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A few simple problems watch

1. Enjoy

Problem 1

Let

and be positive integers (not necessarily distinct)

Prove that

is divisible by

without using the remainder theorem

Problem 2

Find

Problem 3

Let be a function satisfying

for real

You are given that

Find the value of

Problem 4

Find

Problem 5

A rational function is defined as follows

Find the value of

2. Will demotivate myself with these after the Arsenal game
3. (Original post by drandy76)
Will demotivate myself with these after the Arsenal game
Me too after the rivals game ....
4. (Original post by Indeterminate)
Enjoy

Problem 1

Let

and be positive integers (not necessarily distinct)

Prove that

is divisible by

I'm tired and it's time to hit this with an ugly hammer, I think there's a nice geometrical series proof?

Claim:

The base case () holds, since .

Assuming , we then multiply by on both sides to get:

So, adding to both sides, we have:

and we are done!

Hence, and is hence definitely divisible by .
5. Simple? What level would you call this?
6. Surely this is trivial, as you can just set p=1 -> p^q-1=0, so by the Remainder Theorem it divides (p-1) as required.

(Original post by Indeterminate)
Enjoy

Let

and be positive integers (not necessarily distinct)

Prove that

is divisible by
7. (Original post by TeeEm)
Me too after the rivals game ....
I fear a positive result for tottenham will mean a postpone of my attempts until tomorrow
8. (Original post by Zacken)
I'm tired and it's time to hit this with an ugly hammer, I think there's a nice geometrical series proof?

Claim:

The base case () holds, since .

Assuming , we then multiply by on both sides to get:

So, adding to both sides, we have:

and we are done!

Hence, and is hence definitely divisible by .
No need for all that - see my post above.
9. (Original post by drandy76)
I fear a positive result for tottenham will mean a postpone of my attempts until tomorrow
West ham has been a boggie team for us for years ...
The odds are in your favour tonight
10. (Original post by constellarknight)
No need for all that - see my post above.
11. (Original post by constellarknight)
Surely this is trivial, as you can just set p=1 -> p^q-1=0, so by the Remainder Theorem it divides (p-1) as required.
I initially thought of that as well - but then I figured that it was too trivial, so I figured assuming the remainder theorem wasn't allowed.
12. (Original post by TeeEm)
West ham has been a boggie team for us for years ...
The odds are in your favour tonight
hopefully they chip away at your shiny goal difference as well
13. (Original post by drandy76)
hopefully they chip away at your shiny goal difference as well
14. we just took the lead , do i dare hope?

(Original post by TeeEm)
15. (Original post by undercxver)
Simple? What level would you call this?
A-level
16. (Original post by Indeterminate)
This proves that is divisible by for all when

What happens when ?
x=1 is a root of x^q-1, regardless of the value of q. Thus (x-1) is a factor, by the Remainder Theorem. I don't really understand your point??
17. (Original post by Indeterminate)
This proves that is divisible by for all when

What happens when ?
The remainder theorem implies the other factor is always a polynomial in p, no?
18. (Original post by constellarknight)
x=1 is a root of x^q-1, regardless of the value of q. Thus (x-1) is a factor, by the Remainder Theorem. I don't really understand your point??
(Original post by EricPiphany)
The remainder theorem implies the other factor is always a polynomial in p, no?
Ignore me

I'll edit the OP
19. (Original post by EricPiphany)
The remainder theorem implies the other factor is always a polynomial in p, no?
Exactly. Like I said in my previous post, Indeterminate is not making much sense, which is apt given his username.

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