Question
The sum of the exponents of the prime factors in the
prime factorization of 108 isSolution
ATQ, Letβs break 108 into prime factors: 108 = 2 Γ 2 Γ 3 Γ 3 Γ 3 So, 108 = 2Β² Γ 3Β³ Then, add the exponents: The exponent of 2 is 2. The exponent of 3 is 3. Total = 2 + 3 = 5. Hence, 5 is the answer.
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