Hint 1

This maze is very difficult if one tries to solve it using only trial and error. However, with a little bit of mathematics, the maze becomes much easier. Because of this special property of the maze, I will not give you the solution immediately. Rather, I will give you one hint at a time. Try to solve the maze using the fewest number of hints.

Let us label the 9 positions in the square as follows:

A B C D E F G H I

The point B is a crucial point. If you are at B, which way can you go? If x is the value in the square B, you cannot go left to A unless x+1 is a multiple of 3. The reason is that, after multiplying by 2 and adding 5, you will be at an odd number, so you will not be able to go right to E. Rather, you must go down to G, which involves dividing by 3. This will only be possible if we started with a number that was 1 less than a multiple of 3.

If x+1 is a multiple of 3 at B, we will be able to get to G by going through A and D. But you will have an odd number at G, so you will be forced to go to H, adding 8. Since the number will still be odd, going to I will be impossible, so you will be forced to go to E, adding 8 again. Once again the number will be odd, so you will be forced to go back to B.

After doing some algebra, you will find that if we started at B with a value of x, then going to the left will force you back to B with the value of (2x+80)/3.

Can you find a similar result for going to the right from the point B? This is the key to understanding how the maze is solved.

If we are at B, when can we go to the right to C? We will then have a multiple of 3, and will be forced to go down to F. But after adding 2, we will not have a multiple of 3 anymore, so we cannot go to I, but must divide by 4 to get to E. If we start with an odd number at B, we will be stuck with an odd number at F. On the other hand, if we start with a multiple of 4 at B, then we will no longer have a multiple of 4 after adding 2, so again we will be stuck. Hence, to go to the right from B, one must start with an even number that is not a multiple of 4. In other words, if x is the value at B, x+2 must be a multiple of 4 to go to C. You will wind up at E with the value (3x+2)/4.

Thus, if you are at B, you can only go to the left if x+1 is a multiple of 3, and can only go to the right if x+2 is a multiple of 4. Can you find any restrictions for when you can go down?

Still need another hint? Click here.