# Calendar Formulas

## Formulas To Solve Calendars Problem in Aptitude

On this page, you will find out all the Calendar Formula for Solving Questions in Competitive Exams. Go through this page to get useful Insights to solve Calendar Questions.

### Calendar Formulas & Definitions

• A calendar is a method for systematically  organizing days for  religious, social, commercial or administrative purposes.It is done by giving names to periods of time, such as  days, weeks, months and years.
• A date is single, specific day within such a system
• The year is classified into two type:
1. Ordinary years (365 days)
2. Leap year (366 days.)z
• The calendar starts on 1st January and ends on 31st December.

### Calendar Formulas for Odd days

• Odd days are the number of days that are more than the number of days in a complete week.
• For example:         Calculate odd days for 10 and 14 days

10 days = 1 week (7 days) + 3 days. Here, 3 days are odd days

14 days = 2 weeks (14 days) + 0 day (0 odd day).

### Formulas of Calendar For Number of Odd Days

• 1 ordinary year has 1 odd day

Explanation :  In an ordinary year, there are 365 days, which means 52 x 7 + 1, or 52 weeks and one day. This additional day is called an odd day.

• 1 leap year has 2 odd days

Explanation:  A leap year has 366 days. There are 29 days in February in a leap year. There are 52 weeks and 2 odd days in a leap year.

• 100 years has 5 odd days

Expalnation :  Odd days in a leap year = (52 weeks +2) days  .In 100 years , there will be 24 leap years and 76 non-leap years. So odd days in 100 years will be (76 x 1 + 24 x 2) which is 124 odd days. This can also be written as 17 weeks + 5 days. So every 100 years will have 5 odd days.

• 200 years has 3 odd days

Expalnation: 100 years give us 5 odd days as calculated above. 200 years give us 5 x 2 = 10 .Hence , 7 days (one week) = 3 odd days.

• 300 years has 1 odd day

Expalnation: 300 years give us 5 x 3 = 15.Hence, 14 days (two weeks) = 1 odd day.

• 400 years has 0 odd day

Expalnation:  The number of odd days in 400 years will be ( 5 x 4 + 1) because 400 is itself a leap year and that is why it has one odd day extra. Thus odd days in 400 will be 0.

• Similarly, all the 4th centuries 800 years, 1200 years, 1600 years, 2000 years etc. have 0 odd day.
• Mapping of the number of odd day to the day of the week
DaysSundayMondayTuesdayWednesdayThursdayFridaySaturday

No. of Odd Days

0123456

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### Reference Chart that gives odd days for the given months

• To determine whether a year is a leap year, follow these steps
1. If the year is evenly divisible by 4, go to step 2. Otherwise, go to step 5.
2. If the year is evenly divisible by 100, go to step 3. Otherwise, go to step 4.
3. If the year is evenly divisible by 400, go to step 4. Otherwise, go to step 5.
4. The year is a leap year (it has 366 days).
5. The year is not a leap year (it has 365 days).
• For example:

1. Check for years not ending with “00”.

Year 1997 is not a leap year because it is not divisible by 4.

Year 2016 is a leap year because it is divisible by 4.

2. Now, check for years ending with “00”.

Year 2000 is a leap year because it is divisible by 4,100 and 400.

Year 1900 is not a leap year because it is divisible by 4 and 100 but not 400.

Year 1600 is a leap year because it is divisible by 4, 100, and 400.

• For this reason, the following years are not leap years:

1700, 1800, 1900, 2100, 2200, 2300, 2500, 2600

This is because they are evenly divisible by 100 but not by 400.

• The following years are leap years:

1600, 2000, 2400

This is because they are evenly divisible by both 100 and 400.

### Points to remember

• Last day of a century cannot be Tuesday or Thursday or Saturday because of the number of odd days.
• For the calendars of two different years to be the same, the following conditions must be satisfied.

a.   Both years must be of the same type. i.e., both years must be ordinary years or both years must be leap years.

b.   1stJanuary of both the years must be the same day of the week.

### Sample Calendar Questions with Solution

Question:1 How many odd days are there in 3000 days?

Solution:

To find the number of odd days in 3000 days, divide 3000 by 7 (the number of days in a week)

= 3000 ÷ 7 = 428 with a remainder of 4

So, there are 428 weeks with no odd days and 4 odd days remaining.

Therefore, the number of odd days in 3000 days is 4.

Question: 2 Is the year 2100 a leap year?

Solution:

To determine if the year 2100 is a leap year, we need to check if it satisfies the leap year rule.

A year is a leap year if it is divisible by 4, except for years that are divisible by 100 but not divisible by 400.

Year = 2100

Step 1: Check if it is divisible by 4. 2100 ÷ 4 = 525 with no remainder. So, it is divisible by 4.

Step 2: Check if it is divisible by 100. 2100 ÷ 100 = 21 with no remainder. So, it is divisible by 100.

Step 3: Check if it is divisible by 400. 2100 ÷ 400 = 5 with no remainder. So, it is divisible by 400.

Since the year 2100 satisfies all three conditions, it is a leap year.

Question: 3 In a leap year, how many days are there between February 15th and September 10th inclusive?

Solution:

In a leap year, February has 29 days, and all other months have their usual number of days.

Number of days from February 15th to February 29th = 29 – 15 + 1 = 15 days

Number of days from March 1st to August 31st (inclusive) = 31 + 30 + 31 + 30 + 31 + 31 = 184 days

Number of days from September 1st to September 10th = 10 days

Total days = 15 + 184 + 10 = 209 days

Question: 4 What is the next year that will have the same calendar as the year 2023?

Solution:

The calendar repeats itself every 28 years because the days of the week and the number of days in each month follow a 28-year cycle.

2023 + 28 = 2051

Thus, the year 2051 will have the same calendar as the year 2023.

Question: 5 If today is Sunday, what day of the week will it be 10 days from today?

Solution:

Since there are 7 days in a week, to find the day of the week 10 days from today,

We can divide 10 by 7 to get the number of complete weeks and then find the day of the week for the remaining days.

10 ÷ 7 = 1 week with a remainder of 3 days.

Counting forward 3 days from Sunday, we determine the day of the week.

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