The number of zeros at the end of the product 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 is :

Answer: Option C

Let N = 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50
= 5¹⁰ × ( 1 × 2 × 3 × 4 × ..... × 10) = 5¹⁰ × 10!
Highest power of 2 in 10!
$$=\left[\frac{10}{2}\right]+\left[\frac{10}{2^2}\right]+\left[\frac{10}{2^3}\right]$$
= 5 + 2 + 1
= 8
Highest power of 5 in 10! = $$\left[\frac{10}{5}\right]$$ = 2
∴ N = 2⁸ × 5¹² × k
Since highest power of 2 is less than that of 5,
So, required number of zeros = 8