5 Best Ways to Check Whether Sum of Digits at Odd Places of a Number is Divisible by k in Python

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💡 Problem Formulation: We need to analyze a given integer to determine if the sum of digits at odd indices (when counting from the right, starting at 1) is divisible by a specified integer k. For example, if the number is 123456 and k is 2, we calculate the sum of digits in odd places (1 + 3 + 5) and check if 9 is divisible by 2 (which in this case, it is not).

Method 1: Using Loops and Conditional Statements

This method entails converting the number to a string, iterating through characters at odd indices, summing them, and checking for divisibility by k. It’s straightforward, easy to understand for beginners, and allows for additional logic during iteration if needed.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.
def is_sum_odd_divisible_by_k(number, k):
  num_str = str(number)
  sum_odd = sum(int(num_str[i]) for i in range(0, len(num_str), 2))
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k accepts a number and a divisor k, then sums the digits at odd indices of the number and checks divisibility by k. This snippet demonstrates the function on number 123456 with k being 2, and correctly returns False.

Method 2: Using String Slicing

This method slices the string representation of the number to fetch digits at odd places and proceeds with summing and divisibility check. It is concise and leverages Python’s powerful slicing capabilities.

Here’s an example:

def is_sum_odd_divisible_by_k_slicing(number, k):
  num_str = str(number)[::2]
  sum_odd = sum(int(char) for char in num_str)
  return sum_odd % k == 0

print(is_sum_odd_divisible_by_k_slicing(123456, 2))

Output: False

By slicing the number string with a step of 2, the function is_sum_odd_divisible_by_k_slicing grabs every other digit starting from the first position (the odd positions), sums them up, and checks if the sum is divisible by k.

Method 3: Using List Comprehension

This approach makes use of list comprehension within Python, offering a compact and idiomatic way to perform operations on a list or string. It is essentially a more concise version of Method 1.

Here’s an example:

def is_sum_odd_divisible_by_k_list_comp(number, k):
  return sum(int(digit) for index, digit in enumerate(str(number)) if index % 2 == 0) % k == 0

print(is_sum_odd_divisible_by_k_list_comp(123456, 2))

Output: False

The function is_sum_odd_divisible_by_k_list_comp combines list comprehension with enumeration to sum digits at odd indices and then performs the modulo operation to check for divisibility by k.

Method 4: Using Recursion

This method applies recursion to sum digits at odd places and check divisibility. It’s an elegant solution that, while not being the most efficient for large numbers, demonstrates an important programming concept.

Here’s an example:

def is_sum_odd_divisible_by_k_recursive(number, k, is_odd=True):
  if number == 0:
    return 0
  remainder = is_sum_odd_divisible_by_k_recursive(number // 10, k, not is_odd)
  return remainder + (number % 10 if is_odd else 0) % k == 0

print(is_sum_odd_divisible_by_k_recursive(123456, 2))

Output: False

The recursive function is_sum_odd_divisible_by_k_recursive reduces the number by a factor of 10 each call, summing the last digit when the current level represents an odd position, and then checks divisibility by k.

Bonus One-Liner Method 5: Using the Reduce Function

This one-liner uses the functools.reduce function to iterate and sum the digits at odd places in a functional programming style. It’s the least readable for those unfamiliar with functional paradigms but is very expressive.

Here’s an example:

from functools import reduce

def is_sum_odd_divisible_by_k_reduce(number, k):
  return reduce(lambda acc, val: acc + int(val[1]) if val[0] % 2 == 0 else acc, enumerate(str(number)), 0) % k == 0

print(is_sum_odd_divisible_by_k_reduce(123456, 2))

Output: False

The is_sum_odd_divisible_by_k_reduce function effectively performs a fold operation, starting with an accumulator of 0, adding the value of the digit at an odd index each time through the enumeration of the string representation of the number.

Summary/Discussion

  • Method 1: Using Loops and Conditional Statements. Allows detailed control over the flow of the program. Can be a bit verbose.
  • Method 2: Using String Slicing. Offers a clean and concise solution utilizing Python’s string manipulation strengths. It is less explicit and might confuse those new to slicing.
  • Method 3: Using List Comprehension. Elegant and Pythonic, it makes for short readable code. Performance-wise, it is similar to a loop but may be harder to understand at a glance.
  • Method 4: Using Recursion. An academic approach that is not the most efficient for large numbers due to possible stack overflow but teaches important programming concepts.
  • Bonus One-Liner Method 5: Using the Reduce Function. Concise and functional, but can be obscure for those not used to functional programming.