In https://bombshelterradioblog.azurewebsites.net/2022/11/28/the-train-puzzle/ I mentioned a nifty puzzle that was shared on Mastodon.
Here is how I solved it:
- Assign each car a letter (A-K). Assign the number of passengers per car to each letter. The Ninth car would be I.
- “in any 3 consecutive cars there are 99 passengers”, so…
- A+B+C=99, B+C+D=99, through to I+J+K=99
- By combining two sets of cars (e.g. A+B+C=B+C+D) and reducing we find that A=D=G=J, B=E=H=K, C=F=I
- “There are 381 passengers total”, so…
- A+B+C+D+E+F+G+H+I+J+K=381
- A+B+C+A+B+C+A+B+C+A+B=381
- 4A+4B+3C=381
- “in any 3 consecutive cars there are 99 passengers”, so…
- A+B+C=99
- 4A+4B+4C=396
- 4A+4B+4C-15 is equal to 4A+4B+3C, solve for C
- 4C-15=3C
- 1C-15=0
- C=15
- The ninth car (I) contains the same number of people as third car (C) and the sixth car (F). Thus the answer is 15.
I do not believe there is a way to solve for the other cars given the information provided. We do know there are 84 (99-C) people between the car pairs of A+B, D+E, G+H, and J+K. But we do not know the relationship between A+B.
About the picture: An Amtrak GG-1 pulls the Broadway Limited. Picture taken from https://history.amtrak.com/blogs/blog/digging-into-the-archives-the-amazing-gg-1