Calendar Generator worksheet

1. Days in August 2800: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 2799, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 399\right)\text{mod 7}$$=\mathrm{2,806}\text{mod 7}$$=6$

2. A remainder of 6 means January 1, 2800 is a Saturday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Saturday}$

3. 2800 is a century year that divides evenly by 400, so it is a leap year with 366 days

   $\text{days in February 2800}$$=29$

4. Add up the days in the months before August

   $31+29+31+30+31+30+31$$=213$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(6+213\right)\text{mod 7}$$=2\text{= Tuesday}$

6. August 1 is day 214 of the year, and August 31 is day 244

   $\text{days of the year}$$=\text{214–244}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Tue, Wed, Thu, adding 3 more

   $4\times 5+3$$=23$

8. So August 2800 has 31 days, from Tuesday the 1st to Thursday the 31st

   $\text{days in August}$$=31$

2. Days in August 2400: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 2399, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 399\right)\text{mod 7}$$=\mathrm{2,806}\text{mod 7}$$=6$

2. A remainder of 6 means January 1, 2400 is a Saturday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Saturday}$

3. 2400 is a century year that divides evenly by 400, so it is a leap year with 366 days

   $\text{days in February 2400}$$=29$

4. Add up the days in the months before August

   $31+29+31+30+31+30+31$$=213$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(6+213\right)\text{mod 7}$$=2\text{= Tuesday}$

6. August 1 is day 214 of the year, and August 31 is day 244

   $\text{days of the year}$$=\text{214–244}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Tue, Wed, Thu, adding 3 more

   $4\times 5+3$$=23$

8. So August 2400 has 31 days, from Tuesday the 1st to Thursday the 31st

   $\text{days in August}$$=31$

3. Days in February 1600: 29

1. Find the weekday of January 1 with Gauss's rule. It divides 1599, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 399\right)\text{mod 7}$$=\mathrm{2,806}\text{mod 7}$$=6$

2. A remainder of 6 means January 1, 1600 is a Saturday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Saturday}$

3. 1600 is a century year that divides evenly by 400, so it is a leap year with 366 days

   $\text{days in February 1600}$$=29$

4. Add up the days in the months before February

   $\text{days in January}$$=31$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(6+31\right)\text{mod 7}$$=2\text{= Tuesday}$

6. February 1 is day 32 of the year, and February 29 is day 60

   $\text{days of the year}$$=\text{32–60}$

7. Four full weeks hold 5 weekdays each. The 1 extra day is Tue, adding 1 more

   $4\times 5+1$$=21$

8. So February 1600 has 29 days, from Tuesday the 1st to Tuesday the 29th

   $\text{days in February}$$=29$

4. Days in May 2200: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 2199, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 199\right)\text{mod 7}$$=\mathrm{1,606}\text{mod 7}$$=3$

2. A remainder of 3 means January 1, 2200 is a Wednesday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Wednesday}$

3. 2200 is a century year that doesn't divide evenly by 400, so it is a common year with 365 days

   $\text{days in February 2200}$$=28$

4. Add up the days in the months before May

   $31+28+31+30$$=120$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(3+120\right)\text{mod 7}$$=4\text{= Thursday}$

6. May 1 is day 121 of the year, and May 31 is day 151

   $\text{days of the year}$$=\text{121–151}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Thu, Fri, Sat, adding 2 more

   $4\times 5+2$$=22$

8. So May 2200 has 31 days, from Thursday the 1st to Saturday the 31st

   $\text{days in May}$$=31$

5. Days in August 2200: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 2199, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 199\right)\text{mod 7}$$=\mathrm{1,606}\text{mod 7}$$=3$

2. A remainder of 3 means January 1, 2200 is a Wednesday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Wednesday}$

3. 2200 is a century year that doesn't divide evenly by 400, so it is a common year with 365 days

   $\text{days in February 2200}$$=28$

4. Add up the days in the months before August

   $31+28+31+30+31+30+31$$=212$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(3+212\right)\text{mod 7}$$=5\text{= Friday}$

6. August 1 is day 213 of the year, and August 31 is day 243

   $\text{days of the year}$$=\text{213–243}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Fri, Sat, Sun, adding 1 more

   $4\times 5+1$$=21$

8. So August 2200 has 31 days, from Friday the 1st to Sunday the 31st

   $\text{days in August}$$=31$

6. Days in October 1800: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 1799, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 199\right)\text{mod 7}$$=\mathrm{1,606}\text{mod 7}$$=3$

2. A remainder of 3 means January 1, 1800 is a Wednesday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Wednesday}$

3. 1800 is a century year that doesn't divide evenly by 400, so it is a common year with 365 days

   $\text{days in February 1800}$$=28$

4. Add up the days in the months before October

   $31+28+31+30+31+30+31+31+30$$=273$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(3+273\right)\text{mod 7}$$=3\text{= Wednesday}$

6. October 1 is day 274 of the year, and October 31 is day 304

   $\text{days of the year}$$=\text{274–304}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Wed, Thu, Fri, adding 3 more

   $4\times 5+3$$=23$

8. So October 1800 has 31 days, from Wednesday the 1st to Friday the 31st

   $\text{days in October}$$=31$

7. Days in May 2000: 31

1. Find the weekday of January 1 with Gauss's rule. It divides 1999, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 399\right)\text{mod 7}$$=\mathrm{2,806}\text{mod 7}$$=6$

2. A remainder of 6 means January 1, 2000 is a Saturday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Saturday}$

3. 2000 is a century year that divides evenly by 400, so it is a leap year with 366 days

   $\text{days in February 2000}$$=29$

4. Add up the days in the months before May

   $31+29+31+30$$=121$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(6+121\right)\text{mod 7}$$=1\text{= Monday}$

6. May 1 is day 122 of the year, and May 31 is day 152

   $\text{days of the year}$$=\text{122–152}$

7. Four full weeks hold 5 weekdays each. The 3 extra days are Mon, Tue, Wed, adding 3 more

   $4\times 5+3$$=23$

8. So May 2000 has 31 days, from Monday the 1st to Wednesday the 31st

   $\text{days in May}$$=31$

8. Days in April 2250: 30

1. Find the weekday of January 1 with Gauss's rule. It divides 2249, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 1+4\times 49+6\times 249\right)\text{mod 7}$$=\mathrm{1,696}\text{mod 7}$$=2$

2. A remainder of 2 means January 1, 2250 is a Tuesday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Tuesday}$

3. 2250 doesn't divide evenly by 4, so it is a common year with 365 days

   $\text{days in February 2250}$$=28$

4. Add up the days in the months before April

   $31+28+31$$=90$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(2+90\right)\text{mod 7}$$=1\text{= Monday}$

6. April 1 is day 91 of the year, and April 30 is day 120

   $\text{days of the year}$$=\text{91–120}$

7. Four full weeks hold 5 weekdays each. The 2 extra days are Mon, Tue, adding 2 more

   $4\times 5+2$$=22$

8. So April 2250 has 30 days, from Monday the 1st to Tuesday the 30th

   $\text{days in April}$$=30$

9. Days in September 2500: 30

1. Find the weekday of January 1 with Gauss's rule. It divides 2499, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 3+4\times 99+6\times 99\right)\text{mod 7}$$=\mathrm{1,006}\text{mod 7}$$=5$

2. A remainder of 5 means January 1, 2500 is a Friday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Friday}$

3. 2500 is a century year that doesn't divide evenly by 400, so it is a common year with 365 days

   $\text{days in February 2500}$$=28$

4. Add up the days in the months before September

   $31+28+31+30+31+30+31+31$$=243$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(5+243\right)\text{mod 7}$$=3\text{= Wednesday}$

6. September 1 is day 244 of the year, and September 30 is day 273

   $\text{days of the year}$$=\text{244–273}$

7. Four full weeks hold 5 weekdays each. The 2 extra days are Wed, Thu, adding 2 more

   $4\times 5+2$$=22$

8. So September 2500 has 30 days, from Wednesday the 1st to Thursday the 30th

   $\text{days in September}$$=30$

10. Days in September 2350: 30

1. Find the weekday of January 1 with Gauss's rule. It divides 2349, the year before, by 4, 100 and 400. The remainders track how many leap days came before

   $\left(1+5\times 1+4\times 49+6\times 349\right)\text{mod 7}$$=\mathrm{2,296}\text{mod 7}$$=0$

2. A remainder of 0 means January 1, 2350 is a Sunday (0 is Sunday, 1 is Monday, up to 6 for Saturday)

   $\text{January 1}$$=\text{Sunday}$

3. 2350 doesn't divide evenly by 4, so it is a common year with 365 days

   $\text{days in February 2350}$$=28$

4. Add up the days in the months before September

   $31+28+31+30+31+30+31+31$$=243$

5. Move on that many days from January 1. Only the remainder after dividing by 7 matters, since the week repeats

   $\left(0+243\right)\text{mod 7}$$=5\text{= Friday}$

6. September 1 is day 244 of the year, and September 30 is day 273

   $\text{days of the year}$$=\text{244–273}$

7. Four full weeks hold 5 weekdays each. The 2 extra days are Fri, Sat, adding 1 more

   $4\times 5+1$$=21$

8. So September 2350 has 30 days, from Friday the 1st to Saturday the 30th

   $\text{days in September}$$=30$