Substitute each of the capital letters in bold by a different digit from 0 to 9 such that the sum of digits of each of POIRE, OPERA, PORTE, TAPIS, ASTRE, PATRE, LOTUS, LISTE, LOUPE and, OUTRE (in this order) constitutes ten consecutive terms of an arithmetic sequence in ascending order. It is known that none of P, O, T, A and L can be zero.

This does not define the words as well as Charlie has with his but it works.

OPEN "poire.txt" FOR OUTPUT AS #1
CLS


FOR a = 1 TO 9
IF used(a) = 0 THEN
used(a) = 1
FOR o = 1 TO 9
IF used(o) = 0 THEN
used(o) = 1
FOR l = 1 TO 9
IF used(l) = 0 THEN
used(l) = 1
FOR p = 1 TO 9
IF used(p) = 0 THEN
used(p) = 1
FOR t = 1 TO 9
IF used(t) = 0 THEN
used(t) = 1
FOR e = 0 TO 9
IF used(e) = 0 THEN
used(e) = 1
FOR i = 0 TO 9
IF used(i) = 0 THEN
used(i) = 1
FOR u = 0 TO 9
IF used(u) = 0 THEN
used(u) = 1
FOR r = 0 TO 9
IF used(r) = 0 THEN
used(r) = 1
FOR s = 0 TO 9
IF used(s) = 0 THEN
used(s) = 1

a1 = p + o + i + r + e
a2 = o + p + e + r + a
a3 = p + o + r + t + e
a4 = t + a + p + i + s
a5 = a + s + t + r + e
a6 = p + a + t + r + e
a7 = l + o + t + u + s
a8 = l + i + s + t + e
a9 = l + o + u + p + e
a10 = o + u + t + r + e

IF (a1 < a2 AND a2 < a3 AND a3 < a4 AND a4 < a5 AND a5 < a6 AND a6 < a7 AND a7 < a8 AND a8 < a9 AND a9 < a10) THEN
PRINT a; e; i; o; u; l; p; r; s; t
PRINT a1; a2; a3; a4; a5; a6; a7; a8; a9; a10

PRINT #1, a; e; i; o; u; l; p; r; s; t
PRINT #1, a1; a2; a3; a4; a5; a6; a7; a8; a9; a10
END IF






used(s) = 0
END IF
NEXT
used(r) = 0
END IF
NEXT
used(u) = 0
END IF
NEXT
used(i) = 0
END IF
NEXT
used(e) = 0
END IF
NEXT
used(t) = 0
END IF
NEXT
used(p) = 0
END IF
NEXT
used(l) = 0
END IF
NEXT
used(o) = 0
END IF
NEXT
used(a) = 0
END IF
NEXT
CLOSE 1

Yields:
 a    e     i     o    u   l     p    r    s    t
 6    8     5    3    9   4    2    0    1   7
Sum of Digits
 18  19  20  21  22  23  24  25  26  27

Posted by brianjn on 2009-11-21 19:15:47