A natural number divisible by ‘3’ is

A natural number divisible by 3, if and only if sum of the digits is divisible by 3. Now,

A) $129\to 1+2+9=12\to$  divisible by 3

B) $12345\to 1+2+3+4+5=15\to$  divisible by 3

C) $61392\to 6+1+3+9+2=21\to$  divisible by 3

$\therefore 129,12345,61392$  are exactly divisible by ‘3’