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Thread: Remainder

  1. #1
    wps
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    Remainder

    Let m be a natural number. If m^{53} has remainder 31 when divided by 143, what is the remainder of m when divided by 143 ?
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    Re: Remainder

    This is an RSA encryption problem with n=143=11*13.

    $n=143=11\cdot13$. So $\phi(n)=10\cdot12=120$. Now 53 is prime to 120 and so has an inverse mod 120. With some computer help, $1=19\cdot120+(-43)\cdot53$. So the inverse of 53 mod 120 is 120-43=77. Hence $m\equiv m^{53\cdot 77}\equiv 31^{77}\equiv 70 (mod\,\, 143)$. Again, the computer helped.
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