Question

if the prime factorization of a natural number n is 2³×3²×5²×7, write the number of consecutive zeros in which 'n' ends.​

Answer 1

Answer:

2

Step-by-step explanation:

n = (2^3) × (3^2) × (5^2) × 7

= 8 × 9 × 25 × 7

= 12600

therefore,

number of consecutive zeros in which n ends = 2