Originally posted by bugzpodder
right most non-zero digit? easy.

of course you need to use a program to do it.

loop from 1 to N {i}
lets say 'i' is divisible by 2^a, where a is the largest integer such that 2^a<i
and 'i' is divisible by 2^b, where b is the largest integer such that 2^b<i
k=(k*i/(2^a*5^b)) mod 10
numberOfTwo+=a
numberOfFive+=b
}

loop from 1 to numberOfTwo-numberOfFive
k=(k*2)mod 10

and k would be the right-most non-zero digit
if you wrote ...and 'i' is divisible by 5^b, where b is the largest integer s.t. 5^b<i there may be less confusion