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 \(2964\).

- across

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- anonymous

not enough :)

- across

Okay, let us do this in steps. Who can find me the prime-power factorization of that number?

- anonymous

including 34 and 1? (with the respective number of zeroes after their numbers)

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## More answers

- TuringTest

17*2*10^(let me count..)=-1*17*2*5^(...)*2(...)

- anonymous

there is a really easy way to do this but i forgot, finding primes of really really large numbers

- 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

10^36 * 34 * (-10)
17*2*10^(36) * (-10)

- anonymous

152 factors?

- anonymous

wait, itd be greater than that.

- shadowfiend

What am I saying, I'm a programmer...
Shadowfiend@al-farabi ~> ruby -e "puts '0000000000000000000000000000000000000'.length"
37

- TuringTest

17*2^(37)*5^(36)
If I trust that Ishaan counted correctly

- shadowfiend

^ Ruby counts better. It's 37 0s, so should be 38 and 37, if I'm not mistaken?

- anonymous

I used python interpreter to count
17*2^38* 5^37*(-1)

- TuringTest

(-1)*17*2^(38)*5^(37)
keep forgetting the minus
now what?

- shadowfiend

Combinations of the 2s, 5s, and -1?

- shadowfiend

Why do I feel like Pascal's triangle is about to pop out of my screen and bite me?

- 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

oh I always forget that theorem...
probably because I never use it

- anonymous

disincluding the possibility of being dividing by itself, i got 300 divisors. and im probs mega wrong. :P

- shadowfiend

Wow. I haven't seen that theorem in about five years haha. I had forgotten it... How you call it... Existed.

- TuringTest

what's it called again?

- anonymous

how does it work?

- anonymous

Interesting.

- 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

http://en.wikipedia.org/wiki/Divisor_function

- TuringTest

ahhhh

- shadowfiend

Or, as across updated the question to read, the τ function.

- 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

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

@funinabox that's not how the theorem works...

- anonymous

instead of multiply* dunno why i said subtract

- 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

ok, who else started seeing "math processing error" all of a sudden?

- TuringTest

it was just me I guess
it's gone now

- 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

I forgot to close the parenthesis up there.

- TuringTest

oh wow, and now I think I actually understand the theorem
thanks again across

- 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

I noticed anoter typo. Oh well, you guys get the idea. ;P

- shadowfiend

♥ elegant proofs, elegant theorems.

- 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

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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