(In reply to
5/6 of computer solution by Charlie)
Looking for minimums and patterns for A3 and B3.
In order for both sod(P1) and sod(P+1) to be divisible by the given integer, sod(P1)  sod(P+1) must also be divisible by the integer.
1/13 = 0.[076923]...
The first occurance for
sod(P1)  sod(P+1) =
7 @ P = 3×13 = 39
16 @ P = 23×13 = 299
25 @ P = 923×13 = 11999
34 @ P = 6923×13 = 89999
52 @ P = 76923×13 = 999999
43 @ P = 100000×13 = 1300000
...
52 is the first multiple of 13 that appears, thus P > 999999.
As Charlie has found, P = 39000000.
1/11 = 0.[09]...
The first occurance for
sod(P1)  sod(P+1) =
16 @ P = 9×11 = 99
7 @ P = 10×11 = 110
34 @ P = 909×11 = 9999
25 @ P = 1000×11 = 11000
52 @ P = 90909×11 = 999999
43 @ P = 100000×11 = 1100000
70 @ P = 9090909×11 = 99999999
61 @ P = 10000000×11 = 110000000
88 @ P = 909090909×11 = 9999999999
...
88 is the first multiple of 11 that appears, thus P > 9999999999.
It appears Charlie's program looked only up to 11×99999999, thus it was unable to reveal the solution for B3.
Edited on March 19, 2013, 9:11 am

Posted by Dej Mar
on 20130319 08:18:32 