S2T2

Jim Randell 10:15 am on 26 September 2025 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

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  Jim Randell 10:15 am on 26 September 2025 Permalink | Reply

  The following Python program runs in 61ms. (Internal runtime is 135µs).

  from enigma import (irange, div, subsets, union, printf)
  
  # collect radius^2 -> coin value
  d = dict()
  for v in irange(1, 99):
    k = (v if v % 2 == 1 else 2 * v)
    d[k] = v
  
  # now look for k values where k/9 and k/4 are also values
  ss = list()
  for (k, v) in d.items():
    (v3, v2) = (d.get(div(k, 9)), d.get(div(k, 4)))
    if v3 and v2:
      printf("[{v} covers 6x {v3} or 2x {v2}]")
      ss.append((v, v3, v2))
  
  # find 2 sets that use exactly 5 coins between them
  for (s1, s2) in subsets(ss, size=2):
    rs = union([s1, s2])
    if len(rs) == 5:
      printf("{s1} & {s2} -> {rs}", rs=sorted(rs))
  

  Solution: The coin values are: 2, 8, 9, 18, 72 cents.

  There are 4 candidate configurations of coins where the largest can exactly cover 6 and 2 of the smaller coins.

  18 cent covers 6× 2 cent or 2× 9 cent = (18, 2, 9)
  54 cent covers 6× 6 cent or 2× 27 cent = (54, 6, 27)
  72 cent covers 6× 8 cent or 2× 18 cent = (72, 8, 18)
  90 cent covers 6× 10 cent or 2× 45 cent = (90, 10, 45)

  And we need to find 5 coins that can make two of these sets (i.e. one of the coins is shared between sets).

  The solution comes from:

  (18, 2, 9) and (72, 8, 18)

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