Solution:Any number will be divisible by '3' if the sum of the digits of that number is completely divisible by '3'From option
59670135+9+6+7+0 1 + 3 = '31' not divisible by 3 1111 1+1+1+1=4, '4' is nt divisible by '3' 541326-5+4+1+3+2+6 = '21', is divisible by '3'