Question

If the expansion of 126! is completely divisible by 3"k, then what is the maximum value of k? (N! is the product of all the positive integers from N to 1)

• A. 56
• B. 61
• C. 19
• D. 72

Answer 1

The Correct Option is B

Finding the Maximum Value of \( k \) in \( 126! \) Divisible by \( 3^k \): 

To determine the maximum value of \( k \) such that \( 126! \) is divisible by \( 3^k \), we need to calculate the exponent of \( 3 \) in the prime factorization of \( 126! \). This is done using the formula:

\[ \text{Exponent of } p \text{ in } N! = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \dots \]

Step-by-Step Calculation:

Here, \( p = 3 \) and \( N = 126 \). Let’s compute the exponent step by step:

• First term: \[ \left\lfloor \frac{126}{3} \right\rfloor = 42 \]
• Second term: \[ \left\lfloor \frac{126}{9} \right\rfloor = 14 \]
• Third term: \[ \left\lfloor \frac{126}{27} \right\rfloor = 4 \]
• Fourth term: \[ \left\lfloor \frac{126}{81} \right\rfloor = 1 \]
• Fifth term: \[ \left\lfloor \frac{126}{243} \right\rfloor = 0 \] Since this term is 0, we stop here.

Total Exponent:

\[ 42 + 14 + 4 + 1 = 61 \]

Final Answer:

Thus, the maximum value of \( k \) is 61 (Option B).