Question

Write a number n in usual form whose prime factorization is ,n=2⅞×5⅝×13

Answer 1

The required number is 5200000 and number of zero the number has 5.

Explanation:

Given : Prime factorization of n,  $2^7 \times5^5 \times 13$

To find : Write a number n in usual form and the number of zeroes it contains ?

Solution :

Let  $x=2^7 \times5^5 \times 13$

Solve the equation,

$x=2^5 \times5^5 \times 13\times 2^2$

$x=(2\times5)^5 \times 13\times 4$

$x=(10)^5 \times 52$

$x=5200000$

The number x contains 5 zeroes.

#Learn more

Write the number x in usual form, whose prime factorisation is given below:n=2^7×5^6×13. Also, find the number of zeroes it contains.

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Answer 2

Answer:

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