Solution for the week of January 5, 2015:

The first thing you should notice is that every instance of a multiple

of 10 in the factorial expansion will yield a trailing zero. Next,

notice that multiples of 5 and 2 will yield trailing zeroes, including

the same yielded by the multiples of 10.

In the factorization of 1000000, there are far more 2s than there are

5s. Thus, in order to count the number of trailing zeroes, we need

only count the number of 5s that occur as factors.

This can be done simply with:

floor(1000000/5) + floor(1000000/5^2) + floor(1000000/5^3) +

floor(1000000/5^4) + floor(1000000/5^5) + floor(1000000/5^6) +

floor(1000000/5^7) + floor(1000000/5^8)

(And since 5^9 = 1953125 > 1000000, no further terms are needed past 5^8.)

The result:

200000 + 40000 + 8000 + 1600 + 320 + 64 + 12 + 2 = 249998

Two other related problems from Project Euler can be found athttps://projecteuler.net/index.php?section=problems&id=20

and https://projecteuler.net/index.php?section=problems&id=97.

http://hpcuniversity.org/students/weeklyChallenge/10