Calendar problems are based on the relationship between dates, days, weeks, months and years. The most important idea is to determine the day of the week for a particular date.
| Number | Day |
|---|---|
| 0 | Sunday |
| 1 | Monday |
| 2 | Tuesday |
| 3 | Wednesday |
| 4 | Thursday |
| 5 | Friday |
| 6 | Saturday |
An odd day is the number of days left after complete weeks are removed from a given number of days.
Odd Days = Total Days mod 7
100 = 14 × 7 + 2
Therefore, odd days = 2.
An ordinary year has 365 days.
Therefore, an ordinary year has 1 odd day.
A leap year normally contains 366 days, with February having 29 days.
Therefore, a leap year has 2 odd days.
However, century years must be divisible by 400 to be leap years.
Examples:
2024 → Leap year
2025 → Ordinary year
1900 → Not a leap year
2000 → Leap year
| Month | Days |
|---|---|
| January | 31 |
| February | 28 / 29 |
| March | 31 |
| April | 30 |
| May | 31 |
| June | 30 |
| July | 31 |
| August | 31 |
| September | 30 |
| October | 31 |
| November | 30 |
| December | 31 |
For a group of years, count the odd days contributed by ordinary and leap years.
Leap Year → 2 odd days
Total Odd Days = Number of ordinary years \+ 2 × Number of leap years
Odd days = 5 × 1
= 5 odd days
In the Gregorian calendar, a 100-year period generally contains either 24 or 25 leap years depending on the exact century interval. For standard aptitude calculations, carefully count century years using the divisibility-by-400 rule.
Ordinary years = 76
Odd days = 76 + (24 × 2)
= 124
124 mod 7 = 5 odd days
Count leap years carefully and then reduce the total number of odd days modulo 7.
To find the day corresponding to a date, calculate the total odd days from the chosen reference point up to the required date.
10 mod 7 = 3
Day number 3 = Wednesday
Another common aptitude method uses month codes together with the year, century and date. Different formula tables may use different code conventions, so the complete code system must be kept consistent.
When a question asks which day will occur after a certain number of days, divide the number by 7 and use only the remainder.
45 mod 7 = 3
Three days after Monday = Thursday.
To find the day difference between two dates, count the remaining days of the starting month, all complete months in between, and the days of the ending month.
March has 31 days.
From 10 March to 10 April = 31 days.
From a date in one year to the same date in the next year, the shift depends on whether a leap day occurs during the interval.
Leap-year interval → usually 2-day shift when Feb 29 lies within the interval
Shift = 1 day.
Add the odd days contributed by each year in the interval. The final remainder after division by 7 gives the weekday shift.
Required Shift = Total Shift mod 7
A calendar repeats when the accumulated weekday shift becomes a multiple of 7 and the leap-year structure is compatible.
Total accumulated odd days ≡ 0 (mod 7)
and the leap-year arrangement must match.
7 mod 7 = 0
Therefore, the weekday shift over those seven ordinary years is zero days.
Note: Actual calendar repetition also depends on leap-year placement, so a complete Gregorian calendar should be checked rather than assuming every 7-year interval repeats.
A month may contain 28, 29, 30 or 31 days. Divide the number of days by 7 to determine how many complete weeks occur.
29 days → one weekday occurs 5 times
30 days → two weekdays occur 5 times
31 days → three weekdays occur 5 times
Wednesday, Thursday and Friday.
In a 30-day month, at most 2 weekdays occur five times.
In February with 28 days, every weekday occurs exactly 4 times.
| Concept | Formula / Rule |
|---|---|
| Ordinary Year | 365 days = 52 weeks + 1 odd day |
| Leap Year | 366 days = 52 weeks + 2 odd days |
| Odd Days | Total days mod 7 |
| Leap Year Rule | Divisible by 4, except century years must be divisible by 400 |
| 28-Day Month | Each weekday occurs 4 times |
| 29-Day Month | One weekday occurs 5 times |
| 30-Day Month | Two weekdays occur 5 times |
| 31-Day Month | Three weekdays occur 5 times |
| Future Day | Shift by N mod 7 |
100 mod 7 = 2
Two days after Friday = Sunday.
2100 is divisible by 100 but not by 400.
Therefore, 2100 is not a leap year.
30 = 4 weeks + 2 days.
Therefore, the first two weekdays occur five times:
Monday and Tuesday.
31 = 4 weeks + 3 days.
Therefore:
Saturday, Sunday and Monday occur five times.
``` 1. 1 odd day
2. 2 odd days
3. Yes
4. No
5. Thursday
6. Friday
7. Thursday, Friday and Saturday
8. Friday and Saturday
9. Monday = 4, Tuesday = 4, Sunday = 4
10. Monday ```
``` 2. Ordinary year → 1 odd day.
3. Leap year → 2 odd days.
4. Remember the leap-year exception: century years must be divisible by 400.
5. For 31 days, the first three weekdays occur five times.
6. For 30 days, the first two weekdays occur five times.
7. For 28 days, every weekday occurs exactly four times.
8. Always check whether February has 28 or 29 days before calculating a date interval. ```
