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

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