Solution
= Solution
By the time `print(n)` is called, `record` has evaluated `divisors(i)` once for every $1\leq i\leq n$. In the <unit-cost model>, the total number of remainder tests is
$$
\sum_{i=1}^{n}i=\frac{n(n+1)}2.
$$
The <record values> comparisons add only $O(n)$ work, and output is cheap by assumption. \b[The elapsed time is $\Theta(n^2)$.] This measures time to reach a given <highly composite number>, rather than time as a function of the number of printed records.