\( \begin{array}{ll}\text { (c) } 13068 & \text { (d) } 1084 \\ \text { Find the greatest number of } 5 \text {-digits which when } \\ \text { divided by } 3,5,8 \text { and } 12 \text { will have } 2 \text { as remainder }\end{array} \)

To find the greatest 5-digit number that, when divided by 3, 5, 8, and 12, leaves a remainder of 2, we can follow these steps: 1. **Identify the conditions**: We need a number \( N \) such that: \[ N \equiv 2 \mod 3 \] \[ N \equiv 2 \mod 5 \] \[ N \equiv 2 \mod 8 \] \[ N \equiv 2 \mod 12 \] 2. **Simplify the problem**: Since all these congruences have the same remainder (2), we can rewrite \( N \) as: \[ N = k - 2 \] where \( k \) is a number that is divisible by the least common multiple (LCM) of 3, 5, 8, and 12. 3. **Calculate the LCM**: - The prime factorization of the numbers is: - \( 3 = 3^1 \) - \( 5 = 5^1 \) - \( 8 = 2^3 \) - \( 12 = 2^2 \cdot 3^1 \) - The LCM takes the highest power of each prime: - For \( 2 \): \( 2^3 \) - For \( 3 \): \( 3^1 \) - For \( 5 \): \( 5^1 \) Thus, the LCM is: \[ \text{LCM} = 2^3 \cdot 3^1 \cdot 5^1 = 8 \cdot 3 \cdot 5 = 120 \] 4. **Express \( N \)**: Since \( N \equiv 2 \mod 120 \), we can express \( N \) as: \[ N = 120m + 2 \] for some integer \( m \). 5. **Find the greatest 5-digit number**: The largest 5-digit number is 99999. We need to find the largest \( m \) such that: \[ 120m + 2 \leq 99999 \] Subtracting 2 from both sides gives: \[ 120m \leq 99997 \] Dividing by 120: \[ m \leq \frac{99997}{120} \approx 833.3083 \] The largest integer \( m \) is 833. 6. **Calculate \( N \)**: Now substituting \( m = 833 \) back into the equation for \( N \): \[ N = 120 \cdot 833 + 2 = 99960 + 2 = 99962 \] 7. **Conclusion**: The greatest 5-digit number that, when divided by 3, 5, 8, and 12, leaves a remainder of 2 is: \[ \boxed{99962} \]