 Solution

ID: 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 form

d = p_{1}^{e_{1}} \dots p_{k}^{e_{k}}, 0 \leq e_{j} \leq i_{j} .

There are

i_{j} + 1

independent choices for each prime exponent; different choices give different divisors, and every divisor occurs. The multiplication principle therefore gives $τ (n) = j = 1 \prod k (i_{j} + 1)$ . For

n = 1

, the empty product is

1

, giving its one positive divisor.

 New to topics? Read the docs here!