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anonymous

  • one year ago

List all the positive integer divisors of 3^s5^t, where s,t are integers and s,t >0. If r,s,t are positive integers, how many positive divisors does 2^r3^s5^t have? Please explain the answer, I am not sure how to get the result.

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  1. zepdrix
    • one year ago
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    Oh boy.. what's a nice compact way to list all of these divisors.. hmm :d

  2. zepdrix
    • one year ago
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    take for example the number: \(\large\rm 16875=3^3\cdot5^4\) This number is divisible by: \(\large\rm 3^1,3^2,3^3\) \(\large\rm 5^1,5^2,5^3,5^4\) \(\large\rm 3^1\cdot5^1,~3^1\cdot5^2,~3^1\cdot5^3,~3^1\cdot5^4\) \(\large\rm 3^2\cdot5^1,~3^2\cdot5^2,~3^2\cdot5^3,~3^2\cdot5^4\) \(\large\rm 3^3\cdot5^1,~3^3\cdot5^2,~3^3\cdot5^3\) You can see it gets kind of crazy ^ We need a nice way to write this.. hmm

  3. zepdrix
    • one year ago
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    Is this confusing? Maybe I can do a smaller number as an example:

  4. zepdrix
    • one year ago
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    \(\large\rm 45=5\cdot9=5\cdot3\cdot3\) So this number is divisible by 3, 5, 3*3, 3*5, and 3*3*5. So the divisors are 3,5,9,15, and 45.

  5. zepdrix
    • one year ago
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    Hmm sorry, I'm not sure :c

  6. zepdrix
    • one year ago
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    @ganeshie8 @Kainui @dan815 any ideas? 0_o

  7. dan815
    • one year ago
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    |dw:1444634846449:dw|

  8. Kainui
    • one year ago
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    The amount of ways to combine all the divisors together is quite nice, for instance, s and t could be 3 and 4 then you have: |dw:1444634910261:dw|

  9. dan815
    • one year ago
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    o thats nice

  10. Kainui
    • one year ago
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    Actually I should have included \(3^0\) and \(5^0\) so it should be (3+1)*(5+1). This actually all stems from the fact that the divisor counting function is multiplicative, but there's no reason to get into that right now I think.

  11. dan815
    • one year ago
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    oh i remember doing this on a peuler question actually

  12. Kainui
    • one year ago
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    So if you have \[n = \prod_k p_k^{\alpha_k}\] \[\tau(n) = \prod_k (1+\alpha_k)\]

  13. dan815
    • one year ago
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    |dw:1444635166020:dw|

  14. zepdrix
    • one year ago
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    Well the problem stated that s,t>0. So you don't really need to worry about that.

  15. dan815
    • one year ago
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    then just s*t

  16. zepdrix
    • one year ago
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    The first part of the question actually wanted you to `list them out`. I wasn't sure of an economical way to do that XD Maybe like the chart thing that Kai was doing?

  17. dan815
    • one year ago
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    oh u can write a nice forloop for kais chart

  18. dan815
    • one year ago
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    for i = 1:s for j=1:t print "3"+"^" + \i * + "*5"+"^" + \j * end end

  19. zepdrix
    • one year ago
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    Is there a math symbol for `for loop`? >.< Like.. Hey .. thing... I want to do this set of operations.. from this lower bound to this upper bound... I don't want you to add the results or multiply them... just list them -_- pls

  20. dan815
    • one year ago
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    |dw:1444635697073:dw|

  21. dan815
    • one year ago
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    im not really familiar with set notation

  22. dan815
    • one year ago
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    but something like this all the products for all the possible i , js, you will have a product for every different combinations of i and j

  23. zepdrix
    • one year ago
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    interesting :o

  24. Kainui
    • one year ago
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    @zepdrix No we actually do have to worry about it, see you made the same mistake as me.

  25. Kainui
    • one year ago
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    "positive integer divisors of 3^s5^t, where s,t are integers and s,t >0" 1, 3, and 5 are positive integer divisors of 15.

  26. dan815
    • one year ago
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    oh true it is (s+1)(t+1) -1 but subtract 1 because we only dont allow both to have 0 exponent

  27. Kainui
    • one year ago
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    No wrong dan

  28. dan815
    • one year ago
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    i thought they dont like 1, since both s,t have to be more than 0

  29. Kainui
    • one year ago
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    The divisors don't obey the same rule that s and t obey

  30. dan815
    • one year ago
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    ohh

  31. dan815
    • one year ago
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    i gotcha now

  32. dan815
    • one year ago
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    ya that was silly lol xD why were we cofusing the question for the actual divisors

  33. Kainui
    • one year ago
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    Yeah actually it's kinda cool that the divisor function is a specific case of this: \[\sigma_k(n) = \sum_{d|n} d^k\] with the divisor counting function when k=0. You can see that you will get a geometric series so you have to use L'Hopital's rule to derive it from the general case if you're curious.

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