Copyright © University of Cambridge. All rights reserved.
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.
Josh captured the different stages here
Patrick from Woodbridge School wrote:The fastest solution I have found uses Strategy 1 (the one above):
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.
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.
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