in how many ways we can form a $8$ digit numbers from $1,2,3,4,5$ with repetition allowed & divisible by $8$.
MY APPROACH :
to be divisible by 8 : last 3 digit of the no. must be divisible by 8 like $152 , 112$(as repetition allowed),.......[ let total no. of these 3 digit numbers are x ]
then total ways of other 5 diigit = 5^5
now for each way of first 5^5 combinations there will be further x combination for last three
so it becomes (5^5) $*$ (x^5)
is it correct???i dont think it is correct???how to solve it???