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Math Help - More congruence trouble

  1. #1
    Junior Member Nerdfighter's Avatar
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    More congruence trouble

    for which integers c with 0 \leq c < 1001 does the congruence 154x \equiv c (\bmod 1001) have solutions? For each case when there are solutions, find how many solutions are not congruent modulo 1001.

    Originally, I thought this problem might have something to do with multiplicative inverses, but I didn't get anywhere on that front.

    If it matters, I calculated \gcd (154,1001) to be 77.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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     ax \equiv b \mod n has a solution iff  (a,n) \mid b . So  154x \equiv c \mod 1001 is solvable iff  (154,1001)=77 \mid c .
    Therefore  c = 77n, \; n \in \mathbb{N} .
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  3. #3
    Moo
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    Quote Originally Posted by chiph588@ View Post
     ax \equiv b \mod n has a solution iff  (a,n) \mid b . So  154x \equiv c \mod 1001 is solvable iff  (154,1001)=77 \mid c .
    Therefore  c = 77n, \; n \in \mathbb{N} .
    Since we're working with modulos, I would say that it's 77n mod 1001.

    And since 1001/77=13, there are 13 different values for c...
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