Correct Option: A

We find LCM of 18, 21 and 24

LCM of 18, 21 and 24 = 2 × 3 × 3 × 7 × 4 = 504
Now compare the divisors with their respective remainders. We observe that in all the cases the remainder is just 11 less than their respective divisor. So the number can be given by 504K – 11. Where K is a positive integer
Since 23 × 21 = 483
We can write 504K – 11 = (483 + 21)K – 11
⇒ 504K – 11 = 483 K + (21K – 11)
483 K is multiple of 23, since 483 is divisible by 23.
So, for (504K – 11) to be multiple of 23, the remainder (21K – 11) must be divisible by 23.
Put the value of K = 1, 2, 3, 4, 5, 6, ..... and so on successively. We find that the minimum value of K for which (21K – 11) is divisible by 23. is 6, (21 × 6 – 11)
= 115 which is divisible by 23.
Therefore, the required least number = 504 × 6 – 11 = 3013