# A long race

crazyblue
Posers and Puzzles 23 Jun '05 23:28
1. 23 Jun '05 23:28
Long time ago I came up with a puzzle which I could never solve somehow. Now I remembered it again, so maybe someone here knows how to do it? I think I'll give it another try too...

On a racing track (6.9 km length) three vehicles run with constant speed. Michael Schumachers Ferrari runs with constant 290 km/h, a Ford KA runs with 130 km/h and I ride with my Moped with 70 km/h.

We all start with our constant speed (no acceleration or so) at the start/finish line at 0:00 h (midnight) of Day 1.

a) When (time and lap) will we all be at the same point (how many km from start/finish line?) somewhere on the track again?
b) When (time and lap) will we all be at the start/finish line at the same time again?
2. sonhouse
Fast and Curious
24 Jun '05 16:48
Originally posted by crazyblue
Long time ago I came up with a puzzle which I could never solve somehow. Now I remembered it again, so maybe someone here knows how to do it? I think I'll give it another try too...

On a racing track (6.9 km length) three vehicles run with constant speed. Michael Schumachers Ferrari runs with constant 290 km/h, a Ford KA runs with 130 km/h and I ride ...[text shortened]... again?
b) When (time and lap) will we all be at the start/finish line at the same time again?
well the fastest car, 290 kliks makes 1008.695652 laps/day
the B car 130 kliks makes 452.173913 L/D
and the moped makes 243.4782608 L/D.
you multiplly all three L/D #'s to get 111,051,861.5 days
which is 304,043 years, 5 months, 2 days, 50 minutes and 0.384 seconds which seems to be when they will all line up at the starting
line again. Of course we are ignoring the differance in inner track
length vs. track outer edge, which are differant lengths.
3. 03 Jul '05 21:391 edit
Yes, we can ignore inner track lengths and that stuff.
I'm sure though there is a faster time, but I dont know how to explain it in English why there is lol
It's like when you have to add the numbers 1/24 + 1/36....you change it to x/72 and not to x/(24*36) because that number would be too large, even though its not wrong. This is somehow what you did by multiplying all times. Or am i wrong? Hope you know what I mean, I dont know how else to explain it with my English :-)
4. 03 Jul '05 23:073 edits
For meeting at the starting point we are looking for distances a,b,c such that a,b,c = 6.9k and a/29 = b/13 = c/7

I believe the smallest such solution is
a = 6.9 * 29 * 29 * 13 * 7 = 528063.9 km
b = 6.9 * 29 * 13 * 13 * 7 = 236718.3 km
c = 6.9 * 29 * 13 * 7 * 7 = 127463.7 km

Ferrari makes 76531 laps.
Ford makes 34307 laps.
Moped makes 18473 laps.

And a/29 = b/13 = c/7 = 18209.1 hours = 758.7125 days.

I don't think it's possible for them to meet elsewhere as 69 is relatively prime to 29, 13, and 7.

5. 04 Jul '05 09:552 edits
ðŸ˜€
6. 04 Jul '05 10:42
Originally posted by THUDandBLUNDER
ðŸ˜€
lol do you need two edits for a smiley??ðŸ˜‰ðŸ˜›ðŸ˜€
7. 04 Jul '05 11:29
Originally posted by Florrat
lol do you need two edits for a smiley??ðŸ˜‰ðŸ˜›ðŸ˜€
I change my mind a lot.

Actually, I thought my answer was wrong - but now I don't think it is.

8. 05 Jul '05 10:44
Originally posted by THUDandBLUNDER
I change my mind a lot.
Oops, they first meet after 0.345 hours.
Ferrari has done 14.5 laps.
Ford has done 6.5 laps.
Moped has done 3.5 laps.

Double those numbers to find when they first meet at the starting point.