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Math Help - How do you determine the total amount of positive factors of a number

  1. #1
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    How do you determine the total amount of positive factors of a number

    Part of a homework assignment I'm stuck on from my discrete math class. I suspect this may belong in another forum, but I wasn't sure which one.

    Part a.) Find the prime factorization of 7056. okay.. no problem
    7056 = 2 * 3528 = 2^2 * 1764 = 2^3 * 882 = 2^4 * 441 = 2^4 * 3 * 147 = 2^4 * 3^2 * 49 = 2^4 * 3^2 * 7^2

    Part b.) How many positive factors are there of 7056?
    I'm stuck on this, I used an example of a previous question from the homework where I discovered that:
    48 = 2^4 * 3 and that it has 10 positive factors. (namely 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48) and 10 negative identical factors.

    I can't seem to derive the method to determine total factors based on the powers of each individual factor. I certainly don't want to write them all out. My best guess is 36, which i figured by adding each exponent, then adding each exponent multiplied to the other exponents, in turn, then adding all three exponents multiplied. (1+4+2+2+1)+(4*2+4*2+2*2)+(4*2*2)=10+20+16=36

    something seems fishey about my methodology though..

    Thanks for the help!
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  2. #2
    Super Member Gamma's Avatar
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    just a counting argument. any factor of that number is going to share some or all of the prime factors.

    Let n=p_1^{e_1}\cdot... \cdot p_n^{e_1}
    Then any factor of n will have 0 factors of p_1, 1 factor of p_1, ... or e_1 factors of p_1.

    Similarly for each prime.

    Thus the total number of possible combinations of factors will be
    (e_1+1)\cdot ... \cdot (e_n+1)

    So in your example you have n=2^43^1
    so it has (4+1)(1+1)=5\cdot2=10 factors
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  3. #3
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    Hello, TravisH82!

    Part (a) Find the prime factorization of 7056.

    okay.. no problem . . . 7056 \:=\: 2^4\cdot3^2\cdot7^2

    Part (b) How many positive factors are there of 7056?
    Let's construct a factor of 7056 . . .

    We have 5 choices for twos: . \begin{Bmatrix}\text{0 twos} \\ \text{1 two} \\ \text{2 twos} \\ \text{3 two's} \\ \text{4 two's} \end{Bmatrix}

    We have 3 choices for threes: . \begin{Bmatrix}\text{0 threes} \\ \text{1 three} \\ \text{2 threes} \end{Bmatrix}

    We have 3 choices for sevens: . \begin{Bmatrix}\text{0 sevens} \\ \text{1 seven} \\ \text{2 sevens} \end{Bmatrix}


    Hence, there are: . 5\cdot3\cdot3 \:=\:45 choices for a factor of 7056.

    . . Therefore, 7056 has \boxed{45} factors.

    . . . This includes 1 and 7056 itself.



    Now you can see why Gamma's formula works.


    Find the prime factorization: . 7056 \;=\;2^4\cdot3^2\cdot7^2

    Add 1 to each exponent and multiply:

    . . (4+1)(2+1)(2+1) \;=\;5\cdot3\cdot3 \;=\;45 factors . . . see?

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  4. #4
    Super Member Gamma's Avatar
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    Exactly. Wow very nice looking solution Soroban, exactly what I was going for, but I suck at TeX, lol. Thanks for taking the time to clean it up.
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  5. #5
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    Ahh.. I see where I was going wrong.. i was neglecting to consider 2^0 as a choice. Thank you all for your help.. this makes it much clearer to me!

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