We only need to examine numbers from 1 up to ⌊√n⌋. Because any divisor d larger than √n pairs with a quotient n/d smaller than √n, testing up to √n finds them all.
Class 9 Chapter 11 The World of Algorithms solutions
Class 9 Maths Ganita Manjari Part 2 Book and Solutions
We only need to examine numbers from 1 up to √n (specifically, the integer part ⌊√n⌋). If a number d divides n, then n/d is also a divisor. One factor in the pair must always be less than or equal to √n, so finding all smaller divisors automatically discovers the paired larger ones without scanning all the way to n.
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