Three consecutive years!

Let us first try to simplify the fraction a bit.

In $\frac { 2014!2015! }{ 2013! }$ ,
$\frac { 2014! }{ 2013! } =\frac { 2014 \times 2013! }{ 2013! } =2014$

$Therefore,\quad it\quad is\quad simplified\quad into\quad 2014(2015!).\\ \\ Now,\quad to\quad find\quad the\quad number\quad of\quad trailing\quad zeroes\quad in\quad this\quad expression,\\ \\ By\quad de\quad Polignac's\quad formula,\\ \\ 2015/5\quad =\quad 403\quad \quad \\ \\ 2015/25\quad =\quad 80\quad (Ignore\quad the\quad remainder)\\ \\ 2015/125\quad =\quad 16\quad (Ignore\quad the\quad remainder)\\ \\ 2015/625\quad =\quad 3\quad (Again,\quad ignore\quad the\quad remainder)\\ \\ Therefore,\quad number\quad of\quad trailing\quad zeroes\quad is\quad 502\quad (by\quad adding).\\ \\ Hence,\quad 1000\quad -\quad x\quad =\quad 498.\\ \\$