Substitute each capital letter in bold by a different base ten digit from 0 to 9, such that (TAU)^.BETA when rounded off to the nearest integer is equal to PI, and the absolute difference of (TAU)^.BETA and PI is the minimum. Each of A, P and T is nonzero.

Notes:

(i) PI does not represent the quantity π and, .BETA denotes a decimal fraction.

(ii) Relevant alphametic rules are applicable for this problem.

I used the following Qbasic code
DEF fnRound# (num#)
 x1# = INT(num#)
 x2# = x1# + 1
 IF num# - x1# < x2# - num# THEN
  fnRound# = x1#
 ELSE
  fnRound# = x2#
 END IF
END DEF
DEF fnAbs# (num#)
 IF num# >= 0 THEN
  fnAbs# = num#
 ELSE
  fnAbs# = -num#
 END IF
END DEF
CLS 0
min# = 1000000
FOR t# = 1 TO 9
 FOR a# = 0 TO 9
  IF a# <> t# THEN
   FOR u# = 0 TO 9
    IF u# <> t# AND u# <> a# THEN
     FOR b# = 0 TO 9
      IF b# <> t# AND b# <> a# AND b# <> u# THEN
       FOR e# = 0 TO 9
        IF e# <> t# AND e# <> a# AND e# <> u# AND e# <> b# THEN
         tau# = 100 * t# + 10 * a# + u#
         beta# = (b# / 10) + (e# / 100) + (t# / 1000) + (a# / 10000)
         v1# = tau# ^ beta#
         v1f# = fnRound#(v1#)
         IF v1f# >= 10 AND v1f# < 100 THEN
          i# = v1f# MOD 10
          IF i# <> t# AND i# <> a# AND i# <> u# AND i# <> b# AND i# <> e# THEN
           p# = (v1f# - i#) / 10
           IF p# <> t# AND p# <> a# AND p# <> u# AND p# <> b# AND p# <> e# AND p#<>i# THEN
            df# = fnAbs#(v1f# - v1#)
            IF df# < min# THEN
             min# = df#
             PRINT tau#; beta#; v1f#; v1#; df#
            END IF
           END IF
          END IF
         END IF
        END IF
       NEXT e#
      END IF
     NEXT b#
    END IF
   NEXT u#
  END IF
 NEXT a#
NEXT t#

and go the following final minimum
tau=659 beta=0.4365 and pi=17
tau^beta=16.99978714015282
rouned that gives 17=pi
with a minimal absolute difference of 0.000212859848

Posted by Daniel on 2009-09-05 12:58:55