 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/imperative-programming-resit/1/d/solution

p_{1}^{e_{1}} \dots p_{k}^{e_{k}}

with

0 \leq e_{j} \leq i_{j}

. Conversely every such exponent choice gives a divisor. The exponent choices are independent, so the product rule gives

τ (n) = j = 1 \prod k (i_{j} + 1) .

For

n = 1

the factorization is empty, and the empty product is one, corresponding to the divisor one.

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