Hey, 👲
you're probably here searching for the solution to an interestingly difficult task of counting the number of trailing 0's in the factorial of a number.
Straightaway, here's the solution:
f(n) = [n/5] + [n/52] + [n/53] + ... [<1]
[ where [ ] is the greatest integer function (also called box or floor function) that truncates the fractional part of a real number & returns an integer value. ]
> Examples:
i) for n = 23:
f(n) = f(23) = [23/5] + [23/52] = [4.6] + [0.92] = 4 + 0 = 4
Hence, 23! has 4 trailing 0s.
👉 Reason:
How many multiples of 5 are between 1 and 23?
We have: 5, 10, 15, and 20 - four multiples of 5 which paired with 2's from the even factors, makes for 4 factors of 10.
Thus, 23! has four trailing zeroes.
ii) Find the number of trailing zeroes in the expansion of 1000!
for n = 1000:
f(n) = f(1000) = [1000/5] + [1000/52] + [1000/53] + [1000/54] + [1000/55] = [200] + [40] + [8] + [1.6] + [0.32] = 200 + 40 + 8 + 1 = 249
Hence, 1000! has 249 trailing 0s.
👉 Reason:
There are 1000 ÷ 5 = 200 multiples of 5 between 1 and 1000.
The next power of 5 i.e. 25 has 1000 ÷ 25 = 40 multiples between 1 and 1000.
The next power of 5 i.e. 125 has 1000 ÷ 125 = 8 multiples between 1 and 1000.
The next power of 5 i.e. 625 fits in the expansion too, and has 1000 ÷ 625 = 1.6 multiples implying 1 complete multiple between 1 and 1000. (the remaining = 1000-625 accounts for that '0.6' excess which is hence truncated)
Thus, in total, we have 200 + 40 + 8 + 1 = 249 occurrences of the factor 5 in the expansion of 1000!
