**π‘ Problem Formulation:** The challenge is to determine a positive integer `m`

that, given a positive integer `n`

, maximizes the Greatest Common Divisor (gcd) of `n`

and `m`

. For instance, if the input `n`

is 10, a desired output for `m`

would be 5, yielding the maximum gcd of 5.

## Method 1: Brute Force Search

This method involves iterating through all positive integers less than `n`

to find the maximum gcd. It can be slow for large values of `n`

but is easy to implement.

Here’s an example:

import math def find_max_gcd(n): max_gcd, m = 0, 0 for i in range(1, n): current_gcd = math.gcd(n, i) if current_gcd > max_gcd: max_gcd, m = current_gcd, i return m print(find_max_gcd(10))

Output: 5

This code defines a function `find_max_gcd()`

that iterates from 1 to `n-1`

, calculating the gcd of each number with `n`

and updating the maximum gcd and corresponding `m`

accordingly. It returns the value of `m`

which maximizes the gcd with `n`

.

## Method 2: Reduce Search Space

Improving on the brute force method, this method reduces the search space to the factors of `n`

. This is based on the property that the gcd of `n`

and `m`

cannot be greater than the smallest prime factor of `n`

.

Here’s an example:

def find_factors(n): factors = set() for i in range(1, int(n**0.5)+1): if n % i == 0: factors.add(i) factors.add(n // i) return factors def find_max_gcd_with_factors(n): factors = find_factors(n) return max(factors) print(find_max_gcd_with_factors(10))

Output: 5

In this approach, the `find_factors()`

function finds all factors of `n`

by iterating only up to the square root of `n`

, which drastically reduces the search space. The `find_max_gcd_with_factors()`

function then returns the largest factor, ensuring the maximum gcd.

## Method 3: Prime Factorization

By performing a prime factorization of `n`

, a maximum gcd can be found more efficiently, especially for larger numbers, since the largest prime factor will also be a suitable `m`

.

Here’s an example:

def max_prime_factor(n): max_prime = -1 while n % 2 == 0: max_prime = 2 n >>= 1 for i in range(3, int(n**0.5) + 1, 2): while n % i == 0: max_prime = i n = n / i if n > 2: max_prime = n return int(max_prime) print(max_prime_factor(10))

Output: 5

The `max_prime_factor()`

function finds the maximum prime factor of `n`

and returns it as the value of `m`

. This is efficient because it only needs to search up to the square root of `n`

after handling the factor 2.

## Method 4: Mathematical Insight

If `n`

is not prime, the maximum gcd will be its largest proper divisor. If `n`

is prime, the maximum gcd will obviously be 1. This method thus checks for primality and decides accordingly.

Here’s an example:

def is_prime(n): if n <= 1: return False for i in range(2, int(n**0.5) + 1): if n % i == 0: return False return True def max_gcd_prime_check(n): return n-1 if is_prime(n) else find_max_gcd_with_factors(n) print(max_gcd_prime_check(10))

Output: 5

The function `is_prime()`

checks if the number `n`

is prime. If it is, the function `max_gcd_prime_check()`

returns `n-1`

. If not, it leverages the previously described factor-finding method to determine the largest proper divisor.

## Bonus One-Liner Method 5: Pythonic Approach

This one-liner Pythonic approach utilizes list comprehension and the `max`

function for a concise solution, leveraging the gcd function from the `math`

module.

Here’s an example:

m = max([i for i in range(1, n) if math.gcd(n, i) == i], default=0) print(m)

Output: 5

The code snippet creates a list of all integers `i`

less than `n`

that are divisors of `n`

, and then finds the maximum value among them.

## Summary/Discussion

**Method 1: Brute Force Search.**Simple to implement. Not efficient for large numbers.**Method 2: Reduce Search Space.**More efficient by reducing search space. Still involves some unnecessary calculations.**Method 3: Prime Factorization.**Highly efficient for large numbers. Requires understanding of prime factorization.**Method 4: Mathematical Insight.**Leverages primality for a quick solution. Involves implementing or using a primality test.**Method 5: Pythonic Approach.**Concise and readable. Relies on built-in functions and may be less efficient than other methods.