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The sum of the powers of the prime factors in $108 \times 192$ is:
A. 5
B. 7
C. 8
D. 12

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Last updated date: 20th Jun 2024
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Answer
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Hint: First of all find the prime factorization of $108$ and$192$ individually.(example: prime factorization of $48$ is $2 \times 2 \times 2 \times 2 \times 3$. Add the powers of prime numbers with the same base(example: $48 = {2^4} \times {3^1}$). Then multiply the prime factorization of both given numbers.

Complete step-by-step answer:
Step-1
Prime factorization of $108$ is $2 \times 2 \times 3 \times 3 \times 3$
${a^b} \times {a^c} = {a^{b + c}}$
${a^1} \times {a^1} \times ... \times {a^1} = {a^{1 + 1 + ... + 1}} = {a^n}$ when $a$ is multiplied by itself $n$ times
Use the above formula for adding powers of numbers having same base
On adding powers of primes having same base we can rewrite prime factorization of $108$ as ${2^2} \times {3^3}$
Step-2
Prime factorization of $192$ is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$
Again use the formula for adding powers given in step-1
We can rewrite prime factorization of $192$ as ${2^6} \times {3^1}$
Step-3
From step-1 and step-2 write the prime factorization of $108 \times 192$
$ \Rightarrow 108 \times 192 = {2^2} \times {3^3} \times {2^6} \times {3^1}$
$ \Rightarrow 108 \times 192 = {2^2} \times {2^6} \times {3^3} \times {3^1}$
${a^b} \times {a^c} = {a^{b + c}}$
Use above formula for adding powers of numbers having same base
 $ \Rightarrow 108 \times 192 = {2^{2 + 6}} \times {3^{3 + 1}}$
 $ = {2^8} \times {3^4}$
Step-4
Sum of the powers of the prime factors of $182 \times 192$ is
(powers of prime number $2$) + (powers of prime number $3$ )
$ = 8 + 4$
$ = 12$
So the sum of the powers of the primes of $108 \times 192$ is $12$ .
Hence option (D) is the correct answer.


Note: Don’t multiply the given numbers. find the prime factorization of the numbers individually.
Verify the prime factorization of a number by multiplying the prime factors so that it is equal to the required number.
Or
 From step-1 add the powers of prime factors of $108$
As $108 = {2^2} \times {3^3}$
$ \Rightarrow $ sum of powers of prime factors of $108$ is $2 + 3 = 5$
 Now, from step-2 add the powers of prime factors of $192$
As $192 = {2^6} \times {3^1}$
$ \Rightarrow $ sum of the powers of prime factors of $192$ is $6 + 1 = 7$
Now, the sum of powers of prime factors of $108 \times 192 = $
(sum of powers of prime factors of $108$) $ + $ (sum of powers of prime factors of $192$)
$ = 5 + 7$
$ = 12$.