This is a generalization of Rupees and Paise.

Stan entered a departmental store with A dollars and B cents. When he exited the store, he had B/p dollars and A cents, where B/p is an integer. It was observed that when Stan came out, he had precisely 1/p times the money he had when he came in.

Given that each of A, B and p is a positive integer, with 2 ≤ p ≤ 99, determine the values of p for which this is possible. What values of p generate more than one solution?

DEFDBL A-Z
CLS
FOR p = 2 TO 99
  FOR a = 1 TO 99
    FOR b = p TO 99 STEP p
       IF ((b / p) * 100 + a) * p = a * 100 + b THEN
          PRINT p, a; b, b / p; a
       END IF
    NEXT b
  NEXT a
NEXT p

finds all the solutions:

              entered        left
                with         with
 p              a   b       b/p  a
 
 2             99  98        49  99
 
 4             33  32        8  33
 4             66  64        16  66
 4             99  96        24  99
 
 5             99  95        19  99
 
 10            11  10        1  11
 10            22  20        2  22
 10            33  30        3  33
 10            44  40        4  44
 10            55  50        5  55
 10            66  60        6  66
 10            77  70        7  77
 10            88  80        8  88
 10            99  90        9  99
 
 12            27  24        2  27
 12            54  48        4  54
 12            81  72        6  81
 
 20            99  80        4  99
 
 25            33  25        1  33
 25            66  50        2  66
 25            99  75        3  99
 
 28            77  56        2  77
 
 34            51  34        1  51
 
 40            66  40        1  66
 
 45            81  45        1  81
 
 50            99  50        1  99
 
(blank lines inserted manually between p values)
 
So p can be 2, 4, 5, 10, 12, 20, 25, 28, 34, 40, 45 or 50

among which

p = 4, 10, 12 and 25 lead to multiple solutions.

Posted by Charlie on 2012-08-13 12:43:47