Discrete Maths Last Minute questions (2 Viewers)

[Drsoccerball]

In that case, I now have

Why divide by 2! ? Theres no more repeats.

EDIT: lel the word rat repeats...

 

[4025808]

Would anyone know how to start off this one?

Firstly I did (n+k)|n^2 but I'm not sure how to proceed from there.

 

[Drsoccerball]

Would anyone know how to start off this one?

Firstly I did (n+k)|n^2 but I'm not sure how to proceed from there.

Answers on moodle

Edit: omg i know who you are

 

[leehuan]

Sorry I know I can do this myself but I don't have time to right now. Can someone prove this?

 

[InteGrand]

Sorry I know I can do this myself but I don't have time to right now. Can someone prove this?

 

[Drsoccerball]

Would anyone know how to start off this one?

Firstly I did (n+k)|n^2 but I'm not sure how to proceed from there.

Okay so (n + k)|n^2

n^2 = (n + k)M for some M integer.

Take note: n^2 = (n - k)(n + k) + k^2

(n + k)M = (n - k)(n + k) + k^2

(n + k)(M + k - n) = k^2

Therefore (n + k)|k^2

If k <= sqrt(n) that means n + k > k^2 and thus a contradiction (Take note k, n are positive integers)

Therefore k > sqrt(n)

 

[leehuan]

Oops. Well this was the question and the let us assume a result (potentially true upon the basis of primality though)

 

[leehuan]

Okay so (n + k)|n^2

n^2 = (n + k)M for some M integer.

Take note: n^2 = (n - k)(n + k) + k^2

(n + k)M = (n - k)(n + k) + k^2

(n + k)(M + k - n) = k^2

Therefore (n + k)|k^2

If k <= sqrt(n) that means n + k > k^2 and thus a contradiction (Take note k, n are positive integers)

Therefore k > sqrt(n)

Man... need Peter Brown though... cause that's so not obvious to induce the diff. of two squares and the add/deduct trick

 

[Drsoccerball]

Man... need Peter Brown though... cause that's so not obvious to induce the diff. of two squares and the add/deduct trick

Yeah I would've never got this in the exam...

 

[leehuan]

With these perms and combs I'd appreciate anyone checking but it's also (hopefully) for Drsoccerball's convenience

 

[Paradoxica]

Answers on moodle

Edit: omg i know who you are

I thought everyone knew.

 

[Paradoxica]

Man... need Peter Brown though... cause that's so not obvious to induce the diff. of two squares and the add/deduct trick

You didn't try long division?

gosh you people.

 

[leehuan]

You didn't try long division?

gosh you people.

I thought you were the one that hated long division

Though yeah that never crosses my mind with algebra anymore unless I explicitly see polynomials

 

[InteGrand]

Oops. Well this was the question and the let us assume a result (potentially true upon the basis of primality though)

Yeah if p is a prime, then that's a true result.

 

[Paradoxica]

I thought you were the one that hated long division

Though yeah that never crosses my mind with algebra anymore unless I explicitly see polynomials

I wouldn't long divide. Just saying you might have.

 

[leehuan]

The question isn't hard but I just want to see it get done because I want to see how you'd set out your working



I'll put down the relevant Euclidean algorithm to save a bit of work

 

[Drsoccerball]

The question isn't hard but I just want to see it get done because I want to see how you'd set out your working



I'll put down the relevant Euclidean algorithm to save a bit of work

If your gcd != 1 and your desired number in this case being 8 is not equal to the gcd then you divide everything by the gcd(including the modulus) and multiply everything by 8 except the modulus.

 

[leehuan]

Woohoo no more first year math again

Shoutout to everyone who helped esp InteGrand

 

[Drsoccerball]

Woohoo no more first year math again

Shoutout to everyone who helped esp InteGrand

nek minute fails...

Thanks integrand you've saved my wam

 

[leehuan]

nek minute fails...

Thanks integrand you've saved my wam

Can you like, not jinx it ples