Sunday Times Teaser 3198 – The Sixth Element

by BRG on January 5, 2024

by Stephen Hogg

Published Friday January 05 2024 (link)

Agent J discovered that S.P.A.M.’s IQ booster drug comprises five elements with atomic numbers Z under 92. Some information had been coded as follows: Z1[4;6], Z2[1;4], Z3[4;6], Z4[0;6] and Z5[7;2], where Z5 is lowest and Z3 is highest. Code key: [Remainder after dividing Z by 8; Total number of factors of Z (including 1 and Z)]. The drug’s name is the elements’ chemical symbols concatenated in encoded list order.

J subsequently discovered the Z3 and Z5 values. Now MI6 had just fifteen possible sets of atomic numbers to consider. Finally, J sent a sixth element’s prime Z value (a catalyst in the drug’s production, not part of it). She’d heard it was below Z1 and above Z2, without knowing these. MI6 boffins were now sure of the drug’s name.

Give the catalyst’s atomic number.

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BRG permalink


from itertools import product

# primes less than 100
pr = {3, 5, 7}
pr = {2} | pr | set(n for n in range(11, 100, 2) if all(n % p for p in pr))

# the number of divisors of n 
div_nbrs = lambda n: sum(n % i == 0 for i in range(1, n + 1))

# the element's codes
codes = [(4, 6), (1, 4), (4, 6), (0, 6), (7, 2)]

# find possible atomic numbers for each code
z = dict()
for i in range(1, 92):
  if (k := (i % 8, div_nbrs(i))):
    z[k] = z.get(k, []) + [i]

# now collect possible atomic number sequences for
# different z3 and z5 pairs
d35 = dict()
for zz in product(*(z[c] for c in codes)):
  z1, z2, z3, z4, z5 = zz
  if z5 < min(zz[:-1]) and z3 > max(z1, z2, z4, z5):
    d35[k] = d35.get((k := (z3, z5)), []) + [zz]

# look for lists of atomic number sequences of length 15 
for zseqs in d35.values():
  if len(zseqs) == 15:
    # collect atomic number sequences where
    # there is a prime between z2 and z1
    p2zs = dict()
    for zs in zseqs:
      for n in range(zs[1] + 1, zs[0]):
        if n in pr:
          p2zs[n] = p2zs.get(n, []) + [zs]
    # find a prime that gives a unique result
    for p, l_zs in p2zs.items():
      if len(l_zs) == 1:
        print(F"Atomic numbers: Catalyst: {p}; 5 elements: {l_zs[0]}")

from itertools import product

# primes less than 100

pr = {3, 5, 7}

pr = {2} | pr | set(n for n in range(11, 100, 2) if all(n % p for p in pr))

# the number of divisors of n

div_nbrs = lambda n: sum(n % i == 0 for i in range(1, n + 1))

# the element's codes

codes = [(4, 6), (1, 4), (4, 6), (0, 6), (7, 2)]

# find possible atomic numbers for each code

z = dict()

for i in range(1, 92):

if (k := (i % 8, div_nbrs(i))):

z[k] = z.get(k, []) + [i]

# now collect possible atomic number sequences for

# different z3 and z5 pairs

d35 = dict()

for zz in product(*(z[c] for c in codes)):

z1, z2, z3, z4, z5 = zz

if z5 < min(zz[:-1]) and z3 > max(z1, z2, z4, z5):

d35[k] = d35.get((k := (z3, z5)), []) + [zz]

# look for lists of atomic number sequences of length 15

for zseqs in d35.values():

if len(zseqs) == 15:

# collect atomic number sequences where

# there is a prime between z2 and z1

p2zs = dict()

for zs in zseqs:

for n in range(zs[1] + 1, zs[0]):

if n in pr:

p2zs[n] = p2zs.get(n, []) + [zs]

# find a prime that gives a unique result

for p, l_zs in p2zs.items():

if len(l_zs) == 1:

print(F"Atomic numbers: Catalyst: {p}; 5 elements: {l_zs[0]}")

Frits permalink

With one extra condition (“not part of it”)?


from number_theory import divisors
from itertools import product

# primes less than 90 (as 2 numbers will be lower and higher)
P = [3, 5, 7]
P += [n for n in range(11, 90, 2) if all(n % p for p in P)]

info = [(0, 0), (4, 6), (1, 4), (4, 6), (0, 6), (7, 2)]
# build dictionary of key usage
d = {key: [Z for Z in range(1, 92) if (Z % 8, len(divisors(Z))) == key] 
           for key in set(info[1:])}
           
# pick Z5 and Z3 with Z5 < Z3 
for Z5 in d[info[5]]:
  for Z3 in d[info[3]][::-1]:
    if not(Z5 < Z3): break
    # now MI6 had just fifteen possible sets of atomic numbers to consider
    # so pn(Z1) * pn(Z2) * pn(Z4) == 15 where pn is number of possible values 
    pn = [len([x for x in d[info[i]] if Z5 < x < Z3]) for i in [1, 2, 4]]
    if pn[0] * pn[1] * pn[2] != 15: 
      continue

    # find prime Z6 with Z5 < Z2 < Z6 < Z1 < Z3
    for Z6 in [p for p in P if Z5 < p < Z3]:
      # we should now know Z1 and Z2 in order to know the drug's name
      if len([x for x in d[info[1]] if Z6 < x < Z3]) != 1: continue
      if len([x for x in d[info[2]] if Z5 < x < Z6]) != 1: continue

      # we need to be able to pick a Z4 unequal to Z6
      if not set(d[info[4]]).difference([Z6]): continue

      print("answer:", Z6)

from number_theory import divisors

from itertools import product

# primes less than 90 (as 2 numbers will be lower and higher)

P = [3, 5, 7]

P += [n for n in range(11, 90, 2) if all(n % p for p in P)]

info = [(0, 0), (4, 6), (1, 4), (4, 6), (0, 6), (7, 2)]

# build dictionary of key usage

d = {key: [Z for Z in range(1, 92) if (Z % 8, len(divisors(Z))) == key]

for key in set(info[1:])}

# pick Z5 and Z3 with Z5 < Z3

for Z5 in d[info[5]]:

for Z3 in d[info[3]][::-1]:

if not(Z5 < Z3): break

# now MI6 had just fifteen possible sets of atomic numbers to consider

# so pn(Z1) * pn(Z2) * pn(Z4) == 15 where pn is number of possible values

pn = [len([x for x in d[info[i]] if Z5 < x < Z3]) for i in [1, 2, 4]]

if pn[0] * pn[1] * pn[2] != 15:

continue

# find prime Z6 with Z5 < Z2 < Z6 < Z1 < Z3

for Z6 in [p for p in P if Z5 < p < Z3]:

# we should now know Z1 and Z2 in order to know the drug's name

if len([x for x in d[info[1]] if Z6 < x < Z3]) != 1: continue

if len([x for x in d[info[2]] if Z5 < x < Z6]) != 1: continue

# we need to be able to pick a Z4 unequal to Z6

if not set(d[info[4]]).difference([Z6]): continue

print("answer:", Z6)

JohnZ permalink


from itertools import product

#number of factors for numbers 1 to 91
fs = [sum(n % m == 0 for m in range(1, n + 1)) for n in range(1, 92)]

# prime numbers up to 91
pr = (2, 3, 5, 7)
pr = pr + tuple(n for n in range(12, 92) if all(n % p for p in pr))

Z1, Z2, Z3, Z4, Z5 = 1, 2, 3, 4, 5

# initial knowledge
Zs = ((Z1, 4, 6), (Z2, 1, 4), (Z3, 4, 6), (Z4, 0, 6), (Z5, 7, 2))

# create dictionary of possible Z values based on the initial knowledge
Z = {k: [n for n in range(1, 92) if (n % 8 == m) and (fs[n - 1] == f)] for
     k, m, f in Zs}

Zh = dict() # trial hypothetical values of Z1...Z5

# find key of Zn with only one possible value if any
u = [i for i, x in Z.items() if len(x) == 1][0]

# J subsequently discovered the Z3 and Z5 values. 
# take a value from Z5 and Z3 and build dictionary Zh of hypothetical Z's
for Zminh, Zmaxh in product(Z[Z5], Z[Z3]):
  if Zmaxh > Zminh:
    Zh[Z5], Zh[Z3] = [Zminh], [Zmaxh]
    for p in (Z1, Z2, Z4):
      Zh[p] = [x for x in Z[p] if Zmaxh > x > Zminh]
    
    # Now MI6 had just fifteen possible sets of atomic numbers to consider.
    if len(Zh[Z1]) * len(Zh[Z2]) * len(Zh[Z4]) == 15:
      
      # J sent a sixth element’s prime Z value (a catalyst)
      # She’d heard it was below Z1 and above Z2, without knowing these.
      # so Z6 < max(Z1) and Z6 > min(Z2)

      for Z6 in range(min(Zh[Z2]), max(Zh[Z1])):
        
        if Z6 in pr:
          # Z6 is < than only the maximum possible Z1 and > than only the
          # minimum possible Z2 it is possible to determine Z1 and Z2
          # Z3 and Z5 are known as is Z4 because has only one possible value
          if sum(x > Z6 for x in Zh[1]) == 1 == sum(Z6 > x for x in Zh[2]):
            print(f'Catalyst {Z6=}; Z1 -- Z5:', end = ' ')
            print(max(Zh[Z1]), min(Zh[Z2]), *Zh[Z3], *Zh[Z4], *Zh[Z5])

from itertools import product

#number of factors for numbers 1 to 91

fs = [sum(n % m == 0 for m in range(1, n + 1)) for n in range(1, 92)]

# prime numbers up to 91

pr = (2, 3, 5, 7)

pr = pr + tuple(n for n in range(12, 92) if all(n % p for p in pr))

Z1, Z2, Z3, Z4, Z5 = 1, 2, 3, 4, 5

# initial knowledge

Zs = ((Z1, 4, 6), (Z2, 1, 4), (Z3, 4, 6), (Z4, 0, 6), (Z5, 7, 2))

# create dictionary of possible Z values based on the initial knowledge

Z = {k: [n for n in range(1, 92) if (n % 8 == m) and (fs[n - 1] == f)] for

k, m, f in Zs}

Zh = dict() # trial hypothetical values of Z1...Z5

# find key of Zn with only one possible value if any

u = [i for i, x in Z.items() if len(x) == 1][0]

# J subsequently discovered the Z3 and Z5 values.

# take a value from Z5 and Z3 and build dictionary Zh of hypothetical Z's

for Zminh, Zmaxh in product(Z[Z5], Z[Z3]):

if Zmaxh > Zminh:

Zh[Z5], Zh[Z3] = [Zminh], [Zmaxh]

for p in (Z1, Z2, Z4):

Zh[p] = [x for x in Z[p] if Zmaxh > x > Zminh]

# Now MI6 had just fifteen possible sets of atomic numbers to consider.

if len(Zh[Z1]) * len(Zh[Z2]) * len(Zh[Z4]) == 15:

# J sent a sixth element’s prime Z value (a catalyst)

# She’d heard it was below Z1 and above Z2, without knowing these.

# so Z6 < max(Z1) and Z6 > min(Z2)

for Z6 in range(min(Zh[Z2]), max(Zh[Z1])):

if Z6 in pr:

# Z6 is < than only the maximum possible Z1 and > than only the

# minimum possible Z2 it is possible to determine Z1 and Z2

# Z3 and Z5 are known as is Z4 because has only one possible value

if sum(x > Z6 for x in Zh[1]) == 1 == sum(Z6 > x for x in Zh[2]):

print(f'Catalyst {Z6=}; Z1 -- Z5:', end = ' ')

print(max(Zh[Z1]), min(Zh[Z2]), *Zh[Z3], *Zh[Z4], *Zh[Z5])

Sunday Times Teaser 3198 – The Sixth Element

by Stephen Hogg

Published Friday January 05 2024 (link)

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