Why not 2015?

To make a zero, we need the factors $2$ and $5$ . Since we will have more factors of $2$ than $5$ in any $n!$ calculation we need focus only on the factors of $5$ .

Now in general there are $x_{k} = \lfloor \frac{n}{k} \rfloor$ multiples of $k$ less than or equal to $n$ for any integer $k.$ To find all the factors of $5$ in $n!$ we will need to find

$\sum_{m=1}^{\infty} x_{5^{m}}$ , i.e., $x_{5} + x_{25} + x_{125} + ......$ .

We thus need to find all $n$ such that this sum comes to $2016.$

Next, note that for $n = 3125$ this sum comes to $781$ and for $n = 15625$ this sum comes to $3906.$ So we are looking for values of $n$ between these two values. Thus our sum will only involve a calculation of $x_{5} + x_{25} + x_{125} + x_{625} + x_{3125}.$

For a first estimate, look at the sum without the floor function, i.e.,

$\dfrac{n}{5} + \dfrac{n}{25} + \dfrac{n}{125} + \dfrac{n}{625} + \dfrac{n}{3125} = 2016$

$\Longrightarrow \dfrac{781*n}{3125} = 2016 \Longrightarrow n = 8066.58...$

As the least $n$ we are looking for will be a multiple of $5$ , we will need to make the sum calculations for multiples of $5$ near to this initial estimate. For $n = 8070$ we get a sum of $2014$ , and for $n = 8075$ we get the desired sum of $2016.$ As the smallest $k$ -value in our sum is $5$ , we know that the next four integers after $8075$ will yield a sum of $2016$ as well.

Thus the desired solution is $8075 + 8076 + 8077 + 8078 + 8079 = \boxed{40385}.$

Note that there are no $n$ such that $n!$ ends in precisely $2015$ zeros, hence the title of this question.

The highest power of 5 thats contained in this number has to be the answer. Now with the number approaching 2016 that becomes a little imposing initially. What I did was some approximation to land close to the number As this is a sum of increasingly restrictive values, we are looking for a number which leads us close to the vicinity of 2016. Since this method focuses on the trial and error bit, thus a reasonable number to start with was 8000 as 8000/5 = 1600 and 8000/25 = 320 which together adds upto 1920 which gets us within reasonable proximity of 2016. A common technique, though not foolproof that I have observed which helps in such cases would be to focus on the numbers which are essentially multiples of 5(again not a fail safe method, it has glaring inadequacies). Proceeding this way we can arrive at 8075 with maybe 3-4 rounds of calculations.

Now, if any number n is chosen that has k zeroes at the end of n! then the next 4 numbers also will be ending with the same number of zeroes because its not until the next multiple of 5 does the number of zeroes change.

Any inputs are welcome

n=0

l=0

for i in range(0,10000,5):

l=i

n=0

while( l > 0):

    n = n + l/5

    l=l/5

if (n == 2016):

     print 5*i+10

python code for the problem

To continue with this solution, I would like to say that this was a handy method for me.

We have to see no of 5's in the given number to be exactly 2016.

Let us consider a smaller number. For instance 5! = 5/5 = 1 which has only 1 0.

This continues but for 25! = 25/5 = 5/5 =1 which has 6 0's.

That is for multiple of 25,we will get a jump.

But for 125 = 125/5 = 25/5 =5/5 =1 has 31 0's.

We need to see the series 125,250,375,...

We know that 31(65) = 2015.

But 1250, 2500, 3250, 5000, 6250, 7500 each carries an extra 0 in them.

65th term = 125 + 64(125) = 65(125) = 8125 will have 8125/5 = 1625/5 =325/5 = 65/5 = 13/5 = 2 will have 2030 0's.

we get 8075 having 2016 0's if we check it.

The next 5 numbers will have exactly 2016 0's...8075...8079

Total is 40385.