Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 20th Jun 2024
Total views: 404.4k
Views today: 10.04k
Verified
404.4k+ views
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.

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 $192$)
$= 5 + 7$
$= 12$.