Find the greatest number of five digits which when divided by 3, 5, 8, 12 leaves 2 as remainder ?

Answer: Option C

LCM of (3, 5, 8, 12)
⇒ 3 × 5 × 4 × 2 = 120
⇒ Now greatest five digits number is 99999
On dividing 99999 by = 120 (LCM)
We get remainder = $$\frac{99999}{120},$$  remainder = 39
⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)
∴ 99999 - 39 = 99960
⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.
⇒ add 2 in the 99960
⇒ 99960 + 2
⇒ 99962