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In the prime factorization of 109!, what is the exponent of 3? (Reminder: The number n! is the product of the integers from 1 to n. For example, 5!=5* 4*3*2*1= 120$.)

In the prime factorization of 109!, what is the exponent of 3? (Reminder: The number n! is the product of the integers from 1 to n. For example, 5!=5* 4*3*2*1= 120$.)
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In the prime factorization of 109!, what is the exponent of 3? (Reminder: The number n! is the product of the integers from 1 to n. For example, 5!=5* 4*3*2*1= 120$.)

User Nithin Baby
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1 Answer

2 votes

Answer:

53 is the exponent

Explanation:

User Gian Segato
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