base = 3
22 = 11 + 4*1
base = 4
11 = 1 + 4*1
base = 6
24 = 12 + 4*2
base = 8
12 = 2 + 4*2
base = 9
26 = 13 + 4*3
base = 10
39 = 27 + 4*3
base = 12
13 = 3 + 4*3
28 = 14 + 4*4
base = 15
2A = 15 + 4*5
base = 16
14 = 4 + 4*4
base = 18
2C = 16 + 4*6
base = 20
15 = 5 + 4*5
3i = 2E + 4*6
base = 21
2E = 17 + 4*7
base = 24
16 = 6 + 4*6
2G = 18 + 4*8
base = 27
2i = 19 + 4*9
base = 28
17 = 7 + 4*7
base = 30
2K = 1A + 4*A
3R = 2L + 4*9
base = 32
18 = 8 + 4*8
base = 33
2M = 1B + 4*B
base = 36
19 = 9 + 4*9
2o = 1C + 4*C
base = 39
2Q = 1D + 4*D
base = 40
1A = A + 4*A
3 = 2S + 4*C
base = 42
2S = 1E + 4*E
base = 44
1B = B + 4*B
base = 45
2U = 1F + 4*F
base = 48
1C = C + 4*C
2W = 1G + 4*G
where A represents 10, B represents 11, etc., and i and o have been lowercased to avoid confusion with 1 and 0.
Something must have gone awry with Ady's analysis as base 10 has only one solution. The first base with more than one solution is base 12.
DECLARE FUNCTION dig$ (n#)
DECLARE FUNCTION cvb$ (n#)
DEFDBL AZ
DIM SHARED b
PRINT
FOR b = 2 TO 48
done = 0
FOR n1 = 1 TO b  1
FOR n2 = 1 TO b  1
IF b * n1 + n2 = n1 * n2 + 4 * n2 / n1 THEN
IF done = 0 THEN
PRINT
PRINT "base ="; b
done = 1
END IF
PRINT " "; dig$(n1); dig$(n2); " = "; cvb$(n1 * n2); " + 4*"; cvb$(n2 / n1)
END IF
NEXT
NEXT
NEXT b
FUNCTION cvb$ (n)
build$ = ""
r = n
DO
d = r MOD b
r = r \ b
build$ = dig$(d) + build$
LOOP UNTIL r = 0
cvb$ = build$
END FUNCTION
FUNCTION dig$ (n)
dig$ = MID$("0123456789ABCDEFGHiJKLMNoPQRSTUVWXYZ", n + 1, 1)
END FUNCTION
Edited on September 17, 2007, 12:09 pm

Posted by Charlie
on 20070917 11:53:36 