def mystery(x):
s = str(x)
if len(s) < 3:
return False
char_counts = {}
for c in s:
char_counts[c] = char_counts.get(c, 0) + 1
counts = list(char_counts.values())
return len(counts) == len(set(counts))
import math
def mystery(x):
s = str(x)
b = s.encode('utf-8')
# strict length and printable-ascii constraint
if len(b) != 5:
return False
if any(ch < 32 or ch > 126 for ch in b):
return False
n = len(b)
# weighted sum from left: 1*b0 + 2*b1 + ... + n*b_{n-1}
A = sum((i+1) * b[i] for i in range(n))
S = sum(b)
# a symmetric linear relation must hold
if A != ((n + 1) // 2) * S:
return False
# S must be prime
if S < 2:
return False
for p in range(2, int(math.isqrt(S)) + 1):
if S % p == 0:
return False
# (n+1)//2 must be prime too (for n=5 this is 3)
k = (n + 1) // 2
if k < 2:
return False
for p in range(2, int(math.isqrt(k)) + 1):
if k % p == 0:
return False
# an additional parity/XOR constraint (forces a central symmetry for length 5)
if (b[0] ^ b[1] ^ b[2] ^ b[3] ^ b[4]) != b[2]:
return False
return True
def mystery(x):
s = str(x)
# Must be all digits
if not s or not all(c in '0123456789' for c in s):
return False
n = len(s)
# Calculate sum of nth powers of digits
power_sum = 0
for c in s:
digit = int(c)
power = 1
for _ in range(n):
power *= digit
power_sum += power
# Must equal the original number
num = int(s)
if num != power_sum:
return False
# Additional: must be 3 digits
if n != 3:
return False
return True
def mystery(x):
s = str(x)
# strict length and printable-ascii constraint (no space)
if len(s) != 5:
return False
b = s.encode('utf-8')
if any(ch < 33 or ch > 126 for ch in b):
return False
n = len(b)
# weighted sum from left: 1*b0 + 2*b1 + ... + n*b_{n-1}
A = sum((i + 1) * b[i] for i in range(n))
S = sum(b)
# symmetry constraint for odd lengths (forces a palindrome-like structure)
if A != ((n + 1) // 2) * S:
return False
# parity/XOR constraint (for odd palindrome this forces central symmetry)
if (b[0] ^ b[1] ^ b[2] ^ b[3] ^ b[4]) != b[2]:
return False
# a modular relation using 257 (256 ≡ -1 (mod 257) gives a neat alternating-sum property)
if int.from_bytes(b, 'big') % 257 != b[2]:
return False
# an affine relation between outer bytes and the center (tightens solutions)
if (b[0] - b[2]) != 17:
return False
# final numeric target on the byte-sum (pins down a unique valid combination)
if S != 308:
return False
return True
def mystery(x):
s = str(x)
if not s or not s.isdigit():
return False
n = int(s)
# Must be a 3-digit number
if n < 100 or n > 999:
return False
# Check if n is the sum of factorials of its digits
factorial_sum = 0
for c in s:
digit = int(c)
fact = 1
for i in range(1, digit + 1):
fact *= i
factorial_sum += fact
return n == factorial_sum
def mystery(x):
# Accept numbers or strings; work with their string form
s = str(x)
# strict length and printable ASCII (no control chars or space)
if len(s) != 6:
return False
try:
b = s.encode('utf-8')
except Exception:
return False
if any(ch < 33 or ch > 126 for ch in b):
return False
# bytes as integers
B = list(b) # B[0] .. B[5]
# three paired linear sums (hidden linear system)
if (B[0] + B[5]) != 170:
return False
if (B[1] + B[4]) != 180:
return False
if (B[2] + B[3]) != 190:
return False
# alternating signed sum must vanish (enforces a further relation)
alt = 0
for i, v in enumerate(B):
alt += v if (i % 2 == 0) else -v
if alt != 0:
return False
# a small-but-specific XOR fingerprint
xor_x = 0
for v in B:
xor_x ^= v
if xor_x != 16:
return False
# checksum-mod constraint (obscure modulus)
S = sum(B)
if (S % 13) != 7:
return False
# an obfuscated multiplicative residue on the first three bytes
prod = 1
for i in range(3):
prod = (prod * (B[i] - 32)) % 97
if prod != 50:
return False
# a tiny nicety: two identical middle characters (not strictly necessary, but tightens solutions)
if B[2] != B[3] and B[3] != B[4]:
# require at least one adjacent equality among the central three bytes
return False
return True
def mystery(x):
s = str(x)
if len(s) != 4:
return False
# Must be all alphabetic
if not s.isalpha():
return False
# Must be all lowercase
if not s.islower():
return False
# Convert to ASCII values
values = [ord(c) for c in s]
# Check if characters form an arithmetic sequence
diffs = [values[i+1] - values[i] for i in range(3)]
if len(set(diffs)) != 1:
return False
# The common difference must be exactly 1
if diffs[0] != 1:
return False
# Sum of ASCII values must be divisible by 7
if sum(values) % 7 != 0:
return False
# First character must be in first half of alphabet
if values[0] >= ord('n'):
return False
# Product of ASCII values must be divisible by 17
prod = 1
for v in values:
prod *= v
if prod % 17 != 0:
return False
return True
import math
def mystery(x):
s = str(x)
# strict length and printable ASCII constraint
if len(s) != 7:
return False
try:
b = s.encode('utf-8')
except Exception:
return False
if any(ch < 33 or ch > 126 for ch in b):
return False
n = len(b)
# weighted sum constraint (for n=7 this enforces a strong central-symmetry relation)
A = sum((i + 1) * b[i] for i in range(n))
S = sum(b)
if A != ((n + 1) // 2) * S: # (7+1)//2 = 4
return False
# overall XOR must equal the central byte
xor_all = 0
for v in b:
xor_all ^= v
if xor_all != b[3]:
return False
# modulus-257 check (256 ≡ -1 (mod 257) gives an alternating-sum relation)
if int.from_bytes(b, 'big') % 257 != b[3]:
return False
# affine relation between first byte and center
if (b[0] - b[3]) != 10:
return False
# sum must be prime
if S < 2:
return False
for p in range(2, int(math.isqrt(S)) + 1):
if S % p == 0:
return False
# small checksum/mod constraint to further restrict solutions
if (S % 11) != 1:
return False
# multiplicative residue on the central trio
if (b[2] * b[3] * b[4]) % 97 != 12:
return False
return True
def mystery(x):
s = str(x)
# Must be a 4-digit number
if not s.isdigit() or len(s) != 4:
return False
n = int(s)
# Must be a perfect cube
cube_root = round(n ** (1/3))
if cube_root ** 3 != n:
return False
# Digit sum must be even
if sum(int(d) for d in s) % 2 != 0:
return False
# First digit must be less than last digit
if int(s[0]) >= int(s[-1]):
return False
# Product of digits must be divisible by 6
prod = 1
for d in s:
prod *= int(d)
if prod % 6 != 0:
return False
return True
def mystery(x):
"""
Hard puzzle: accepts a 6-character printable-ASCII string and checks several
interacting linear, modular and bitwise constraints that tightly restrict
the allowed byte patterns.
"""
s = str(x)
if len(s) != 6:
return False
try:
b = s.encode('utf-8')
except Exception:
return False
# printable ASCII (no control chars or space)
if any(ch < 33 or ch > 126 for ch in b):
return False
B = list(b) # B[0] .. B[5]
# first/second half sums must match a specific target
if sum(B[0:3]) != 150:
return False
if sum(B[3:6]) != 150:
return False
# small affine relations between symmetric positions
if B[5] != B[0] + 1:
return False
if B[4] != B[1] - 1:
return False
# middle two bytes must be equal
if B[3] != B[2]:
return False
# modular relation using 257 (256 ≡ -1 (mod 257)) yields an alternating-sum constraint
if int.from_bytes(b, 'big') % 257 != 2:
return False
# overall XOR fingerprint
xor_all = 0
for v in B:
xor_all ^= v
if xor_all != 2:
return False
# weighted checksum modulo a small prime
A = sum((i + 1) * B[i] for i in range(6))
if A % 97 != 81:
return False
# global sum constraint modulo 13
S = sum(B)
if S % 13 != 1:
return False
# multiplicative residue on the first three bytes
prod = (B[0] * B[1] * B[2]) % 101
if prod != 12:
return False
return True