**Calendar Problem Tips**

In competitive exam always questions are asked on calendar tricks and maximum students are get confused about the answer as concepts are not cleared. Here we provide simplified steps from basic level to advanced level to solve calendar reasoning problems.

**Basic level:**

In ordinary year there are 365 days means 52 weeks and 1 extra day**. We can call odd day to this extra day. **

If 1^{st} January of this year is Sunday, then 1^{st} January of next year will be Monday.

**Means in a year same day is shifted by one day ahead because of this extra odd day. **

But the concept is changed for leap year.

**Leap year: Leap year is a year that contains 366 days with February 29 as an additional day. **

Leap year is identified as if last two digits of a year are perfectly divisible by 4, then this year is a leap year. **But this rule is not applicable for century year**.

A century is a leap year if its first 2 digits are divisible by 4.

For example: Year 2000, 2400,3200 are leap year as first two digits are divisible by 4, whereas 1100, 1700, 1900 are not a leap year as first 2 digits are not divisible by 4.

**However as 1 day is extra in leap year so in a year same date is exceed by 2 days if it passes the February month. **

**e.g**. If 26 January 1995 is Thursday then 26 January 1996 will Friday. Even though 1996 is a leap year then also **same date is exceeded by one day only as we do not cross the February month.**

But if 15 April 1995 is Saturday then 15 April 1996 will be Monday , **same date is exceeded by 2 days as we are crossing the February month.**

**Solved Examples: **

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**Ex. i)** **If 22 April 1945 was Sunday, what was the day on 22 April 1946?**

**Ans: **As given year is not a leap year so same date is exceeded by one day only for one year.

So, if 22 April 1945 was Sunday, then the day on 22 April 1946 will be Monday.

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**Ex. ii)** **If 13 August 1987 was Thursday, what was the day on 13 August 1988?**

**Ans:** As Year 1988 is a leap year, so year exceed by two days for the same date as it is crossing the February Month.

So, if 13 August 1987 was Thursday then 13 August 1988 will be Saturday.

**Ex. ii)** **19 ^{th} September was the 3^{rd} Saturday. On which day will 12^{th} October fall in that same year?**

**Ans:** Count the number of days from 19^{th} September to 12^{th} October .

We know that September have total 30 days, So, 30 -19 = 11 days left from September & 12 days from October.

Total days = 11 + 12 =23

As there are 7 days in a week, so we divide 23 by 7, Remainder is 2.

So, we go 2 days ahead by Saturday, means 12^{th} October will be Monday.

**Advanced Leve**l:

**i) Single year Calendar-**

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We can identify any day of a given date of any particular year for that we should know the 12 digits key of the particular year.

Here 12 digits key of the year 2014 is given, by using this key we can find day of any date of 2014 year.

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**Key: 522 641 637 527**

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**Above 12 digits number represent the month key of year 2014.**

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Here number 5 is for month January, 2 is given for month February, next 2 is for March and so on…

**Basically, each of those numbers represents first Sunday of that month occurred on corresponding date**.

So, from above number key we can say that,

5^{th} January is Sunday

^{ }2^{nd} February is Sunday

2^{nd} March is Sunday

So, if we remember this 12-digit key we can identify day of any date of the corresponding year.

**Solved Example:**

**Ex i) What is the day on 8 ^{th} March?**

**Ans: **As March is 3^{rd} month so we check the 3^{rd} key and it is 2. Means 2^{nd} March is Sunday, So, 8^{th} March will be Saturday.

**Ex. ii) What is the day on 15 August?**

**Ans:** As August is 8^{th} Month in a year, we look at 8^{th} digit key which is 3. It means 3^{rd} August is Sunday. Hence the next Sunday is on 3+7= 10. Means 10^{th} August is Sunday so 15^{th} August will be on Friday.

**ii) Calendar of Century: **

Now we will study a system for finding the day on which any date falls for the entire century. After using this technique, you will be able to find out the day of any date falls from 1^{st} January 1901 to 31^{st} December 2000.

Before proceeding for the method, we will see the key of the months. This key will remain same for the century.

In this technique, final answer is obtained in the form of remainder. To predicate the day based on remainder we should know the following key.

As we can see this key is very simple, so no need to remember. Now we will do an overview of the steps needed to predicate the day.

**Steps:**

1) Very first step is take the last 2-digits of the year.

e.g. if year is 1987 then take 87.

2) Add the number of leap year from 1901 to the year asked to find the day.

3) Add the month key

4) Add the date

5) Divide the total by 7

6) Now last step is taking the remainder & verify it with the day key.

These six steps are needed to find the day of any date.

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**Solved Examples:**

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**e.g. i) What day is 8 ^{th} March 1985?**

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**Step 1:** First we take the last two digits of the year = **85**

**Step 2:** We add the number of leap years from 1901 to 1985

It will be **21**

**Step 3: **We add the month key for the March =** 4**

(Refer to Month key table)

**Step 4: **Now add the date = **08**

Total = 85 + 21 + 4 + 8= 118

Now divide 118 by 7, we get Q =16, R = 6

From the day key table number 6 represent Friday.

**So, 8 ^{th} March 1985 was a Friday.**

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**e.g. ii) What is the day on 30 November 1978?**

**Ans:**

** **Last two digit of the year** = 78**

** **Number of leap years from 1901 to 1978** = 19**

Month key of November** = 4**

** **Date =** 30**

Total = 78 + 19 + 4 + 30 = 131

Now, 131 ÷ 7, Q =19, R = 5

From the day key 5 represents Thursday.

**So, 30 ^{th} November 1978 was Thursday.**

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