Question

what will be the least number which is divided by 3, 4, 6and 9 leaves remainder 2, but when divided by 11 leaves no remainder?

Answer 1

First find the lcm of the numbers that the required number is divisible with, i.e., 3,5,6,8,10,12 which is 120

Therefore the required number is of the form 120x+2 (since it leaves a reminder of 2)

Now the number is a multiple of 13. 
so 120x+2 is a multiple of 13
=> 117x + 3x+2 is a multiple of 13

The least value of x for which this holds true is 8 (3*8+2 = 26)
Therefore the least such number is 120*8+2 =962