Question

Let \[ G = \{ n \in \mathbb{N} : n \leq 55, \, \gcd(n, 55) = 1 \} \] be the group under multiplication modulo 55. Let \( x \in G \) be such that \( x^2 = 26 \) and \( x>30 \). Then \( x \) is equal to ................

Hint:

When solving quadratic congruences, try checking potential solutions directly by squaring numbers modulo the modulus.

Answer 1

Correct Answer: 31

Step 1: List elements of \( G \).
The group \( G \) consists of integers less than or equal to 55 that are coprime with 55. The prime factorization of 55 is: \[ 55 = 5 \times 11. \] Thus, we need to find numbers less than or equal to 55 that are not divisible by 5 or 11.
Step 2: Solve \( x^2 = 26 \mod 55 \).
We are given that \( x^2 \equiv 26 \pmod{55} \), and we need to find \( x \). First, check numbers greater than 30 that satisfy this condition: - \( x = 31 \), then \( 31^2 = 961 \equiv 26 \pmod{55} \). Thus, \( x = 31 \) is a solution.
Step 3: Conclusion.
Therefore, the value of \( x \) is: \[ \boxed{31}. \]