Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming (resit) / 1 / a

Test each positive candidate divisor and count the successful remainder tests:The loop invariant is $1\le d\le n+1$ and $count=|\{j:1\le j<d,j\mid n\}|$, where $d$ denotes candidate. A successful test adds exactly the new divisor; an unsuccessful test leaves the count unchanged. The loop variant is $n+1-d$, which decreases by one. At termination $d=n+1$, so the returned value is $\tau (n)$. A Long loop counter prevents the increment after the last positive Int candidate from wrapping around.