Sometimes an arithmetic procedural error produces a correct answer.
An accidental cancellation is reducing a fraction by canceling individual digits in the numerator and denominator.

In some cases the result is equal to that obtained by a correct process.
Examples:
16/64 = 16/64 = 1/4;
26/65 = 26/65 = 2/5;
266/665 = 266/665 = 2/5;
49/98 = 4/8 = 1/2 etc.

Try to find two-digit cases in bases other than base 10,(e.g. 13/32 = 1/2 ( the only solution in base 4).

(In reply to re: computer solutions -a partial errata by Ady TZIDON)

Ah yes, you're so right. I neglected the very important fact that the canceled digits had to be equal to each other!

I've fixed the program, and extended to base 20 as there are fewer results. Sorry about that.

13/32=1/2   4
15/53=1/3   6
25/54=2/4   6
17/74=1/4   8
37/76=3/6   8
14/43=1/3   9
28/86=2/6   9
16/64=1/4   10
19/95=1/5   10
26/65=2/5   10
49/98=4/8   10
1B/B6=1/6   12
2B/B8=2/8   12
3B/B9=3/9   12
5B/BA=5/A   12
1D/D7=1/7   14
6D/DC=6/C   14
17/75=1/5   15
27/76=2/6   15
2C/C9=2/9   15
2E/EA=2/A   15
3C/CA=3/A   15
4E/EC=4/C   15
15/54=1/4   16
19/96=1/6   16
1F/F8=1/8   16
2A/A8=2/8   16
39/98=3/8   16
3F/FC=3/C   16
7F/FE=7/E   16
1H/H9=1/9   18
2H/HC=2/C   18
5H/HF=5/F   18
8H/HG=8/G   18
1J/JA=1/A   20
3J/JF=3/F   20
4J/JG=4/G   20
9J/JI=9/I   20

The revised program:

DECLARE FUNCTION alp$ (x!)
CLS
OPEN "cancfall.txt" FOR OUTPUT AS #2
FOR b = 2 TO 20

FOR a1 = 1 TO b - 1
 FOR a2 = 1 TO b - 1
  FOR b1 = 1 TO b - 1
    FOR b2 = 1 TO b - 1
      IF a1 <> a2 OR b1 <> b2 THEN
       IF a1 <> b1 OR a2 <> b2 THEN
        num = b * a1 + a2: den = b * b1 + b2
         IF den >= num THEN
              IF a2 * den = b1 * num AND a1 = b2 THEN
                PRINT alp$(a1); alp$(a2); "/"; alp$(b1); alp$(b2); "="; alp$(a2); "/"; alp$(b1); TAB(12); b
                PRINT #2, alp$(a1); alp$(a2); "/"; alp$(b1); alp$(b2); "="; alp$(a2); "/"; alp$(b1); TAB(12); b
                DO: LOOP UNTIL INKEY$ > ""
              END IF
              IF a1 * den = b2 * num AND a2 = b1 THEN
                PRINT alp$(a1); alp$(a2); "/"; alp$(b1); alp$(b2); "="; alp$(a1); "/"; alp$(b2); TAB(12); b
                PRINT #2, alp$(a1); alp$(a2); "/"; alp$(b1); alp$(b2); "="; alp$(a1); "/"; alp$(b2); TAB(12); b
              END IF
           END IF
       END IF
      END IF
    NEXT
  NEXT
 NEXT
NEXT

NEXT b

CLOSE

END

FUNCTION alp$ (x)
  alp$ = MID$("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ", x + 1, 1)
END FUNCTION

 

Posted by Charlie on 2013-01-18 20:51:57