The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is 

Given,

Number is divisible by 7 and when divided by 5, 10, 12 and 15 leaves remainder 2 in each case

∴ Smallest number which is divisible by 5, 10, 12 and 15

= LCM of 5, 10, 12 and 15

Finding LCM

5 = 1× 5

10 = 2 × 5

12 = 2 × 2 × 3

15 = 3 × 5

∴ LCM = 2 × 2 × 3 × 5 = 60

∴ The number which when divided by 5, 10, 12 and 15 leaves remainder 2 in each case is (60n + 2) where, n is an integer.

Now, we have to find the smallest value of n for which (60n + 2) is divisible by 7.

For, n = 1, 60n + 2 = 62, not divisible by 7

For, n = 2, 60n + 2 = 122, not divisible by 7

For, n = 3, 60n + 2 = 182, which is divisible by 7

∴ Required number is 182