## Wednesday, January 11, 2006

### Ticket to Ride

Each of the railroad stations in a certain area sells tickets to every other station on the line. This practice was continued when several new stations were added, and 52 additional sets of tickets had to be printed. How many stations were there originally, and how many new ones had to be added?

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#### 5 comments:

1. It's an interesting question, but I don't think I understand it. Does each station also sell tickets for it's own station? What exactly is a set of tickets? A stack of tickets for 1 other station? Or a stack of tickets for all other stations?

2. A ticket is sold to go to one other station. Each station does not sell tickets to itself.

So if there were 3 stations originally, each station would have 2 tickets to sell.

I'm sorry the question wasn't clear. I hope this explanation helps.

3. There were five stations and they added 4 to make nine. The five stations need 4 tickets each making a total of 20. Nine stations need 8 tickets each making a total of 72. The difference makes the 52 additional tickets needed.

4. Bentley92 got it right. Now here's the math...

If F = the number of stations and N = the number of new ones. We must have 2FN + N(N-1) = 52. This can be rewritten as N^2 + (2F -1)N - 52 = 0. The possible factorization of -52 is 1x52, 2x26 and 4x13. The sum of the roots must equal 51, -51, 24, -24, 9 or -9. But this sum must equal 2F-1 (and F is a positive integer). The only possibility is 2F-1 = 9, so F = 5. Therefore N = 4.

You may have noted that 2F-1 = 51 works, but then F = 26, N = 1, but the problem stated several new stations were added, not just 1.

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