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March 20th, 2018, 08:59 PM   #11
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Well if you are complaining from over East...

Yeah as the title suggests you need to experience Perth to truly know terror.

Have found a few peeps that are fantastic though, couple of the graduates from UWA I met over facebook somehow. Really clever stuff and they have been immeasurably helpful with some of the problems I ask for help with, so yes we do live in a cultural shithole but its not all doom and gloom.

You should apply for those semesters abroad in Budapest, that's what ive got on the agenda for the next 12 months. Should be good although I still need to find someone to review me in one more subject. (You can submit letters of recommendation by faculty members in your application)
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March 20th, 2018, 09:32 PM   #12
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Sorry I need to also share another proof for this one, this time from a local in Perth named Andre Rhine Davis from UWA, showing that there is indeed *some* culture in here, as is very characteristic of this chap the proof is direct and well worthy of the prize money, which has already been spent on coca cola but none the less:

Let's try considering one prime factor at a time.
Let v_p(x) be the function that gives the highest power of prime p that divides x. Note that v_p is completely additive.

So, n = 0 (mod k^3) means k^3 divides n, that is:
v_p(k^3) = 3 v_p(k) ≤ v_p(n)
for every prime p

Now what rad does is effectively set the power of every prime factor to 1. That is:
v_p(rad(x)) = min(1,v_p(x))

So, let r := rad[n / rad(n)] · rad(n) · k
v_p(r)
= v_p[rad[n / rad(n)] · rad(n) · k]
= v_p[rad[n / rad(n)]] + v_p[rad(n)] + v_p[k]
= min(1,v_p[n / rad(n)]) + min(1,v_p(n)) + v_p(k)
= min(1,v_p(n) – v_p[rad(n)]) + min(1,v_p(n)) + v_p(k)
= min(1,v_p(n) – min(1,v_p(n))) + min(1,v_p(n)) + v_p(k)

We want to prove that r divides n, that is v_p(r) ≤ v_p(n) for all primes p.
We can split this into several cases.

• v_p(k) = 0 and v_p(n) = 0
v_p(r) obviously must be ≤ 0 = v_p(n)

• v_p(k) = 0 and v_p(n) = 1
v_p(r) = min(1,1 – min(1,1)) + min(1,1) + 0
= min(1,0) + min(1,1)
= 1
= v_p(n)

• v_p(k) = 0 and v_p(n) ≥ 2
v_p(r)
= min(1,v_p(n) – min(1,v_p(n))) + min(1,v_p(n)) + 0
= min(1,v_p(n) – 1) + 1
= 2
≤ v_p(n)

• v_p(k) ≥ 1, and remember we also have the condition 3 v_p(k) ≤ v_p(n)
v_p(r)
= min(1,v_p(n) – min(1,v_p(n))) + min(1,v_p(n)) + v_p(k)
= min(1,v_p(n) – 1) + 1 + v_p(k)
= 2 + v_p(k)
≤ 3 v_p(k)
≤ v_p(n)

Therefore in every case, we have v_p(r) ≤ v_p(n), and no restrictions were put on p, so this holds for every p. Therefore, r divides n.

As for proving the converse is false, just take v_p(k) = 2 and v_p(n) = 4.
Then v_p(r) = 2 + v_p(k) ≤ v_p(n), but 3 v_p(k) is not ≤ v_p(n)

e.g.
n = 16, k = 4
We have:
r = rad[n / rad(n)] · rad(n) · k
= rad[16 / rad(16)] · rad(16) · 4
= rad[16 / 2] · 2 · 4
= rad[8] · 2 · 4
= 2 · 2 · 4
= 16

So r=16 is a divisor of n=16, but k^3=64 is not a divisor of n=16


So there you go, they both had me back in my box with 30 minutes of posting the question I might add, thought I had at least a few days up my sleave but as again I am the tortoise it seems...
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March 20th, 2018, 11:17 PM   #13
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Quote:
Originally Posted by Adam Ledger View Post
i get quite low on carbs without "mi goreng"
You eat Indomie? How much does it cost there?
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March 21st, 2018, 02:04 AM   #14
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Originally Posted by Monox D. I-Fly View Post
You eat Indomie? How much does it cost there?
Lived on the mi goreng for ages.. You can get them pretty cheap in bulk at the Asian food stores here. But individually in the main supermarkets is like 75c each or something.
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March 21st, 2018, 02:06 AM   #15
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Yeah as the title suggests you need to experience Perth to truly know terror.
Try the NT, @$$ crack of the world right there . But Perth is weird from what I remember.
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March 21st, 2018, 04:42 AM   #16
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oh....NT

Well naturally I didn't regard that region as pertaining to an actual population of sentient life forms but mi goreng is approx. 35 cents each if you buy them in the boxes at coles or iga.
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March 21st, 2018, 05:26 AM   #17
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what... what did you mean by that exactly I'm not sure I like your sarcasm dennis
WHY are you "not sure"?

Oh...it's Denis, not Dennis:
my dad was Irish, my mom French AND THE BOSS!!
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March 21st, 2018, 06:07 AM   #18
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Ah ok well I don't have any Irish blood, as far as I know I am pure anyway, but my father did conduct a pretty detailed study into the family's ancestry, and was able to trace us back to the French revolution, where we were protestants that fled persecution, run away screaming like a girl actually I am a total coward when confronted with physical violence so it totally makes sense I have French lineage.

and yes I'm not sure if someone is being facetious or sarcastic with me I need them to elaborate because it's ok if they do it well, I only get angry or insulted if I am made funny of with poor wit, if it's well done then ill be laughing myself to sleep about it that night
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March 21st, 2018, 06:13 AM   #19
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With a last name Ledger, I'm surprised you're not an Accountant

Or a General in the army !
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March 21st, 2018, 06:26 AM   #20
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Well my dad did always want me to be an accountant actually, he was a college drop from an accounting degree so I suppose its one of those things whereby one generation attempts to for fill their failed dreams in the next generation, I was meant to go back to university and do an accounting and be the accountant for his little engineering company but it's just too painful hearing him spout on about his stupid boom doogle, its like "OMG the genius of it all you applied high school Newtonian physics to an industrial process please do go on sir I am SO intrigued"
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