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Math Help - code breaking modular encryptions

  1. #1
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    code breaking modular encryptions

    I can not figure out this problem and it is starting to annoy me!

    So far I have got (may be wrong)
    1.

    L encrypted as J implies 12 a + b ≅ 10 mod 26

    G encrypted as C implies 7a + b 3 mod 26


    my set of notes then say to subtract the first congruence from the 2nd which gave me:

    -5 a ≅ -7 mod 26

    keep getting mixed up after here when trying to find the value of a and b

    any help would be greatly appreciated
    question is as follows:



    Last edited by darego; November 30th 2012 at 07:26 PM.
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  2. #2
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    Re: code breaking modular encryptions

    tried it again and got it wrong. for the first encrypted word "PTU" i got JVY and the last time i checked Jvy is not a word a normal person would use in a sentence
    Last edited by darego; November 30th 2012 at 07:12 PM.
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  3. #3
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    Re: code breaking modular encryptions

    Unless you have miswritten one of the equations, subtracting the second equation from the first gives you -7a= -5 (mod 26). There is no 'b' in the equation- the whole point of subtracting is to eliminate b. That is the same as 7a= 5 (mod 26). We can write that as 7a= 5+ 26n and then as 7a- 26n= 5.

    Now, 7 divides into 26 three times with remainder 5: 26- 3(7)= 7(-3)- 26(-1)= 5. So we can take a= -3. If you want it between 0 and 26, -3 is congruent to 26-3= 23 (mod 26).

    That is, 7(23)= 161= 6(26)+ 5. We can take a= 23.

    Now, go back to 3a+ b= 7 (mod 26). With a= 23, that is 3(23)= 69+ b= 17+ b= 7(mod 26) so b= 7- 17= -10= 16 (mod 26).
    Thanks from darego
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  4. #4
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    Re: code breaking modular encryptions

    i tried to make the decryption map work as y ≅ (23x + 16) mod 26: (using your a = 23 and b = 16)

    23x + 16 ≅ y mod 26
    -> 23x ≅ (y - 16) mod 26

    but when i tried to find the inverse of 23 i couldn't

    23z ≅ 1 mod 26
    23z ≅ 27 mod 26
    23z ≅ 53 mod 26
    23z ≅ 79 mod 26
    etc

    by the way, the example i was following said to subtract the first congruence from the 2nd:

    L encrypted as J implies 12 a + b ≅ 10 mod 26

    G encrypted as C implies 7a + b 3 mod 26

    =

    -5 a ≅ -7 mod 26

    thanks a lot for your reply, you are a gentleman!
    Last edited by darego; November 30th 2012 at 07:36 PM.
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