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Q.

The sum of the exponents of the prime factors in the prime factorization of 108 is

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a

5

b

1

c

6

d

4

answer is A.

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Detailed Solution

$\begin{matrix} 2108 \\ 254 \\ 327 \\ 39 \\ 23 \\ 1 \end{matrix}$

$108 = 2^{2} \times 3^{3}$

Sum of the exponent $= 2 + 3 = 5$

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The sum of the exponents of the prime factors in the prime factorization of 108 is