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Math Help - Number of odd divisors

  1. #1
    Junior Member NoFace's Avatar
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    Number of odd divisors

    Compute the number of odd divisors of 112,000

    My book did an example similar to this, but it was very vague. I didn't really understand where it came up with the numbers.

    I think this one is giving me a hard time because I am using the trial and error algorithm--if there is such a thing--for getting a prime factorization. Does anyone have any tricks for getting a prime factorization?

    Thanks.
    Last edited by NoFace; September 17th 2008 at 07:04 AM. Reason: Spelling
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  2. #2
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    Well I don't know any short cut method to find prime factorization, but I guess you can look at the number in hand an at least figure out some of the prime factors(this is offcourse not possible always)
    about the number 112,000
    has got three tens.....
    that is let 112,000 = n * 1000
    now in 1000 how many odd prime factors you get?
    3 odd prime factors, namely 5.... since 10 = 5 * 2.....

    now in the residue, 112, this is obviously divisible by 2, well if you check for divisibility test of 3, it fails, obviously this is not divisible by 5(since 5/0 is not the last digit), try 7.... the next prime..... 7 * 16 = 112.....
    now 16 is all having even prime factor namely 2....

    therefore, the number of odd prime factor = three 5s + one 7 = 4 odd prime factor which is 1 less than number of total prime factors....
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  3. #3
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    Quote Originally Posted by natarajchakraborty View Post
    the number of odd prime factor = three 5s + one 7 = 4 odd prime factor which is 1 less than number of total prime factors....
    Just to modify that answer slightly: you have found the prime factorisation of 112000 as 2^7\times5^3\times7. So you can get odd divisors of the number by taking any product of the 5s and the 7. There will be seven such divisors, namely

    5,\ 5^2,\ 5^3,\ 7,\ 7\times5,\ 7\times5^2,\ 7\times5^3.
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  4. #4
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    Opalg... Ahha you are right... I misread the questions as number of odd prime factors, and so I said that its 1 less than the total! my mistake!
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  5. #5
    Junior Member NoFace's Avatar
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    Quote Originally Posted by Opalg View Post
    Just to modify that answer slightly: you have found the prime factorisation of 112000 as 2^7\times5^3\times7. So you can get odd divisors of the number by taking any product of the 5s and the 7. There will be seven such divisors, namely

    5,\ 5^2,\ 5^3,\ 7,\ 7\times5,\ 7\times5^2,\ 7\times5^3.
    Great. So I'm assuming it's the same procedure for even divisors?

    How would you works that for distinct divisors? Just use some sort of subtraction?

    EDIT: Never mind, I figured it out.
    Last edited by NoFace; September 17th 2008 at 08:07 AM. Reason: Dumb question
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