Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. Neither P nor O can be zero.
PANE = [OUTSHOUT/ PIANO], where the remainder obtained upon dividing OUTSHOUT by PIANO is less than 500.
Note: [x] denotes the greatest integer ≤ x.
using Qbasic I found
PANE=1670
OUTSHOUT=32854328
PIANO=19673
and this gives a remainder of 418
My Qbasic program follows
CLS 0
FOR p = 1 TO 9
FOR a = 0 TO 9
IF a <> p THEN
FOR n = 0 TO 9
IF n <> p AND n <> a THEN
FOR e = 0 TO 9
IF e <> p AND e <> a AND e <> n THEN
v1# = 1000 * p + 100 * a + 10 * n + e
FOR o = 1 TO 9
IF o <> p AND o <> a AND o <> n AND o <> e THEN
FOR u = 0 TO 9
IF u <> p AND u <> a AND u <> n AND u <> e AND u <> o THEN
FOR t = 0 TO 9
IF t <> p AND t <> a AND t <> n AND t <> e AND t <> o AND t <> u THEN
FOR s = 0 TO 9
IF s <> p AND s <> a AND s <> n AND s <> e AND s <> o AND s <> u AND s <> t THEN
FOR h = 0 TO 9
IF h <> p AND h <> a AND h <> n AND h <> e AND h <> o AND h <> u AND h <> t AND h <> s THEN
v2# = (10 ^ 7) * o + (10 ^ 6) * u + (10 ^ 5) * t + (10 ^ 4) * s + (10 ^ 3) * h + (10 ^ 2) * o + 10 * u + t
FOR i = 0 TO 9
IF i <> p AND i <> a AND i <> n AND i <> e AND i <> o AND i <> u AND i <> t AND i <> s AND i <> h THEN
v3# = o + 10 * n + 100 * a + 1000 * i + 10000 * p
fl# = INT(v2# / v3#)
rm# = v2# - fl# * v3#
IF fl# = v1# AND rm# <= 500 THEN
PRINT v1#, v2#, v3#, rm#
END IF
END IF
NEXT i
END IF
NEXT h
END IF
NEXT s
END IF
NEXT t
END IF
NEXT u
END IF
NEXT o
END IF
NEXT e
END IF
NEXT n
END IF
NEXT a
NEXT p
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Posted by Daniel
on 2009-08-12 11:40:33 |
basic solution
