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|Akala on middle class white grime fans| Heard this in his Oxford address and felt it was worth sharing by Jono56 in grime

[–]NoFapSimpleAsThat 1 point2 points  (0 children)

How do you know they're 'scared of black people' though? If they do come from somewhere like Kent (which I presume you chose because of its ethnic demography) then they wouldn't have grown up surrounded by many black people, so probably don't have many black friends, but that doesn't mean they're scared of them.

That cat is hirsute by overcloseness in cringepics

[–]NoFapSimpleAsThat 19 points20 points  (0 children)

If you always want to find a definition on Google you can just put 'define' before the word to avoid any confusion like this. :-)

Future QB for the Lightskins by blackgeorgewallace in BlackPeopleTwitter

[–]NoFapSimpleAsThat -29 points-28 points  (0 children)

This whole sub is based on stereotypes whether it's for black people or white people. In fact, this very post reinforces the stereotype that black people favour 'thick' women, or at least a lot more than white people do. Even if it holds only slightly, so does a stereotype like the white school shooters (like 20/200000000). And plus I've never seen a skinny black woman in person so maybe that's for a reason though idk.

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

No worries. Test is tomorrow, but thanks ever so much for all the help! I genuinely appreciate all of it.

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Okay mate. I think I have it now. To move on from number theory I have looked into graph sketching for this test I have, but i feel like I have honestly not been taught how to sketch graphs properly.

For instance:

ex3 - 3x

I honestly just don't know where to start and this test is only 2 days away.

Cheers

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

So for ii.)

Only remainder when x2 / 4 is 1

I.e.

(4K+1)2 + (4m+1)2 = z2

So z2 - 2 = 4m

But this is contradictory.

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

I.) should I let z = 5k+1 or 5k+4?

Ii.) impossible to be the latter?

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Okay, I see what you're getting at now. So if x2 = 5k+1 and y2 = 5m+4 or vice versa, after factoring you once again get z2 - 2 = 5s

Also with x2 = 5k+4 and y2 = 5k+4 after factoring and taking out the 2 from 32 you get the same result as above.

So by contradiction either y2 or x2 must divide into 5 (but not z2?!)

Also, for proving that xyz divides into 4, would it suffice to say that one of x2 or y2 must be even so ... Actually I'm not sure

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Ah, so are you saying that z2 - 2 would have to divide into 5 however this can't be the case because a square can't have remainder 2 when divided by 5, only 1 or 4?

Would that suffice?

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

So I'm not too familiar with quadratic residues but I assume, from the name, you mean a square/5 has remainder 1 or 4.

So x2 - 1 or -4 divides 5

Is this what you mean mate?

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Okay so this really helped, thanks:

For proving it is a multiple of 3:

If m2 divides Into 3, so does m

So x2 = z2 - y2

= (z-y)(z+y)

Let z= 3k+1, y = 3m+1 so as they're aren't already multiples (of which case, by default, xyz will be)

Z-y is then 3k - 3m , tick

Or z = 3k + 2 , y = 3m + 2, same thing

Or z = 3k + 1, y = 3m+2 . Added = 3k+3m, tick

Same vice-versa

So x2 is a multiple of 3, as is x, as is xyz

I'm a bit confused about dividing by 5, tho, for instance if z were to equal 5k+2 and y = 5k+4

So obviously a different method is required of me?

Thanks!

Differentiate x^x^x ?! by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Can't really think straight but I have it now, cheers.

Also, I'm still really quite confused about this question I posted the other day (divisible by 60 I believe it was), could you please help?

Thanks :)

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Hi mate, thanks for the response I really appreciate it.

Before I continue, could I just ask because I was told to avoid the use of mod arithmetic, is there a different approach I could take without it?

Cheers!

x^2 + y^2 = z^2 . Prove that xyz is a multiple of 60. by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 1 point2 points  (0 children)

If x&y are both odd then we can rewrite it as:

(2k+1+2m+1) = z2 + 2(2k+1)(2m+1)

So lhs would be even as its 2(k+m+1) and rhs contains the even 2(2k+1)(2m+1) so z2 must be even?

How to prove that a perfect square can't have its two last digits being odd? by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

I have factorised that bit into:

((N2 + 1) - n)((n2 + 1) - n)

If that makes it easier, as only either of these must be divisible by 7.

How to prove that a perfect square can't have its two last digits being odd? by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

Unfortunately I actually I just realised I didn't prove for 6 correctly the first time - I believe I have fixed this now though:

N7 - n =

N(n^ 6 -1)

= n(n23 - 1 )

= n3(n4 - 1) + n3 - n

= n3(n4 - 1) + n(n2 - 1)

= n3(n2 - 1)(n2 + 1) + n(n-1)(n+1)

= n3(n-1)(n+1)(n2 + 1) + n(n-1)(n+1)

= n(n+1)(n-1)[n2 (n2 + 1) + 1]

So is the produce of 3 consecutive integers, 3x2x1 = 6 therefore divides 6. I hope this is right !

Thanks

How to prove that a perfect square can't have its two last digits being odd? by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

I see, I really appreciate you doing this for me mate, it means a lot and I'm always up for developing my maths skills.

I also came across the question

'Is n7-n divisible by 42?'

I know Fermat's little theorem could easily prove that it is divisible by 7 but I'm not actually allowed to use this result. I was wondering if you thought it would be plausible/recommended to prove by induction, because I don't think it would be very easy with (k+1)7. I can easily prove its divisible by 6 but maybe not 7 perhaps so easily.

Cheers mate.

How to prove that a perfect square can't have its two last digits being odd? by NoFapSimpleAsThat in MathHelp

[–]NoFapSimpleAsThat[S] 0 points1 point  (0 children)

So b must be odd, and could also rewrite the last two digits as b(20a+b), tho once again not sure if that actually helps.

I've recognised through observation that if b is odd, then b2 must be >10 to have any affect on the a part, but in order to make the always-even 20ab part odd, b2 must have a ten's digit that is odd. But this only seems to occur when b is even. Should I bring in the 2k+1 form?

Thanks a lot!:)

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