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Math Help - Divisibility

  1. #1
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    Divisibility

    For a = 238000 = 2^4 * 5^3 * 7 * 17 and b = 299880 = 2^3 * 3^2 * 5 * 7^2 * 17

    (a) Is there an integer k so that a divides b^k? If so, what is the smallest possibility for k? If not, why not.

    (b) Is there an integer k so that b divides a^k? If so, what is the smallest possibility for k? If not, why not.
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  2. #2
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    Hello, kiddopop!

    \begin{array}{ccccc}a &=& 238,\!000 &=& 2^4\cdot5^3\cdot7\cdot17 \\<br />
b &=&  299,\!880 &=&  2^3\cdot3^2\cdot5\cdot7^2\cdot17 \end{array}

    \text{(a) Is there an integer }k\text{ so that }a\text{ divides }b^k\,?
    . . . . \text{If so, what is the smallest possibility for }k\,?\:\text{ If not, why not?}

    Can this be an integer? . \dfrac{b^k}{a} \;=\;\dfrac{(2^3\cdot3^2\cdot5\cdot7^2\cdot17)^k}{  2^4\cdot5^3\cdot7\cdot17}

    We have: . \displaystyle \frac{2^{3k}\cdot3^{2k}\cdot5^k\cdot7^{2k}\cdot17^  i}{2^4\cdot5^3\cdot7\cdot17} \;=\; 2^{3k-4}\cdot3^{2k}\cdot5^{k-3}\cdot7^{2k-1}\cdot17^{k-1}


    The answer is yes, for k \ge 3.




    \text{(b) Is there an integer }k\text{ so that }b\text{ divides }a^k\,?
    . . . . \text{ If so, what is the smallest possibility for }k\,?\:\text{ If not, why not?}

    Can this be an integer? . \dfrac{a^k}{b} \;=\;\dfrac{(2^4\cdot5^3\cdot7\cdot17)^k}{2^3\cdot  3^2\cdot5\cdot7^2\cdot17}

    We have: . \displaystyle \frac{2^{4k}\cdot5^{3k}\cdot7^k\cdot17^k}{2^3\cdot  3^2\cdot5\cdot7^2\cdot17} \;=\;\dfrac{2^{4k-3}\cdot5^{3k-1}\cdot7^{k-2}\cdot 17^{k-1}}{3^2}


    The answer is no.
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