Christmas Streak 31/88: A Bit Farther

There are a total of 9 possible jumps, listed below in order of shortest to longest:

• Length $1$ : $(1,0)$
• Length $\sqrt{2}$ : $(1,1)$
• Length $2$ : $(2,0)$
• Length $\sqrt{5}$ : $(2,1)$
• Length $2\sqrt{2}$ : $(2,2)$
• Length $3$ : $(3,0)$
• Length $\sqrt{10}$ : $(3,1)$
• Length $\sqrt{13}$ : $(3,2)$
• Length $3\sqrt{2}$ : $(3,3)$

It turns out they are all possible in sequence. I found it easier to work backwards from the longest possible jump to the shortest possible jump. The number of ways to jump this way is the same as what is asked in the problem.

The longest possible jump is from opposite corners. Since there are 4 corners, there are 4 choices for placing the start of the longest jump. Then, for each of these choices, there is only one circle to jump to.

Now there are 2 possible choices for the next longest jump (either circle adjacent to the 1 circle).

It may seem that there are now two choices for the next longest jump, but placing the 4 in the bottom left corner will not allow us to move exactly 3 spaces in the jump afterwards. Thus, there is only one choice for the next jump, and then only one choice for the jump afterwards.

For the next several jumps, there is only one possible location for each.

Now, there are 3 places the last jump can go to (any of the open circles adjacent to the 9 circle).

By rule of product, there are $4\times 2 \times 3 = 24$ ways to jump this sequence. Thus, the answer is $10 \times 24 = \boxed{240}.$