import math
def is_prime(n):
if n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
def sum_digits(n):
return sum(int(d) for d in str(n))
def mystery(x):
# Condition 1: x must be an integer
if not isinstance(x, int):
return False
# Condition 2: x must be positive
if x <= 0:
return False
# Condition 3: x must be a perfect square
try:
sqrt_x = math.isqrt(x)
except ValueError:
return False
if sqrt_x * sqrt_x != x:
return False
# Condition 4: The sum of the digits of x must be a prime number
s_digits = sum_digits(x)
if not is_prime(s_digits):
return False
# Condition 5: The string representation of x must contain the digit '5' exactly once
str_x = str(x)
if str_x.count('5') != 1:
return False
# Condition 6: x must be greater than 1000 and less than 10000
if not (1000 < x < 10000):
return False
return True
import datetime
import zlib
def is_prime(n):
if n < 2: return False
# A simple primality test is sufficient here.
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
def mystery(x):
# Condition 1: The input must be a string.
if not isinstance(x, str):
return False
# Condition 2: It must be parseable as a date and time in a specific format.
# This immediately narrows the search space from all possible strings.
try:
dt = datetime.datetime.strptime(x, "%Y-%m-%d %H:%M:%S")
except ValueError:
return False
# Condition 3: The event must have occurred on a Wednesday.
# In Python's datetime library, Monday is 0 and Sunday is 6.
if dt.weekday() != 2:
return False
# Condition 4: The hour of the event must be a prime number.
if not is_prime(dt.hour):
return False
# The next conditions depend on the POSIX timestamp, a non-obvious transformation.
try:
# Note: timestamp() can fail for dates before the epoch on some systems.
timestamp = int(dt.timestamp())
except (ValueError, OSError):
return False
# Condition 5: The sum of the digits of the integer timestamp must be divisible by 9.
# This is a classic divisibility rule.
if sum(int(digit) for digit in str(timestamp)) % 9 != 0:
return False
# Condition 6: The CRC32 checksum of the timestamp's string representation
# must match a specific "magic" value. This is the lock on the puzzle box.
# It is computationally infeasible to reverse this. The only way to find the
# correct timestamp is to use the other conditions to generate likely
# candidates and test them.
checksum = zlib.crc32(str(timestamp).encode('ascii'))
if checksum != 217983273:
return False
return True
import math
def is_prime(n):
if n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
def mystery(x):
# Condition 1: The input must be a string.
if not isinstance(x, str):
return False
# Condition 2: The string length must be exactly 7 characters.
if len(x) != 7:
return False
# Condition 3: The string must represent a valid hexadecimal number.
# We attempt to convert it to an integer (base 16).
try:
n = int(x, 16)
except ValueError:
return False
# Condition 4: The 13th bit (0-indexed) of the integer 'n' must be set.
# This means (n // 2^13) % 2 == 1, or more efficiently, (n >> 13) & 1 == 1.
if not ((n >> 13) & 1):
return False
# Condition 5: The count of set bits (1s) in the binary representation of 'n' must be a prime number.
set_bits_count = bin(n).count('1')
if not is_prime(set_bits_count):
return False
# Condition 6: The sum of the ASCII values of all characters in the original string 'x' must be divisible by 11.
ascii_sum = sum(ord(c) for c in x)
if ascii_sum % 11 != 0:
return False
# Condition 7: The string 'x' must contain at least one uppercase hexadecimal digit (A-F).
if not any(c.isupper() for c in x):
return False
return True
import hashlib
import math
def is_prime(n):
"""A reasonably fast primality test."""
if not isinstance(n, int) or n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
def mystery(x):
# Condition 1: Input must be a string.
if not isinstance(x, str):
return False
# Condition 2: String must contain only lowercase ASCII letters.
# This significantly constrains the character set.
if not x.isascii() or not x.islower():
return False
# Condition 3: The length of the string must be a prime number.
# This is a structural clue about the solution.
if not is_prime(len(x)):
return False
# Condition 4: The number of vowels in the string must also be a prime number.
# This provides another filter on the content of the string.
vowel_count = sum(1 for char in x if char in 'aeiou')
if not is_prime(vowel_count):
return False
# The next conditions are based on a cryptographic hash of the input,
# making them hard to reverse. This is the core of the puzzle.
try:
h = hashlib.sha256(x.encode('utf-8')).hexdigest()
n = int(h, 16)
except (ValueError, TypeError):
return False
# Condition 5: The popcount (number of set bits) of the hash value
# must be a prime number. This is a non-obvious property that is
# quick to check but hard to reverse.
popcount = bin(n).count('1')
if not is_prime(popcount):
return False
# Condition 6: The "lock". The hash value, treated as an integer,
# must satisfy a specific modular arithmetic congruence. This is
# computationally infeasible to reverse. A solver must find a
# candidate string using the other clues and test it against this.
MODULUS = 999999937 # A large prime number
REMAINDER = 703940130
if n % MODULUS != REMAINDER:
return False
return True
import math
import base64
def is_perfect_square(num):
"""Checks if a number is a perfect square."""
if num < 0:
return False
if num == 0:
return True
sqrt_num = int(num**0.5)
return sqrt_num * sqrt_num == num
def is_perfect_cube(num):
"""Checks if a number is a perfect cube."""
if num < 0:
return False
if num == 0:
return True
# Use binary search to find integer cube root for robustness with large numbers
low, high = 1, num
while low <= high:
mid = (low + high) // 2
cube = mid * mid * mid
if cube == num:
return True
elif cube < num:
low = mid + 1
else: # cube > num
high = mid - 1
return False
def sum_digits(n):
"""Calculates the sum of the digits of a number."""
return sum(int(d) for d in str(n))
def is_power_of_two(n):
"""Checks if a number is a positive power of two."""
return n > 0 and (n & (n - 1) == 0)
def mystery(x):
# Condition 1: The input must be a string.
if not isinstance(x, str):
return False
# Condition 2: The string must be a valid base64 encoded string.
# This immediately constrains the search space to valid base64 formats.
try:
decoded_bytes = base64.b64decode(x, validate=True)
except (base64.binascii.Error, TypeError):
return False
# Condition 3: The decoded bytes, when interpreted as ASCII, must form a string
# representing a positive integer 'n'.
try:
decoded_str = decoded_bytes.decode('ascii')
n = int(decoded_str)
except (UnicodeDecodeError, ValueError):
return False
if n <= 0:
return False
# Condition 4: The integer 'n' must be a perfect cube.
# This significantly reduces the number of integers to check.
if not is_perfect_cube(n):
return False
# Condition 5: The sum of the digits of 'n' must be a perfect square.
# This adds another mathematical constraint on the numerical value.
s_digits = sum_digits(n)
if not is_perfect_square(s_digits):
return False
# Condition 6: The original base64 string 'x' must contain at least two different uppercase letters.
# This is a property of the encoded string itself, not the decoded number.
upper_chars = set(c for c in x if 'A' <= c <= 'Z')
if len(upper_chars) < 2:
return False
# Condition 7: The length of the original base64 string 'x' must be a power of two.
# Another structural constraint on the encoded string.
if not is_power_of_two(len(x)):
return False
# Condition 8: The integer 'n' must be greater than 1,000,000,000 (1 billion).
# Ensures the number is large enough, making brute force harder.
if n <= 1000000000:
return False
return True
import math
def mystery(x):
# Condition 1: The input must be a float.
if not isinstance(x, float):
return False
# Condition 2: Use the canonical string representation of the float.
# This avoids issues with precision and makes the puzzle deterministic.
# We also disallow scientific notation for simplicity.
s = repr(x)
if 'e' in s.lower():
return False
# Condition 3: The string representation must be in the form "3.f..."
try:
int_part, frac_part = s.split('.')
except ValueError:
# This happens if there is no decimal point.
return False
if int_part != '3':
return False
# Condition 4: The fractional part must have an even number of digits.
# This is necessary for the de-interleaving step.
if len(frac_part) % 2 != 0:
return False
# Condition 5: The fractional digits must come from a restricted set.
# This significantly prunes the search space.
if any(c in '09' for c in frac_part):
return False
# Condition 6: De-interleave the fractional digits to form two numbers, A and B.
# For example, if frac_part is "1234", a_str becomes "13" and b_str becomes "24".
a_str = frac_part[0::2]
b_str = frac_part[1::2]
try:
A = int(a_str)
B = int(b_str)
except ValueError:
return False # Should not happen with the digit check, but good practice.
# Condition 7: A and B must be coprime. This is a subtle but strong constraint.
if math.gcd(A, B) != 1:
return False
# Condition 8: (A, B, C) must form a Pythagorean triple.
# This is the main mathematical core of the puzzle.
c_squared = A*A + B*B
C = math.isqrt(c_squared)
if C*C != c_squared:
return False
# Condition 9: The "lock". The sum of the digits of the hypotenuse C
# must equal a specific magic number.
s_digits_C = sum(int(d) for d in str(C))
if s_digits_C != 11:
return False
return True
import math
def is_prime(n):
"""Checks if a number is prime."""
if n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
def mystery(x):
# Condition 1: The input must be a string.
if not isinstance(x, str):
return False
# Condition 2: The string must be exactly 5 characters long.
# This immediately limits the search space for the solver.
if len(x) != 5:
return False
# Condition 3: The string must contain only '0' or '1' characters.
# This implies it's a binary string.
if not all(c in '01' for c in x):
return False
# Condition 4: The string must be a palindrome.
# This adds a structural constraint on the binary string.
if x != x[::-1]:
return False
# Condition 5: When the string 'x' is interpreted as a binary number 'n',
# 'n' must be a prime number.
# This is the core mathematical constraint and the lock for the puzzle.
try:
n = int(x, 2)
except ValueError:
# This should theoretically not happen if Condition 3 is met.
return False
# Optimization: For a number to be prime (and > 1), its binary representation
# cannot start with '0' if it's longer than a single digit, as that would imply
# a leading zero, which means it's effectively a shorter number.
# For a 5-character binary string, '0xxxx' implies a value less than 16.
# Primes like 2, 3, 5, 7, 11, 13 are possible, but for a 5-char string,
# '00010' (2), '00011' (3) would not be considered "5-char binary numbers".
# The smallest prime 5-char binary number is 10001 (17).
if x[0] == '0':
return False
if not is_prime(n):
return False
return True
import math
def is_prime(n):
"""A reasonably fast primality test."""
if not isinstance(n, int) or n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
def mystery(x):
# Condition 1: Input must be a string containing exactly one underscore.
# This suggests the input is composed of two parts.
if not isinstance(x, str) or x.count('_') != 1:
return False
try:
name_part, num_part_str = x.split('_')
except ValueError:
return False
# Condition 2: The part before the underscore must be purely alphabetic
# and not empty.
if not name_part.isalpha():
return False
# Condition 3: The part after the underscore must represent a valid integer.
try:
num = int(num_part_str)
except ValueError:
return False
# Condition 4: This number must be prime.
# This is a strong mathematical constraint on the second part.
if not is_prime(num):
return False
# Condition 5: A critical link between the two parts. The length of the
# name must be exactly one less than the prime number. This is the main
# clue that allows for a structured, rather than random, search.
if len(name_part) != num - 1:
return False
# Condition 6: A property of the name part. It must contain strictly
# more vowels than consonants. Vowels are 'aeiou', case-insensitive.
vowels = "aeiou"
vowel_count = sum(1 for char in name_part.lower() if char in vowels)
consonant_count = len(name_part) - vowel_count
if vowel_count <= consonant_count:
return False
# Condition 7: The "lock". A checksum is calculated from the name part
# by summing the squares of the alphabetical positions of its letters
# (a=1, b=2, ...). This checksum must satisfy a specific modular congruence
# with the prime number. This is computationally difficult to reverse.
try:
# Using a generator expression for efficiency.
checksum = sum((ord(c) - ord('a') + 1)**2 for c in name_part.lower())
except TypeError:
return False
if checksum % num != 6:
return False
return True
import math
def is_prime(n):
"""Checks if a number is prime."""
if not isinstance(n, int) or n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
def is_perfect_square(num):
"""Checks if a non-negative integer is a perfect square."""
if not isinstance(num, int) or num < 0: return False
if num == 0: return True
sqrt_num = int(math.isqrt(num))
return sqrt_num * sqrt_num == num
def sum_digits(n):
"""Calculates the sum of the digits of a non-negative integer."""
if not isinstance(n, int) or n < 0: return -1 # Indicate error or invalid input
return sum(int(d) for d in str(n))
def mystery(x):
# Condition 1: The input must be a string.
if not isinstance(x, str):
return False
# Condition 2: The string must be a comma-separated list of exactly three positive integers.
parts = x.split(',')
if len(parts) != 3:
return False
numbers = []
for p in parts:
try:
num = int(p)
if num <= 0: # Numbers must be positive
return False
numbers.append(num)
except ValueError: # Not a valid integer string
return False
# Prepare to store properties for each number in the list
num_properties = []
for num in numbers:
# Condition 3a: Each integer in the list must be a perfect square.
if not is_perfect_square(num):
return False
sqrt_num = int(math.isqrt(num))
# Condition 3b: The sum of digits of each number must be greater than 1.
sd = sum_digits(num)
if sd <= 1:
return False
# Determine the properties of the sum of digits (sd) for each number
is_prime_sd = is_prime(sd)
is_ps_sd = is_perfect_square(sd)
is_neither_sd = not is_prime_sd and not is_ps_sd # sd is neither prime nor perfect square
num_properties.append({
'num': num,
'sqrt_num': sqrt_num,
'sd': sd,
'is_prime_sd': is_prime_sd,
'is_ps_sd': is_ps_sd,
'is_neither_sd': is_neither_sd
})
# Condition 4: The three sum-of-digits properties (prime, perfect square, neither)
# must be uniquely distributed among the three numbers.
# Specifically, exactly one number's sum of digits must be prime (and not a perfect square),
# exactly one number's sum of digits must be a perfect square (and not prime),
# and exactly one number's sum of digits must be neither prime nor a perfect square.
# Collect the boolean flags for each property across all three numbers
props_prime_counts = [p['is_prime_sd'] for p in num_properties]
props_ps_counts = [p['is_ps_sd'] for p in num_properties]
props_neither_counts = [p['is_neither_sd'] for p in num_properties]
# Check that each property count is exactly 1
if sum(props_prime_counts) != 1 or \
sum(props_ps_counts) != 1 or \
sum(props_neither_counts) != 1:
return False
# Condition 5: The square roots of the three numbers, when sorted, must form a Pythagorean triple.
# i.e., a^2 + b^2 = c^2 for a < b < c.
sqrt_numbers = sorted([p['sqrt_num'] for p in num_properties])
a, b, c = sqrt_numbers[0], sqrt_numbers[1], sqrt_numbers[2]
if a*a + b*b != c*c:
return False
# Condition 6 (The "lock"): The length of the original input string 'x' must be a prime number.
if not is_prime(len(x)):
return False
return True
import math
def mystery(x):
# Condition 1: Input must be a positive integer.
if not isinstance(x, int) or x <= 0:
return False
# Condition 2: The integer's decimal representation must be a palindrome.
s = str(x)
if s != s[::-1]:
return False
# Helper function to convert an integer to its standard Roman numeral representation.
# Handles numbers from 1 to 3999.
def to_roman(num):
if not (0 < num < 4000):
return "" # Out of standard range
val = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
syb = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman_num = ''
i = 0
while num > 0:
for _ in range(num // val[i]):
roman_num += syb[i]
num -= val[i]
i += 1
return roman_num
# Condition 3: The Roman numeral representation of x must also be a palindrome.
# This is a key constraint, requiring symmetry in two different numeral systems.
roman_s = to_roman(x)
if not roman_s or roman_s != roman_s[::-1]:
return False
# Helper function for primality test.
def is_prime(n):
if n < 2: return False
if n == 2: return True
if n % 2 == 0: return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
# Condition 4: The number of unique symbols used in the Roman numeral
# must be a prime number. (e.g., 'CCC' uses 1 symbol, 'XIX' uses 2).
unique_symbols = set(c for c in roman_s)
if not is_prime(len(unique_symbols)):
return False
# Condition 5 (The "lock"): The integer x must be divisible by 3.
# This is a simple but effective final filter.
if x % 3 != 0:
return False
return True