Question

write the number of zeros in the end of a number whoes prime factors is 2^2×5^3×3^2×17​

Answer 1

Answer:

If we expand it, it will be 2*2*5*5*5*3*3*17

Step-by-step explanation:

2*5=10 the no. of times the above expansion satisfy this condition that no . of zeros it will have at the end

Therefore it will have 2 zeros