Math
posted by Ali on .
The year 1999 contains 365 days and begins on Friday. Use deductive reasoning to figure out what day of the week the year 2000 begins on. Show your method. The year 2000 is a leap year; 366 days. What day of the week does 2001 begin on? Show method. How many different calanders are possible? Show method. I am confused. Please help.

This site shows the 1999 calendar. You can also get calendars for other years at this site. On which day does each year begin? Can you figure out what the pattern is?
http://www.timeanddate.com/calendar/index.html?year=1999&country=1 
1999 starts on Friday, so the first week ends on Thursday. 7 days in a week x 52 weeks in a year = 364 days + 1 + 365 so the year 2000 starts on a Saturday.
2000 starts on Saturday, so the first week ends on Friday. 7 x 52 = 364 + 2 (for leap year) = 366 so the year 2001 starts on Monday
Is this correct?
Now how do reason out how many calendars are possible? Is it 7, because it's possible for the year to start on each day of the week?
Thank you for the help! 
There are 7 days in a week, so it's the same day of the week 7, 14, 21, 28,.. days from now. E.g., to find out what day of the week it is 30 days from now, you can use that 28 days from now it's the same day, so it will be two days later in the week (e.g. Friday>Sunday).
So, as far as days in the week are concerned, 30 = 2, in mathematics we write:
30 Mod 7 = 2
(30 Modulo 7 = 2), which means that 30 equals 2 up to a multiple of 7.
If
X Mod 7 = A
and
Y Mod 7 = B
then
X*Y Mod 7 = A*B Mod 7
Proof:
X Mod 7 = A >
X = A + 7 n for some integer n
Y Mod 7 = A >
Y = B + 7 m for some integer m
Therefore:
X*Y = (A + 7 n)*(B + 7 m) = A*B + some multiple of 7
We also have that (X+Y)Mod 7 = (A + B) Mod 7, of course.
To compute 365 Mod 7, you can use that:
7 Mod 7 = 0 >
49 Mod 7 = 7^2 Mod 7 = 0^2 = 0 >
50 Mod 7 = 1 >
350 Mod 7 = 7*50 Mod 7 = 0*1 = 0 >
365 Mod 7 = (15 + 350)Mod 7 = 15 Mod 7 = 1
So, we find that 2000 begins on Saturday and that 2001 begins on Monday.