Here is the labeling of the 9 positions from before.
A B C D E F G H I
If one starts with an odd number at E, we saw last time that one must go up to B. After adding 5, we will have an even number, so if we don't have a multiple of 4 we can go to C. But what if we go to A instead? In the first hint, we saw we can do this as long as x+1 is a multiple of 3, and we will end up back at B with the value of (2x+80)/3. But if x is an even number, then (2x+80)/3 will be a multiple of 4. This means we will not be able to go to C, but are forced to go back to A. Once again, we will be in a loop, which will eventually kill us (unless x=80). Hence, if we have an odd number x in the center, we must not only go up to B, but then go to the right to C, down to F, and then left back to E. This will give us (3x+29)/4, so we must start with a number that is 1 more than a multiple of 4.
At this point, it is useful to introduce modular arithmetic notation. We say that a number is 1 (mod 4) if it is one more than a multiple of 4. That is, if we divide by 4, we will have 1 as a remainder. Thus, if we have a number x that is 1 (mod 4) in the center, we must go up to B, right to C, down to F, and left back to E, giving us (3x+29)/4.
What happens if we have a number that is 3 (mod 4) in the center? We will be doomed to fail. We have already seen that we must go up, but after adding 5 we will have a multiple of 4, so we can't go to the right. Sooner or later, we will be stuck.
From this information, can you determine the first couple of moves in the maze?
Still need another hint? Click here.