Legendre's Formula with a sample problem - pending

Q. Find the number of values of \( n \in \mathbb{N} \), such that that the largest exponent of 2 dividing \( n! \) equals \( n \).

1. The Formula
Legendre's Formula states that the largest exponent of a prime \( p \) dividing \( n! \) (denoted as \( v_p(n!) \)) is given by:
\[ v_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
An alternative and very useful form of this formula relates the exponent to the sum of the digits of \( n \) when written in base \( p \):
\[ v_p(n!) = \frac{n - s_p(n)}{p - 1} \]
where \( s_p(n) \) is the sum of the digits of \( n \) in base \( p \).
2. Solving the Problem
The problem asks for the number of values of \( n \in \mathbb{N} \) such that \( v_2(n!) = n \).
Step 1: Apply the formula for \( p = 2 \)
Using the second form of Legendre's Formula:
\[ v_2(n!) = \frac{n - s_2(n)}{2 - 1} = n - s_2(n) \]
Step 2: Set up the equation
According to the problem statement, we need \( v_2(n!) = n \). Substituting our formula:
\[ n - s_2(n) = n \]
Step 3: Simplify and analyze
Subtracting \( n \) from both sides gives:
\[ - s_2(n) = 0 \Rightarrow s_2(n) = 0 \]
Step 4: Conclusion
In base 2 (binary), \( s_2(n) \) represents the sum of the bits of \( n \). For any natural number \( n \in \mathbb{N} \) (\( n \ge 1 \)):
- \( n \) must have at least one digit equal to 1 in its binary representation.
- Therefore, \( s_2(n) \ge 1 \) for all \( n \in \mathbb{N} \).

Since \( s_2(n) \) can never be 0 for a natural number, there are zero values of \( n \) that satisfy this condition. Final Answer: The number of values of \( n \) is \( 0 \).