Question

A number less than 100, when divided by 4, 5 and 6 leaves remainder 2 in each case. The number is:

• A. \( 62 \)
• B. \( 54 \)
• C. \( 38 \)
• D. \( 44 \)

Hint:

When a number leaves the same remainder with multiple divisors, subtract the remainder and find the LCM of the divisors to simplify the problem.

Answer 1

The Correct Option is A

Step 1: Let the number be \( x \).
Given: \[ x \equiv 2 \pmod{4}, \quad x \equiv 2 \pmod{5}, \quad x \equiv 2 \pmod{6} \] This implies: \[ x - 2 \text{ is divisible by } \text{LCM}(4, 5, 6) \] \[ \text{LCM}(4, 5, 6) = 60 \Rightarrow x - 2 = 60k \Rightarrow x = 60k + 2 \] Step 2: Find value of \( k \) such that \( x<100 \).
\[ 60k + 2<100 \Rightarrow 60k<98 \Rightarrow k<1.63 \] So the only integer value for \( k \) is \( k = 1 \) \[ x = 60(1) + 2 = 62 \]