Last Whole Piece

I started drawing successive stages to get a feel for how the pieces moved. Even though I had it figured out before drawing the whole thing, I thought it would be a nice picture to share. So I finished it and here it is.

After each stage I recolored the pieces before shuffling and removing a bite of chocolate.

On stage $n=8$ we remove the piece $x=8$ so $nx=\boxed{64}$

We can see how many times we can cut out a piece by looking at how much each rearrangement shortens the height. We will label the amount that the diagonal line rises (per square) $x$ as shown in the first picture.

As we can see from the second picture, in order for the resulting rearrangement to form a rectangle we must have $1+2-x+3-2x=1+x+3+x \implies x=\frac 25$ . We can also see that the section on the top of the right edge changes from $2-x$ to $1+x$ , a loss of $\frac 15$ each rearrangement. We started with an available height (on the top section of the right edge) of $\frac 85$ , and we lose $\frac 15$ each rearrangement, so we can rearrange $8$ times. The third picture is a graph of what happens to each piece every rearrangement. Thus we just need to count backwards around the graph 8 steps (starting with the first piece) to find out which piece is the last one removed. The pieces are removed in this order: $1,2,3,6,4,7,5,8$ so we have square $8$ removed after $8$ steps giving an answer of $\boxed{64}$

The sequence of whole pieces that can be cut from this is $1,2,3,6,4,7,5,8$ and still leave behind a solid rectangle.

If the dimensions of one of the chocolate pieces are $1$ wide by $\dfrac{3}{2}$ high, then the slope of the cut in black has to be $\dfrac{3}{5}$ .

Notice that corner $b$ of one of the pieces is going to move to corner $a$ , while the top of that piece must drop by $\dfrac{3}{2}+\dfrac{3}{10}=\dfrac{9}{5}$ in order to drop the height of the reassembled bar by $\dfrac{3}{10}$ which results an area deficit of $\dfrac{3}{2}$ , the area of one piece. This is the reason why the slope must be $\dfrac{3}{5}$ . See the following figures, each time the height of the reassembled bar dropping by $\dfrac{3}{10}$ . None of the whole pieces and the partial pieces at the start are ever themselves divided in subsequent cuttings, they are just rearranged. The partial pieces are rotated cyclically by $2$ with each reassembly.

Based on the question, "a solid rectangle without holes", we know:

1.It can't have holes (which the animation gives us a good way to avoid that)

2.It has to be a rectangle.

So when does it stop being a rectangle?

It will seem good UNTIL the Right most side of the top section is Gone and the Top Line of the top section is below the top right corner of the bottom section, like the stage 9 picture that another good answer illustrates.

One Key thing in order to solve the problem: When the you reassemble it, notice that the left edge of the left big chunk and right side of the right big chunk has to match, otherwise they will have a height dismatch, therefore not a rectangle any more.

Therefore the left most of red dotted line and the right has exactly a two-row-of-chocolate difference,

Therefore the intersection between red line and the chocolate row line should be in the about-center of the chocolate horizontally, AKA the about-center of the third column WHEN the top section becomes a triangle or almost-triangle,

Therefore there would be 2 bars left(the top left two)

Then you can count the order of eating.

OVER