Question

if the prime factorization of a natural number n is 2^3 x 3^2 x 5^2 x 7, then the number of consecutive zeroes are?

Answer 1

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.

$\begin{array} { c } { n = 2 ^ { 3 } \times 3 ^ { 2 } \times 5 ^ { 2 } \times 7 } \\\\ { n = 2 \times 2 ^ { 2 } \times 3 ^ { 2 } \times 5 ^ { 2 } \times 7 } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 2 ^ { 2 } \times 5 ^ { 2 } } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 10 ^ { 2 } } \\\\ { n = 2 \times 3 ^ { 2 } \times 7 \times 100 } \end{array}$

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

Answer 2

Answer:

The number of consecutive zeroes are two.

i.e.,

n = 2³×3²×5²×7

= 2×3²×7×(2×5)²

= 12600

Explanation:

Given a natural number n is

2³×3²×5²×7

To get the number consecutive zeroes we must find pairs of (2and 5 ).

Now , Rearranging the prime factors, we get

n = 2³×3²×5²×7

= 2×3²×7×(2²×5²)

= 2×3²×7× (2×5)²

= 2×3²×7×(10)²

= 2×3²×7×(100)

Or

= 12600

Therefore,

The number of consecutive zeroes are two.

i.e.,

n = 2³×3²×5²×7

= 12600

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