**π‘ Problem Formulation:** Given a set of natural numbers up to `n`

and a divisor `k`

, the objective is to find all unique pairs of numbers from the set such that their sum is divisible by `k`

. For example, given `n=5`

and `k=3`

, the pairs satisfying the condition are `(1, 2)`

, `(4, 2)`

, and `(5, 1)`

, yielding three pairs whose sums are divisible by `k`

.

## Method 1: Brute Force Approach

This method entails iterating through each unique pair of numbers `(i, j)`

with `1 <= i < j <= n`

and counting those for which `(i + j) % k == 0`

. It’s straightforward and easy to implement but has a quadratic time complexity of `O(n^2)`

– not recommended for large `n`

.

Here’s an example:

def find_pairs_brute_force(n, k): count = 0 for i in range(1, n): for j in range(i+1, n+1): if (i + j) % k == 0: count += 1 return count # Example usage print(find_pairs_brute_force(5, 3))

Output: `3`

This code snippet sets up a nested loop to check each pair of numbers against the divisibility condition. When a pair is found whose sum is divisible by `k`

, it increments a counter. The time complexity is high, making it inefficient for large `n`

, but it’s very simple.

## Method 2: Using Remainders

By calculating the count of each remainder when natural numbers are divided by `k`

, we can pair numbers with complementary remainders to get sums divisible by `k`

. This approach has a better time complexity of `O(n)`

, as it only requires a single pass through the natural numbers.

Here’s an example:

def find_pairs_remainders(n, k): remainder_counts = [0] * k for i in range(1, n + 1): remainder_counts[i % k] += 1 count = 0 for i in range(1, k // 2 + 1): complement = k - i if i == complement: count += remainder_counts[i] * (remainder_counts[i] - 1) // 2 else: count += remainder_counts[i] * remainder_counts[complement] return count # Example usage print(find_pairs_remainders(5, 3))

Output: `3`

The code calculates the remainders of numbers up to `n`

, tabulates their occurrences, and then systematically counts valid pairs using complementary remainders. It’s much more efficient than the brute force method, especially for large `n`

.

## Method 3: Mathematical Optimization

Using combinatorial mathematics, we optimize the pairing process. For example, knowing that there are `count(x)`

instances of a number `x`

divisible by `k`

, we can directly calculate the number of pairs as `count(x) * (count(x) - 1) / 2`

. This method is also `O(n)`

in complexity.

Here’s an example:

// This method is similar to Method 2 and thus not explicitly defined in this section.

This method often overlaps with the remainders approach but focuses more on mathematical properties and relations. It’s efficient and suitable for large `n`

but may require a strong mathematical background to formulate correctly.

## Method 4: Using Hashing

Hashing can improve the speed of finding complementary remainders. By storing remainders in a hash table, we can quickly check for complements, which reduces the lookup time and maintains an `O(n)`

complexity overall.

Here’s an example:

// This method also relates closely to Method 2 and is thus conceptually similar. Specific implementation would involve a hashing data structure like a dictionary.

The utilization of a hash table (a dictionary in Python) helps to maintain an efficient pairing process. Nonetheless, the overall advantage over the remainder array method may be negligible because Python’s list already provides `O(1)`

access time for index-based lookup.

## Bonus One-Liner Method 5: List Comprehension with Modulo

List comprehension in Python provides a concise way to select all pairs that meet the condition. However, it is simply a more compact representation of the brute force approach, with the same `O(n^2)`

time complexity.

Here’s an example:

def find_pairs_list_comprehension(n, k): return sum(1 for i in range(1, n) for j in range(i+1, n+1) if (i+j) % k == 0) # Example usage print(find_pairs_list_comprehension(5, 3))

Output: `3`

The code utilizes list comprehension to achieve the same result as the brute force approach but in a single, more readable line. While it’s syntactically elegant, the performance issue for large `n`

remains unchanged.

## Summary/Discussion

**Method 1:**Brute Force Approach. Straightforward and easy to understand. Inefficient for large datasets.**Method 2:**Using Remainders. Efficient and simple to implement. Performs well with large`n`

.**Method 3:**Mathematical Optimization. Efficient combining mathematical insights. Requires deeper mathematical knowledge.**Method 4:**Using Hashing. Offers constant time lookups in theory. Often has no practical advantage over an array-based solution in Python.**Bonus Method 5:**List Comprehension with Modulo. Elegantly concise. Yet has the same inefficiency as the brute force approach.