Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Imperative programming / 2 / d / Solution

By unique prime factorization, a positive divisor has precisely the formThere are $i_j+1$ independent choices for each prime exponent; different choices give different divisors, and every divisor occurs. The multiplication principle therefore gives $\boxed{\tau (n)=\prod _{j=1}^k(i_j+1)}$. For $n=1$, the empty product is $1$, giving its one positive divisor.