Science:Math Exam Resources/Courses/MATH312/December 2009/Question 04 (a)

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MATH312 December 2009

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Question 04 (a)
Decrypt OAPB, which was encrypted by the affine transformation $\displaystyle C\equiv 7P+11\mod {26}$

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
It might help to refer to the following table A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Note also that you are given the values of C and ask to compute P.

Hint 2
If needed, you can compute inverses modulo 26 via 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. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

First, we isolate for P in the given formula. To do this we need to compute the inverse of 7 modulo 26. We use the Euclidean algorithm to do this. $\displaystyle {\begin{aligned}26&=7(3)+5\\7&=5(1)+2\\5&=2(2)+1\end{aligned}}$ and back substituting $\displaystyle {\begin{aligned}1&=5-2(2)\\1&=5-(7-5(1))(2)\\1&=5(3)-7(2)\\1&=(26-7(3))(3)-7(2)\\1&=26(3)-7(11)\end{aligned}}$ and so the inverse of 7 modulo 26 is (-11) which is equal to 15 modulo 26. Hence $\displaystyle P=15(7P)=15(C-11)\mod {26}$ We plug in the four values of C given by $\displaystyle OAPB=(14)(0)(15)(1)$ and see that $\displaystyle {\begin{aligned}P_{1}&\equiv 15(14-11)\equiv 15(3)\equiv 45\equiv 19\equiv T\mod {26}\\P_{2}&\equiv 15(0-11)\equiv 15(-11)\equiv -165\equiv 17\equiv R\mod {26}\\P_{3}&\equiv 15(15-11)\equiv 15(4)\equiv 60\equiv 8\equiv I\mod {26}\\P_{4}&\equiv 15(1-11)\equiv 15(-10)\equiv -150\equiv 6\equiv G\mod {26}\end{aligned}}$ Thus, the plaintext was the word TRIG.

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