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across Group Title

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 \(2964\).

  • 2 years ago
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  1. masc1124 Group Title
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    not enough :)

    • 2 years ago
  2. across Group Title
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    Okay, let us do this in steps. Who can find me the prime-power factorization of that number?

    • 2 years ago
  3. Dumb_as_a_Duck Group Title
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    including 34 and 1? (with the respective number of zeroes after their numbers)

    • 2 years ago
  4. TuringTest Group Title
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    17*2*10^(let me count..)=-1*17*2*5^(...)*2(...)

    • 2 years ago
  5. funinabox Group Title
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    there is a really easy way to do this but i forgot, finding primes of really really large numbers

    • 2 years ago
  6. shadowfiend Group Title
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    \(17\cdot2\cdot 5^n\cdot 2^n\) where \(n\) is the number of 0s that I'm too lazy to count? :p

    • 2 years ago
  7. Ishaan94 Group Title
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    10^36 * 34 * (-10) 17*2*10^(36) * (-10)

    • 2 years ago
  8. Dumb_as_a_Duck Group Title
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    152 factors?

    • 2 years ago
  9. Dumb_as_a_Duck Group Title
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    wait, itd be greater than that.

    • 2 years ago
  10. shadowfiend Group Title
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    What am I saying, I'm a programmer... Shadowfiend@al-farabi ~> ruby -e "puts '0000000000000000000000000000000000000'.length" 37

    • 2 years ago
  11. TuringTest Group Title
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    17*2^(37)*5^(36) If I trust that Ishaan counted correctly

    • 2 years ago
  12. shadowfiend Group Title
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    ^ Ruby counts better. It's 37 0s, so should be 38 and 37, if I'm not mistaken?

    • 2 years ago
  13. Ishaan94 Group Title
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    I used python interpreter to count 17*2^38* 5^37*(-1)

    • 2 years ago
  14. TuringTest Group Title
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    (-1)*17*2^(38)*5^(37) keep forgetting the minus now what?

    • 2 years ago
  15. shadowfiend Group Title
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    Combinations of the 2s, 5s, and -1?

    • 2 years ago
  16. shadowfiend Group Title
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    Why do I feel like Pascal's triangle is about to pop out of my screen and bite me?

    • 2 years ago
  17. across Group Title
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    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: \((38+1)(37+1)(1+1)\).

    • 2 years ago
  18. TuringTest Group Title
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    oh I always forget that theorem... probably because I never use it

    • 2 years ago
  19. Dumb_as_a_Duck Group Title
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    disincluding the possibility of being dividing by itself, i got 300 divisors. and im probs mega wrong. :P

    • 2 years ago
  20. shadowfiend Group Title
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    Wow. I haven't seen that theorem in about five years haha. I had forgotten it... How you call it... Existed.

    • 2 years ago
  21. TuringTest Group Title
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    what's it called again?

    • 2 years ago
  22. Dumb_as_a_Duck Group Title
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    how does it work?

    • 2 years ago
  23. fewscrewsmissing Group Title
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    Interesting.

    • 2 years ago
  24. TuringTest Group Title
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    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

    • 2 years ago
  25. shadowfiend Group Title
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    http://en.wikipedia.org/wiki/Divisor_function

    • 2 years ago
  26. TuringTest Group Title
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    ahhhh

    • 2 years ago
  27. shadowfiend Group Title
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    Or, as across updated the question to read, the τ function.

    • 2 years ago
  28. funinabox Group Title
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    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..

    • 2 years ago
  29. Dumb_as_a_Duck Group Title
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    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.

    • 2 years ago
  30. TuringTest Group Title
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    @funinabox that's not how the theorem works...

    • 2 years ago
  31. funinabox Group Title
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    instead of multiply* dunno why i said subtract

    • 2 years ago
  32. TuringTest Group Title
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    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

    • 2 years ago
  33. TuringTest Group Title
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    ok, who else started seeing "math processing error" all of a sudden?

    • 2 years ago
  34. TuringTest Group Title
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    it was just me I guess it's gone now

    • 2 years ago
  35. across Group Title
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    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. :)

    • 2 years ago
  36. across Group Title
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    I forgot to close the parenthesis up there.

    • 2 years ago
  37. TuringTest Group Title
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    oh wow, and now I think I actually understand the theorem thanks again across

    • 2 years ago
  38. funinabox Group Title
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    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)

    • 2 years ago
  39. across Group Title
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    I noticed anoter typo. Oh well, you guys get the idea. ;P

    • 2 years ago
  40. shadowfiend Group Title
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    ♥ elegant proofs, elegant theorems.

    • 2 years ago
  41. ParthKohli Group Title
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    \[-1 \times 17 \times 2 \times 10000000000000000000000000000000000000 \]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 \((3 + 1)(1 + 1)=8\).

    • one year ago
  42. ParthKohli Group Title
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    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 :-)

    • one year ago
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