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Math Help - positive integral divisors

  1. #1
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    positive integral divisors

    The question is how many positive divisors do each of the following numbers have:

    1) 6
    2) 48
    3) 100
    4) 2^{n}
    5) 10^{n}
    6) 0

    I understand how to do #1, #2, and #3. For example:

    6 = 2^1 * 3^1 = (1 + 1) * (1 + 1) = 4 positive divisors. (which happen to be 1, 2, 3, and 6)

    I used the same method (converting each number to a factor of primes in exponential notation, adding 1 to each exponent, and multiplying the exponents) to solve #2 and #3.

    I'm not sure where to start on 4, 5, or 6. Could anyone shed some light on the correct method for doing these problems?
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  2. #2
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    Quote Originally Posted by absvalue View Post
    The question is how many positive divisors do each of the following numbers have:
    4) 2^{n}
    5) 10^{n}
    6) 0
    I'm not sure where to start on 4, 5, or 6. Could anyone shed some light on the correct method for doing these problems?
    For #4, there are (n+1) factors. Why?

    For #5, there are (n+1)(n+1) factors. Why?

    For #6, how many positive integers divide zero?
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  3. #3
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    Quote Originally Posted by absvalue View Post
    The question is how many positive divisors do each of the following numbers have:


    4) 2^{n}
    I think the answer is n+1

    You should be able to count them, try n = 1,2,3,4, and 5.

    You'll see the pattern.

    Try the same for 10^{n}
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  4. #4
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    Quote Originally Posted by Plato View Post
    For #4, there are (n+1) factors. Why?

    For #5, there are (n+1)(n+1) factors. Why?

    For #6, how many positive integers divide zero?
    Thanks, I think I see the pattern. That's beautiful.

    For #4 there are n + 1 factors because:

    2^{n} = (n + 1)
    2^{3} = (3 + 1) = 4 (there are 4 positive divisors of 8)
    2^{4} = (4 + 1) = 4 (there are 5 positive divisors of 16)

    For #5 there are (n + 1)^2 factors because:

    10^{n} = (n + 1)^2
    10^2 = (2 + 1)^2 = 9 (there are 9 positive divisors of 100)
    10^3 = (3 + 1)^2 = 16 (there are 16 positive divisors of 1000)

    I understand the pattern above. But what if the question was a number other than 2^{n} or 10^{n}? Is there a way to find the number of positive divisors for, say, 8^{n}?

    For #6: Zero divided by any number (except zero, because that is undefined) should be zero. So the answer would be the set of all numbers not equal to zero?
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  5. #5
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    Here is what you need to know.
    The number 2^43^25^67 has (5)(3)(7)(2) divisors. Do you see why?

    The number 2^a3^b5^c7^d has (a+1)(b+1)(c+1)(d+1) divisors. Do you see why?
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  6. #6
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    Quote Originally Posted by Plato View Post
    Here is what you need to know.
    The number 2^43^25^67 has (5)(3)(7)(2) divisors. Do you see why?

    The number 2^a3^b5^c7^d has (a+1)(b+1)(c+1)(d+1) divisors. Do you see why?
    Yes, that makes sense. There are (5)(3)(7)(2) positive divisors because 2^4 has (4 + 1) divisors, 3^2 has (2 + 1) divisors, 5^6 has (6 + 1) divisors, and 7^1 has (1 + 1) divisors.

    So, does this only work with prime numbers? Is it not possible to do this with 16^{n} or 8^{n} or 20^{n}?

    Because 8^{n} does not have n + 1 factors... For example:

    8^2 = 64 = 2^6 = 6 + 1 = 7 positive divisors.
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  7. #7
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    Quote Originally Posted by absvalue View Post
    So, does this only work with prime numbers? Is it not possible to do this with 16^{n} or 8^{n} or 20^{n}?
    But every positive integer has a prime factorization.
    Thus 20^{n}=2^{2n}\cdot 5^n. So the number has (2n+1)(n+1) factors.
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  8. #8
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    Thanks so much for your explanation. I think I understand it now.
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