Question 1111102: The Mayan used several different calendar systems one system used 365 days another system used 260 days. Suppose the first day of both calendars occurred on the same day. After how many days would the first day of the year again occur on the same day? About how long is this in years? (Assume 1 year has 365 days)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! to understand what this means, make a much smaller calendar so you can see what's going on.
i chose 2 calenders, the first one being 3 days and the second one being 5 days.
i then strung out the days to see when the first day of each calendar would coincide again.
my string looks like this:
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1.........
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1.........
* . . . . . . . . . . . . . . * . . . . . . . . . . . . . . *.........
it turns out that the first day of each calendar coincides again after 15 days have passed.
15 is the least common multiple of 3 and 5.
take this concept and apply it to your problem.
you have 365 day calendars and 260 day calendars.
a common multiple would be 365 * 260 = 94900.
this, however, is not the least common multiple.
you can divide 365 by 5 and 260 by 5 and 94900 by 5.
you will get 73 * 52 = 18980.
there are no other common divisors between 73 and 52, so 18980 is your least common multiple.
18980 / 365 = 52
18980 / 260 = 73
since the least common multiple will be the next time the first day of each calendar coincide, that should be your solution.
the 365 and 260 day calendars will have their first day fall on the same day in 18,980 days.
that's what i think your answer will be.
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