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Math Help - How many factors does a million have?

  1. #1
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    How many factors does a million have?

    How many factors does a million have?

    1 000 000
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  2. #2
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    Quote Originally Posted by Natasha1
    How many factors does a million have?

    1 000 000
    In number theorem there is a number-theoretic function,
    \tau (n). It is defined as the number of distinct divisors of n. For example,
    \tau (3)=2 cuz 3 only got 2 factors 1 and 3.


    There is a formula, if
    n=p_1^{a_1}p_2^{a_2}....p_k^{a_k}
    is the prime-factorization of n
    Then,
    \tau (n)=(a_1+1)(a_2+1)...(a_k+1)

    Thus, since,
    1,000,000=10^6=(2\cdot 5)^6=2^6\cdot 5^6
    Thus, (6+1)(6+1)=7\cdot 7=49
    Last edited by ThePerfectHacker; May 8th 2006 at 02:55 PM.
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  3. #3
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    ns formula!

    what is the proof?
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  4. #4
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    Quote Originally Posted by srulikbd
    what is the proof?
    Let

    <br />
n=p_1^{a_1}p_2^{a_2}....p_k^{a_k}<br />

    Then any factor r of n is of the form:

    <br />
r=p_1^{b_1}p_2^{b_2}....p_k^{b_k}<br />

    where b_i \in \{0,\ 1, \ ..,\ a_i\}, and each number of this
    form is a factor of n.

    So how many r's are there? Well the first exponent can take
    any one of a_1+1 values the second can ....

    Hence the number of factors is:

    <br />
\prod_{i=1}^n (a_i+1)<br />

    RonL
    Last edited by CaptainBlack; May 23rd 2006 at 11:49 AM.
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  5. #5
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    tnx a lot

    easy indeed
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  6. #6
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    Quote Originally Posted by srulikbd
    what is the proof?
    One more.

    This one is more advanced but more elegant

    Theorem:
    If f is a number theoretic function which is multiplicative. Then, the function F definied as, F(n)=\sum_{d|n}f(d) is also.
    ---------
    Note that,
    \tau (n)=\sum_{d|n}1
    Since, f(n)=1 is multiplicative since,
    f(ab)=f(a)f(b) whenever \gcd(a,b)=1 we have that,
    \tau(n) is mulitplicative.

    Since,
    \tau(p^n)=n+1
    You have,
    \tau(p_1^{k_1}\cdot ....\cdot p_s^{k_s})=(k_1+1)\cdot ...\cdot (k_s+1)
    because or multiplicativity.
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