Calendars

Day of week, odd days, leap years

Calendar problems involve determining the day of the week for a given date, calculating odd days, and understanding leap year rules. These problems test your ability to work with the Gregorian calendar system and apply modular arithmetic concepts. Mastering calendar calculations is essential for solving date-based logical reasoning questions efficiently.

Key Formulas / Rules

\text{Odd Days = Remainder when total days / 7}

\text{Leap Year: Divisible by 4, but not by 100 unless also divisible by 400}

\text{Century Odd Days: 100 years = 5 odd days, 200 years = 3 odd days, 300 years = 1 odd day, 400 years = 0 odd days}

\text{Month Days: Jan(31), Feb(28/29), Mar(31), Apr(30), May(31), Jun(30), Jul(31), Aug(31), Sep(30), Oct(31), Nov(30), Dec(31)}

\text{Day Codes: Sunday=0, Monday=1, Tuesday=2, Wednesday=3, Thursday=4, Friday=5, Saturday=6}

\text{Zeller's Congruence: h = (q + 13(m+1)/5 + K + K/4 + J/4 - 2J) mod 7}

\text{Same Date Same Day: After 400 years, calendar repeats exactly}

Key Concepts

Odd Days Concept

Odd days are the extra days left after complete weeks are removed from any given period. A normal year has 365 days = 52 weeks + 1 day, giving 1 odd day. A leap year has 366 days = 52 weeks + 2 days, giving 2 odd days. To find the day of week, calculate total odd days and apply modulo 7. For example, if today is Monday and we add 3 odd days, the result day is Thursday.

Leap Year Rules

A year is a leap year if: (1) divisible by 4 but not by 100, OR (2) divisible by 400. Years like 1900, 2100 are NOT leap years despite being divisible by 4. Years like 2000, 2400 ARE leap years. February has 29 days in leap years and 28 in normal years. This distinction is crucial for calculating odd days across February boundaries.

Century Calculations

Every 100 years, the calendar shifts by 5 odd days (since 76 normal years x 1 + 24 leap years x 2 = 124 odd days = 5 mod 7). For multiple centuries: 100 years = 5 odd days, 200 years = 3 odd days (5+5=10, 10 mod 7 = 3), 300 years = 1 odd day, 400 years = 0 odd days (accounting for the leap century). This pattern repeats every 400 years.

Month Code System

To quickly calculate day shifts: January=0 (or 6 if preceding year was leap), February=3, March=3, April=6, May=1, June=4, July=6, August=2, September=5, October=0, November=3, December=5. These codes represent the cumulative odd days from the start of the year. Add the date number and apply modulo 7 to find the day code.

Same Day Date Patterns

In a normal year, the calendar for a month repeats after 6 years (adding 6+1=7 or 7+2=9 odd days). In a leap year cycle, the pattern shifts. Key patterns: Last day of Feb, Mar, Nov is same weekday. Same for Jan, Oct in non-leap years. Apr, Jul in all years. Sep, Dec in all years. These shortcuts help verify calculations.

Solved Examples

Problem 1:

What day of the week was 15th August 1947?

Solution:

Step 1: Break into centuries - 1600 years + 300 years + 47 years (Jan 1 to Aug 15).
Step 2: 1600 years have 0 odd days (complete 400-year cycle).
Step 3: 300 years have 1 odd day (from century table).
Step 4: In 47 years: 11 leap years (1940-1944-1948 excluded, so 1940,1944 + 9 others) and 36 normal years.
Step 5: Odd days from 47 years = (11x2) + (36x1) = 22 + 36 = 58 = 58 mod 7 = 2 odd days.
Step 6: Days from Jan 1 to Aug 15, 1947: Jan(31) + Feb(28) + Mar(31) + Apr(30) + May(31) + Jun(30) + Jul(31) + Aug(15) = 227 days = 227 mod 7 = 3 odd days.
Step 7: Total odd days = 0 + 1 + 2 + 3 = 6.
Step 8: Day code 6 = Friday.
Answer: Friday

Problem 2:

If 1st January 2001 was Monday, what day was 1st January 2010?

Solution:

Step 1: Calculate years from 2001 to 2009 (9 years, since we're finding Jan 1, 2010).
Step 2: Identify leap years: 2004, 2008 are leap years (2000 is already passed).
Step 3: Normal years: 7 years x 1 odd day = 7 odd days.
Step 4: Leap years: 2 years x 2 odd days = 4 odd days.
Step 5: Total odd days = 7 + 4 = 11 = 11 mod 7 = 4 odd days.
Step 6: Starting from Monday (1), add 4 days: Tuesday(2), Wednesday(3), Thursday(4), Friday(5).
Answer: Friday

Problem 3:

How many times does the 29th of February occur in 400 years?

Solution:

Step 1: In 400 years, count leap years using the complete rule.
Step 2: Years divisible by 4: 400/4 = 100 candidates.
Step 3: Subtract years divisible by 100 (but not 400): 100, 200, 300 = 3 years.
Step 4: Add back years divisible by 400: 400 = 1 year.
Step 5: Total leap years = 100 - 3 + 1 = 97 leap years.
Step 6: Each leap year has one 29th February.
Answer: 97 times

Problem 4:

What was the day on 26th January 1950 (Republic Day of India)?

Solution:

Step 1: Break 1950 into 1600 + 300 + 50 years, but we need up to Jan 26.
Step 2: 1600 years = 0 odd days.
Step 3: 300 years = 1 odd day.
Step 4: In 49 complete years (1901-1949): 12 leap years (1904,08,12,16,20,24,28,32,36,40,44,48).
Step 5: Odd days from 49 years = (12x2) + (37x1) = 24 + 37 = 61 = 61 mod 7 = 5 odd days.
Step 6: Days in 1950 until Jan 26 = 26 days = 26 mod 7 = 5 odd days.
Step 7: Total odd days = 0 + 1 + 5 + 5 = 11 = 11 mod 7 = 4.
Step 8: Day code 4 = Thursday.
Answer: Thursday

Tips & Strategies

Always verify leap years carefully: century years (1900, 2100) are NOT leap years unless divisible by 400 (2000, 2400 are leap years).
Use the century odd days pattern: 100=5, 200=3, 300=1, 400=0. This repeats every 400 years, making calendar calculations cyclical.
For quick month calculations, memorize that Jan+Feb combined have 59 days (60 in leap), which is 3 or 4 odd days respectively.
When finding a past date's day, subtract odd days instead of adding, or add (7 - odd days) to move backward equivalently.
Same day of week falls on: 1 Jan (non-leap) = 1 Oct; 1 Jan (leap) = 1 Jul; 29 Feb only exists in leap years.
Use reference dates you know: Independence Day 1947 = Friday, Republic Day 1950 = Thursday, or any known date as anchor point.

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