Zeros and the IMO!

According to Legendre 's formula :

$\lfloor x/5 \rfloor + \lfloor x/5^2 \rfloor + \lfloor x/5^3 \rfloor + ... \infty = 1987$

Suppose we treat this temporarily as a geometric progression by removing the GIF sign.Then using infinite formula we get $x(\dfrac{ \dfrac{1}{5} }{ 1 - \dfrac{1}{5}}) = 1987$ thus $x \approx 7948$ . However putting this in actual formula we find that answer comes $1983$ and we are off by $4$ zeroes.

If we induce some small changes in value of x this will only change the first and second term that is $\lfloor x/5 \rfloor$ and $\lfloor x/5^2 \rfloor$ will change.

By Trial and error slowly increasing value of $x$ from $7948$ to $7950$ to $7955$ to $7960$ we find we get $3$ extra zeroes from first term and one extra zero from second term thus total we have obtained our $4$ needed zeroes.