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I am not sure where to start to solve this problem:

  • Smith, Jones, and Rodriguez are the engineer, brakeman, and fireman on a train, not necessarily in that order.

  • Riding on the train are three passengers with the same last names as the crewmembers, identified as passenger Smith, passenger Jones, and passenger Rodriguez.

  • The brakeman lives in Denver.

  • Passenger Rodriguez lives in San Francisco.

  • Passenger Jones long ago forgot all the algebra that he learned in high school.

  • The passenger with the same name as the brakeman lives in New York.

  • The brakeman and one of the passengers, a professor of mathematical physics, attend the same health club.

  • Smith beat the fireman in a game of tennis at a court near their homes.

Can you discover a theorem that tells the name of the engineer, the brakeman, and the fireman? I am lost on where to start. Thank you

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A teeny-tiny note on language: what you want to do is to derive a theorem from the given postulates. You don't solve postulates; they're the things you're supposed to take for granted as being true. –  J. M. May 12 '12 at 5:10
2  
What does this have to do with geometry? –  Neal May 12 '12 at 5:19
    
@Neal: Probably the idea of establishing postulates and deriving conclusions from them. I can't say that this problem seems a very good example, though. –  Brian M. Scott May 12 '12 at 5:36
    
This type of puzzle is sometimes called zebra puzzle, see Wikipedia. –  Martin Sleziak May 12 '12 at 5:55

1 Answer 1

This kind of problem is often solved most easily by making a table whose rows are the names and whose columns are the occupations and working through the postulates, marking the combinations $1$ (correct) and $0$ (impossible). Here's the empty table.

$$\begin{array}{r|c} &\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline \text{Smith}\\ \hline \text{Jones}\\ \hline \text{Rodriguez} \end{array}$$

Now number the postulates that give specific information:

  1. The brakeman lives in Denver.
  2. Passenger Rodriguez lives in San Francisco.
  3. Passenger Jones long ago forgot all the algebra that he learned in high school.
  4. The passenger with the same name as the brakeman lives in New York.
  5. The brakeman and one of the passengers, a professor of mathematical physics, attend the same health club.
  6. Smith beat the fireman in a game of tennis at a court near their homes.

(6) clearly implies that Smith is not the fireman. From (2) and (4) you can infer that the brakeman is not Rodriguez: if he were, passenger Rodriguez would lie in New York, and he doesn't. At this point you can enter a couple of $0$'s in the table:

$$\begin{array}{r|c} &\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline \text{Smith}&&&0\\ \hline \text{Jones}\\ \hline \text{Rodriguez}&&0 \end{array}$$

Now (1) and (5) imply that the mathematical physicist lives in or near Denver, so by (2) he can't be passenger Rodriguez. By (3) he also can't be passenger Jones, since he obviously hasn't forgotten high school algebra, so he must be passenger Smith. (4) tells us that the passenger with the same name as the brakeman lives in New York, so the brakeman isn't Smith, and we can enter another $0$ into our table. But Smith has to have one of the occupations, so he must be the engineer: it's the only remaining possibility. And of course nobody else can be the engineer, so we now have:

$$\begin{array}{r|c} &\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline \text{Smith}&1&0&0\\ \hline \text{Jones}&0\\ \hline \text{Rodriguez}&0&0 \end{array}$$

And at this point it all falls into place; we must have

$$\begin{array}{r|c} &\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline \text{Smith}&1&0&0\\ \hline \text{Jones}&0&1&0\\ \hline \text{Rodriguez}&0&0&1 \end{array}$$

Smith is the engineer, Jones is the brakeman, and Rodriguez is the fireman.

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OP said he didn't know where to start. Now he knows where to start, and where to go next, and where to go after that, all the way to where to finish. –  Gerry Myerson May 12 '12 at 6:30
    
@Gerry: I debated with myself about this one but finally decided that a fully-worked example might be more helpful in the long run if the OP had never encountered anything like this before. I'm not entirely happy with that choice, but I also wasn't entirely happy with any of the possible stopping points. –  Brian M. Scott May 12 '12 at 6:35

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