Numbers2 is the successor of an older challenge presented at N00bzCtf 2023 with a few differences:

• It outputs and additional welcome message
• It asks for three different challenge questions
• successful solving attempts are being answered with different responses

We have to solve several programming challenges here:

1. Challenge inputs do not end on EOL but on ': ' instead. So we cannot simply use the input() function to read the questions from stdin
2. We have to parse two integers from within a line of text
3. Questions for the least common multiple or the greatest common divisor can be answered using the builtin methods from the math module math.lcm() / math.gcd(). But to answer the question asking for the largest prime factor we must implement an algorithm or an appropriate library from pip must be used

1. can easily be achieved using the re.findall() method:

[x] = re.findall(r'\d+', "Give me the greatest prime factor of 42")
# x = 42

x, y = re.findall(r'\d+', "Give me the least common multiple of 42 and 23")
# x = "42"
# x = "23"

To solve 2. we need to write an own implementation of an input method that stops reading single characters once it sees a ':'. We simply call sys.stdin.read(1) in a loop and decide whether to continue reading chars or not.

def input_until_char(sep):
    input_read = ""

    c = sys.stdin.read(1)
    input_read += c
    while c != "" and c != "\n" and c != sep:
        c = sys.stdin.read(1)
        input_read += c
    return input_read

For 3. a naive implementation of using a loop over all prime factor candidates to find out whether a candidate factor is prime is fast enough to find the largest prime factor:

largest_prime_factor = 1
for n in range(1,int(x)+1):
    is_prime = True
    for factor in range(2, n):
        if n%factor == 0:
            is_prime = False
            break
    if is_prime and int(x)%n == 0 and n > largest_prime_factor:
        largest_prime_factor = n