As there are always more factors of 2 than 5 in $n!$ ($n \geq 2$), we need to find $n!$ with exactly $1987$ factors of 5 Now, amongst integers $\leq 8000$, there are- $1600$ multiples of $5$ $320$ multiples of $5^2=25$ $64$ multiples of $5^3=125$ $12$ multiples of $5^4=625$ $2$ multiples of $5^5=3215$. This totals 1998, which is 11 too many. So we have $n$ slightly less than 8000- if we reduce it to $7960$, then we have one fewer multiple of 125, two fewer multiples of 25 and 8 fewer multiples of 5, giving us $1987$ zeros on the end of $7960!$. We can't reduce this any further, because $7959!$ only has $1986$ zeros on the end, so the answer is $\boxed{7960}$

we must find n such that $\sum_{i=1}^{k} [\frac {n}{5^{i}}] =1987$ with $5^{k}\leq n$ use that $[x] \leq x$ we have that $1987\leq \frac {n}{4}.(1- \frac{1}{5^k}) \leq \frac {n-1}{4}$ we have that $n\geq 7949$ then easy find n=7960.

power of 2(say a) in n! is always less than or equal to power of 5(say b) in n! and 2*5=10 so the number n! will end with exactly b many zeros. so we need to find n least such that $\ s(n)=[\frac{n}{5}]+[\frac{n}{5^{2}}]+.....=1987$ if at all such n exist. but then see if $\ n=5^{6},s(n)=\frac{5^{6}-1}{4}=3906 $ and $\ n=5^{5},s(n)=\frac{5^{5}-1}{4}=781 $ so if such n exists it is $\ 5^{5} < n < 5^{6} $ so let we take base 5 expansion for n to get $\ n=a_{5}5^{5}+a_{4}5^{4}+...a_{0} $ where $\ 0\leq a_{i} \leq 4 $ then see $\ s(n)=781a_{5}+156a_{4}+31a_{3}+6a_{2}+a_{1} =1987 $ if possible to make n least we try to keep a_{i} 's with the larger i's as small as possible. see a_{5} is at least 1. if we set a_{5}=1 then s(n) is at max 1557 so a_{5} have to be atleast 2. for a_{5}=2 we see $\ 156a_{4}+31a_{3}+6a_{2}+a_{1} =425 $ see 156*3=468>425 so a_{4} have to be less than 3 but if a_{4}=1 then see s(n) can be atmost 308 so a_{4}=2 so $ \ 31a_{3}+6a_{2}+a_{1} =113 $ then by similar arguement $ \ a_{3}=3$ so $\ 6a_{2}+a_{1} =20 $ see by same arguement $\ a_{2}=3 $ then $\ a_{1}=2 $ as we want the least no so we set $\ a_{0}=0$ so the desired no is $\ n=2.5^{5}+2.5^{4}+3.5^{3}+3.5^{2}+2.5+0=7960 $ $ \ n=7960$