Computation Workshop Solution Checker

Counting Divisors

For any positive integer \(x\), let \(\tau(x)\) denote the number of positive integer factors of \(x\) Including \(1\) and \(x\) itself. For example \(\tau(3)=2\) and \(\tau(12)=6\). Given two numbers \(X\) and \(Y\), your task is to find the smallest positive integer \(N\) such that \(\tau(N+1)=X\) and \(\tau(N-1)=Y\).

Part A \(X = 18\) and \(Y = 8\)
Part B \(X = 108\) and \(Y = 3\)