Math Problem Statement

Solve

Encrypt and decrypt using shift cipher, and perform modular arithmetic operations.

Solution

Let's go through each of the questions one by one:

1. Encrypt and decrypt using the shift cipher algorithm.

a) Encrypt the plaintext "COMPUTER SCIENCE" using the key “3”:

The shift cipher works by shifting each letter in the alphabet by a fixed number of positions. Here, we need to shift each letter by 3 positions.

  • "C" → "F"
  • "O" → "R"
  • "M" → "P"
  • "P" → "S"
  • "U" → "X"
  • "T" → "W"
  • "E" → "H"
  • "R" → "U"

For the second word:

  • "S" → "V"
  • "C" → "F"
  • "I" → "L"
  • "E" → "H"
  • "N" → "Q"
  • "C" → "F"
  • "E" → "H"

So, the encrypted text is: "FRPSXWHU VFLHQFH".

b) Decrypt the cipher text "YLAYPLCLMPSLZ" using the key “7”:

For decryption, we shift each letter backward by 7 positions.

  • "Y" → "R"
  • "L" → "E"
  • "A" → "T"
  • "Y" → "R"
  • "P" → "I"
  • "L" → "E"
  • "C" → "V"
  • "L" → "E"
  • "M" → "F"
  • "P" → "I"
  • "S" → "L"
  • "L" → "E"
  • "Z" → "S"

So, the decrypted text is: "RETRIEVES FILES".


2. Modular Arithmetic

We use the formula a=mk+ra = m \cdot k + r, where rr is the remainder (which is the result of amodma \mod m).

a) What is 236mod9-236 \mod 9?

Perform division: 236÷9=26-236 \div 9 = -26 remainder 2. So, 236mod9=2-236 \mod 9 = 2.

b) What is 1025mod10-1025 \mod 10?

Perform division: 1025÷10=103-1025 \div 10 = -103 remainder 5. So, 1025mod10=5-1025 \mod 10 = 5.

c) What is 553mod3-553 \mod 3?

Perform division: 553÷3=185-553 \div 3 = -185 remainder 2. So, 553mod3=2-553 \mod 3 = 2.

d) What is 27mod127 \mod 1?

Any number modulo 1 is 0, so 27mod1=027 \mod 1 = 0.


3. State the residue class of modulo 9 and state 6 integers in each class: 3 negative and 3 positive integers.

The residue classes modulo 9 are the set of possible remainders when dividing by 9. These are: {0,1,2,3,4,5,6,7,8}\{0, 1, 2, 3, 4, 5, 6, 7, 8\}.

Class for 0:

Negative: 9,18,27-9, -18, -27 Positive: 0,9,180, 9, 18

Class for 1:

Negative: 8,17,26-8, -17, -26 Positive: 1,10,191, 10, 19

Class for 2:

Negative: 7,16,25-7, -16, -25 Positive: 2,11,202, 11, 20

Class for 3:

Negative: 6,15,24-6, -15, -24 Positive: 3,12,213, 12, 21

Class for 4:

Negative: 5,14,23-5, -14, -23 Positive: 4,13,224, 13, 22

Class for 5:

Negative: 4,13,22-4, -13, -22 Positive: 5,14,235, 14, 23

Do you need further details or explanations?


Here are 5 related questions to extend your knowledge:

  1. How would you modify the shift cipher if the key was larger than 26?
  2. Can the shift cipher be applied to non-alphabetic characters like numbers?
  3. How can we efficiently compute modular arithmetic for very large numbers?
  4. How is the shift cipher related to modern encryption methods like RSA?
  5. How would modular arithmetic apply to cryptography beyond basic ciphers?

Tip: Always ensure when calculating amodma \mod m for negative numbers, you find the remainder as a non-negative integer between 0 and m1m-1.

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Math Problem Analysis

Mathematical Concepts

Shift cipher
Modular arithmetic
Residue classes

Formulas

a = m * k + r (modular arithmetic formula)
Shift cipher encryption: letter shift

Theorems

Division Algorithm

Suitable Grade Level

Grades 10-12