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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.
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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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh boy.. what's a nice compact way to list all of these divisors.. hmm :d

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1take 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Is this confusing? Maybe I can do a smaller number as an example:

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\rm 45=5\cdot9=5\cdot3\cdot3\) 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm sorry, I'm not sure :c

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 @Kainui @dan815 any ideas? 0_o

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3The 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Actually 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh i remember doing this on a peuler question actually

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3So if you have \[n = \prod_k p_k^{\alpha_k}\] \[\tau(n) = \prod_k (1+\alpha_k)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Well the problem stated that s,t>0. So you don't really need to worry about that.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The 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?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh u can write a nice forloop for kais chart

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for i = 1:s for j=1:t print "3"+"^" + \i * + "*5"+"^" + \j * end end

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Is 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.2im not really familiar with set notation

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but 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

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3@zepdrix No we actually do have to worry about it, see you made the same mistake as me.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3"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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh true it is (s+1)(t+1) 1 but subtract 1 because we only dont allow both to have 0 exponent

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i thought they dont like 1, since both s,t have to be more than 0

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3The divisors don't obey the same rule that s and t obey

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ya that was silly lol xD why were we cofusing the question for the actual divisors

Kainui
 one year ago
Best ResponseYou've already chosen the best response.3Yeah actually it's kinda cool that the divisor function is a specific case of this: \[\sigma_k(n) = \sum_{dn} d^k\] 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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