Calendar Problems in Aptitude Tests: Derive, Don't Memorise

The odd-days method reduces any day-of-the-week question to modular arithmetic across four components: century offset, year offset, month offset, and date offset.

Add those four remainders, take mod 7, and read off the day. No lookup charts. No tables to memorise in the test room. This article derives each component from scratch so you understand why the method works, then applies it in six verified examples.

For speed-focused competitive-exam shortcuts and pattern-match tricks with fewer derivations, see the companion article: solving calendar problems for competitive exams.

What Calendar Problems Test

Calendar problems appear in the quantitative aptitude sections of most Indian placement tests. The TCS NQT Numerical Ability section and AMCAT Quantitative Aptitude both include date-based questions. Infosys InfyTQ and Wipro NLTH also cover this topic.

The core question type: given a known day for one date, find the day for another date. Variations include finding all occurrences of a specific weekday in a month, determining the last day of a century, or calculating the day after n days from a given reference.

Every variation reduces to counting total days, dividing by 7, and reading the remainder.

Foundations: Leap Years and Odd Days

Leap Year Rules

The Gregorian calendar defines three rules:

• Divisible by 4: leap year (366 days).
• Also divisible by 100: not a leap year.
• Also divisible by 400: leap year (overrides the 100 rule).

So 2024 is a leap year (divisible by 4, not by 100). Year 1900 is not a leap year (divisible by 100, not by 400). Year 2000 is a leap year (divisible by 400).

What Is an Odd Day?

“Odd days” is the name for the remainder when a period’s total days is divided by 7. Since the weekly cycle repeats every 7 days, the remainder tells you how many days the weekday has shifted from the starting reference.

Odd Days for Each Year Type

• Non-leap year: 365 = 52 x 7 + 1 → 1 odd day.
• Leap year: 366 = 52 x 7 + 2 → 2 odd days.

The Century Table: Derived, Not Memorised

The centennial odd-days table is the part most students memorise. Here is where it comes from.

In any 100-year block within the Gregorian calendar (using a block where the centennial year is not a 400-year anniversary, for example 1901–2000):

• Leap years: every year divisible by 4, except the centennial. That gives 24 leap years (e.g., 1904, 1908, …, 1996 for the 1901–2000 block; year 2000 is the next centennial and 1900 is the previous one, which was not a leap year).
• Non-leap years: 76.
• Total days: 24 x 366 + 76 x 365 = 8784 + 27740 = 36524.
• Odd days: 36524 divided by 7 gives remainder 5.

For longer periods:

• 200 years: 36524 x 2 = 73048; 73048 mod 7 = 3.
• 300 years: 36524 x 3 = 109572; 109572 mod 7 = 1.
• 400 years: 97 leap years + 303 non-leap years = 97 x 366 + 303 x 365 = 35502 + 110595 = 146097; 146097 mod 7 = 0.

Why does 400 years break the multiplication pattern? Every 400-year cycle has one more leap year than 4 x 100-year blocks would imply. The centennial years divisible by 400 (like 400, 800, 1200, 1600, 2000) are leap years, but the 100-year block calculation excluded them. The 400-year total of 146097 days happens to be a perfect multiple of 7.

Period	Total Days	Odd Days
100 years	36524	5
200 years	73048	3
300 years	109572	1
400 years	146097	0

Because 400 years = 0 odd days, any span of 800, 1200, or 1600 years also yields 0 odd days.

The Standard Method: Four Steps

To find the day of the week for any date, apply these steps in order:

1. Count the odd days for all complete 400-year blocks before the target date (each gives 0).
2. Count the odd days for remaining complete centuries within the current 400-year span using the table above.
3. Count the odd days for each complete year in the current century (1 per non-leap year, 2 per leap year; take the total mod 7).
4. Count the odd days within the current year: add the odd days for each complete month, then add the day-of-month number; take the total mod 7.
5. Sum all four results, take mod 7, and map to the day-of-week key.

Month Odd Days Reference

Month	Days (non-leap)	Odd Days	Days (leap)	Odd Days
January	31	3	31	3
February	28	0	29	1
March	31	3	31	3
April	30	2	30	2
May	31	3	31	3
June	30	2	30	2
July	31	3	31	3
August	31	3	31	3
September	30	2	30	2
October	31	3	31	3
November	30	2	30	2
December	31	3	31	3

Day-of-Week Key

Odd Days	Day
0	Sunday
1	Monday
2	Tuesday
3	Wednesday
4	Thursday
5	Friday
6	Saturday

Six Worked Examples

All examples below are re-derived from scratch.

Example 1: Counting Forward Across Years

• Q: If January 1, 2006 was a Sunday, what day was January 1, 2010?
• Count odd days for each complete year from 2006 through 2009:
  • 2006 (non-leap): 1 odd day.
  • 2007 (non-leap): 1 odd day.
  • 2008 (leap): 2 odd days.
  • 2009 (non-leap): 1 odd day.
  • Total: 5 odd days.
• Sunday + 5 = Friday.
• Answer: January 1, 2010 was a Friday.

Example 2: Finding the Day for Any Date

• Q: What day of the week was May 28, 2006?
• Step 1: Years 1 through 2000 = 5 x 400-year cycles = 0 odd days.
• Step 2: Years 2001 through 2005 (five complete years):
  • 2001: 1, 2002: 1, 2003: 1, 2004 (leap): 2, 2005: 1. Total: 6 odd days.
• Step 3: Within 2006, Jan 1 through May 28 (148 days total):
  • Jan (31) + Feb (28) + Mar (31) + Apr (30) + May 1–28 (28) = 148 days.
  • 148 mod 7 = 1 odd day.
• Grand total: 0 + 6 + 1 = 7 ≡ 0 odd days.
• Answer: 0 odd days = Sunday. May 28, 2006 was a Sunday.
• Cross-check: Jan 1, 2006 = Sunday (from Example 1 working backwards). Days from Jan 1 to May 28 = 147 days elapsed; 147 mod 7 = 0. Same day as Jan 1 = Sunday. ✓

Example 3: The n-Days-From-Now Shortcut

• Q: Today is Monday. What day will it be 61 days from now?
• 61 mod 7 = 5 (since 56 = 8 x 7, remainder 5).
• Monday + 5 days = Saturday.
• Answer: Saturday.

Example 4: Counting Backwards

• Q: If March 6, 2005 was a Monday, what day was March 6, 2004?
• The span from March 6, 2004 to March 6, 2005 = 365 days.
• Although 2004 was a leap year, February 29, 2004 falls before March 6 and is not inside this span.
• 365 mod 7 = 1 odd day.
• Going backwards: Monday minus 1 day = Sunday.
• Answer: March 6, 2004 was a Sunday.

Example 5: Finding All Occurrences of a Weekday

• Q: On what dates did Wednesday fall in April 2001?
• First, find April 1, 2001:
  • Years 1–2000: 0 odd days.
  • January 1, 2001 = 0 + 1 (Jan 1 itself) = 1 odd day = Monday.
  • January + February + March = 31 + 28 + 31 = 90 days; 90 mod 7 = 6 odd days.
  • April 1, 2001 = Monday + 6 = Sunday.
• Sunday is offset 0. Wednesday is offset 3. First Wednesday = April 1 + 3 = April 4.
• Subsequent Wednesdays: April 11, April 18, April 25.
• Answer: Wednesdays in April 2001 were on the 4th, 11th, 18th, and 25th.

Example 6: The Last Day of a Century

• Q: Which day of the week can never be the last day of a century?
• From the century table, the last day of each century type:
  • 100 years = 5 odd days from reference → Friday.
  • 200 years = 3 odd days from reference → Wednesday.
  • 300 years = 1 odd day from reference → Monday.
  • 400 years = 0 odd days from reference → Sunday.
• This four-step pattern repeats exactly because each 400-year cycle resets to 0.
• Possible last days of a century: Sunday, Monday, Wednesday, Friday.
• Answer: The last day of a century cannot be Tuesday, Thursday, or Saturday.

Where These Questions Appear in Placement Tests

Calendar problems appear in the Numerical Ability section of the TCS NQT and in the Quantitative Aptitude module of AMCAT. Infosys and Wipro placement tests include them at similar difficulty.

In these tests, difficulty concentrates in two areas: century crossings, where the 100-year exception to the leap rule matters; and backwards counting, where students subtract rather than add and get an off-by-one result. Example 4 above directly illustrates both pitfalls.

Problems in campus placement tests rarely go beyond three steps. The method in this article handles all three-step problems without any special-case memorisation.

Derivation vs. Speed Tricks: Knowing Which Mode to Use

This article teaches derivation. Once you can re-derive the century table in two minutes, you do not need to carry it as a memorised fact.

In a 60-minute aptitude section, you may encounter one or two calendar problems. For a simple n-days problem like Example 3, apply the mod-7 trick directly. For a date spanning a century boundary, run the full four-step method.

If you are preparing for a competitive exam with back-to-back time pressure on 15-plus quantitative problems, the speed-tricks companion article covers rapid pattern-matching with fewer derivations and more drill-format examples. That article is the right mode for exam-day speed; this one is the right mode for understanding what those shortcuts are actually doing.

The same systematic reasoning applies across coding-decoding questions in aptitude tests, where understanding the underlying substitution rule beats pattern-guessing every time.

The core argument of this article is that derivation is more durable than memorisation in any rule-based domain, calendar maths included. That reasoning transfers directly to AI work: when a model API or library changes, engineers who understand the mechanics adapt faster than those who memorised the syntax. TinkerLLM is a self-paced LLM playground at ₹299 where that first-principles approach to AI tools starts, with build-first exercises that treat the model as a system to understand rather than a syntax to recall.