Henry Ernest Dudeney/Puzzles and Curious Problems/345 - The Egg Cabinet/Solution

by : $345$

 * The Egg Cabinet

Solution
Let the drawer number be $n$.

Then there will be $2 n - 1$ strips one way and $2 n - 3$ strips the other way.

This gives $4 n^2 - 4 n$ cells and $4 n - 4$ strips.

So in the $12$th drawer we get $23$ and $21$ strips, or $44$ altogether, and hence $528$ cells.

This applies to all drawers except the second, where we may have any number of strips one way, and one strip the other way.

So $1$ and $1$ will serve, but because all cells have intersecting strips, it cannot have just one.

Hence there are $262$ strips in all, and $2284$ cells.