The numbers 2, 4, 6, 8 ..... 98, 100 are multiplied together. The number of zeros at the end of the product must be :

Solution (By Examveda Team)

N = 2 × 4 × 6 × 8 × ..... × 98 × 100
    = 2⁵⁰ (1 × 2 × 3 × ..... × 49 × 50)
    = 2⁵⁰ × 50!
Clearly, the highest power of 2 in N is much higher than that of 5
∴ Number of zeros in N = Highest power of 5 in N
= $$\left[ {\frac{{50}}{5}} \right] + \left[ {\frac{{50}}{{{5^2}}}} \right]$$
= 10 + 2
= 12