Solution:Any number will be divisible by '3' if the sum of the digits of that number is completely divisible by '3'From option
5967013→ 5 + 9 + 6 + 7 + 0 + 1 + 3 ='31' not divisible by 3
1111→1 + 1 + 1 + 1 = 4,'4' is not divisible by '3'
5413265→5 + 4 + 1 + 3 + 2 + 6 = '21', is divisible by '3'