Question

Maximum value n such that (66)! is divisible by 3n

Answer 1

We are asked to find the maximum value of \( n \) such that \( (66)! \) is divisible by \( 3^n \). 

Step 1: Use Legendre's formula 
Legendre's formula gives the highest power of a prime \( p \) that divides \( n! \). It is given by: \[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor. \] In this case, we want to find the power of 3 that divides \( 66! \), so we use \( p = 3 \) and \( n = 66 \). 

Step 2: Calculate the sum for \( p = 3 \) 
We calculate each term in the sum: \[ \left\lfloor \frac{66}{3} \right\rfloor = \left\lfloor 22 \right\rfloor = 22, \] \[ \left\lfloor \frac{66}{3^2} \right\rfloor = \left\lfloor \frac{66}{9} \right\rfloor = \left\lfloor 7.33 \right\rfloor = 7, \] \[ \left\lfloor \frac{66}{3^3} \right\rfloor = \left\lfloor \frac{66}{27} \right\rfloor = \left\lfloor 2.44 \right\rfloor = 2, \] \[ \left\lfloor \frac{66}{3^4} \right\rfloor = \left\lfloor \frac{66}{81} \right\rfloor = \left\lfloor 0.8148 \right\rfloor = 0. \] Since \( 3^k > 66 \) for \( k \geq 5 \), all higher terms will be 0. 

Step 3: Add the terms 
The total power of 3 dividing \( 66! \) is the sum of the terms: \[ 22 + 7 + 2 = 31. \] 

Final Answer: \[ \boxed{31}. \]