This was a very popular problem - we received over 200 correct solutions including many from Oxgangs Primary School in Edinburgh, Red Hill Field Primary School, Culford Prep School, Jebel Ali Primary School in Dubai, Greenfields Junior School, Howell's School in Cardiff,Archbishop Temple High School, Stewart County Middle School, Hassall Grove P.S. in Australia, St.Paul's Catholic College, Cardiff High School,Casterton Preparatory School, Outwoods Edge Primary School, Thomas Deacon Academy...

Danielle from Howell's School in Cardiff explained the strategy for getting the four friends across in the shortest time:

The trick to this problem is to get the slowest people across together because otherwise you are wasting too much time. But once you've got them across how do you make one of them not walk back. The answer to that one is to get the fastest people across first so that when the slow people are over the fastest of the group can go quickly back with the flashlight. The two fastest can then run back together.



Then 7 and 10 cross over.

Then 2 crosses back.

Then 2 and 1 cross together.

Josh captured the different stages here

Patrick from Woodbridge School wrote:

The fastest solution I have found uses Strategy 1 (the one above):
Two shortest times cross together, either comes back.
Two longest times cross together, shortest (left there from last step) comes back.
Two shortest cross again.

The second fastest solution uses Strategy 2 :
Shortest crosses with longest, shortest comes back.
Shortest crosses with second longest, shortest comes back.
Two cross together.

The only speeds I can find to make both strategies the same time is if all the people go at the same speed.

Rhea and Macy from Mason Middle School found another set of timings when both Strategies would give identical results:

For the numbers 3, 4, 5, and 6 both Strategies give a shortest crossing of 21 minutes.

Harry from Dumpton School compared the times involved in each strategy:

If A is quickest time to cross then B, then C, then D.

Strategy 1:
A+B cross. B goes back.
C+D cross. A goes back.
A+B cross.

Strategy 2:
A+B cross. A goes back.
A+C cross. A goes back.
A+D cross.

Time Strategy 1 - Time Strategy 2 = 2B - (A+C)

2B > (A+C) Time 1 > Time 2 Strategy 2 best .
2B < (A+C) Time 1 < Time 2 Strategy 1 best .
2B = (A+C) Time 1 = Time 2 Both strategies equal.

Someone calling themselves "a very old person" used the same reasoning to explain how to choose between the strategies:

If we line up the times, fastest to slowest A, B, C, D.

Strategy 1:
(A and B) -->
< -- (A or B, it doesn't matter)
(C and D) -->
< -- (A or B, whatever you didn't use last time)
(A and B) -->


Strategy 2:
(A and B) -->
< -- (A)
(A and C) -->
< -- (A)
(A and D) -->

TAKING ONLY THE LONGEST TIME OF EACH PAIR we find that
Strategy 1 has a total crossing time of B + B + D + A + B
Strategy 2 has a total crossing time of B + A + C + A + D

The time for the slowest to cross (D) is incidental in both strategies.

The difference in time between the two strategies is therefore A+C vs 2B.

So when 2B = A+C they will take the same time.
Wherever A+C is greater than 2B, Strategy 1 will give us the fastest crossing.
Wherever A+C is less than 2B, Strategy 2will give us the fastest crossing.

Here are examples of each case:

2B = A+C (use either Strategy) when the times are 4, 5, 6, and 7

A+C is greater than 2B (use Strategy 1) when the times are 1, 2, 7 and 10 (as in the original problem)

A+C is less than 2B (use Strategy 2) when the times are 1, 8, 9 and 10