How Many Positive Divisors for 2^n and 30? | Discrete Math Question

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SUMMARY

The discussion focuses on determining the number of positive divisors for the expressions 2n and 30. For 30, the correct positive divisors are 1, 2, 3, 5, 6, 10, and 15, while the prime factors are 2, 3, and 5. For 2n, it is established that it has n-1 positive divisors, as it only has one prime divisor, which is 2. The participants emphasize the importance of understanding the divisor count and the role of prime factorization in solving such problems.

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cragar
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Homework Statement


How many positive divisors does each of the following have?

2^n where n is a positive integer.
and 30

The Attempt at a Solution


for 30 i get 2 , 5 , 3 , 10
but my book says 2 ,3 ,5 I don't understand why 10 isn't a divisor.
and for 2^n I am trying to look for a pattern if n=1 i get no divisors
and n=2 i get 1 divisor and n=3 i get 2 divisors so would it be
2^n has n-1 divisors?
 
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hi cragar! :smile:

(try using the X2 tag just above the Reply box, and write "itex" rather than "tex", and it won't keep starting a new line :wink:)
cragar said:
How many positive divisors does each of the following have?

for 30 i get 2 , 5 , 3 , 10
but my book says 2 ,3 ,5

i suspect that that's just a hint, and they're telling you those are the prime divisors, and leaving you to carry on from there

(btw, you've missed out two more)
and for 2^n I am trying to look for a pattern if n=1 i get no divisors
and n=2 i get 1 divisor and n=3 i get 2 divisors so would it be
2^n has n-1 divisors?

yes :smile:

(though you should be able to prove it more rigorously than that! :wink:)
 
ok thanks for your post. so would all the divisors of 30 be 1 , 2 ,5,6,10 ,15 . is one a divisor. for 2^n to have divisors it has to be a multiple of 2 so would I divide it by 2 and then i would get 2^{n-1}
then could i say it has n-1 divisors
 
hi cragar! :wink:
cragar said:
would all the divisors of 30 be 1 , 2 ,5,6,10 ,15 .

yes :smile: (except i don't know whether 1 counts as a divisor :redface:)
for 2^n to have divisors it has to be a multiple of 2 so would I divide it by 2 and then i would get 2^{n-1}
then could i say it has n-1 divisors

better would be …

2n has only one prime divisor, 2 …

so its only divisors are 2k for 0 < k < n, of which there are n - 1 :wink:

(and now try a similar proof for 30 :biggrin:)
 

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