We get the following code:

So basically we can invert the encryption and create a decryptor:

from sympy import mod_inverse
 
# Obtain modular inverse of 123 % 256
INV_123 = mod_inverse(123, 256)  # Calculate 123^(-1) % 256
 
def decrypt(ciphertext):
    pt = []
    for char in bytes.fromhex(ciphertext.strip()):  # Convert from hex to bytes
        pt.append(chr((INV_123 * (char - 18)) % 256))  # Apply ecuation
    return ''.join(pt)  # Convert list in string
 
# Read the encrypted flag
with open('./msg.enc', 'r') as f:
    ciphertext = f.read()
 
# Decrypt and show flag
print(decrypt(ciphertext))
 
# HTB{l00k_47_y0u_r3v3rs1ng_3qu4710n5_c0ngr475}

Challenge completed!