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across
 3 years ago
Funsie: How many positive divisors does\[340000000000000000000000000000000000000\]have? :P
This is actually doable.
Step 1: Find the primepower factorization of that number.
Step 2: Use the \(\tau\) function.
The answer is \(2964\).
across
 3 years ago
Funsie: How many positive divisors does\[340000000000000000000000000000000000000\]have? :P This is actually doable. Step 1: Find the primepower factorization of that number. Step 2: Use the \(\tau\) function. The answer is \(2964\).

This Question is Closed

across
 3 years ago
Best ResponseYou've already chosen the best response.8Okay, let us do this in steps. Who can find me the primepower factorization of that number?

Dumb_as_a_Duck
 3 years ago
Best ResponseYou've already chosen the best response.0including 34 and 1? (with the respective number of zeroes after their numbers)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.017*2*10^(let me count..)=1*17*2*5^(...)*2(...)

funinabox
 3 years ago
Best ResponseYou've already chosen the best response.0there is a really easy way to do this but i forgot, finding primes of really really large numbers

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2\(17\cdot2\cdot 5^n\cdot 2^n\) where \(n\) is the number of 0s that I'm too lazy to count? :p

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.010^36 * 34 * (10) 17*2*10^(36) * (10)

Dumb_as_a_Duck
 3 years ago
Best ResponseYou've already chosen the best response.0wait, itd be greater than that.

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2What am I saying, I'm a programmer... Shadowfiend@alfarabi ~> ruby e "puts '0000000000000000000000000000000000000'.length" 37

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.017*2^(37)*5^(36) If I trust that Ishaan counted correctly

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2^ Ruby counts better. It's 37 0s, so should be 38 and 37, if I'm not mistaken?

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0I used python interpreter to count 17*2^38* 5^37*(1)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0(1)*17*2^(38)*5^(37) keep forgetting the minus now what?

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2Combinations of the 2s, 5s, and 1?

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2Why do I feel like Pascal's triangle is about to pop out of my screen and bite me?

across
 3 years ago
Best ResponseYou've already chosen the best response.8So 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh I always forget that theorem... probably because I never use it

Dumb_as_a_Duck
 3 years ago
Best ResponseYou've already chosen the best response.0disincluding the possibility of being dividing by itself, i got 300 divisors. and im probs mega wrong. :P

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2Wow. I haven't seen that theorem in about five years haha. I had forgotten it... How you call it... Existed.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0what's it called again?

Dumb_as_a_Duck
 3 years ago
Best ResponseYou've already chosen the best response.0how does it work?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0it'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
 3 years ago
Best ResponseYou've already chosen the best response.2Or, as across updated the question to read, the τ function.

funinabox
 3 years ago
Best ResponseYou've already chosen the best response.0wouldn'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..

Dumb_as_a_Duck
 3 years ago
Best ResponseYou've already chosen the best response.0could 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
 3 years ago
Best ResponseYou've already chosen the best response.0@funinabox that's not how the theorem works...

funinabox
 3 years ago
Best ResponseYou've already chosen the best response.0instead of multiply* dunno why i said subtract

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok, who else started seeing "math processing error" all of a sudden?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0it was just me I guess it's gone now

across
 3 years ago
Best ResponseYou've already chosen the best response.8Let 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 \(1,p,p^2,\dots,p^{a1}\). 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 primepower 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
 3 years ago
Best ResponseYou've already chosen the best response.8I forgot to close the parenthesis up there.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0oh wow, and now I think I actually understand the theorem thanks again across

funinabox
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i saw that :) thanks for the better explanation, i just was gonna post it clicked in my head. the wiki was adding everything up onebyone 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
 3 years ago
Best ResponseYou've already chosen the best response.8I noticed anoter typo. Oh well, you guys get the idea. ;P

shadowfiend
 3 years ago
Best ResponseYou've already chosen the best response.2♥ elegant proofs, elegant theorems.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0\[1 \times 17 \times 2 \times 10000000000000000000000000000000000000 \]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
 2 years ago
Best ResponseYou've already chosen the best response.0Challenge #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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