For the start of the semester, I have decided to give everyone a quick brain warm-up. Feel free to e-mail me your solutions (my SIUE address starts with â€Ã...“rlamoni@â€Ã,Â) I will post a reply to this message with the names (or pseudonyms if desired) of the correct submitters in a day or two or when a sufficient number of people have answered the question. At that point, I will add a new question. Each question will have a title that could be a clue or just an inside joke. :-)
Please DO NOT reply to the tread with the answer. E-mail it instead.
Brain warm-up A (Astro-Math) [Read the rules in the previous post FIRST]
The Russians are shuttling astronauts to the International Space Station (ISS) while the USA’s Space Shuttle Program is under review. Unfortunately, the Russian Space Agency (RKA) has fallen upon hard times itself and so must place the following restrictions on its mission. Your task is to plan a series of steps that will get all four astronauts to the ISS.
1) The RKA has only one shuttle but can use it many times for this mission.
2) The RKA’s shuttle can carry only two astronauts at a time (a driver and a passenger.)
3) The shuttle has no auto-pilot and so must have a driver for each part of the mission.
4) The RKA has only 14 units of fuel allocated for this mission.
5) Contrary to any physical explanation, the shuttle must consume one unit of fuel for each unit of weight of the HEAVIEST of the two astronauts. Thus if a 5 unit astronaut and a 4 unit astronaut travel then the trip will take 5 units of fuel.
6) Inexplicably, rule 4 also applies to return trips. However, you will most likely send only one astronaut back with the shuttle each trip so his/her weight will determine the fuel consumed.
7) The four astronauts weigh 1, 2, 5, and 7 units and their names are 1, 2, 5, and 7 respectively ;-)
Example incorrect answer.
14 units remain.
Astronauts 1 and 5 travel up. 9 units remain.
Astronaut 5 brings the shuttle back. 4 units remain.
Astronauts 2 and 5 travel up. -1 units remain. Bummer.
It was Professor Plum, in the Kitchen, with the candlestick!!!
42 :-?
Two people (not the two clowns who posted here) have submitted correct answers. I was hoping for about five (or so) submissions in the first 24 hours. There is still 6 hours left. Puzzle B is called â€Ã...“Boom.â€Ã,Â
:-( a single tear rolls down my cheek
he called me a clown..
...I don't like clowns they scare me
Oops.
Rule 6 should read
6) Inexplicably, rule 5 also applies to return â€Ã,¦
Not
6) Inexplicably, rule 4 also applies to return â€Ã,¦
Thank you to Bill Corcoran for pointing this out.
I’ll leave this puzzle up one more day to give anyone who was confused by this typo a chance to try again.
Solvers: :-)
1. sunil (solved: 1-10-2005 10:14am)
2. tom von kercey (solved:1-10-2005 5:53pm)
Answer: :-o
14 Units Remain
1 & 2 Travel Up 12 Units Remain
2 Travels Back 10 Units Remain
5 & 7 Travel Up 3 Units Remain
1 Travels Back 2 Units Remain
1 & 2 Travel Up 0 Units Remain
Explanation: ;-)
The trick to this problem is recognizing that since only the heaviest astronaut counts you should find a way to send the heaviest and second heaviest astronauts in the same trip. If you try and do this in the first trip then one of them two will have to bring the shuttle back to the surface and will defeat any gain you got from sending them both at once. You cannot send them both in the last trip because one of the people involved in the last trip must have been on the station already to bring the shuttle back and this is just as bad as sending them both up at the start. Thus, you must send the two heaviest after one light astronaut is already on the station to make a return trip.