You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Enfy****Guest**

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.

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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

Offline

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

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

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

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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)

Offline

**Enfy****Guest**

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.

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

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)*

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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)*

Offline

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

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)*

Offline

**Enfy****Guest**

Thanks Ricky, I got it now :]

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

**Enfy****Guest**

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.

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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

Offline

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

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.

Offline

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

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

Offline

Pages: **1**