Problems & Puzzles: Puzzles

Puzzle 152.  The Prime Gears Cars Race

Sudipta Das, sent the following nice puzzle:

Two friends , Tom and Dick , start simultaneously from Puzzledom to reach Primeland . Each of their cars have 4 different front gears ( no back gears ) and each gear makes the car run at a constant prime speed ( in metres / minute ) . And the set of prime values of the 4 gears for one car is completely different from the set for the second one .

Both the cars reach Primeland at the same time , and , believe it or not , each car runs on each gear for an integer number of minutes, and no two integer time values are identical .

Assume that the gear changing is instantaneous , and each car can pick up or drop speed immediately.

1) What is the maximum road distance between Puzzledom and Primeland , if all gear values are less than 1000 metres / minute ?
2) What is the
minimum road distance between Puzzledom and Primeland ?
3) What is the
minimum time in which the race can be finished ?
4) Repeat (1) to (3) , if there are
3 persons participating in the car race, each with a 4-gear car, and all finish simultaneously.
 


Jim Fougeron wrote (24/1/02):

I think I have question 3 of puzzle 152 done. If all gears must be different between the 2 cars, and if all times in the gears must be different, then here is an answer.

First, minimum time is (1+2+3+4+5+6+7+8)/2 minutes == 18 minutes.

Here is a working gear set and working integer minutes which satisfies the 18 minute mark.

Car 1

g1=7 for 1 minute (7m distance)
g2=17 for 8 minutes (136m distance)
g3=53 for 4 minutes (212m distance)
g4=83 for 5 minutes (415m distance)
1+8+4+5 minutes or 18 minutes total time, 7+136+212+415 for 770m distance.

Car 2

g3=5 for 2 minutes (10m distance)
g2=19 for 7 minutes (133m distance)
g3=67 for 3 minutes (201m distance)
g4=71 for 6 minutes (426m distance)
2+7+3+6 minutes or 18 minutes total time, 10+133+201+426 for 770m distance.

Both cars pass the same place at the same time. No 2 gears and no minutes in any one gear are the same. No time less than 18 minutes can be done if minutes must all be unique integers (and none can be 0), so this is a minimal (probably not the ONLY minimal) arrangement to produce the minimal time.

***

Regarding the question 2), Jim wrote (2/2/02):

I have a solution which IS the smallest. I have an iterative method which finds all "possible" valid results. I have found a distance 122, time 18 solution.

3 x 6 + 5 x 7 + 13 x 4 + 17 x 1 t=18 d=122

2 x 8 + 7 x 5 + 11 x 3 + 19 x 2 t=18 d=122

This result IS the minimum. There may be other solutions for d=122, (I still have 4 possible pathways to search), but there is no solution for the distance 120, and there can be no minimal time solution for an odd distance.

***

 

 


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