The smallest number of four digits which on division by 4, 6, 10 and 15 leaves a remainder 2 in each case is:
Options:
A . 1020
B . 1022
C . 1024
D . 1040
Answer: Option B First of all,we find the LCM of 4, 6, 10 and 15 LCM of 4, 6, 10, 10 = 60Now, 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
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