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Math Help - inverse by mod

  1. #1
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    inverse by mod

    You have intercepted a long sequence of numbers and noticed that the
    most frequently occurring numbers were 13 and 7, in this order. You
    know that the ciphertext was obtained with an affine enciphering
    transformation with modulus 35, on single letter message blocks of
    the ordinary English alphabet A - Z and the letter b for “space”. Find
    the deciphering transformation.

    I have done so far:

    Deciphering transformation of the form: g(y) = cx + d mod 35

    Space will be the most frequently occuring, with e the second most frequently occuring.

    So b or 26 ---> 13
    e or 4 ----> 7

    26 = g(13) = 13c + d
    4 = g(7) = 7c + d

    So 6c = 22 mod 35

    But I have no idea where to go from here?!

    Could anyone please explain?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by hunkydory19 View Post
    You have intercepted a long sequence of numbers and noticed that the
    most frequently occurring numbers were 13 and 7, in this order. You
    know that the ciphertext was obtained with an affine enciphering
    transformation with modulus 35, on single letter message blocks of
    the ordinary English alphabet A - Z and the letter b for “space”. Find
    the deciphering transformation.

    I have done so far:

    Deciphering transformation of the form: g(y) = cx + d mod 35

    Space will be the most frequently occuring, with e the second most frequently occuring.

    So b or 26 ---> 13
    e or 4 ----> 7

    26 = g(13) = 13c + d
    4 = g(7) = 7c + d

    So 6c = 22 mod 35

    But I have no idea where to go from here?!

    Could anyone please explain?

    Thanks in advance!
    Multiply both sides of the equation by 6.
    Then you have 36c = 132 mod 35
    or c = 27 mod 35.
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  3. #3
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    Thank you so much for the reply icemanfan, I understand what you have done... except how do you know to multiply by 6?
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  4. #4
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    Quote Originally Posted by hunkydory19 View Post
    Thank you so much for the reply icemanfan, I understand what you have done... except how do you know to multiply by 6?
    In this example it's fairly easy to see that mod 35, 6 is its own inverse, because when you multiply 6 times 6 you get 36 which is congruent to 1. If that number had been 2, for instance, you would find the inverse of 2 mod 35 which is 18, and multiply by that. Sometimes finding an inverse isn't as easy, but it was in this case.
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  5. #5
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    Thanks icemanfan, I just have one more past exam question I'm a little stuck on...

    Explain why,

    h(x) = 7x + 5 mod 35 is not a suitable enciphering transformation.

    This is worth 6 marks, yet the only reason I can think of is because 35 is divisble by 3, and hence when rearranged to y - 5 = 7x + mod 35 it is not possible to find a number x that gives 1 mod 35...could anyone please explain what else I have missed out/need to say?

    Thanks in advance!

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  6. #6
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    Quote Originally Posted by hunkydory19 View Post
    Explain why,

    h(x) = 7x + 5 mod 35 is not a suitable enciphering transformation.

    This is worth 6 marks, yet the only reason I can think of is because 35 is divisble by 3, and hence when rearranged to y - 5 = 7x + mod 35 it is not possible to find a number x that gives 1 mod 35...could anyone please explain what else I have missed out/need to say?

    Thanks in advance!

    I think I have an explanation.... but I am not sure.
    The utility of such a cipher is that 7x+5 should map to as many different numbers as possible for different x.That is we need to have a unique 7x+5 mod 35 for each x.

    But from this thread, we know that in any collection of more than 6 integers, we will have some two whose difference will be divisible by 5. Lets call it a and b. So 5|(a-b) \Rightarrow 35|7(a-b) \Rightarrow 7a = 7b\, mod \,35 \Rightarrow h(a) = 7a + 5 = 7b + 5 = h(b)\, mod \,35.

    So this means we can have a bijection for h, for at most 5 values of x. Thus only 5 characters can be coded successfully using this cipher. Hence it is not a good cipher, by virtue of low capacity.
    Last edited by Isomorphism; May 14th 2008 at 11:35 AM.
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