Bus stops 2
Now there are three buses. And one bus stop.
Bus A arrives every 53 minutes, starting from 0700.
Bus B arrives every 47 minutes, starting from 0701.
Bus C arrives every 31 minutes, starting from 0700.
This time the buses carry on across days so if Bus A finished the day at 2347, then it would arrive at 0040 the next day. How long until they all arrive together, in minutes.
The answer is 37789.
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1 solution
Bus A 0700 53a
Bus B 0701 47b
Bus C 0700 31c
0700+53a = 0701+47b = 0700+31c
53a = 47b+1 = 31c
First solve for a and b,
53a - 47b =1 Using extended euclidean algorithm...
a = 47p + 8 ---- b = 53p + 9
53 * 47p + 53 * 8 = 47 * 53p + 47 * 9 + 1
2491p + 424 = 2491p +423+1
Now solve p and c,
53a = 31c but a=47p+8 so,
2491p + 424 = 31c
31c - 2491p = 424 Again extended euclidean algorithm...
c = 2491q + 1219 ----- p= 31q + 15
a = 47p + 8 b = 53p + 9 c = 2491q + 1219 p=31q + 15
a = 1457q + 713 ---- b = 1643q + 804 ---- c=2419q + 1219
53a = 77221q+37789 = 47b+1 = 77221q+37788+1 = 31c = 77221q + 37789
Therefore the time taken is 37789 min.
The time when this happens is 1249, 26 days later