
Math question
This is a bonus question, only question I was unable to solve.
An ant wanders over the coordinate plane. Whenever it finds itself stationary at a point with coordinates(xy), it walks without stopping to the point with coordinates (3xy,7x+ky), where k is an integer and stops there. What is the value of k if wherever you place the ant on the plane, it returns to that point after a finite number of stops?
Thanks for the help

Let me see if I understand this. If you set the ant down at (x,y), it immediately walks to (3xy, 7x+ ky) and those then become the "(x,y)" for the next step it now walks to (3(3xy) (7x+ ky), 7(3xy)+ k(7x+ky))= (2x+(3k)y, (21+ 7k)x+ (7+k^2)y)
I think we can best do this as a matrix problem: From point (x,y) the ant goes to A(x,y) where A is the matrix
$\displaystyle \begin{bmatrix}3 & 1 \\ 7 & k\end{bmatrix}$
In order that the ant eventually return to the original point, no matter what that point is, we must have k such that, for some finite n, $\displaystyle A^n= I$ or
$\displaystyle \begin{bmatrix}3 & 1 \\ 7 & k\end{bmatrix}^n= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$
Since the product of a two matrices has determinant equal to the product of the determinants of the two matrices, If A had determinant larger than 1, it would get bigger and bigger and could never be 1, as it must in order to be the identity matrix. If A had determinant less than 1, it would get smaller and smaller. A must have determinant 1 or 1.
The determinant of A is 3k+ 7 so either 3k+ 7= 1, which gives 3k= 6 and k= 2, or 3k+ 7= 1 so 3k= 8 and k= 8/3. I think I would take k= 2.
Sure enough, with k= 2, [tex]A^3= I/[tex] so starting at any (x, y), the ant would return to (x,y) after going around a triangle.