Sunday Times Teaser 2512 – No Title

by Graham Smithers

Published November 14 2010 (link)

Using each of the digits 0 to 9 exactly once, I have written down two three-figure numbers and a four- figure number.

In no particular order, one of them is a perfect square, one is a perfect cube, and the other is a prime.

If I told you which of the square, the cube and the prime was the smallest then it would be possible for you to work out what those three numbers are.

What are they?

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from collections import defaultdict
from number_theory import Primes

# return the digits in a number
n2d = lambda n:tuple(int(c) for c in str(n))

# map three and four digit squares to their digits, omitting
# those with duplicate digits
sqrs = [(n * n, n2d(n * n)) for n in range(10, 100)]
sqrs = {k:set(v) for k, v in sqrs if len(v) == len(set(v))}

# map three and four digit cubes to their digits, omitting
# those with duplicate digits
cbes = [(n ** 3, n2d(n ** 3)) for n in range(5, 22)]
cbes = {k:set(v) for k, v in cbes if len(v) == len(set(v))}

# map the digit sets of three and four digit primes to the 
# primes, omitting those with duplicate digits
d2p = defaultdict(list)
for pp in Primes().range(100, 10000):
  pd = n2d(pp)
  if len(pd) == len(set(pd)):
    d2p[frozenset(pd)].append(pp)

# map the type (0 = square, 1 = cube, 2 = prime) of the 
# smallest value in triples to triples
d = defaultdict(list)
# for all square values
for ss, sd in sqrs.items():
  # find cube values that do not share digits with them 
  for cc, cd in cbes.items():
    if not (sd & cd):
      # find primes that use the remaining digits
      for pp in d2p[frozenset(set(range(10)).difference(sd | cd))]:
        scp = (ss, cc, pp)
        d[scp.index(min(scp))].append(scp)

# find the type of the smallest value that gives a unique solution
for k, v in d.items():
  if len(v) == 1:
    (s, c, p), = v
    print(f'square = {s}, cube = {c}, prime = {p}')

from collections import defaultdict

from number_theory import Primes

# return the digits in a number

n2d = lambda n:tuple(int(c) for c in str(n))

# map three and four digit squares to their digits, omitting

# those with duplicate digits

sqrs = [(n * n, n2d(n * n)) for n in range(10, 100)]

sqrs = {k:set(v) for k, v in sqrs if len(v) == len(set(v))}

# map three and four digit cubes to their digits, omitting

# those with duplicate digits

cbes = [(n ** 3, n2d(n ** 3)) for n in range(5, 22)]

cbes = {k:set(v) for k, v in cbes if len(v) == len(set(v))}

# map the digit sets of three and four digit primes to the

# primes, omitting those with duplicate digits

d2p = defaultdict(list)

for pp in Primes().range(100, 10000):

pd = n2d(pp)

if len(pd) == len(set(pd)):

d2p[frozenset(pd)].append(pp)

# map the type (0 = square, 1 = cube, 2 = prime) of the

# smallest value in triples to triples

d = defaultdict(list)

# for all square values

for ss, sd in sqrs.items():

# find cube values that do not share digits with them

for cc, cd in cbes.items():

if not (sd & cd):

# find primes that use the remaining digits

for pp in d2p[frozenset(set(range(10)).difference(sd | cd))]:

scp = (ss, cc, pp)

d[scp.index(min(scp))].append(scp)

# find the type of the smallest value that gives a unique solution

for k, v in d.items():

if len(v) == 1:

(s, c, p), = v

print(f'square = {s}, cube = {c}, prime = {p}')

Sunday Times Teaser 2512 – No Title

by Graham Smithers

Published November 14 2010 (link)

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