S2T2

Jim Randell 12:09 pm on 21 February 2025 Permalink Reply
Tags: by: Victor Bryant ( 139 )   

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  Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply

  The following Python runs in 59ms. (Internal runtime is 367µs).

  from enigma import (irange, multiset, repdigit, subsets, nsplit, map2str, printf)
  
  word = 'EASTER'
  m = multiset.from_seq(word)
  
  # calculate multiplier for sum of letters
  n = m.size()
  p = repdigit(n) * m.multinomial() // n
  
  # even/odd digits
  evens = set(irange(0, 9, step=2))
  odds = irange(1, 9, step=2)
  
  # now assign odd digits to each letter
  rs = set()
  ks = sorted(m.keys())
  for vs in subsets(odds, size=len(ks), select='P'):
    d = dict(zip(ks, vs))
    # calculate the sum of the letters in the word
    s = sum(d[x] for x in word)
    # calculate the sum of all the permutations
    t = s * p
    # check the total contains only even digits
    if not evens.issuperset(nsplit(t)): continue
    # answer is sum of the letters in word
    printf("[sum = {s}; {d} -> {t}]", d=map2str(d))
    rs.add(s)
  
  printf("answer = {rs}", r=sorted(rs))
  

  Solution: E+A+S+T+E+R = 34.

  E is 9, and A, R, S, T are 1, 3, 5, 7 in some order.

  And the sum of all possible arrangements is 226666440.

  ---

  For EASTER there are 360 different arrangements of the letters (there are 2 E‘s that are not distinguishable).

  There are 6 possible positions each letter can appear in. And each of the 6 letters (including repeats) appears in each position 60 times.

  So each letter contributes 60 × 111111 = 6666660 times its own value to the total sum.

  With the sum of the letters in EASTER = 1 + 3 + 5 + 7 + 2×9 = 34 we arrive at a total sum of:

  total = 34 × 6666660 = 226666440

  which consists entirely of even digits.

  And we see that reordering the values of the non-repeated letters does not change the total.

  Using two copies of any odd digit other than 9 gives a total with at least one odd digit.

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