Question

If the prime factorisation of a rational number 'N' is 2^5×3^2×5^4×7,. write the number ​

Answer 1

Answer: There will be 3 consecutive zeroes in N.

Step-by-step explanation:

Given : If the prime factorization of a natural number N is

To find : Write the number of consecutive zeros in N?

Solution :

Prime factorization of a natural number N is

The consecutive zeros are the power of multiple of 2 and 5

So, There will be 3 consecutive zeroes in N.

Answer 2

Answer:

The given expression 'n' has only 2 consecutive zeroes.

Solution:

Given that

The prime factorization of the given natural number n is 2^3× 3^2×5^2× 7

Prime factorization: by using prime factorization we can find out which prime numbers multiple together to make the original number.

To find the consecutive zeroes in a number, split out the 2’s and 5’s which can finally sum up together to give 10.

\begin{lgathered}\begin{array} { c } { n = 2 ^ { 3 } \times 3 ^ { 2 } \times 5 ^ { 2 } \times 7 } \\\\ { n = 2 \times 2 ^ { 2 } \times 3 ^ { 2 } \times 5 ^ { 2 } \times 7 } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 2 ^ { 2 } \times 5 ^ { 2 } } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 10 ^ { 2 } } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 100 } \end{array}\end{lgathered}n=23×32×52×7n=2×22×32×52×7n=2×32×7×22×52n=2×32×7×102n=2×32×7×100

Thus, in the given natural number 'n' there are 2 consecutive zeroes.