**π‘ Problem Formulation:** In this article, we tackle the problem of finding the first element in an arithmetic progression (AP) that is also a multiple of a given prime number. This type of problem is commonly encountered in algorithmic challenges and mathematics-centered programming tasks. For instance, given an AP starting at 10 with a common difference of 5, and a prime number 3, the desired output would be the first number in the sequence divisible by 3.

## Method 1: Brute Force Iteration

The brute force method employs a simple iteration over the arithmetic progression, checking each term for divisibility by the given prime number until the first multiple is found. This method uses a loop to generate terms sequentially until the condition is met.

Here’s an example:

def find_first_multiple(start, difference, prime): element = start while element % prime != 0: element += difference return element first_multiple = find_first_multiple(10, 5, 3) print(first_multiple)

Output:

`15`

In the snippet above, `find_first_multiple()`

iterates through the AP beginning at 10 with a difference of 5 until it finds a number that’s a multiple of the prime 3. The function returns 15, which is the first number in the AP that meets the condition.

## Method 2: Modulo-Based Calculation

This approach reduces the number of iterations needed by using modulo arithmetic to directly calculate the first multiple of the prime in the arithmetic progression. It is more efficient than brute force when dealing with large numbers.

Here’s an example:

def find_first_multiple_mod(start, difference, prime): offset = (-start) % prime return start + (offset // difference) * difference + (prime if offset % difference != 0 else 0) first_multiple_mod = find_first_multiple_mod(10, 5, 3) print(first_multiple_mod)

Output:

`15`

The function `find_first_multiple_mod()`

calculates the necessary offset to adjust the starting term so it becomes a multiple of the given prime. The expression then effectively jumps straight to the first valid term in the sequence.

## Method 3: Use of Arithmetic Progression Properties

This method applies mathematical insights about arithmetic progressions and divisibility to deduce the right term. By understanding the properties of an AP, the calculations become straightforward and computationally inexpensive.

Here’s an example:

def find_first_multiple_properties(start, difference, prime): nth_term = (prime - start % prime) % prime return start + nth_term * difference first_multiple_properties = find_first_multiple_properties(10, 5, 3) print(first_multiple_properties)

Output:

`15`

Here, `find_first_multiple_properties()`

directly computes the nth term of the AP that would be the multiple of the prime. By calculating the positional offset based on the modulus with the prime and adding that to the start, it yields the same accurate answer with fewer calculations.

## Method 4: Algebraic Simplification and Computation

By setting up an algebraic equation for the AP term and the prime, and solving for the term, you can find the solution in an algebraic fashion. This approach shines in clarity and mathematical elegance.

Here’s an example:

def find_first_multiple_algebraic(start, difference, prime): return prime * ((start // prime) + (1 if start % prime else 0)) first_multiple_algebraic = find_first_multiple_algebraic(10, 5, 3) print(first_multiple_algebraic)

Output:

`15`

The function `find_first_multiple_algebraic()`

constructs the first valid term using a simple algebraic expression. The equation takes the integer part of the AP start term divided by the prime, adds 1 if the start isn’t already a multiple, and multiplies by the prime to get to the next biggest multiple.

## Bonus One-Liner Method 5: Pythonic One-Liner with Comprehension

This succinct method utilizes Python’s list comprehension and generator expression features to find the first element in the sequence. It’s a compact and ‘Pythonic’ way to condense the logic into a single line of code.

Here’s an example:

first_multiple_oneliner = next(x for x in range(10, 100, 5) if x % 3 == 0) print(first_multiple_oneliner)

Output:

`15`

The one-liner uses a generator expression along with the `next()`

function to quickly scan through a range representing our AP, immediately stopping and returning the first value that satisfies the modulus check for divisibility by the prime.

## Summary/Discussion

**Method 1: Brute Force Iteration.**Simple to understand. Could be slow for large numbers or large differences in the AP.**Method 2: Modulo-Based Calculation.**More efficient. Requires understanding of modulo arithmetic. May not be as instantly clear to beginners.**Method 3: Use of Arithmetic Progression Properties.**Efficient and mathematically elegant. Relies on the reader’s knowledge of AP properties.**Method 4: Algebraic Simplification and Computation.**Very efficient and clear. Especially useful when dealing with very large sequences or numbers.**Bonus One-Liner Method 5: Pythonic One-Liner with Comprehension.**Concise and elegant. However, less readable and potentially slower due to temporary range generation.