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Math Help - find its moduluo 19

  1. #1
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    find its moduluo 19

    17! . 3 ^16 = a (mod 19)

    what is a
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  2. #2
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    Hello,

    Quote Originally Posted by szpengchao View Post
    17! . 3 ^16 = a (mod 19)

    what is a
    I'd do it straightforward, because I don't know any property letting us simplify this...

    17!=2.3.4....16.17

    (mod 19)
    17=-2
    16=-3
    15=-4
    14=-5
    .
    .
    .

    --> 17! \equiv 2^2 3^2 4^2 5^2 6^2 7^2 8^2 9^2 \mod 19


    ----> \boxed{17! \equiv ({\color{blue}2}\cdot{\color{red}3}\cdot4\cdot5\cd  ot{\color{red}6}\cdot7\cdot8\cdot{\color{blue}9})^  2  \mod 19}

    3x6= -1 mod 19

    2x9= -1 mod 19

    ----> 17! \equiv ({\color{red}4}\cdot{\color{red}5}\cdot{\color{blu  e}7}\cdot{\color{blue}8})^2  \mod 19

    4x5=19+1= 1 mod 19

    7x8=56=57-1=3x19-1= -1 mod 19

    ----> 17! \equiv (-1)^2 \mod 19

    \boxed{17! \cdot 3^{16} \equiv 3^{16} \mod 19}


    Now, calculate a in 3^{16} \equiv a \mod 19
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  3. #3
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    i do this a simple way

    by wilson theorem
    18! = -1 mod 19
    by euler fermat
    3^18=1 mod 19

    times them together

    18! * 3^18= -1 mod 19

    by know, we have
    3^16*17! =a mod 19

    times both sides by 9*18, we have 162 a = 18 mod 19
    divides both sides by 18, 9a=1 mod 19
    9 has inverse 17, we then solve this get
    a= 17 mod 19

    is it right?
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  4. #4
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    Quote Originally Posted by szpengchao View Post
    by wilson theorem
    18! = -1 mod 19
    by euler fermat
    3^18=1 mod 19

    times them together

    18! * 3^18= -1 mod 19

    by know, we have
    3^16*17! =a mod 19

    times both sides by 9*18, we have 162 a = 18 mod 19
    divides both sides by 18, 9a=1 mod 19 << SiMoon says : yep, it's possible because 18 is prime with 19
    9 has inverse 17, we then solve this get
    a= 17 mod 19

    is it right?
    I find the same result with my (veeeeery long method) ^^

    I didn't know about Wilson's theorem, thanks
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  5. #5
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    another problem

    i try to put it into my calculator. but it doesnt work...


    i hav another problem about solving simutalneous congruences

    if we hav two congruences:

    x= 1 mod 2
    x= 3 mod 4

    how to solve that?
    i mean the general case if (mod n) and (mod m) and (m,n)>1 ???
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  6. #6
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    Quote Originally Posted by szpengchao View Post
    i try to put it into my calculator. but it doesnt work...
    My calculator gives this result too

    i hav another problem about solving simutalneous congruences

    if we hav two congruences:

    x= 1 mod 2
    x= 3 mod 4

    how to solve that?
    i mean the general case if (mod n) and (mod m) and (m,n)>1 ???
    Well, here I know that if x=3 mod 4, then x=1 mod 2.
    So the sets of x such that x= 3 mod 4 is included in the sets of x such that x= 1 mod 2.
    Hence, the solutions are every x such that x=3 mod 4.


    More generally, there is the CNT if m and n are coprime. When they're not, I'd do it according to the situation...
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  7. #7
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    cheers

    cheers mate!
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