Number Theory Problem No.6

Since the factors 5 and 2 are needed to make a trailing zero, we can take the floor function of how many times the powers of 5 divide $n$ (since they are less common than powers of 2):

$\displaystyle 250000000 = \sum_{i=1}^{⌊log_{5}n⌋} ⌊\frac{n}{5^{i}}⌋$

Solving the above equation (which is very difficult to me) results in $1000000009 < n < 1000000014$ with $n \in \N$ . So the answer is $n_{min}=\boxed{1000000010}$ .

A good way to approximate the answer is using:

$\displaystyle A = \sum_{i=1}^{⌊log_{5}n⌋} ⌊\frac{n}{5^{i}}⌋ \approx \frac{n}{4}$

With $n=10^9$ , $A=249999998 < 250000000$ , so you would start counting upward.