**π‘ Problem Formulation:** We’re tasked with creating a program in Python that identifies a number `m`

which, when written in base 10, has exactly `n`

zeroes at the end. For instance, if `n`

is 3, a valid `m`

might be 1000 because it ends in three zeroes.

## Method 1: Brute Force Incrementation

This method involves beginning at a number (say 10^n) and incrementally checking each number for the desired number of trailing zeroes. This can be computationally intensive as n increases, but it is straightforward to understand and implement.

Here’s an example:

def find_number_with_zeroes(n): candidate = 10 ** n while str(candidate)[-n:] != '0' * n: candidate += 1 return candidate print(find_number_with_zeroes(3))

Output: 1000

This snippet defines a function `find_number_with_zeroes(n)`

which looks for the smallest number with `n`

zeroes at the end. It increments from an initial estimate until the correct number is found.

## Method 2: Using String Formatting

The string formatting approach leverages Python’s inherent string manipulation abilities, allowing us to format a given base number by appending the needed zeros at the end, effectively building the number from a string.

Here’s an example:

def find_number_with_zeroes(n): return int('1' + '0' * n) print(find_number_with_zeroes(3))

Output: 1000

The function `find_number_with_zeroes(n)`

constructs a string representing the desired number and then converts it back to an integer. The advantage of this method is its simplicity and efficiency.

## Method 3: Mathematical Calculation

Instead of iterating or manipulating strings, this method uses direct mathematical computation to find the number. This is efficient and quick, as it bypasses any need for iteration or string handling.

Here’s an example:

def find_number_with_zeroes(n): return 10 ** n print(find_number_with_zeroes(3))

Output: 1000

This code defines `find_number_with_zeroes(n)`

, which uses exponentiation to calculate the number containing `n`

trailing zeroes directly, making it both fast and simple.

## Method 4: Utilizing Regular Expressions

We can use Python’s regular expressions module to match patterns in strings. Here, we increment numbers and use a regex to check for the requisite number of trailing zeroes.

Here’s an example:

import re def find_number_with_zeroes(n): candidate = 10 ** n while not re.fullmatch(r'\d*0{' + str(n) + '}', str(candidate)): candidate += 1 return candidate print(find_number_with_zeroes(3))

Output: 1000

The function `find_number_with_zeroes(n)`

checks each incremented number against a regex pattern that validates the trailing zeroes. This method is more complex and less efficient than the previous ones.

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

This compact one-liner combines list comprehension and a condition check. It’s elegant and utilizes Python’s concise syntax, making it great for small values of n.

Here’s an example:

print([m for m in range(10**(n-1), 10**n) if str(m).endswith('0' * n)][0])

Output: 1000

Here we create a list of numbers that meet our condition and then extract the first element. This one-liner assumes that `n`

is defined in the code and is a clean solution provided n isn’t excessively large.

## Summary/Discussion

**Method 1:**Brute Force Incrementation. Simple, but inefficient. Best for small n values.**Method 2:**Using String Formatting. Efficient and very quick for all n values. Limited by integer size in Python.**Method 3:**Mathematical Calculation. Direct and fast computation. Best approach when performance matters.**Method 4:**Utilizing Regular Expressions. More complex, not as efficient, but can be useful for pattern matching tasks.**Method 5:**Bonus One-Liner List Comprehension. Elegant and concise. It lacks efficiency for large n.