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this is my first time asking a question here, so sorry in advance if there's anything wrong with the format or place this is posted in.

The problem I need to solve is written as the following: "Four people want to cross a bridge on a very dark night. They all begin on the same side and want to do the crossing as quickly as possible. A maximum of two persons can cross the bridge at one time. Any group that crosses, either 1 or 2 people, must have a flashlight with them. There is only one flashlight. The flashlight must be walked back and forth, it cannot be thrown. Each person walks at a different speed. A pair must walk together at the rate of the slower person’s pace, no one can be carried. Person A takes 1 minute to cross, person B takes 2 minutes, person C takes 5 minutes, and person D takes 10 minutes. What is the least amount of time needed to get all four people across the bridge?"

Here's how I solved it:

Since each person inevitably has to cross the bridge, the only fact we can manipulate is that someone has to come back each time to get the next person.

A and D cross first, then A goes back. A and B cross second, then A goes back. A and C cross third. In total, that gives me (10+1=11) + (2+1=3) + (5) 19 minutes.

The T.A. in class, though, said 19 was an incorrect answer.

Thoughts?

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    $\begingroup$ This would be better on puzzling.stackexchange, a similar question with different constants is puzzling.stackexchange.com/questions/287/… The point is that you want the two slow ones to cross together, so the second slowest doesn't count. $\endgroup$ – Ross Millikan Sep 16 '14 at 2:55
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Your solution is this:

$$\begin{align} ABCD&|&0\\ BC&|AD&10\\ ABC&|D&11\\ B&|ACD&16\\ AB&|CD&17\\ &|ABCD&19 \end{align} $$

whereas you can do this:

$$\begin{align} ABCD&|&0\\ CD&|AB&2\\ ACD&|B&3\\ A&|BCD&13\\ AB&|CD&15\\ &|ABCD&17 \end{align}$$

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AB over for 2, A back for 1, CD over for 10, B back for 2, AB over for 2, total 17.

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TedEx solved this!

Not my solution, but this illustrates it.

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