LCM of 18, 21 and 24
LCM = 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 504 K –
11. Where K is a positive integer
Since 23 × 21 = 483
We can write 504 K – 11
= (483 + 21) K – 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