The numbers 1, 2, 3, 4, ......, 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be :

Answer: Option D

Let N = 1 × 2 × 3 × 4 × ..... × 1000 = 1000!
Clearly, the highest power of 2 in N very high as compared to that of 5.
So, the number of zeros in N will be equal to the highest power of 5 in N.
∴ Required number of zeros
= $$\left[\frac{1000}{5}\right]+\left[\frac{1000}{5^2}\right]+\left[\frac{1000}{5^3}\right]$$      $$+\left[\frac{1000}{5^4}\right]$$
= 200 + 40 + 8 + 1
= 249