New Scientist Enigma 936 – Shift of Fortune
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”?
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Brian Gladman permalink1234567891011121314151617181920212223242526272829303132333435from itertools import combinations# primes below fiftypr = {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 digitfor tens in range(1, 5):# combine this with all combinations of six units digitsfor u6 in combinations(range(10), 6):lns = set(10 * tens + u for u in u6)# now find offsets from this selection to one without primesfor 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 Alo = -4 if i < 0 else i - 4for j in range(lo, lo + 4 - abs(i)):# form the five selectionsa2e = 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 otherif any(len(a.difference(b)) < 4 or len(b.difference(a)) < 4for a, b in combinations(a2e, 2)):continue# all pairs of selections must share at least two numbersif any(len(a.intersection(b)) < 2 for a, b in combinations(a2e, 2)):continueprint(tuple(sorted(a2e[0])))
