Introduction
Have you ever thought about how a calendar app or even an old almanac can instantly tell you the day of the week for a date in the past or future? The secret lies in mathematical calendar calculations, and one of the simplest and most exam-friendly methods is the Odd Days Method.
This method is widely used in SSC, UPSC, Banking, and other competitive exams to solve “Calendar Questions” within seconds. In this article, we’ll break it down from scratch, so even if you’ve never seen it before, you’ll be able to calculate any weekday confidently.
🌍 Step 1: Understanding Odd Days
- A week has 7 days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
- When dividing the total days by 7, the remainder is called the odd days.
👉 Example:
- 15 days = 2 weeks + 1 odd day.
- That means, after 15 days from Sunday, it will be Monday (since 1 odd day = Monday).
So, odd days are just the “leftover” days after making full weeks.
📆 Step 2: Odd Days from Years
Every year contributes odd days. But we must separate ordinary years and leap years.
- Ordinary Year = 365 days = 52 weeks + 1 day → 1 odd day
- Leap Year = 366 days = 52 weeks + 2 days → 2 odd days
So:
- 100 years = 76 ordinary years + 24 leap years = 76×1 + 24×2 = 124 odd days = 124 ÷ 7 = 5 odd days
- 200 years = 3 odd days
- 300 years = 1 odd day
- 400 years = 0 odd days
This cycle repeats every 400 years.
📅 Step 3: Odd Days from Months
We also need to know how many days each month contributes (up to the given date).
For a normal year:
- Jan → 31 = 3 odd days
- Feb → 28 = 0 odd days
- Mar → 31 = 3 odd days
- Apr → 30 = 2 odd days
- May → 31 = 3 odd days
- Jun → 30 = 2 odd days
- Jul → 31 = 3 odd days
- Aug → 31 = 3 odd days
- Sep → 30 = 2 odd days
- Oct → 31 = 3 odd days
- Nov → 30 = 2 odd days
- Dec → 31 = 3 odd days
For a leap year, February has 1 odd day.
📌 Step 4: Final Table of Odd Days → Days of Week
| Odd Days | Day |
|---|---|
| 0 | Sunday |
| 1 | Monday |
| 2 | Tuesday |
| 3 | Wednesday |
| 4 | Thursday |
| 5 | Friday |
| 6 | Saturday |
🔎 Step 5: Full Example (Explained in Detail)
Problem: Find the day on 12th January 1979.
Step A: Break into parts
1978 years completed + 12 days of January 1979.
Step B: Count Odd Days for Years
- 1600 years → 0 odd days
- Next 300 years (1601–1900) → 1 odd day
- Remaining 78 years → 59 ordinary + 19 leap = 59×1 + 19×2 = 97 odd days = 13 weeks + 6 odd days
So total odd days from 1978 years = 0 + 1 + 6 = 7 = 0 odd days.
Step C: Odd Days from 12th Jan 1979
- January (till 12th) → 12 days → 12 ÷ 7 = 1 week + 5 odd days
Step D: Total Odd Days
0 (years) + 5 (days) = 5 odd days → Friday ✅
💡 Sample Questions with Solutions
Q1: Find the day on 26th January 1950.
- 1949 years + 26 days
- 1600 → 0 odd days
- 300 → 1 odd day
- 49 years = 12 leap + 37 ordinary = 12×2 + 37×1 = 61 odd days = 5 odd days
- Total from years = 1 + 5 = 6 odd days
- January 26 → 26 ÷ 7 = 3 odd days + remainder 5
- Final = 6 + 5 = 11 = 4 odd days = Thursday ✅
Q2: Find the day on 15th August 1947.
- 1946 years + (days till 15th Aug 1947)
- 1600 → 0 odd days
- 300 → 1 odd day
- 46 years = 11 leap + 35 ordinary = 57 odd days = 1 odd day
- Year total = 1 + 1 = 2 odd days
- Days from Jan–July + Aug 15 = (Jan 3 + Feb 0 + Mar 3 + Apr 2 + May 3 + Jun 2 + Jul 3 + 15) = 31 days = 31 ÷ 7 = 3 odd days + 3
- Total = 2 + 3 = 5 odd days = Friday ✅
Q3: Find the day on 1st January 2000.
- 1999 years completed
- 1600 → 0 odd days
- 300 → 1 odd day
- 99 years = 24 leap + 75 ordinary = 123 odd days = 4 odd days
- Total = 5 odd days → Saturday ✅
Q4: Find the day on 2nd October 1869 (Mahatma Gandhi’s birth date).
- Years till 1868 + days till 2nd Oct 1869
- After calculation → Saturday ✅
Q5: Find the day on 14th November 1889 (Jawaharlal Nehru’s birth date).
- After calculation → Thursday ✅
Q6: Find the day on 4th July 1776 (US Independence Day).
- After calculation → Thursday ✅
🎯 Conclusion
The Odd Days Method is an elegant way to calculate weekdays for any date without a calendar. By understanding ordinary/leap years, century rules, and monthly odd days, you can easily break down problems step by step. With practice, you’ll solve such questions in seconds — a huge advantage in competitive exams.
