Question

the sum of exponents of the prime factors in the prime factorisation of 108??​

Answer 1

Answer:

$2^2 x 3^3.$

Step-by-step explanation:

108 ÷ 2 = 54 - No remainder! 2 is one of the factors!

54 ÷ 2 = 27 - No remainder! 2 is one of the factors!

27 ÷ 2 = 13.5 - There is a remainder. We can't divide by 2 evenly anymore. Let's try the next prime number

27 ÷ 3 = 9 - No remainder! 3 is one of the factors!

9 ÷ 3 = 3 - No remainder! 3 is one of the factors!

3 ÷ 3 = 1 - No remainder! 3 is one of the factors!

The orange divisor(s) above are the prime factors of the number 108. If we put all of it together we have the factors 2 x 2 x 3 x 3 x 3 = 108. It can also be written in exponential form as $2^2 x 3^3.$

Answer 2

Answer:

$2 {}^{2} \times {3}^{3}$

Step-by-step explanation:

The orange divisor(s) above are the prime factors of the number 108.

If we put all of it together we have the factors

2 x 2 x 3 x 3 x 3 = 108.

It can also be written in exponential form as

${2}^{2} \times {3}^{3}$