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.