Question

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

• A. only one solution
• B. no solution
• C. \(5\) solutions
• D. \(10\) solutions

Hint:

For \(ax\equiv b\pmod m\), first calculate \(\gcd(a,m)\). If it divides \(b\), the number of solutions is \(\gcd(a,m)\).

Answer 1

The Correct Option is C

Concept:
A linear congruence \[ ax \equiv b \pmod m \] has solutions if and only if \[ \gcd(a,m)\mid b \] If this condition is satisfied, then the number of incongruent solutions modulo \(m\) is \[ \gcd(a,m) \]

Step 1: Compare with the standard form. The given congruence is \[ 10x \equiv 15 \pmod{35} \] So, \[ a=10,\qquad b=15,\qquad m=35 \]

Step 2: Find \(\gcd(10,35)\). \[ 10=2\times 5 \] \[ 35=7\times 5 \] Therefore, \[ \gcd(10,35)=5 \]

Step 3: Check the existence condition. The congruence has solutions if \[ \gcd(10,35)\mid 15 \] That means we check whether \[ 5\mid 15 \] Since \[ 15=5\times 3 \] the condition is satisfied.

Step 4: Find the number of solutions. Since the condition is satisfied, the number of incongruent solutions modulo \(35\) is \[ \gcd(10,35)=5 \]

Step 5: Final answer. Hence, the congruence has \[ \boxed{5\text{ solutions}} \]