across
  • across
Funsie: How many positive divisors does\[-340000000000000000000000000000000000000\]have? :P This is actually doable. Step 1: Find the prime-power factorization of that number. Step 2: Use the \(\tau\) function. The answer is \(2964\).
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
not enough :)
across
  • across
Okay, let us do this in steps. Who can find me the prime-power factorization of that number?
anonymous
  • anonymous
including 34 and 1? (with the respective number of zeroes after their numbers)

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TuringTest
  • TuringTest
17*2*10^(let me count..)=-1*17*2*5^(...)*2(...)
anonymous
  • anonymous
there is a really easy way to do this but i forgot, finding primes of really really large numbers
shadowfiend
  • shadowfiend
\(17\cdot2\cdot 5^n\cdot 2^n\) where \(n\) is the number of 0s that I'm too lazy to count? :p
anonymous
  • anonymous
10^36 * 34 * (-10) 17*2*10^(36) * (-10)
anonymous
  • anonymous
152 factors?
anonymous
  • anonymous
wait, itd be greater than that.
shadowfiend
  • shadowfiend
What am I saying, I'm a programmer... Shadowfiend@al-farabi ~> ruby -e "puts '0000000000000000000000000000000000000'.length" 37
TuringTest
  • TuringTest
17*2^(37)*5^(36) If I trust that Ishaan counted correctly
shadowfiend
  • shadowfiend
^ Ruby counts better. It's 37 0s, so should be 38 and 37, if I'm not mistaken?
anonymous
  • anonymous
I used python interpreter to count 17*2^38* 5^37*(-1)
TuringTest
  • TuringTest
(-1)*17*2^(38)*5^(37) keep forgetting the minus now what?
shadowfiend
  • shadowfiend
Combinations of the 2s, 5s, and -1?
shadowfiend
  • shadowfiend
Why do I feel like Pascal's triangle is about to pop out of my screen and bite me?
across
  • across
So we found that \((-1)\cdot2^{38}\cdot5^{37}\cdot17^1\). :) It follows that the number of factors is the product of the powers plus one: \((38+1)(37+1)(1+1)\).
TuringTest
  • TuringTest
oh I always forget that theorem... probably because I never use it
anonymous
  • anonymous
disincluding the possibility of being dividing by itself, i got 300 divisors. and im probs mega wrong. :P
shadowfiend
  • shadowfiend
Wow. I haven't seen that theorem in about five years haha. I had forgotten it... How you call it... Existed.
TuringTest
  • TuringTest
what's it called again?
anonymous
  • anonymous
how does it work?
anonymous
  • anonymous
Interesting.
TuringTest
  • TuringTest
it's this dealie http://answers.yahoo.com/question/index?qid=20080716092006AAWNZNu what is it friggin' called though? I don't think I ever knew that
shadowfiend
  • shadowfiend
http://en.wikipedia.org/wiki/Divisor_function
TuringTest
  • TuringTest
ahhhh
shadowfiend
  • shadowfiend
Or, as across updated the question to read, the τ function.
anonymous
  • anonymous
wouldn't it make more sense to add all the exponents together instead of subtract? sense you are looking for the amount of divisors (or total prime factors), multiplying them all together would give you some strange number it should be (38+1) + (37+1) + (1+1) or 79 i think..
anonymous
  • anonymous
could someone explain the divisor function in simple english? :P I tried reading the wikipedia link, but alll the words were just swirling around my hear.
TuringTest
  • TuringTest
@funinabox that's not how the theorem works...
anonymous
  • anonymous
instead of multiply* dunno why i said subtract
TuringTest
  • TuringTest
I think the best way to see how the divisor function works is to watch carefully how this problem got solved. stare at it for a while, and you'll see the patttern is not so difficult. How that formula is derived though, \(that\) I would like to know
TuringTest
  • TuringTest
ok, who else started seeing "math processing error" all of a sudden?
TuringTest
  • TuringTest
it was just me I guess it's gone now
across
  • across
Let us refer to the number of divisors function as \(\tau\). If \(p\) is a prime number, then \(\tau(p^a)=a+1\), where \(a\) is a positive integer. This is because the divisors of \(p^a\) are \(1,p,p^2,\dots,p^{a-1}\). Therefore, \(p^a\) has exactly \(a+1\) divisors. Since the function is multiplicative, (I do not feel like proving that ;P), we have that for any positive integer \(n\) with prime-power factorization \(n=p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}\), \(\tau(n)=\tau(p_1^{a_1}p_2^{a_2}\cdots p_s^{a_s}=\tau(p_1^{a_1})\tau(p_2^{a_2})\cdots\tau(p_s^{a_s})\), and @funinabox, this is why we multiply. :)
across
  • across
I forgot to close the parenthesis up there.
TuringTest
  • TuringTest
oh wow, and now I think I actually understand the theorem thanks again across
anonymous
  • anonymous
yeah i saw that :) thanks for the better explanation, i just was gonna post it clicked in my head. the wiki was adding everything up one-by-one like (1+1+1+1+1+1) for factors of 12, and here everyone used exponents. it messed me up till i realized it was the same deal cause 2^2 + 3^1 will be (2+1)(1+1) or 6 (same as 1+1+1+1+1+1)
across
  • across
I noticed anoter typo. Oh well, you guys get the idea. ;P
shadowfiend
  • shadowfiend
♥ elegant proofs, elegant theorems.
ParthKohli
  • ParthKohli
\[-1 \times 17 \times 2 \times 10000000000000000000000000000000000000 \]An intuition for the \(\tau\) function is the basic multiplication rule in Combinatorics. the prime factorization of \(24 = 2^3 \times 3\). Here, the divisors are all the numbers we may generate with the powers of your prime factors less than or equal to the current ones. Like a divisor is \(12 = 2^2 \times 3^1\) or \(1 =2^0 \times 3^0 \). So we must find the number of powers \(2\) can take which is \(0,1,2,3\). Notice that because we include \(0\), the number of choices will increase by \(1\). If you have two ties red and blue, then two suits grey and black, what are the number of ways to wear them? \(2 \times 2 = 4\). Here, there are \(3 + 1\) ways to dress up \(2\) and \(1 + 1\) for \(3\). The total number of dresses is \((3 + 1)(1 + 1)=8\).
ParthKohli
  • ParthKohli
Challenge #2: What is the least positive number such that the number of divisors of the number of divisors of the number of divisors of the number of divisors is \(3\)? @across @shadowfiend @TuringTest @Ishaan94 :-)

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