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 formu...