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A few simple problems

Enjoy :biggrin:

1 Problem




2 Problem




3 Problem




4 Problem




5 Problem

(edited 8 years ago)

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Will demotivate myself with these after the Arsenal game:biggrin:
Reply 2
Avatar for TeeEm
TeeEm
19
Original post by drandy76
Will demotivate myself with these after the Arsenal game:biggrin:


Me too after the rivals game ....
Reply 3
Avatar for Zacken
Zacken
22
Original post by Indeterminate
Enjoy :biggrin:

1 Problem



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

Claim: xn−1=(x−1)(xn−1+xn−2+⋯+1)x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)

The base case (n=2n=2) holds, since x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1).

Assuming xn−1=(x−1)(xn−1+xn−2+⋯+1)x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1), we then multiply by xx on both sides to get:

xn+1−x=(x−1)(xn+xn−1+⋯+x)x^{n+1} - x = (x-1)(x^n + x^{n-1} + \cdots + x)

Cleverly adding zero:

xn+1−1−(x−1)=(x−1)(xn+xn−1+⋯+x)x^{n+1} - 1 - (x-1) = (x-1)(x^n + x^{n-1} + \cdots + x)

So, adding x−1x-1 to both sides, we have:

xn+1−1=(x−1)(xn+xn−1+⋯+x+1)x^{n+1} - 1 = (x-1)(x^n + x^{n-1} + \cdots +x + 1) and we are done!

Hence, pq−1=(p−1)(pq−1+⋯+1)p^q - 1 = (p-1)(p^{q-1} + \cdots + 1) and is hence definitely divisible by p−1p-1.
Simple? :redface: What level would you call this? :frown:
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 :biggrin:

Let

pp and qq be positive integers (not necessarily distinct)

Prove that

pq−1p^q - 1 is divisible by (p−1)(p-1)

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
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: xn−1=(x−1)(xn−1+xn−2+⋯+1)x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1)

The base case (n=2n=2) holds, since x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1).

Assuming xn−1=(x−1)(xn−1+xn−2+⋯+1)x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + 1), we then multiply by xx on both sides to get:

xn+1−x=(x−1)(xn+xn−1+⋯+x)x^{n+1} - x = (x-1)(x^n + x^{n-1} + \cdots + x)

Cleverly adding zero:

xn+1−1−(x−1)=(x−1)(xn+xn−1+⋯+x)x^{n+1} - 1 - (x-1) = (x-1)(x^n + x^{n-1} + \cdots + x)

So, adding x−1x-1 to both sides, we have:

xn+1−1=(x−1)(xn+xn−1+⋯+x+1)x^{n+1} - 1 = (x-1)(x^n + x^{n-1} + \cdots +x + 1) and we are done!

Hence, pq−1=(p−1)(pq−1+⋯+1)p^q - 1 = (p-1)(p^{q-1} + \cdots + 1) and is hence definitely divisible by p−1p-1.


No need for all that - see my post above.
Reply 8
Avatar for TeeEm
TeeEm
19
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
Reply 9
Avatar for TeeEm
TeeEm
19
Original post by constellarknight
No need for all that - see my post above.


f2.jpg
Reply 10
Avatar for Zacken
Zacken
22
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.
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
Reply 12
Avatar for TeeEm
TeeEm
19
Original post by drandy76
hopefully they chip away at your shiny goal difference as well


we are losing already
we just took the lead:h: , do i dare hope?

Original post by TeeEm
we are losing already
Capture.PNG
Original post by undercxver
Simple? :redface: What level would you call this? :frown:


A-level :yep:
Original post by Indeterminate
This proves that pq−1p^q - 1 is divisible by (p−1)(p-1) for all qq when p=1p =1

What happens when p≠1p \neq 1 ?

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 Indeterminate
This proves that pq−1p^q - 1 is divisible by (p−1)(p-1) for all qq when p=1p =1

What happens when p≠1p \neq 1 ?


The remainder theorem implies the other factor is always a polynomial in p, no?
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 :facepalm:

I'll edit the OP
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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