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Lesson 1 of 8

Find divisors in pairs

Replace a full scan with an integer-safe square-root boundary.

Understand the idea

For positive n, each divisor d has partner n/d. At least one member of a pair is at most √n, so test d while d ≤ n/d. Process the partner only when it differs from d, avoiding a duplicated square root.

Trace a small example

For 36, the pairs are (1,36), (2,18), (3,12), (4,9), (6,6). Its nine divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. To output them sorted, collect small divisors and large partners separately, then reverse the large list.

Conditions and cost

The scan costs O(√n), plus output work. n/d avoids overflow in d*d; d starts at 1. The input 1 has one positive divisor and no positive proper divisors. Zero does not have a finite positive-divisor list.

Pause & predict

Why process n/d only when it differs from d?

Choose an answer
Try it yourself

Turn the idea into code

List the divisors of 18 in increasing order.

Give me a hint

Find pairs using d = 1 through 4.

Show the explained solution

Pairs (1,18), (2,9), (3,6) yield 1, 2, 3, 6, 9, 18.

Your notes

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