**π‘ Problem Formulation:** We want to know the remainder of a number consisting of `n`

consecutive 1’s when divided by another number `m`

. For example, if `n = 5`

and `m = 2`

, the number is 11111, and the remainder when dividing this number by 2 is 1.

## Method 1: Brute Force String Conversion

This method involves creating a string with `n`

consecutive 1’s and then converting this string into an integer before performing the modulus operation with `m`

. It’s straightforward but can be inefficient for large numbers.

Here’s an example:

def find_remainder(n, m): number_of_ones = '1' * n remainder = int(number_of_ones) % m return remainder print(find_remainder(5, 2))

Output: `1`

This code snippet defines a function `find_remainder`

that generates a string of 1’s with length `n`

, converts it to an integer, and finds the remainder when divided by `m`

. It’s a direct translation of the problem statement to code.

## Method 2: Mathematical Approach

By recognizing that a number composed of `n`

1’s is equivalent to a geometric series, we can use a mathematical formula to calculate the remainder without generating the actual number.

Here’s an example:

def find_remainder(n, m): remainder = (10**n - 1)//9 % m return remainder print(find_remainder(5, 2))

Output: `1`

This code unleashes the power of mathematical insight to compute the remainder of a number made up entirely of `n`

1’s divided by `m`

, using an algebraic formula derived from the sum of a geometric series.

## Method 3: Iterative Remainder Calculation

The iterative method keeps adding 1βs and taking the modulo with `m`

at each step to avoid large number computation.

Here’s an example:

def find_remainder(n, m): remainder = 0 for i in range(n): remainder = (remainder * 10 + 1) % m return remainder print(find_remainder(5, 2))

Output: `1`

Instead of creating a massive number, this code progressively calculates the remainder by iterating `n`

times, appending one more 1 with each iteration, and applying modulo operation.

## Method 4: Use of pow Function

This approach utilizes Python’s built-in `pow`

function for efficient exponentiation in modulo arithmetic, thereby calculating the remainder directly.

Here’s an example:

def find_remainder(n, m): remainder = (pow(10, n, m*9) - 1) // 9 return remainder print(find_remainder(5, 2))

Output: `1`

The code snippet saves processing power by leveraging Python’s `pow`

function that can raise a number to a power and take the modulo simultaneously, which is significantly faster for large numbers.

## Bonus One-Liner Method 5: Lambda Expression

A concise, one-liner solution using a lambda function can also solve the problem efficiently, offering both brevity and speed.

Here’s an example:

find_remainder = lambda n, m: int('1' * n) % m print(find_remainder(5, 2))

Output: `1`

This one-liner involves a lambda expression that encapsulates the logic of creating a string containing `n`

1’s and computing its modulo `m`

as the remainder. This method is elegant but might not be as readable for beginners.

## Summary/Discussion

**Method 1: Brute Force String Conversion.**Easy to understand. Not scalable for very large`n`

.**Method 2: Mathematical Approach.**Efficient for any size of`n`

. Requires understanding of geometric series.**Method 3: Iterative Remainder Calculation.**Avoids large number calculations. More efficient and readable than method 1.**Method 4: Use of pow Function.**Very efficient for very large numbers. Uses in-built Python functionality for heavy lifting.**Method 5: Lambda Expression.**Compact and efficient. Less readable for those unfamiliar with 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.