Question

all positive integers, one after another, are written in a spiral way:

7,8,9,10
6,1,2,11
5,4,3,12
...,15,14,13
where does the number 2009 appear with respect to 1/ FOR EXAMPLE, 12 APPEARS one unit down and two units to the right of 1

Answer 1

Did you notice the pattern that the perfect squares appear alternatively on the bottom left and top right of the current square?
Check the position of 1, 4, 9, 16, 25, 36,...
If you tabulate the coordinates of these corners relative to 1, they would be:
n n² (x,y)
1 1 (0,0)
2 4 (0,-1)
3 9 (1,1)
4 16 (-1,-2)
5 25 (2,2)
6 36 (-2,-3)
....
and in general terms,
m m² ((2-m)/2, -m/2) {mod(m,2)=0}
n n² ((n-1)/2, (n-1)/2) {mod(n,2)=1}

Since 45²=2025,
and mod(45,2)=1,
we conclude that 2025 is on the top right of the number 1, at a position of ((45-1)/2,(45-1)/2) = (22,22)
2009 is situated at 2025-2009=16 positions to the left of 2025, therefore 2009 is at (22-16,22)=(6,22) with respect to 1.

Please check my calculations.