I have multiplied two 3-digit numbers and coded the whole process twice:
In the 1st coding
replaces an odd digit and E
In the 2nd coding S
replaces a digit smaller than 6 and B
a digit bigger than 5.
Try to get my original numbers(digit 9 does not appear in the multiplication) and all other possible solutions, if any.
(In reply to Not that hard (spoiler)
by Steve Herman)
1. Assigning a degree of difficulty is a subjective task, usely controversial.
2. I have "tested" my "double code" by asking two of my math-friends to solve it. It took them over 40 minutes each.
3. Apparently both of them are less gifted or less systematic
or both than you.
4. I agree that the solving is easy , once you grasp the fact that all B&O dtigits are implicitly defined as 7.
5. Why no one rates this problem?
It is original, thought-provoking and leads to a single solution...