For each integer \(n\), \(1<n<500\), there exists a number that is divisible by all positive integers less than or equal to \(n\) except \(m\) and \(m+1\). Find the sum of all possible \(m\).
Solution
Answer: \(447\)
The PDF argues that both omitted numbers \(m\) and \(m+1\) must be prime powers; otherwise, their proper factors would force them to divide the chosen number. Since they have different parity, one must be a power of \(2\).
The solution lists the candidate values \(1,2,3,4,7,16,31,127,256\). Adding them gives \(447\), the answer recorded in the PDF.