Science:Math Exam Resources/Courses/MATH312/December 2010/Question 06
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Question 06 

Recall that an affine cryptosystem with encryption key and decryption key works as follows. Each letter is assigned a numerical value, in alphabetical order:
These numbers are then viewed modulo 26: They are encrypted by the formula
and decrypted by the formula
It is known that the most frequently occurring letter in the English language is E (with numerical value 4) and the second most frequently occurring letter is T (with numerical value 19). Suppose these letters are encoded as F (numerical value 5) and G (numerical value 6), respectively. (a) Find the encryption key (b) Find the decryption key 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

For part (a), we seek to solve the system of congruences given by
and for part (b), we seek to solve

Hint 2 

After isolating for a variable, this problem becomes an issue of finding inverses modulo 26 which can be done using the Euclidean Algorithm. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. For part (a), we seek to solve the system of congruences given by
Subtracting the two equations yields
Thus we need to find the inverse of 15 modulo 26. To do this we use the Euclidean Algorithm.
and back substituting gives
and so an inverse of 15 modulo 26 is given by 7. Thus as , we have that . Plugging back into the original equation gives and isolating for b gives the value of 3 and so the encryption key is For part (b), we seek to solve
Subtracting these two equations gives . Thus as (so d is 7), we have that the decryption key is given by . 