1. ## 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?!

2. Originally Posted by hunkydory19
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?!

Multiply both sides of the equation by 6.
Then you have 36c = 132 mod 35
or c = 27 mod 35.

3. Thank you so much for the reply icemanfan, I understand what you have done... except how do you know to multiply by 6?

4. Originally Posted by hunkydory19
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.

5. 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?

6. Originally Posted by hunkydory19
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?

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 $\displaystyle 5|(a-b) \Rightarrow 35|7(a-b) \Rightarrow 7a = 7b\, mod \,35 \Rightarrow h(a) = 7a + 5 = 7b + 5 = h(b)\, mod \,35$.