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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