Enigmatic Code
Programming Enigma Puzzles
Enigma 660: Bookworm
Posted by on 20 May 2022
From New Scientist #1815, 4th April 1992 [link]
I have ten volumes of Applications of Abstract Algebra together in order on my shelves. Each volume has 100 pages and the page numbering continues through the volumes from 1 to 1000.
A bookworm started to nibble through the even-numbered side of a leaf (whose number was a perfect square): it kept nibbling in a straight line — through covers as well — and when it stopped it had just emerged through a leaf onto its odd-numbered side (whose number was also a perfect square). Ignoring the covers, it had eaten through a number of leaves which was also a perfect square.
What were those three squares?
[enigma660]
Enigmatic Code 
The key thing to note with this kind of puzzle is that when books are placed on shelves, the spine is usually placed facing out, so that the pages of a book numbered 1/2, 3/4, 5/6, …, 99/100 when placed on the shelf would be positioned (left to right): 100/99, 98/97, …, 2/1.
So if all 10 volumes are stacked left (vol 1) to right (vol 10), the pages will go:
The bookworm starts on an even page number and eats through leaves from left right and ends on an odd number in a different volume.
This Python program runs in 60ms. (Internal run time is 311µs).
Run: [ @replit ]
from enigma import (irange, div, is_square_p, printf) # consider starting leaf (even numbered squares) for i in irange(2, 30, step=2): start = i * i (sv, sp) = divmod(start - 1, 100) # consider finishing leaf (odd numbered squares) for j in irange(i + 1, 31, step=2): finish = j * j (fv, fp) = divmod(finish - 1, 100) # worm eats left to right (and ends in a different volume) if not(sv < fv): continue # calculate the number of leaves eaten (must be a square) n = 50 * (fv - sv - 1) + div(sp + 1, 2) + div(100 - fp, 2) if not is_square_p(n): continue # output solution printf("start={start} finish={finish} n={n}")Solution: The three squares are: 16 (start page), 289 (finish page), 64 (number of leaves chomped).