**π‘ Problem Formulation:** Checking if a given integer `n`

is a power of 3 in Python is a common task that can be approached in several different ways. Given an input like `9`

, the function should return `True`

because 9 is 3 squared, whereas for an input like `8`

, the function should return `False`

, as 8 is not a power of 3.

## Method 1: Iterative Method

An intuitive method to determine if a given integer is a power of 3 is to iteratively divide the number by 3 until the result is 1 or the remainder is non-zero. This method is straightforward and can be implemented with a simple loop.

Here’s an example:

def is_power_of_three(n): if n < 1: return False while n % 3 == 0: n //= 3 return n == 1 print(is_power_of_three(9))

**Output:** True

This code snippet defines a function `is_power_of_three`

that checks if an integer `n`

is a power of 3. If `n`

is less than 1, it immediately returns `False`

. Otherwise, it iteratively divides `n`

by 3 until it cannot divide evenly anymore. If the final result is 1, it returns `True`

; otherwise, it returns `False`

.

## Method 2: Logarithmic Check

By using logarithms, we can check if the log base 3 of an integer is an integer itself. This method involves fewer operations than iterative methods and may be faster for large integers.

Here’s an example:

import math def is_power_of_three(n): if n < 1: return False # Calculate the log base 3 by dividing the logarithm of n by the logarithm of 3 log_base3 = math.log(n, 3) # Check if the log_base3 is an integer by comparing it to an integer version of itself return math.isclose(log_base3, round(log_base3)) print(is_power_of_three(27))

**Output:** True

This snippet uses the `math.log`

function to calculate the logarithm of `n`

with base 3. It then checks if the result is an integer using the `round`

function and comparing it with the `math.isclose`

function. The method returns `True`

if the two values are close enough to be considered equal, indicating `n`

is a power of 3.

## Method 3: Using Integer Limits

The maximum integer that is also a power of 3 within the 32-bit signed integer range is 3^19. We can check if the input integer is a power of three by seeing if 3^19 is divisible by it without a remainder.

Here’s an example:

def is_power_of_three(n): return n > 0 and (3**19) % n == 0 print(is_power_of_three(81))

**Output:** True

In this function, `is_power_of_three`

, we check if `n`

is positive and if `3**19`

(3 to the power of 19) is divisible by `n`

without any remainder. If these conditions are true, `n`

is a power of 3.

## Method 4: Recursive Check

Recursion can also be used to solve this problem by repeatedly calling the function with the input divided by 3 until the input becomes 1 or non-divisible by 3.

Here’s an example:

def is_power_of_three(n): return n == 1 if n < 0 else (n % 3 == 0 and is_power_of_three(n / 3)) print(is_power_of_three(243))

**Output:** True

This recursive implementation of `is_power_of_three`

checks if the input `n`

is immediately 1, indicating that the previous call was divisible by 3. If not, it checks divisibility by 3 and makes a recursive call with `n`

divided by 3.

## Bonus One-Liner Method 5: Using Set Membership

Leverage the properties of sets to check if the given integer is in a predefined set of powers of 3. This predefined set can be generated up to a certain limit to optimize the check.

Here’s an example:

powers_of_three = {3**i for i in range(20)} def is_power_of_three(n): return n in powers_of_three print(is_power_of_three(729))

**Output:** True

By creating a set `powers_of_three`

that contains 3 raised to the power of numbers from 0 to 19, we have a quick way to check membership using the `in`

keyword. This one-liner function is efficient for inputs that are within the precalculated range.

## Summary/Discussion

**Method 1:**Iterative Method. Simple and easy to understand. However, it can be slower for very large numbers.**Method 2:**Logarithmic Check. Efficient and concise. Rounding issues may cause inaccuracies with very large or small numbers.**Method 3:**Using Integer Limits. Extremely fast but only works within certain fixed limits, like the 32-bit signed integer range.**Method 4:**Recursive Check. Elegant and simple. Nevertheless, deep recursion can lead to stack overflow with very large numbers.**Method 5:**Using Set Membership. Fast for numbers within a predefined range but requires memory to store the set and is limited to that range.