 16:18:31

Monox D. I-Fly
Member
From: Indonesia
Registered: 2015-12-02
Posts: 2,000

Re: integer proof

2x - 5 < 10
2x < 10 + 5
2x < 15
x < 15/2
x < 7.5

x/3 > 6
x > 6(3)
x > 18


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