tryingtogetbetter

Having an even number of steps actually makes this problem simpler.

The maximum distance traveled by going in any direction is 12. Also note that any pair of coordinates (x, y) in which x + y is odd is not valid because an even number of steps should either cancel out the direction or produce an even sum.

Counting, we see that there are 6(4) + 1 = 25 such points on the x and y axes.

On any line y = n, we should have n + x as an even number so for odd n's we can do ceil(n/2). For quadrant 1 this is:

ceil(11/2) + ceil(9/2) + ceil(7/2) + ceil(5/2) + ceil(3/2) + ceil(1/2) = 5 + 4 + 3 + 2 + 1 = 15.

For even n's, we can just do n/2 which is also, in quadrant 1, equal to 15.

So we should have 4(15+15) + 25 = 145.

Please let me know if I did anything wrong.

We can start by making numbers with just the digits. The only ones that work are 1111111111 and 55. Next, we can check for numbers with just 1 and 3:

3331 has 4 placements of 1.

331111 has 15 arrangements.

31111111 has 8 placements of 3.

Then numbers with 1 and 5:

511111 has 6 arrangements.

We can not have numbers with just 3 and 5 so we check for numbers with all 1, 3, and 5. There are 6 possible arrangements.

The answer should be 2+4+15+8+6+6 = 41 such numbers possible.

Please let me know if I got anything wrong.

The problem can be thought of as placing four rooks on a chessboard in a way that no two rooks can capture another.

We do this by placing the first counter on the first rank for which we have four choices. Then we have 3 choices for the second counter and so on. The answer should be 4! = 24.

Please correct me if I'm wrong.