## 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

Oh boy.. what's a nice compact way to list all of these divisors.. hmm :d

2. zepdrix

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

Is this confusing? Maybe I can do a smaller number as an example:

4. zepdrix

$$\large\rm 45=5\cdot9=5\cdot3\cdot3$$ 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

Hmm sorry, I'm not sure :c

6. zepdrix

@ganeshie8 @Kainui @dan815 any ideas? 0_o

7. dan815

|dw:1444634846449:dw|

8. Kainui

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

o thats nice

10. Kainui

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

oh i remember doing this on a peuler question actually

12. Kainui

So if you have $n = \prod_k p_k^{\alpha_k}$ $\tau(n) = \prod_k (1+\alpha_k)$

13. dan815

|dw:1444635166020:dw|

14. zepdrix

Well the problem stated that s,t>0. So you don't really need to worry about that.

15. dan815

then just s*t

16. zepdrix

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

oh u can write a nice forloop for kais chart

18. dan815

for i = 1:s for j=1:t print "3"+"^" + \i * + "*5"+"^" + \j * end end

19. zepdrix

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

|dw:1444635697073:dw|

21. dan815

im not really familiar with set notation

22. dan815

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

interesting :o

24. Kainui

@zepdrix No we actually do have to worry about it, see you made the same mistake as me.

25. Kainui

"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

oh true it is (s+1)(t+1) -1 but subtract 1 because we only dont allow both to have 0 exponent

27. Kainui

No wrong dan

28. dan815

i thought they dont like 1, since both s,t have to be more than 0

29. Kainui

The divisors don't obey the same rule that s and t obey

30. dan815

ohh

31. dan815

i gotcha now

32. dan815

ya that was silly lol xD why were we cofusing the question for the actual divisors

33. Kainui

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.