OraQA

The following is an interesting puzzle posted on the programming praxis website:

Feynman’s Puzzle

This Puzzles was proposed by Richard Feyman ( An American physicist)

‘A’ can be any digit; each ‘.’ can be any digit from 0 through 9 except ‘A’. There is only one solution.

         ..A.
    ---------
.A. ) ....A..
      ..AA
      ----
       ...A
        ..A
       -----
        ....
        .A..
        -----
         ....
         ....
         ----
            0

COLUMN str FORMAT  A38
COLUMN d_e_A_f FORMAT A8
COLUMN b_A_c  FORMAT  A8

-----------------------------------------------------------SQL Solution----------------------------------------------------------------------------

WITH DATA1 AS
(SELECT LEVEL n FROM DUAL CONNECT BY LEVEL < 10 ),
DATA2 AS
(SELECT LEVEL-1 n FROM DUAL CONNECT BY LEVEL  < 11 )
SELECT d_e_A_f||chr(10)||'     ------------'||chr(10)||b_A_c||'  )'||Div as str, b_A_c, d_e_A_f
FROM
(SELECT b.n|| A.n|| c.n as b_A_c ,  d.n||e.n||A.n||f.n  as d_e_A_f , to_number(b.n|| A.n|| c.n )* to_number(d.n||e.n||A.n||f.n) as Div
 FROM data2 A,  data1 b, data2 c,  data1 d,  data2 e, data2 f
 WHERE b.n != A.n
 AND   c.n != A.n
 AND   d.n != A.n
 AND   e.n != A.n
 AND   f.n != A.n
 AND length ((b.n * 100 + A.n *10 + c.n) *  (d.n *1000 + e.n *100 + A.n *10 + f.n)) =7
 AND to_number(substr((b.n * 100 + A.n *10 + c.n) *  (d.n *1000 + e.n *100 + A.n *10 + f.n), 5, 1 ) ) = A.n
 AND to_number(substr((b.n * 100 + A.n *10 + c.n) * d.n, 3, 2 ))  = to_number(A.n||A.n)
 AND to_number(substr((b.n * 100 + A.n *10 + c.n)*e.n, 3, 1 )) = A.n
 AND to_number(substr((b.n * 100 + A.n *10 + c.n) * A.n , 2, 1 ) ) = A.n
 AND e.n < d.n
);

STR                                    B_A_C      D_E_A_F
---------------------------           --------    --------
         7289                           484        7289
     ------------
484  )3527876

SQL>

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