Don't forget that every year divisible by 4 is a leap year, except that century years are only leap years if divisible by 400 (e.g., 2000 was a leap year, but 1900 was not).

Another question in my homework.. not really sure how to go about solving it... It there some sort of method to solve this using modular arithmetic?


First you need to calculate the number of days of the last $1014$ years.
That is $365\times 1000 +$ the number of leap years over the last $1014$ years (not necessarily 1014/4!)

Once you found that, divide by $7$ and you will obtain a remainder between $0$ and $6$.

Today is Friday/Saturday (depending on your time zone), add the remainder to today's day and you're done.

  • $\begingroup$ 3 leap years this century.. + 24(10) for every previous century + 3 leap years for centuries divisible by 400.. Make 246 leap years in the last 1000 years? - 3 leap years missed in the 11th century, making that 243 leap years $\endgroup$ – user122661 May 3 '14 at 16:16
  • $\begingroup$ There are 25 leap years per century. Also I don't get why 3 leap years for centuries divisible by 400. In total you should have 253 leap years (year 1000 doesn't count as we passed the 29th of February). I get 250 leap years between 1001 and 2000 (last millenum) plus another three this millenum. $\endgroup$ – user88595 May 4 '14 at 11:24
  • $\begingroup$ it states that century years are not leap years unless they are divisible by 400.. hence 24 leap years/century. Also.. there are only 3 century years divisible by 400 since the year 1000.. hence 3 century leap years. 24*10+3=243 $\endgroup$ – user122661 May 4 '14 at 20:27
  • $\begingroup$ Ok I didn't know that. Then yes you are correct $\endgroup$ – user88595 May 5 '14 at 8:20

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