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Upon Reflection (Posted on 2013-11-22) Difficulty: 3 of 5
The Alphametics:
WED + TAB + PIN = X j k Y
and
DEW + BAT + NIP = Y k j X
..... when considered concurrently have 864 solutions.

In the totals X=1 and Y=2 or X=2 and Y=1, and j and k are distinct integers represented in the set of variables.

Why can X and Y never equate to W, T, P, D, B or N? The digits 0-9 are all available but there are no leading zeroes.

See The Solution Submitted by brianjn    
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Some Thoughts Is this it? | Comment 1 of 13

DECLARE SUB permute (a$)
CLS
a$ = "1234567890": h$ = a$
DO
  w = VAL(MID$(a$, 1, 1))
  e = VAL(MID$(a$, 2, 1))
  d = VAL(MID$(a$, 3, 1))
  t = VAL(MID$(a$, 4, 1))
  a = VAL(MID$(a$, 5, 1))
  b = VAL(MID$(a$, 6, 1))
  p = VAL(MID$(a$, 7, 1))
  i = VAL(MID$(a$, 8, 1))
  n = VAL(MID$(a$, 9, 1))

  wedv = VAL(LEFT$(a$, 3))
  tabv = VAL(MID$(a$, 4, 3))
  pinv = VAL(MID$(a$, 7, 3))
  x..y = wedv + tabv + pinv
  x = x..y \ 1000: y = x..y MOD 10
  IF (x = 1 AND y = 2 OR x = 2 AND y = 1) THEN
   IF w <> 0 AND t <> 0 AND p <> 0 AND d <> 0 AND b <> 0 AND n <> 0 THEN
    dew = 100 * d + 10 * e + w
    bat = 100 * b + 10 * a + t
    nip = 100 * n + 10 * i + p
    y..x = dew + bat + nip
    y = y..x \ 1000: x = y..x MOD 10
    IF x = 1 AND y = 2 OR x = 2 AND y = 1 THEN
     xy$ = LTRIM$(STR$(x..y))
     yx$ = LTRIM$(STR$(y..x))
     good = 1
     FOR j = 1 TO 4
       IF MID$(xy$, j, 1) <> MID$(yx$, 5 - j, 1) THEN good = 0: EXIT FOR
     NEXT j
     IF good THEN
       ct = ct + 1
       bad = 0
       IF w = 1 OR t = 1 OR p = 1 OR d = 1 OR b = 1 OR n = 1 THEN bad = 1
       IF w = 2 OR t = 2 OR p = 2 OR d = 2 OR b = 2 OR n = 2 THEN bad = 1
     ' IF w = 3 OR t = 3 OR p = 3 OR d = 3 OR b = 3 OR n = 3 THEN bad = 1
        
       IF bad THEN PRINT a$, xy$, yx$
     END IF
    END IF
   END IF
  END IF
  permute a$
LOOP UNTIL a$ = h$

PRINT ct

(The permute subroutine appears in the solution to the Permutations puzzle on this site.)

does indeed find 864 solutions to the alphametics taken simultaneously.

However we do get solutions of the type

675+308+249 = 1232
576+803+942 = 2321

where P = 2 = Y.

Perhaps what is meant is that in all these solutions none of W, T, P, D, B or N, nor even E, A or I, is a 1, so that X and Y can't both be one of those digits, even though one can be.


  Posted by Charlie on 2013-11-22 12:56:19
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