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 1). 189 2). 182 3). 175 4). 91
LCM of 5, 10, 12, 15
LCM = 2 × 3 × 5 × 2 = 60
\ Number = 60k + 2
Now, the required number should
be divisible by 7.
Now, 60k + 2 = 7 × 8k + 4k + 2
If we put k = 3, (4k + 2) is equal
to 14 which is exactly divisible
by 7.
Required number = 60 × 3 + 2
= 182