Discrete Maths Last Minute questions (3 Viewers)

Drsoccerball · Nov 14, 2016

leehuan said:
In that case, I now have

$8\times\frac{7!}{3!2!2!}-\left(6\times5\times\frac{4!}{2!}-\frac{4!}{2!}\right)$

Why divide by 2! ? Theres no more repeats.

EDIT: lel the word rat repeats...

4025808 · Nov 14, 2016

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 · Nov 14, 2016

4025808 said:
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 · Nov 14, 2016

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

$a \mid b^2 \implies a \mid b$

InteGrand · Nov 14, 2016

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

$a \mid b^2 \implies a \mid b$

$\noindent This is not true. For example, $4$ divides $2^{2}$, but $4$ does not divide $2$.$

Drsoccerball · Nov 14, 2016

4025808 said:
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 · Nov 14, 2016

InteGrand said:
$\noindent This is not true. For example, $4$ divides $2^{2}$, but $4$ does not divide $2$.$

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

leehuan · Nov 14, 2016

Drsoccerball said:
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 · Nov 14, 2016

leehuan said:
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 · Nov 14, 2016

$\text{a) }\binom{13}{1}\binom43\binom11\binom{12}{7}$

$\text{b) }\binom41\binom{13}3\binom{39}7-\binom42\binom{13}{3}^2\binom{26}{4}+\binom{4}{3} \binom{13}{3} ^3\binom{13}{1}$

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

Paradoxica · Nov 14, 2016

Drsoccerball said:
Answers on moodle

Edit: omg i know who you are

I thought everyone knew.

Paradoxica · Nov 14, 2016

leehuan said:
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 · Nov 14, 2016

Paradoxica said:
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 · Nov 14, 2016

leehuan said:
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 · Nov 14, 2016

leehuan said:
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 · Nov 14, 2016

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

$\text{Q: Solve }296x \equiv 8 (\mod 692)$

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

$\begin{align*}692&=2\times 296+100\\ 296&=2\times 100+96\\ 100&=1\times 96+4\\ 96&=24\times 4\end{align*}$

Drsoccerball · Nov 14, 2016

leehuan said:
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

$\text{Q: Solve }296x \equiv 8 (\mod 692)$

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

$\begin{align*}692&=2\times 296+100\\ 296&=2\times 100+96\\ 100&=1\times 96+4\\ 96&=24\times 4\end{align*}$

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 · Nov 15, 2016

Woohoo no more first year math again

Shoutout to everyone who helped esp InteGrand

Drsoccerball · Nov 16, 2016

leehuan said:
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 · Nov 17, 2016

Drsoccerball said:
nek minute fails...

Thanks integrand you've saved my wam

Can you like, not jinx it ples

Discrete Maths Last Minute questions (3 Viewers)

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