Key Concept: Legendre's Formula This problem requires us to find the largest natural number $n$ such that $3^n$ is a divisor of $66!$. This is equivalent to determining the exponent of the prime number $3$ in the prime factorization of $66!$. This can be efficiently found using Legendre's Formula (also known as de Polignac's Formula).

Legendre's Formula states that for a prime number $p$ and a natural number $N$, the exponent of $p$ in the prime factorization of $N!$ is given by: $E_p(N!) = \sum_{k=1}^{\infty} \left\lfloor \frac{N}{p^k} \right\rfloor$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. This formula can be expanded as: $E_p(N!) = \left\lfloor \frac{N}{p} \right\rfloor + \left\lfloor \frac{N}{p^2} \right\rfloor + \left\lfloor \frac{N}{p^3} \right\rfloor + \dots$

Step-by-Step Calculation In this problem, we need to find the exponent of $p=3$ in $N=66!$. Applying Legendre's Formula, we calculate $n = E_3(66!)$: $n = \left\lfloor \frac{66}{3} \right\rfloor + \left\lfloor \frac{66}{3^2} \right\rfloor + \left\lfloor \frac{66}{3^3} \right\rfloor + \left\lfloor \frac{66}{3^4} \right\rfloor + \dots$

1. First Term: Counting Multiples of $3^1$ The first term, $\left\lfloor \frac{66}{3} \right\rfloor$, identifies all natural numbers from $1$ to $66$ that are multiples of $3$. Each of these numbers contributes at least one factor of $3$ to $66!$. $\left\lfloor \frac{66}{3} \right\rfloor = \left\lfloor 22 \right\rfloor = 22$ This means there are $22$ numbers ($3, 6, 9, \dots, 66$) that are multiples of $3$.

2. Second Term: Counting Additional Factors from Multiples of $3^2$ The second term, $\left\lfloor \frac{66}{3^2} \right\rfloor = \left\lfloor \frac{66}{9} \right\rfloor$, counts numbers from $1$ to $66$ that are multiples of $9$. Numbers like $9, 18, \dots, 63$ contain at least two factors of $3$. One factor has already been counted in the first term (as they are also multiples of $3$). This term accounts for the second factor of $3$ that these numbers contribute. $\left\lfloor \frac{66}{9} \right\rfloor = \left\lfloor 7.333 \dots \right\rfloor = 7$ There are $7$ such numbers.

3. Third Term: Counting Additional Factors from Multiples of $3^3$ Similarly, the third term, $\left\lfloor \frac{66}{3^3} \right\rfloor = \left\lfloor \frac{66}{27} \right\rfloor$, counts numbers from $1$ to $66$ that are multiples of $27$. Numbers like $27, 54$ contain at least three factors of $3$. This term accounts for the third factor of $3$ these numbers contribute. $\left\lfloor \frac{66}{27} \right\rfloor = \left\lfloor 2.444 \dots \right\rfloor = 2$ There are $2$ such numbers.

4. Subsequent Terms: Multiples of $3^k$ for $k \ge 4$ We continue this process for higher powers of $3$: $\left\lfloor \frac{66}{3^4} \right\rfloor = \left\lfloor \frac{66}{81} \right\rfloor = \left\lfloor 0.814 \dots \right\rfloor = 0$ Since $3^4 = 81$ is greater than $66$, there are no natural numbers from $1$ to $66$ that are multiples of $81$. Thus, this term and all subsequent terms (for $3^5, 3^6, \dots$) will be $0$. This indicates we have accounted for all possible factors of $3$.

Summing the Exponents To find the total exponent $n$, we sum the values from each calculated term: $n = 22 + 7 + 2 = 31$ Therefore, the largest natural number $n$ such that $3^n$ divides $66!$ is $31$.

Tips and Common Mistakes

• Prime Base Only: Legendre's Formula is designed specifically for finding the exponent of a prime number in a factorial. If the base is a composite number (e.g., $10^n$), you must first find the exponents of its prime factors (e.g., $2^n$ and $5^n$) separately and then determine the minimum of these exponents.
• The Floor Function is Essential: Always apply the greatest integer function ($\lfloor x \rfloor$) to each term. This ensures that you only count whole multiples and correctly reflects the number of factors.
• Don't Stop Prematurely: Continue calculating terms until the argument of the floor function becomes less than $1$ (which means the floor value will be $0$). Stopping too early will lead to an underestimation of the total exponent.
• Understanding Each Term: Each term $\left\lfloor \frac{N}{p^k} \right\rfloor$ represents the count of numbers up to $N$ that are multiples of $p^k$. By summing these, the formula cleverly accounts for each prime factor $p$ exactly once, regardless of how many times it divides a particular number in the factorial.

Summary and Key Takeaway Legendre's Formula provides a precise and systematic method to determine the highest power of any prime number that divides a given factorial. By applying this formula, we found that the highest power of $3$ that divides $66!$ is $3^{31}$. Hence, the largest natural number $n$ satisfying the condition is $31$.