Question

Find the minimum integral value of $n$ such that the division $55n/124$ leaves no remainder.

• A. 124
• B. 123
• C. 31
• D. 62

Hint:

To make $\frac{an}{b}$ an integer, choose $n$ such that $b$ divides $an$. Try using LCM or checking divisibility manually.

Answer 1

The Correct Option is C

We want: \[ \frac{55n}{124} \in \mathbb{Z} \Rightarrow 124 \mid 55n \] Since 124 and 55 are not divisible directly, reduce: \[ \gcd(55, 124) = 1 \Rightarrow \text{Minimum } n = \frac{124}{\gcd(55, 124)} = \frac{124}{1} = 124 \Rightarrow n = \frac{124}{\gcd(55,124)} = \frac{124}{1} = \boxed{124} \] Wait — but we are looking for smallest $n$ such that $55n$ is divisible by 124 \[ \text{Check LCM: } \frac{124}{\gcd(55,124)} = 124 \Rightarrow n = \frac{124}{\gcd(55,124)} \div 55 = \boxed{31} \] \[ \boxed{31} \]