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Math Help - Is this easily solvable without a computer?

  1. #1
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    Is this easily solvable without a computer?

    Find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power.

    Is there a way to do this without using a computer? If so, can anyone give me a hint. It's killing me.
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Find an integer n such that n/2 is a square, n/3 is a cube, and n/5 is a fifth power.

    Is there a way to do this without using a computer? If so, can anyone give me a hint. It's killing me.


    Write n=p_1^{a_1}\cdot\ldots\cdot p_k^{a_k}\,,\,\,p_i\,\,prime\,,\,\,0<a_i\in\mathbb  {Z} , then n is

    an m-th power iff \forall i=1,2,...,k\,,\,\,a_i=0\!\!\pmod m .

    Now solve your problem...

    Tonio
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  3. #3
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    Could you elaborate more? Based on what you're saying, I need some product of primes where each prime is raised to the power 30? (Since 30 mod 2 = 30 mod 3 = 30 mod 2 = 0)

    I'm still not seeing how to solve for all 3 of the conditions at one time...
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  4. #4
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    Hello, paupsers!

    \text{Find an integer }n\text{ so that }\dfrac{n}{2}\text{ is a square, }\dfrac{n}{3}\text{ is a cube, and }\dfrac{n}{5}\text{ is a fifth power.}

    Since \,n is divided by 2, 3, and 5,
    . . we assume that \,n has factors of 2, 3, and 5 (at least).

    Let n \;=\;2^a\cdot 3^b\cdot5^c . for some positive integers a,b,c.


    \dfrac{n}{2} is a square.
    We have: . \dfrac{n}{2} \;=\;2^{a-1}\cdot 3^b\cdot 5^c

    . . where: . a\!-\!1,\;b,\;c are all even. .[1]


    \dfrac{n}{3} is a cube.
    We have: . \dfrac{n}{3} \;=\;2^a\cdot3^{b-1}\cdot 5^c

    . . where: . a,\;b\!-\!1,\;c are all mutiples of 3. .[2]


    \dfrac{n}{5} is a fifth power.
    We have: . \dfrac{n}{5} \;=\;2^a\cdot 3^b \cdot 5^{c-1}

    . . where: . a,\;b,\,c\!-\!1 are all multiples of 5. .[3]



    Combining [1], [2] and [3], we have:

    \begin{Bmatrix}a\1-\!1\text{ is even} \\ a\text{ is a multiple of 3} \\ a\text{ is a multiple of 5} \end{Bmatrix} . This is satisfied by: . a\,=\,15

    \begin{Bmatrix}b\text{ is even} \\ b\!-\!1\text{ is a multiple of 3} \\ b\text{ is a multiple of 5}\end{Bmatrix} .This is satisfied by: . b \,=\,10

    \begin{Bmatrix}c\text{ is even} \\ c\text{ is a multiple of 3} \\ c\!-\!1\text{ is a multiple of 5}\end{Bmatrix} . This is satisfied by: . c \,=\,6


    Therefore, the least positive integer \,n is:

    . . . . n \;=\;2^{15}\!\cdot3^{10}\!\cdot5^6

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