Question

The greatest number of 5 digits which when divided by 4, 5, 6, 7, and 8 leaves 1, 2, 3, 4, and 5, respectively, as remainders is:

• A. 99313
• B. 99117
• C. 99691
• D. 99957

Hint:

Use the concept of LCM to simplify problems involving multiple divisibility conditions with remainders.

Answer 1

The Correct Option is D

Let the required number be \( N \). We are given that: \[ N \equiv 1 \, (\text{mod} \, 4), \quad N \equiv 2 \, (\text{mod} \, 5), \quad N \equiv 3 \, (\text{mod} \, 6), \quad N \equiv 4 \, (\text{mod} \, 7), \quad N \equiv 5 \, (\text{mod} \, 8) \] This implies: \[ N + 3 \equiv 0 \, (\text{mod} \, 4, 5, 6, 7, 8) \] We need to find the least common multiple (LCM) of 4, 5, 6, 7, and 8: \[ \text{LCM}(4, 5, 6, 7, 8) = 840 \] Thus, \( N + 3 = 840k \) for some integer \( k \), and \( N = 840k - 3 \). Now, we find the largest \( N \) that is a 5-digit number: \[ 840k - 3 \leq 99999 \] \[ 840k \leq 100002 \] \[ k \leq \frac{100002}{840} = 119 \] For \( k = 119 \): \[ N = 840 \times 119 - 3 = 99957 \] Thus, the greatest number is 99957.