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Lange Divisionen, Nr. 9 - Lösung

  EAGHBI / CDEF = AB
  EFGFG
  -----
   DCEJI
   DAACD
   -----
    HGJG

Ich [Jim Loy] erhielt eine nette Email von Bill Smith mit Ratschlägen zum Lösen von Alphametics-Aufgaben. Die Hauptaussage ist diese:

  Der Buchstabe, der für die Ziffer 0 steht, kann oft
  gefunden werden, indem man die 0 für die restlichen
  neun Buchstaben ausschließt.

Bill behauptet, daß dies in der Hälfte aller Aufgaben funktioniert, bei denen unmittelbar brauchbare Erkenntnisse fehlen. Diese Aufgabe hier ist ein Beispiel dafür.

  EAGHBI / CDEF = AB
  EFGFG
  -----
   DCEJI
   DAACD
   -----
    HGJG

Bill findet den Buchstaben, der für die 0 steht in nur drei Schritten:

Insgesamt haben wir 9 Ziffern für die 0 ausgeschlossen, also muß I für die 0 stehen. Bills Technik funktioniert hier hervorragend und verkürzt den Lösungsweg signifikant. In anderen Aufgaben sind subtilere Argumente notwendig, um die Nicht-Nullen auszuschließen (wie beispielsweise "aus N<M folgt, das M nicht 0 sein kann"). Selbst wenn nur mehr zwei oder drei Kandidaten für die 0 übrig bleiben, kann dies bereits hilfreich sein.


A second technique of Bill's is to identify pairs of consecutive letters. He points out that (in the same puzzle) EA and DB are consecutive (E is one less than A). This comes from a property of multipiers that begin with 9: A x 9DEF=EFGFG, E will be one less than A. In other puzzles, there are often simpler ways of identifying consecutive digits, like M-N=zero in which M is one greater than N, and there was a borrow. Sometimes you see M-N=1, and maybe you have already deduced that there was no borrow. Anyway, Bill points out that the consecutive letters EA appear in the subtraction D9EJ0 - DAA9D=HGJG (the middle digits of the two 5-digit numbers). Depending on whether there was a borrow or not, that makes G 8 or 9. Of course, C is 9, so G is 8. This gives us D=2, and B=3 (they were consecutive). Consecutive letters can give you a variety of clues in a variety of situations. Bill spells out these rules (assuming AB are consecutive):

  1a.  If xAx - xBx = xCx then C = 8 or 9.
  1b.  If xAx - xCx = xBx then C = 8 or 9.
  2a.  If xBx - xAx = xCx then C = 0 or 1.
  2b.  If xBx - xCx = xAx then C = 0 or 1.

I would like to add that such ideas do not just apply to consecutive digits. Let's say that we know that A-B=5 (or 10+A-B=5), on the right end of some subtraction, and also that xBx-xCx=xAx, then we can deduce that C=5 or 4.


Bill points out that in Dell Math Logic Problems Number 7, Word Arithmetic #42 had two solutions (only one of which spells out a readable phrase). I think I saw something like that, when I was a child, maybe not in a Dell magazine. It has happened every once in a while in Cross Sums and other puzzles. I remember one Cross Sum with about 20 solutions. Cross Sum #4, in the May 1998 Math Puzzles And Logic Problems, has four solutions. I found one puzzle (not Dell) that an author implied had only one solution, which had over 10,000 solutions, that I had my computer print out.