import math
def mystery(x):
"""
Find a value x for which this function returns True.
"""
# Constraint 1: Must be a string of a certain length.
if not isinstance(x, str) or len(x) < 2:
return False
# Constraint 2: Must be a valid hexadecimal number.
try:
val = int(x, 16)
except ValueError:
return False
# Constraint 3: The length of the string must be a prime number.
n = len(x)
if n > 1:
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
else:
return False
# Constraint 4: A specific property of the numerical value.
s = val
while s >= 10:
s = sum(int(d) for d in str(s))
return s == 1
import math
def mystery(x):
"""
Find a value x (string of digits) for which this function returns True.
"""
# Must be a non-empty string of digits
if not isinstance(x, str) or not x.isdigit():
return False
n = int(x)
if n <= 1:
return False
# Condition 1: n ≡ 1 (mod 8)
if n % 8 != 1:
return False
# Condition 2: n is divisible by the sum of its digits
digit_sum = sum(int(d) for d in x)
if digit_sum == 0 or n % digit_sum != 0:
return False
# Condition 3: n is not divisible by 5
if n % 5 == 0:
return False
# Helper: primality test
def is_prime(p):
if p < 2:
return False
if p % 2 == 0:
return p == 2
for d in range(3, int(math.sqrt(p)) + 1, 2):
if p % d == 0:
return False
return True
# Helper: Carmichael test (Korselt criterion)
def is_carmichael(num):
if is_prime(num):
return False
# factorize by trial division
factors = set()
temp = num
for d in range(2, int(math.sqrt(temp)) + 1):
while temp % d == 0:
factors.add(d)
temp //= d
if temp > 1:
factors.add(temp)
# Check squarefree and each p-1 divides num-1
for p in factors:
if not is_prime(p):
return False
if (num - 1) % (p - 1) != 0:
return False
return True
if not is_carmichael(n):
return False
# Condition 4: n is a taxicab number (at least two distinct representations as sum of two positive cubes)
def is_taxicab(num):
limit = int(round(num ** (1/3))) + 1
pairs = []
for i in range(1, limit):
i3 = i * i * i
if i3 > num:
break
for j in range(i + 1, limit):
j3 = j * j * j
s = i3 + j3
if s > num:
break
if s == num:
pairs.append((i, j))
if len(pairs) >= 2:
return True
return False
if not is_taxicab(n):
return False
return True
import hashlib
def mystery(x):
"""
Find a value x for which this function returns True.
"""
# C1: Input must be a string and have a specific format.
if not isinstance(x, str) or ':' not in x:
return False
try:
f_string, nonce_str = x.rsplit(':', 1)
# The first part must look like it's intended for string formatting.
if '{' not in f_string or '}' not in f_string:
return False
nonce = int(nonce_str)
except (ValueError, IndexError):
return False
# C2: The f-string part, when formatted with the nonce, creates the body.
try:
body = f_string.format(nonce)
except (ValueError, IndexError):
# Handles cases like "{} and {}" with a single nonce, which are invalid for this puzzle.
return False
# C3: A cryptographic hash of the body is computed.
h = hashlib.sha1(body.encode()).digest()
# C4: A number is derived from the tail of the hash digest.
# The digest must be long enough.
if len(h) < 2:
return False
val_from_hash = int.from_bytes(h[-2:], 'big')
# C5: The derived number must have a special relationship with the nonce.
return val_from_hash == nonce
import math
def mystery(x):
# Returns True only for a very special number x.
if not isinstance(x, str):
return False
if not x.isdigit():
return False
if len(x) < 2:
return False
n = int(x)
# Helper: primality test
def is_prime(p):
if p < 2:
return False
if p % 2 == 0:
return p == 2
for d in range(3, int(math.sqrt(p)) + 1, 2):
if p % d == 0:
return False
return True
# Helper: happy number test
def is_happy(num):
seen = set()
while num != 1 and num not in seen:
seen.add(num)
num = sum(int(d)**2 for d in str(num))
return num == 1
# Helper: perfect square test
def is_square(m):
if m < 0:
return False
r = int(math.sqrt(m))
return r*r == m
# 1. Prime
if not is_prime(n):
return False
# 2. Palindromic
if x != x[::-1]:
return False
# 3. Circular prime
s = x
for i in range(len(s)):
rot = int(s[i:] + s[:i])
if not is_prime(rot):
return False
# 4. Truncatable prime (both left and right)
for i in range(1, len(s)):
if not is_prime(int(s[i:])): # left-truncation
return False
if not is_prime(int(s[:-i])): # right-truncation
return False
# 5. Happy prime
if not is_happy(n):
return False
# 6. Sum of digits is prime
digit_sum = sum(int(d) for d in x)
if not is_prime(digit_sum):
return False
# 7. Product of digits is a perfect square
digit_prod = 1
for d in x:
digit_prod *= int(d)
if not is_square(digit_prod):
return False
# 8. Pythagorean prime: n ≡ 1 (mod 4)
if n % 4 != 1:
return False
# 9. Centered square number: exists k such that n = k^2 + (k+1)^2
disc = 4 * (2*n - 1)
sqrt_disc = int(math.isqrt(disc))
if sqrt_disc * sqrt_disc != disc:
return False
k = (-2 + sqrt_disc) // 4
if k <= 0:
return False
if n != k*k + (k+1)*(k+1):
return False
return True
import struct
import math
def mystery(x):
"""
Find a value x for which this function returns True.
"""
# C1: Input must be a string of a specific length and format.
if not isinstance(x, str) or len(x) != 12 or '.' not in x:
return False
parts = x.split('.')
if len(parts) != 2:
return False
a_str, b_str = parts
if not a_str.isdigit() or not b_str.isdigit() or not a_str or not b_str:
return False
a = int(a_str)
b = int(b_str)
# C2: The two integer parts must satisfy a linear Diophantine equation.
if 3 * a - 5 * b != 1:
return False
# C3: Convert to float and get its 64-bit integer representation (IEEE 754).
try:
f = float(x)
# Get the 8-byte (64-bit) representation in big-endian format.
packed = struct.pack('>d', f)
# Interpret these 8 bytes as a single 64-bit unsigned integer.
i = int.from_bytes(packed, 'big')
except (ValueError, OverflowError):
return False
# C4: The number of set bits (popcount) in this integer representation...
popcount = bin(i).count('1')
# ...must be a prime number.
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
return is_prime(popcount)
import math
def mystery(x):
# Returns True only for a very special number x.
if not isinstance(x, str):
return False
if not x.isdigit():
return False
if len(x) < 3:
return False
n = int(x)
if n <= 100:
return False
def is_prime(p):
if p < 2:
return False
if p == 2:
return True
if p % 2 == 0:
return False
for i in range(3, int(math.sqrt(p)) + 1, 2):
if p % i == 0:
return False
return True
def is_happy(num):
seen = set()
while num != 1 and num not in seen:
seen.add(num)
num = sum(int(d)**2 for d in str(num))
return num == 1
if not is_prime(n):
return False
if x != x[::-1]:
return False
binary = bin(n)[2:]
if binary != binary[::-1]:
return False
if not is_happy(n):
return False
digit_sum = sum(int(d) for d in x)
if not is_prime(digit_sum):
return False
digit_prod = 1
for d in x:
digit_prod *= int(d)
if digit_prod <= 0:
return False
root = int(math.sqrt(digit_prod))
if root * root != digit_prod:
return False
if n % 4 != 1:
return False
val = 2 * n - 1
sqrt_val = int(math.isqrt(val))
if sqrt_val * sqrt_val != val:
return False
k = ( -1 + sqrt_val ) // 2
if k <= 0:
return False
if n != k*k + (k+1)*(k+1):
return False
return True
import re
def mystery(x):
"""
Find a value x for which this function returns True.
"""
# C1: The input must be a string representation of an integer.
if not isinstance(x, str) or not x.isdigit():
return False
try:
s = int(x)
except ValueError:
return False
# C2: The puzzle uses a hardcoded RSA public key (n, e).
# n is the modulus, e is the public exponent.
n = 933333333333333333333333332158888888888888888888888888888887
e = 65537
# The input 's' is treated as a signature. We verify it by computing
# message = s^e mod n.
m = pow(s, e, n)
# C3: The decrypted message 'm' must be a valid UTF-8 string
# that conforms to a specific poetic structure (a haiku).
try:
# Calculate the number of bytes needed to represent m.
# An empty message (m=0) is invalid.
if m == 0:
return False
byte_length = (m.bit_length() + 7) // 8
message_bytes = m.to_bytes(byte_length, 'big')
message_text = message_bytes.decode('utf-8')
except (UnicodeDecodeError, OverflowError):
return False
def is_haiku(text):
"""
A helper function to check if a text is a 5-7-5 haiku.
Uses a simple heuristic for syllable counting.
"""
def count_syllables(line):
line = line.lower().strip()
line = re.sub(r'[^a-z\s]', '', line) # Keep only letters and spaces
if not line:
return 0
words = line.split()
total_syllables = 0
for word in words:
if len(word) <= 3:
syllable_count = 1
else:
# remove silent 'e' at the end, unless it's like 'le'
if word.endswith('e') and not word.endswith('le'):
word = word[:-1]
# count vowel groups
vowel_groups = re.findall(r'[aeiouy]+', word)
syllable_count = len(vowel_groups)
total_syllables += max(1, syllable_count) # each word has at least one syllable
return total_syllables
lines = text.strip().split('\n')
if len(lines) != 3:
return False
syllables_per_line = [count_syllables(line) for line in lines]
return syllables_per_line == [5, 7, 5]
return is_haiku(message_text)
import math
def mystery(x):
"""
Find a value x for which this function returns True.
"""
if not isinstance(x, str):
return False
x = x.strip()
try:
val = float(x)
except ValueError:
return False
if val <= 1:
return False
# Check golden ratio equation: val^2 == val + 1
return abs(val*val - val - 1) < 1e-14
import hashlib
def mystery(x):
"""
Find a value x for which this function returns True.
The secret lies not in the code you see, but the map it implies.
"""
# C1: Input must be a string composed of 'U', 'D', 'L', 'R'.
if not isinstance(x, str) or any(c not in 'UDLR' for c in x):
return False
# The maze is procedurally generated but deterministic. Its structure
# is derived from a seed, which is subtly hidden within this function.
maze_seed = mystery.__doc__
def is_wall(r1, c1, r2, c2):
"""Determines if a wall exists between two adjacent cells."""
# Create a canonical representation of the wall's location
# to ensure is_wall(p1, p2) == is_wall(p2, p1).
if (r1, c1) > (r2, c2):
r1, c1, r2, c2 = r2, c2, r1, c1
h = hashlib.sha256()
h.update(maze_seed.encode())
h.update(f"wall:({r1},{c1})-({r2},{c2})".encode())
# A wall exists if the first byte of the hash is below a threshold.
# This creates a pseudo-random but fixed maze layout.
return h.digest()[0] < 100
# Maze parameters
SIZE = 16
START = (0, 0)
END = (SIZE - 1, SIZE - 1)
pos = list(START)
# C2: The input string `x` must represent a valid path from START to END.
for move in x:
r, c = pos
# Determine the next position based on the move.
next_r, next_c = r, c
if move == 'U': next_r -= 1
elif move == 'D': next_r += 1
elif move == 'L': next_c -= 1
elif move == 'R': next_c += 1
# Check for boundary violations.
if not (0 <= next_r < SIZE and 0 <= next_c < SIZE):
return False
# Check if the path is blocked by a wall.
if is_wall(r, c, next_r, next_c):
return False
# Update position.
pos = [next_r, next_c]
# C3: The path must end at the correct destination.
return tuple(pos) == END
import math
def mystery(x):
"""
Find a value x for which this function returns True.
"""
# Must be a non-empty string of digits, no leading zero
if not isinstance(x, str) or not x.isdigit():
return False
if len(x) > 1 and x[0] == '0':
return False
n = int(x)
# Helper: primality test
def is_prime(p):
if p < 2:
return False
if p == 2:
return True
if p % 2 == 0:
return False
for i in range(3, int(math.sqrt(p)) + 1, 2):
if p % i == 0:
return False
return True
# 1. n must be prime
if not is_prime(n):
return False
# 2. x must be palindromic
if x != x[::-1]:
return False
# 3. n must be a Fibonacci number
def is_fibonacci(num):
s1 = 5*num*num + 4
s2 = 5*num*num - 4
return int(math.isqrt(s1))**2 == s1 or int(math.isqrt(s2))**2 == s2
if not is_fibonacci(n):
return False
# 4. n must be a Wilson prime: (n-1)! ≡ -1 (mod n^2)
def is_wilson(p):
if p < 2:
return False
fact = 1
for i in range(1, p):
fact = (fact * i) % (p*p)
return fact == (p*p - 1) % (p*p)
if not is_wilson(n):
return False
# 5. n must be a safe prime: (p-1)/2 is also prime
def is_safe(p):
q = (p-1)//2
return is_prime(q)
if not is_safe(n):
return False
# 6. n must be a Chen prime: p+2 is either prime or a product of two primes
def is_chen(p):
r = p + 2
if is_prime(r):
return True
temp = r
factors = 0
d = 2
while d*d <= temp:
while temp % d == 0:
factors += 1
temp //= d
d += 1 if d == 2 else 2
if temp > 1:
factors += 1
return factors == 2
if not is_chen(n):
return False
# 7. n ≡ 1 (mod 4)
if n % 4 != 1:
return False
# 8. n must be a centered square number: n = k^2 + (k+1)^2 for some k
def is_centered_square(num):
disc = 8*num - 4
sqrt_disc = int(math.isqrt(disc))
if sqrt_disc * sqrt_disc != disc:
return False
k = (-2 + sqrt_disc) // 4
if k <= 0:
return False
return num == k*k + (k+1)*(k+1)
if not is_centered_square(n):
return False
# 9. n must be the hypotenuse of a primitive Pythagorean triple
def is_hypotenuse_primitive(h):
h_sq = h*h
limit = int(math.isqrt(h_sq))
for a in range(1, limit):
a_sq = a*a
b_sq = h_sq - a_sq
if b_sq < a_sq:
break
b = int(math.isqrt(b_sq))
if b*b == b_sq and math.gcd(a, b) == 1:
return True
return False
if not is_hypotenuse_primitive(n):
return False
# 10. Sum of digits of x is prime
digit_sum = sum(int(d) for d in x)
if not is_prime(digit_sum):
return False
# 11. Product of digits of x is a Fibonacci number
digit_prod = 1
for d in x:
digit_prod *= int(d)
if not is_fibonacci(digit_prod):
return False
return True