The following alphametic summation puzzle:

   WORDS
  +WORDS
  +WORDS
-----------
  AVOWED

has a unique solution in base 10.

Find it.

Is there a solution in base 8?

DECLARE SUB permute (a$)
CLS
a$ = "1234567890": h$ = a$
DO
    IF INSTR(a$, "0") > 2 THEN
        w = VAL(MID$(a$, 1, 1))
        a = VAL(MID$(a$, 2, 1))
        o = VAL(MID$(a$, 3, 1))
        r = VAL(MID$(a$, 4, 1))
        d = VAL(MID$(a$, 5, 1))
        s = VAL(MID$(a$, 6, 1))
        v = VAL(MID$(a$, 7, 1))
        e = VAL(MID$(a$, 8, 1))
    END IF
    words = 10000 * w + 1000 * o + 100 * r + 10 * d + s
    avowed = 100000 * a + 10000 * v + 1000 * o + 100 * w + 10 * e + d
    IF 3 * words = avowed THEN
        PRINT words, avowed
    END IF
    permute a$
LOOP UNTIL a$ = h$
a$ = "12345670": h$ = a$
DO
    IF INSTR(a$, "0") > 2 THEN
        w = VAL(MID$(a$, 1, 1))
        a = VAL(MID$(a$, 2, 1))
        o = VAL(MID$(a$, 3, 1))
        r = VAL(MID$(a$, 4, 1))
        d = VAL(MID$(a$, 5, 1))
        s = VAL(MID$(a$, 6, 1))
        v = VAL(MID$(a$, 7, 1))
        e = VAL(MID$(a$, 8, 1))
    END IF
    words = 8 * 8 * 8 * 8 * w + 8 * 8 * 8 * o + 8 * 8 * r + 8 * d + s
    avowed = 8 * 8 * 8 * 8 * 8 * a + 8 * 8 * 8 * 8 * v + 8 * 8 * 8 * o + 8 * 8 * w + 8 * e + d
    IF 3 * words = avowed THEN
        PRINT words, avowed
    END IF
    permute a$
LOOP UNTIL a$ = h$

finds

34786         104358

for WORDS and AVOWED respectively in base 10.

None are found for base 8.

Posted by Charlie on 2014-01-07 18:05:03