If x ≡ 3 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solut

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Question: If x ≡ 3 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

Options:

Correct Answer: 26

Solution:

Using the method of successive substitutions or the Chinese Remainder Theorem, the smallest solution is x = 26.

If x ≡ 3 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

If x ≡ 3 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

Step 1: Understand the notation. 'x ≡ 3 (mod 7)' means when x is divided by 7, the remainder is 3. 'x ≡ 5 (mod 11)' means when x is divided by 11, the remainder is 5.
Step 2: Write down the first equation from the first condition: x = 7k + 3, where k is an integer.
Step 3: Substitute the expression for x from Step 2 into the second condition: 7k + 3 ≡ 5 (mod 11).
Step 4: Simplify the equation from Step 3: 7k + 3 - 5 ≡ 0 (mod 11) which simplifies to 7k - 2 ≡ 0 (mod 11).
Step 5: Rearrange the equation: 7k ≡ 2 (mod 11).
Step 6: Find the multiplicative inverse of 7 modulo 11. The inverse is 8 because 7 * 8 = 56 ≡ 1 (mod 11).
Step 7: Multiply both sides of the equation 7k ≡ 2 (mod 11) by 8: k ≡ 16 (mod 11).
Step 8: Simplify k ≡ 16 (mod 11) to k ≡ 5 (mod 11). This means k can be written as k = 11m + 5 for some integer m.
Step 9: Substitute k back into the expression for x: x = 7(11m + 5) + 3 = 77m + 35 + 3 = 77m + 38.
Step 10: The smallest positive integer solution occurs when m = 0: x = 38.
Step 11: Check if x = 38 satisfies both original conditions: 38 mod 7 = 3 and 38 mod 11 = 5.
Step 12: Since both conditions are satisfied, the smallest positive integer solution for x is 38.

Modular Arithmetic – Understanding congruences and how to solve systems of congruences.
Chinese Remainder Theorem – A method for solving simultaneous congruences with different moduli.
Successive Substitution – A technique for finding solutions to congruences by substituting values iteratively.