Question

The number of prime factors in ${\left(\frac{1}{6}\right)}^{12}\times (8{)}^{25}\times {\left(\frac{3}{4}\right)}^{15}$

• A. $33$
• B. $37$
• C. $52$
• D. None of these

Answer 1

The correct option is D None of these

Consider the given expression ${\left(\frac{1}{6}\right)}^{12}\times {\left(8\right)}^{25}\times {\left(\frac{3}{4}\right)}^{15}$ and express it as a product of power of primes as shown below:

${\left(\frac{1}{6}\right)}^{12}\times {\left(8\right)}^{25}\times {\left(\frac{3}{4}\right)}^{15}={\left(\frac{1}{2\times 3}\right)}^{12}\times {(2\times 2\times 2)}^{25}\times {\left(\frac{3}{2\times 2}\right)}^{15}={\left(\frac{1}{2\times 3}\right)}^{12}\times {\left(2^3\right)}^{25}\times {\left(\frac{3}{2^2}\right)}^{15}=\frac{1^{12}}{{(2\times 3)}^{12}}\times {\left(2^3\right)}^{25}\times \frac{3^{15}}{{\left(2^2\right)}^{15}}\{\because {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\}$

$=\frac{1}{2^{12}\times 3^{12}}\times {\left(2\right)}^{3\times 25}\times \frac{3^{15}}{{\left(2\right)}^{2\times 15}}\{\because {(a\times b)}^n=a^n\times b^{n}and{\left(a^m\right)}^n=a^{m\times n}\}={\left(2\right)}^{-12}\times {\left(3\right)}^{-12}\times {\left(2\right)}^{75}\times {\left(3\right)}^{15}\times {\left(2\right)}^{-30}\{\because \frac{1}{a^n}=a^{-n}\}={\left(2\right)}^{-12+75-30}\times {\left(3\right)}^{-12+15}\{\because a^m\times a^n=a^{m+n}\}={\left(2\right)}^{33}\times {\left(3\right)}^3$

Now, the number of prime factors is equals to the sum of the exponents of the primes in the prime factorization of a number, therefore, the number of prime factors are $33+3=36$.

Hence, the number of prime factors in ${\left(\frac{1}{6}\right)}^{12}\times {\left(8\right)}^{25}\times {\left(\frac{3}{4}\right)}^{15}$ are $36$.