**π‘ Problem Formulation:** You’re tasked with determining whether the product of the first *n* natural numbers (also known as the factorial of *n*) is divisible by the sum of these numbers. For instance, if *n = 4*, the factorial `4!`

equals `24`

, and the sum of the first four natural numbers is `10`

. You want a Python function that checks if `24 % 10 == 0`

.

## Method 1: Iterative Approach

For the iterative approach, we define a function that calculates both the sum and the product of the first *n* natural numbers using a loop. Once we have both values, we use the modulo operator to check if the product is divisible by the sum.

Here’s an example:

def is_divisible_sum_product(n): total_sum = 0 total_product = 1 for i in range(1, n+1): total_sum += i total_product *= i return total_product % total_sum == 0 print(is_divisible_sum_product(4))

Output: False

This code snippet defines a function `is_divisible_sum_product()`

that iterates from 1 to *n*, accumulating the sum and product of these numbers. At the end of the loop, it checks the divisibility using the modulo operator and returns the boolean result.

## Method 2: Mathematics Shortcut

This method takes advantage of the mathematical property that a number *n* factorial (`n!`

) is never divisible by *n* for any *n* greater than 2. We’ll use a simple mathematical check to avoid unnecessary calculations.

Here’s an example:

def is_divisible_shortcut(n): return False if n > 2 else True print(is_divisible_shortcut(4))

Output: False

In this example, the function `is_divisible_shortcut()`

immediately returns `False`

if *n* is greater than 2. This takes advantage of the known mathematical fact and provides a simple, efficient check for divisibility in this particular case.

## Method 3: Functional Programming Approach

By employing Python’s functional programming tools such as `reduce()`

and `sum()`

, we can elegantly express the solution on a single line. This is a more Pythonic way, utilizing built-in high-order functions.

Here’s an example:

from functools import reduce from operator import mul def is_divisible_functional(n): numbers = range(1, n+1) product = reduce(mul, numbers) total_sum = sum(numbers) return product % total_sum == 0 print(is_divisible_functional(4))

Output: False

The `is_divisible_functional()`

function creates a range for the natural numbers, then uses the `reduce()`

function with the multiplication operator to find the product. It calculates the sum with `sum()`

and finally checks for divisibility.

## Method 4: Using Factorial and Sum Formulas

Instead of calculating the sum and product through an iterative process, we can use direct mathematical formulas — the factorial of *n* and the sum of an arithmetic series. This approach reduces complexity by avoiding loops.

Here’s an example:

import math def is_divisible_formula(n): product = math.factorial(n) total_sum = n * (n + 1) // 2 return product % total_sum == 0 print(is_divisible_formula(4))

Output: False

The function `is_divisible_formula()`

uses Python’s `math.factorial()`

to compute the factorial of *n*. It then calculates the sum with the arithmetic series sum formula and checks for divisibility.

## Bonus One-Liner Method 5: Compact Lambda Expression

For those who prefer concise Python expressions, lambda functions provide a way to write the entire operation in a single line. This approach is a compact version of the functional approach.

Here’s an example:

from functools import reduce from operator import mul is_divisible = lambda n: reduce(mul, range(1, n+1)) % sum(range(1, n+1)) == 0 print(is_divisible(4))

Output: False

This one-liner consists of a lambda function `is_divisible`

, which computes the factorial with `reduce()`

, calculates the sum with `sum()`

, and then performs the divisibility check, all in a single expression.

## Summary/Discussion

**Method 1:**Iterative Approach. Simple to understand. Time-consuming for large*n*due to the loop.**Method 2:**Mathematics Shortcut. Extremely fast. Limited to the mathematical property that only applies for*n*greater than 2.**Method 3:**Functional Programming Approach. Pythonic and concise. May be less readable to those unfamiliar with functional programming.**Method 4:**Using Factorial and Sum Formulas. Efficient for large*n*. Requires knowledge of the`math`

module.**Method 5:**Compact Lambda Expression. One-liner elegance. Readability might suffer for those not versed in lambda functions.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.