If x ≡ 4 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solut

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Question: If x ≡ 4 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

Options:

Correct Answer: 25

Solution:

Using the method of successive substitutions or the Chinese Remainder Theorem, we find that the smallest positive integer solution is x = 25.

If x ≡ 4 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

If x ≡ 4 (mod 7) and x ≡ 5 (mod 11), what is the smallest positive integer solution for x?

Step 1: Understand the notation. The expression 'x ≡ 4 (mod 7)' means that when x is divided by 7, the remainder is 4. Similarly, 'x ≡ 5 (mod 11)' means that when x is divided by 11, the remainder is 5.
Step 2: Start with the first equation. From 'x ≡ 4 (mod 7)', we can write x as: x = 7k + 4, where k is any integer.
Step 3: Substitute x into the second equation. Replace x in 'x ≡ 5 (mod 11)' with '7k + 4': 7k + 4 ≡ 5 (mod 11).
Step 4: Simplify the equation. Subtract 4 from both sides: 7k ≡ 1 (mod 11).
Step 5: Find the multiplicative inverse of 7 modulo 11. We need to find a number m such that 7m ≡ 1 (mod 11). Testing values, we find that m = 8 works because 7 * 8 = 56, and 56 mod 11 = 1.
Step 6: Multiply both sides of the equation 7k ≡ 1 (mod 11) by 8: k ≡ 8 (mod 11). This means k can be written as k = 11j + 8, where j is any integer.
Step 7: Substitute k back into the equation for x: x = 7(11j + 8) + 4 = 77j + 56 + 4 = 77j + 60.
Step 8: To find the smallest positive integer solution for x, set j = 0: x = 60.
Step 9: Check if x = 60 satisfies both original equations: 60 mod 7 = 4 and 60 mod 11 = 5. Both conditions are satisfied.
Step 10: Therefore, the smallest positive integer solution for x is 60.

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.