combinatorics and divisibilty 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???
 A: A number is divisible by $ 8 $ if the last three digits are divisible by $ 8$.
Now, we can arrange the first $5$ digits of our answer in $ 5^5 $ ways, because each of the position can take $ 1$ of $5$ values.
Now, our problem reduces to the following.
How many three digit numbers formed with $\{1,2,3,4,5\}$ are divisible by $8$?
We can enumerate all the possibilities, preferably in a programming language. In Python,
acc = 0
for x in range(1,6):
    for y in range(1,6): 
        for z in range(1,6):
             if int(str(x)+str(y)+str(z))%8 == 0: acc += 1

print acc

This yields an output of $ 13 $. The favourable numbers are 
112
144
152
224
232
312
344
352
424
432
512
544
552

Thus, since the last three digits can take any of these $ 13 $ values, and the first five digits can take $ 5^5 $ values, our final answer is $ 5^5 \times 13 = \boxed{40625} $
A: Of all the numbers that are formed with 1,2,3,4,5 - the last three digits need to be divisible 
by 8.  There are 5^3 ways you could arrange the five numbers for the last three digits.  Of 
these last three digits that are divisible by 8 are 312, 152, 512, 432, 352, 112, 232, 224, 
144, 424, 344, 552, 544.  A total of 13 of them which I got by brute force in EXCEL by writing 
a small program.  Divisibility by 8 is the screwist one.
With these as the last three digits with the first five from {1,2,3,4,5}.  The total number of ways then will be $13*5^5$
The answer is $13*5^5$
