New Scientist Enigma 936 – Shift of Fortune

by BRG on March 11, 2022

by Susan Denham

From Issue #2091, 19th July 1997 (link)

To enter the lottery I chose six lucky numbers: I called this “selection A”. I have added 1 to each of those six numbers to form a new “selection B”, I’ve added 1 again to each to form “selection C”, and repeated this twice more to form “selection D” and “selection E”. So, for example, the lowest number in selection E is four more than the lowest number in selection A. Of course, as always, each selection is in the range 1-49.

In one of the selections (but not selection A) the six numbers are all two-digit numbers with the same first digit.

In one of the selections (but not selection A) there are no prime numbers.

Each week I decide to have two goes at the lottery and I choose at random two of my five selections. I know that whichever pair I choose, it would be possible for one of the selections to contain four of the six winning numbers and the other selection to contain none. On the other hand, whichever pair I choose, it would also be possible for both selections to contain four of the six winning numbers.

Which six numbers form “selection A”?

From → Uncategorized

3 Comments Leave one →

Brian Gladman permalink


from itertools import combinations

# primes below fifty
pr = {3, 5, 7}
pr |= {2} | {n for n in range(11, 50, 2) if all(n % p for p in pr)} 

# consider the selection having six numbers with the same tens digit
for tens in range(1, 5):
  # combine this with all combinations of six units digits 
  for u6 in combinations(range(10), 6):
    lns = set(10 * tens + u  for u in u6)
    
    # now find offsets from this selection to one without primes
    for i in set(range(-3, 4)).difference([0]):
      if not any(p in pr for p in (x + i for x in lns)):
        
        # now consider possible offsets from the 'equal tens
        # digit' selection to selection A
        lo = -4 if i < 0 else i - 4
        for j in range(lo, lo + 4 - abs(i)):
          # form the five selections
          a2e = tuple(set(x + j + k for x in lns) for k in range(5))
          
          # for all pairs of selections one must have at least four 
          # numbers not in the other
          if any(len(a.difference(b)) < 4 or len(b.difference(a)) < 4 
                      for a, b in combinations(a2e, 2)):
            continue
          # all pairs of selections must share at least two numbers
          if any(len(a.intersection(b)) < 2 for a, b in combinations(a2e, 2)):
            continue
          
          print(tuple(sorted(a2e[0])))

from itertools import combinations

# primes below fifty

pr = {3, 5, 7}

pr |= {2} | {n for n in range(11, 50, 2) if all(n % p for p in pr)}

# consider the selection having six numbers with the same tens digit

for tens in range(1, 5):

# combine this with all combinations of six units digits

for u6 in combinations(range(10), 6):

lns = set(10 * tens + u for u in u6)

# now find offsets from this selection to one without primes

for i in set(range(-3, 4)).difference([0]):

if not any(p in pr for p in (x + i for x in lns)):

# now consider possible offsets from the 'equal tens

# digit' selection to selection A

lo = -4 if i < 0 else i - 4

for j in range(lo, lo + 4 - abs(i)):

# form the five selections

a2e = tuple(set(x + j + k for x in lns) for k in range(5))

# for all pairs of selections one must have at least four

# numbers not in the other

if any(len(a.difference(b)) < 4 or len(b.difference(a)) < 4

for a, b in combinations(a2e, 2)):

continue

# all pairs of selections must share at least two numbers

if any(len(a.intersection(b)) < 2 for a, b in combinations(a2e, 2)):

continue

print(tuple(sorted(a2e[0])))

Frits permalink

You assume that the selection with the same first digits is different from the one without prime numbers. This is not explicitly stated.

Reply
- BRG permalink
  
  I do, but the code is fine without this assumption (leave 0 in on line 15).
  
  Reply

New Scientist Enigma 936 – Shift of Fortune

by Susan Denham

From Issue #2091, 19th July 1997 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta