Anyone solve it plz!!!!!!!!!

Arun Sharma has explained this concept really nicely.

For n! to have x zeroes, it must have x number of 5's in it. Unless (n+1)! is a multiple of 5, (n+1)! will also have x zeroes.

Here, the main concept is to understand that if (n+1)! has 'x' zeroes more than n! then it means that (n+1) is a multiple of $5^x$.

n!	Number of zeros	(n+1)!	Number of zeroes
n = 29	$\left\lfloor \frac{29}{5} \right\rfloor + \left\lfloor \frac{29}{25} \right\rfloor = 6$	n + 1= 30	$\left\lfloor \frac{30}{5} \right\rfloor + \left\lfloor \frac{30}{25} \right\rfloor = 7$
n = 124	$\left\lfloor \frac{124}{5} \right\rfloor + \left\lfloor \frac{124}{25} \right\rfloor + \left\lfloor \frac{124}{125} \right\rfloor = 28$	n + 1 = 125	$\left\lfloor \frac{125}{5} \right\rfloor + \left\lfloor \frac{125}{25} \right\rfloor + \left\lfloor \frac{125}{125} \right\rfloor = 31$
n = 624	$\left\lfloor \frac{624}{5} \right\rfloor + \left\lfloor \frac{624}{25} \right\rfloor + \left\lfloor \frac{624}{125} \right\rfloor = 152$	n + 1 = 625	$\left\lfloor \frac{625}{5} \right\rfloor + \left\lfloor \frac{625}{25} \right\rfloor + \left\lfloor \frac{625}{125} \right\rfloor + \left\lfloor \frac{625}{625} \right\rfloor = 156$

Notice the similarities here. For 30!, 30 will add one 5 or $5^1$ therefore it has one more zero. For 125!, 125 will add $5^3$ therefore it has 3 more zeroes. Similarly for 625!, it will add $5^4$ therefore it was 4 more zeroes.

So to solve the question, if (n+1)! has 3 more zeroes than n!, it means (n+1) is a multiple of 125. Therefore for $1 \le n + 1 \le 1000$, we will have 8 multiples of 125. But this also contains 625 and as illustrated above, it will have 4 more zeroes. Subtracting that: 8 - 1 = 7 which is your final answer