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Thread: divisibily

  1. #1
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    divisibily

    how many posotive integer divide 2^7 x 3^8 such that they are not prime or 1,
    thanks for any help
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  2. #2
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    Quote Originally Posted by hmmmm View Post
    how many posotive integer divide 2^7 x 3^8 such that they are not prime or 1,
    thanks for any help
    How many numbers are there of the form $\displaystyle 2^{\alpha} \times 3^{\beta}$; $\displaystyle \alpha, \beta \in \mathbb{N}$

    where $\displaystyle \alpha+\beta >1$, $\displaystyle 0\le \alpha\le 7$, $\displaystyle 0\le \beta \le 8$ ?

    CB
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  3. #3
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    hmmm really should have spotted that, 72 - 2 primes and 1 so 69, thanks alot for the help
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  4. #4
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    Hello, hmmmm!

    You got it!


    How many positive integer divide $\displaystyle 2^7\cdot3^8$ such that they are not prime or 1?

    There are 8 choices for the 2's: .$\displaystyle 2^0,\:2^1,\;2^2,\;\hdots \:2^7$

    There are 9 choices for the 3's: .$\displaystyle 3^0,\:3^1,\:3^2,\hdots\:3^8$

    Hence, there are: .$\displaystyle 8\times 9 \,=\,72$ choices for numbers of the form $\displaystyle 2^a3^b$


    But this includes: .$\displaystyle 2^03^0 \,=\,1,\;2^13^0 = 2,\;2^03^1 = 3$
    . .
    The rest are all composite.


    Therefore, the answer is: .$\displaystyle 72 - 3 \;=\;69$

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