To be divisible by 55, a number must be divisible by both 5 and 11. Divisibility by 5 is simple enough. To be divisible by 11, A=sum of digits in odd numbered spaces, B=sum of digits in even spaces. The difference AB must be divisible by 11.
In the starting number A=1+3+8+4+5+0=21 and B=2+7+9+5+6=29. So we have AB=8.
Case 1:
One number changed.
Either change A by +8 or 3, or change B by 8 or +3.
Note we can't change the 0 and retain divisibility by 5.
Note the digit 7 cannot be changed by 8 or +3.
Each of the other _9_ digits can be changed. (1 to 9, 2 to 5 etc)
Case 2:
Two numbers changed including the last digit from 0 to 5.
Now A=26, B=29 so change A by +3 or 8 or B by 3 or +8.
This time it's the 2 that can't change.
Each of the other _9_ digits can be changed.
So I'm up to 18 possibilities.
Case 3:
Two numbers changed, not including the last digit.
This is going to be complicated and I haven't worked it out yet.
Ignore the last digit, since it is zero and can't be changed and this case simplifies to "Change two digits of 1237894556 to make the number divisible by 11."
The two 5's make it a little tricky. When I finish I will make a separate post.

Posted by Jer
on 20210208 09:29:25 