If the prime factorization of a natural number `n` is `2^3*3^2*5^2*6,` write the number of consecutive zeros in `ndot`

If the prime factorization of a natural number (n) is 2^2xx3^2xx5^2xx7. How many consecutive zeroes will it have at the end of it? Justify your answer.

If n=2^3xx3^4xx5^4xx7 , where n is a natural number, then find the number of consecutive zeros in n

The prime factorization of a number is 3xx5 . What is the number and what are all of its factors?

Is 2^(3-n) a prime number for all natural numbers n?

The prime factorization of a number is 2xx3xx13 . What is the number and what are all of its factors?

The prime factorization of a number is 2xx5xx7 . What is the number and what are all of its factors?

If 2 appears 3 times in the prime factorization of the number n, then in n^(2) with appear

The number of consecutive zeros in 2^3xx\ 3^4xx\ 5^4xx\ 7, is

The number of factors of a number N=2^(3)xx3^(2)xx5^(3) is

If [x] denotes the integral part of x and n is a natural number such that 1<=n<100 then the number of solutions of the equation [(n)/(2)]+[(n)/(3)]+[(n)/(5)]=(n)/(2)+(n)/(3)+(n)/(5) is