Question

$\huge\red{\tt{Question:-}}$
If the prime factorization of a natural number n is 2³x3⁴x5⁴x7, write the number of
consecutive zeroes in n.​

Answer 1

Answer

CONCEPT

consecutive zeroes are power of 10

make factor of number as pair of 2 and 5

the number of pair is your answer

SOLUTION

given

n = 2³x3⁴x5⁴x7

$= > n \: = ({2}^{3} \times {5}^{3} ) \times 5 \times \times {3}^{4} \times 7 \\ = > n \: = {10}^{3} \times 5 \times {3}^{4} \times 7$

so number of consecutive zeroes are

3 (power of 10 )

Answer 2

HELLO MATE

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.

Hope it helps you mate.

please thank on my answer.

humble request....

@ ANUSHA ❤✌