Multiplying cab*car one gets "fur, ciea & bbec"
as partial results;

Multiplying car*cab we get "car, cotu & bbeu" .

Although the final result is the same, it is you who will decide what is better.

DEFDBL A-Z
CLS
FOR c = 1 TO 9
 used(c) = 1
FOR a = 1 TO 9
 IF used(a) = 0 THEN
   used(a) = 1
FOR b = 0 TO 9
 IF used(b) = 0 THEN
   used(b) = 1
FOR r = 1 TO 9
 IF used(r) = 0 THEN
   used(r) = 1

   cab = 100 * c + 10 * a + b
   pr1 = cab * r
   IF pr1 > 99 AND pr1 < 1000 THEN
     pr2 = cab * a
     IF pr2 > 999 AND pr2 < 10000 THEN
       pr3 = cab * c
       IF pr3 > 999 AND pr3 < 10000 THEN
         f = pr1 \ 100
         IF used(f) = 0 THEN
           used(f) = 1
         u = (pr1 \ 10) MOD 10
         IF used(u) = 0 THEN
           used(u) = 1
         IF r = pr1 MOD 10 THEN
       
         IF pr2 \ 1000 = c AND pr3 \ 100 = 11 * b AND pr3 MOD 10 = c THEN
           IF (pr2 \ 10) MOD 10 = (pr3 \ 10) MOD 10 THEN
           PRINT cab; r, pr1; pr2; pr3
           END IF
         END IF
       
         END IF
           used(u) = 0
         END IF
           used(f) = 0
         END IF
       END IF
     END IF
   END IF

   used(r) = 0
 END IF
NEXT
   used(b) = 0
 END IF
NEXT
   used(a) = 0
 END IF
NEXT
 used(c) = 0
NEXT

PRINT "-------"

FOR c = 1 TO 9
 used(c) = 1
FOR a = 1 TO 9
 IF used(a) = 0 THEN
   used(a) = 1
FOR r = 0 TO 9
 IF used(r) = 0 THEN
   used(r) = 1
FOR b = 1 TO 9
 IF used(b) = 0 THEN
   used(b) = 1

   car = 100 * c + 10 * a + r
   pr1 = car * b
   IF pr1 > 99 AND pr1 < 1000 THEN
     pr2 = car * a
     IF pr2 > 999 AND pr2 < 10000 THEN
       pr3 = car * c
       IF pr3 > 999 AND pr3 < 10000 THEN
         IF pr1 = car THEN
           u = (pr2) MOD 10
           IF used(u) = 0 THEN
             used(u) = 1
           IF u = pr3 MOD 10 AND pr3 \ 100 = 11 * b THEN
            IF used((pr3 \ 10) MOD 10) = 0 THEN
             PRINT car; b, pr1; pr2; pr3
            END IF
           END IF
             used(u) = 0
           END IF
         END IF
       END IF
     END IF
   END IF

   used(b) = 0
 END IF
NEXT
   used(r) = 0
 END IF
NEXT
   used(a) = 0
 END IF
NEXT
 used(c) = 0
NEXT

Finds

 381  2        762  3048  1143
-------
 382  1        382  3056  1146
 384  1        384  3072  1152
 
showing two solutions for the second multiplication, but only one for the first.

In the first, CAR is 381 and B is 2. The three partial products are shown: 762  3048  1143.

Indeed, one of the two solutions for the second, 382*381 does match, in reverse order, the multiplication in the first alphametic.

This makes the first alphametic better, in having only one solution.

Posted by Charlie on 2013-10-12 17:53:20