Question:
If \( \frac{6!}{3^m} \) is an integer, which of the following options are greater than the largest possible value of \( m \)? Indicate all possible values.
If \( \frac{6!}{3^m} \) is an integer, which of the following options are greater than the largest possible value of \( m \)? Indicate all possible values.
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Always use prime factorization to determine the maximum power of a prime dividing a factorial.
Updated On: Sep 30, 2025
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The Correct Option is D
Solution and Explanation
Step 1: Factorize 6!
We know that: \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \] Prime factorization of \( 720 \): \[ 720 = 2^4 \times 3^2 \times 5 \]
Step 2: Condition for \( \frac{6!}{3^m} \) to be an integer.
For the expression to be an integer, \( 3^m \) must divide \( 6! \). Since the highest power of 3 in the factorization is \( 3^2 \), the largest possible value of \( m \) is: \[ m = 2 \]
Step 3: Compare options with the largest \( m \).
Largest possible \( m = 2 \). Options greater than 2 are: 3, 4, and 5.
Step 4: Eliminate carefully.
The question asks for values \emph{greater than the largest possible \( m \)}. Thus the correct answers are: \[ \boxed{4 \text{ and } 5} \]
We know that: \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \] Prime factorization of \( 720 \): \[ 720 = 2^4 \times 3^2 \times 5 \]
Step 2: Condition for \( \frac{6!}{3^m} \) to be an integer.
For the expression to be an integer, \( 3^m \) must divide \( 6! \). Since the highest power of 3 in the factorization is \( 3^2 \), the largest possible value of \( m \) is: \[ m = 2 \]
Step 3: Compare options with the largest \( m \).
Largest possible \( m = 2 \). Options greater than 2 are: 3, 4, and 5.
Step 4: Eliminate carefully.
The question asks for values \emph{greater than the largest possible \( m \)}. Thus the correct answers are: \[ \boxed{4 \text{ and } 5} \]
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