What is the smallest number which when divided by 3, 5, 6, 8, 10 and 12 leaves a remainder of 2 in each case but which when divided by 13 leaves no remainder?

Calculation

LCM of 3, 5, 6, 8, 10 and 12 is 120.

If the remainder in each case is 2 and the number is divisible by 13, then

The number is (120x + 2) divisible by 13, so

(120x + 2)/13

⇒ 9x + (3x + 2)/13

(3x + 2) is divisible by 13 if we put x = 1, 2, 3, 4, ……..

Put x = 8, then

⇒ (3 × 8 + 2)

⇒ 26

As we know, 13 is divisible by 13 so put x = 8

120x + 2

⇒ 120 × 8 + 2

⇒ 962

The answer is 962.