Brainteaser 1757: Enigma variations
From The Sunday Times, 19th May 1996 [link]
I recently joined Enigma, the international club for puzzle-solvers. When I received my membership card I noticed that my membership number was a six-figure number and that its digits could be rearranged to make three consecutive two-figure numbers.
I also noticed that its six digits could be rearranged to form two consecutive three-figure numbers.
Furthermore, no one with a lower six-figure membership number could make the same claims.
What is my membership number?
This puzzle was included in the book Brainteasers (2002, edited by Victor Bryant). The puzzle text above is taken from the book.
[teaser1757]
Jim Randell 8:51 am on 19 November 2019 Permalink |
We can examine sequences of 3 consecutive 2-digit numbers (there are only 88 of them), and look for ones whose digits can be rearranged to form 2 consecutive 3-digit numbers. Each of these produces a collection of 6-digit membership numbers, and we can choose the smallest of these.
This Python program runs in 135ms.
Run: [ @repl.it ]
from enigma import irange, tuples, multiset, nsplit, subsets, nconcat, Accumulator, printf # collect the digits from a sequence of numbers digits = lambda ns: multiset.from_seq(*(nsplit(n) for n in ns)) # turn digits ds into n consecutive k-digit numbers def consecutive(ds, n, k): # extract the first k-digit number for s in subsets(ds, size=k, select="mP"): if s[0] == 0: continue # make the sequence of consecutive numbers s = nconcat(s) ss = tuple(irange(s, s + n - 1)) # does it use the same digits? if digits(ss) == ds: yield ss # find the smallest number r = Accumulator(fn=min) # consider 3 consecutive 2-digit numbers for n2s in tuples(irange(10, 99), 3): ds = digits(n2s) # can the digits be rearranged to form 2 consecutive 3-digit numbers? n3s = list(consecutive(ds, 2, 3)) if not n3s: continue # construct the smallest possible 6 digit number from the digits n = sorted(ds) # but leading zeros aren't allowed if n[0] == 0: # bring the first non-zero digit forward n.insert(0, n.pop(next(i for (i, x) in enumerate(n) if x > 0))) n = nconcat(n) printf("[{n:06d} -> {n2s} -> {n3s}]") r.accumulate(n) # output solution printf("membership number = {r.value:06d}")Solution: Your membership number is: 102222.
The six digits of 102222 can be rearranged as (20, 21, 22) and also (220, 221).
The largest such number is 999987:
There are 19 viable sequences of 3 consecutive 2-digit numbers, which give a total of 1195 candidate membership numbers.
If leading zeros were allowed (which would not be uncommon for membership numbers) the lowest viable membership number would be 012222 (and there would be 1350 candidate numbers).
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