Question:medium

The congruence \(10x \equiv 15 \pmod{35}\) has

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For \(ax\equiv b\pmod m\), first calculate \(\gcd(a,m)\). If it divides \(b\), the number of solutions is \(\gcd(a,m)\).
  • only one solution
  • no solution
  • \(5\) solutions
  • \(10\) solutions
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
This question is about solving a linear congruence of the form \(ax \equiv b \pmod{n}\). The number of solutions to such a congruence depends on the greatest common divisor (GCD) of the coefficient \(a\) and the modulus \(n\).

Step 2: Key Formula or Approach:

A linear congruence \(ax \equiv b \pmod{n}\) has solutions if and only if \(\gcd(a, n)\) divides \(b\). Let \(d = \gcd(a, n)\). - If \(d\) does not divide \(b\), there are no solutions. - If \(d\) divides \(b\), there are exactly \(d\) incongruent solutions modulo \(n\).

Step 3: Detailed Explanation:

The given congruence is \(10x \equiv 15 \pmod{35}\). Here, we have \(a = 10\), \(b = 15\), and \(n = 35\). First, we calculate the greatest common divisor of \(a\) and \(n\): \[ d = \gcd(10, 35) \] The divisors of 10 are {1, 2, 5, 10}. The divisors of 35 are {1, 5, 7, 35}. The greatest common divisor is 5. So, \(d = 5\). Next, we check if \(d\) divides \(b\). We need to check if 5 divides 15. \[ 15 \div 5 = 3 \] Yes, 5 divides 15. Since \(\gcd(10, 35)\) divides 15, the congruence has solutions. The number of incongruent solutions modulo 35 is equal to \(d\). Therefore, there are exactly 5 solutions.

Step 4: Final Answer:

The congruence has \(\gcd(10, 35) = 5\) solutions. This matches option (C).
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