The smallest number of four digits which on division by 4, 6, 10 and 15 leaves a remainder 2 in each case is:

First of all,we find the LCM of 4, 6, 10 and 15
LCM of 4, 6, 10, 10 = 60
Now, The smallest four digit no. is 1000. We divide it by 60
$$\frac{{1000}}{{60}}$$ It leaves remainder 40
Now, the smallest four digit no which is divisible by 4, 6, 10, 15 is,
1000 + (60 - 40) = 1020
So, required number (as it gives remainder 2 always) would be = 1020 + 2 = 1022