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Math Help - number of divisors

  1. #1
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    number of divisors

    let p_{i} be a prime number and a_{i} be a positive integer for all i

    a) list all distinct divisors of 32

    1,2,4,8,16,32

    2^5 5+1 = 6 divisors

    b) how many distinct divisors does p^a have?

    a+1 divisors

    c) how many distinct divisors does p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot\cdot\cdot p_{m}^{a_{m}} have?

    (a_{1}+1)(a_{2}+1)\cdot\cdot\cdot(a_{m}+1) divisors

    d) how many distinct divisors does 60 have?

    using prime factorization 60 = 2^2\cdot3\cdot5

    so it has (2+1)(1+1)(1+1) = 12

    here is the one I don't know how to solve

    e) What is the smallest positive integer with exactly 100 distinct divisors?
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  2. #2
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    Re: number of divisors

    isn't it going to be the product of the first 100 primes?
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  3. #3
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    Re: number of divisors

    Quote Originally Posted by romsek View Post
    isn't it going to be the product of the first 100 primes?

    that seems to makes sense.

    I was thinking I want 100 = 2\cdot 50 divisors

    2 \cdot 50=2\cdot2\cdot5

    so the number would be something like p^1q^1r^4 but that doesn't really tell me what the number is.

    I'm not really sure how to show it would be the product of the first 100 primes
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    Re: number of divisors

    Quote Originally Posted by Jonroberts74 View Post
    that seems to makes sense.

    I was thinking I want 100 = 2\cdot 50 divisors

    2 \cdot 50=2\cdot2\cdot5

    so the number would be something like p^1q^1r^4 but that doesn't really tell me what the number is.

    I'm not really sure how to show it would be the product of the first 100 primes
    ok I read a similar problem. It's not the product of the 1st 100 primes.

    you have

    $n = \displaystyle{\prod_{k=1}^N}p_k^{a_k}$

    You want to find $N$ and the $a_k$ such that

    $\displaystyle{\prod_{k=1}^N}a_k=100$

    and

    $n$ is minimized.

    I'll have to think about how to solve this one.
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  5. #5
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    Re: number of divisors

    Quote Originally Posted by romsek View Post
    ok I read a similar problem. It's not the product of the 1st 100 primes.

    you have

    $n = \displaystyle{\prod_{k=1}^N}p_k^{a_k}$

    You want to find $N$ and the $a_k$ such that

    $\displaystyle{\prod_{k=1}^N}a_k=100$

    and

    $n$ is minimized.

    I'll have to think about how to solve this one.
    yeah I did some experimenting. I did the product of the first 7 primes and got 128 distinct divisors
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  6. #6
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    Re: number of divisors

    I was wrong

    I said 2*2*5 = 50,

    I meant 2*5*5= 50

    so 2*2*5*5 equals 100

    so the number should be

    p^1q^1r^4s^4

    so the first 4 primes

    2^4 \times 3^4 \times 5^1 \times 7^1 = 45360
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