Question

2. If the prime factorization of a natural number n is 2^3 x 3^2 x 5^2 x 7, write the number of
consecutive zeros of n.​

Answer 1

Answer:

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{gathered}\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{gathered}

n=2

3

×3

2

×5

2

×7

n=2×2

2

×3

2

×5

2

×7

n=2×3

2

×7×2

2

×5

2

n=2×3

2

×7×10

2

n=2×3

2

×7×100

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