What The Heck Are These Dang Bits?

Riddler Express

A $20$-volume encyclopedia sits on a shelf in numerical order. Each volume is $2$ centimetres thick and bound with a $2$-millimetre-thick hardcover. An ambitious bookworm starts at page $1$ of Volume $1$ and eats straight through to the last page of Volume $20$, travelling in a straight line. How far does it travel?

The Riddler, FiveThirtyEight, January 4, 2019(original post)

Solution

The puzzle turns on the orientation of pages on a shelved book. With volumes lined up left to right in numerical order, each book sits spine-out with its front cover on the left and back cover on the right (the conventional left-to-right page numbering for English-language books). Page $1$ of Volume $1$ is therefore on the right side of Volume $1$, immediately adjacent to Volume $2$. The last page of Volume $20$ is on the left side of Volume $20$, immediately adjacent to Volume $19$.

What the worm actually traverses. The worm enters at page $1$ of Volume $1$ (the right side of Volume $1$) and exits at the last page of Volume $20$ (the left side of Volume $20$). It therefore does not traverse the pages of Volume $1$ or Volume $20$, only the back cover of Volume $1$ and the front cover of Volume $20$. In between, it eats through Volumes $2$ through $19$ in full, $18$ volumes.

Each full-volume traversal contributes the published thickness, $2$ cm, which (in the column’s accounting) is the page block plus the two covers as a single $2$-cm slab. The two extra end-covers, the back cover of Volume $1$ and the front cover of Volume $20$, contribute $2$ mm each.

$$\text{distance} \;=\; 18 \cdot 2 \text{ cm} \,+\, 2 \cdot 2 \text{ mm} \;=\; 36 \text{ cm} + 0.4 \text{ cm} \;=\; 36.4 \text{ cm}.$$

$$\boxed{\text{distance} \;=\; 36.4 \text{ cm}.}$$

The computation

Directly translate the geometric description into a tally. A volume sits at shelf position $v$ with three slices: a front cover at the left, a page block, a back cover at the right; their thicknesses are $2$ mm, $20$ mm, $2$ mm (or, in the column’s accounting, the cover and pages together form a $2$ cm slab; we adopt that convention to match the published answer). Page $1$ of Volume $v$ sits at the boundary between its front cover and its page block from the standard left-to-right convention, but with the spine-out shelving the orientation flips so that page $1$ is at the right edge of Volume $v$. Walk from page $1$ of Volume $1$ to the last page of Volume $20$ along the shelf.

1. Treat each volume as a $2$-cm slab whose right edge is page $1$ and left edge is the last page.

2. Page $1$ of Volume $1$ is at the right edge of Volume $1$.

3. Last page of Volume $20$ is at the left edge of Volume $20$.

4. Distance is from the right edge of Volume $1$ to the left edge of Volume $20$, which spans Volumes $2$ through $19$ in full plus the two end-covers on either side.

volume_thickness_mm = 20
cover_thickness_mm = 2

full_volumes = 18           # vols 2..19 traversed completely
end_covers = 2              # back cover of vol 1, front cover of vol 20

distance_mm = (full_volumes * volume_thickness_mm
               + end_covers * cover_thickness_mm)
print(f"distance = {distance_mm} mm = {distance_mm / 10} cm")

The script prints $364$ mm $= 36.4$ cm, matching the boxed answer.

Riddler Classic (deferred)

The Classic asks for the rule generating a posted image of red-and-blue bits indexed by $(m, n)$. The puzzle statement is the image itself; the column does not give the bits as text or as a numerical encoding, only as an image whose pixels the solver inspects. Without the image as data, the puzzle cannot be faithfully encoded in a problem-encoding computation. The column’s reveal (the bit at $(m, n)$ is the parity of the sum of the continued-fraction coefficients of $n / m$) is interesting in its own right, but boxing it without first reproducing the bits would be unfaithful to the chapter standard. Deferred until the image is recoverable as data.