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**GETTING STARTED:**

If calendar shows Monday today, what would be the 3rd day starting tomorrow?

If we see the timeline, 3rd day from today is Thursday.

What would be the 10th day from today?

It would also be a **Thursday**.

Reason for that is periodicity. If we start on a Sunday, we arrive at the same day again after 7 days.

Eg. 1st, 8th, 15th, 22nd and 29th of a same month in that year occur on a same day.

**ODD DAYS:**

Remainder after dividing a given set of days by 7.

**Eg.**

1 Normal year with 365 days has 1 odd day.

365 when divided by 7, gives a remainder of 1

1 Leap year with 366 days has 2 odd days.

**LEAP YEARS:**

Every 4^{th} year is a leap year except the years ending in ‘00’.

A leap year has 366 days whereas a normal year has 365 days.

**A year is a leap year if:**

It is a non-century year and divisible by 4.

Eg. 2004, 1964, 1976

It is a century year and divisible by 400.

Eg. 1600, 1200,2000

**Eg. If 1st Feb 2019 is a Friday, what will be the day on 19th September 2019?**

Soln:

Number of days between 1st Feb 2019 and 19th Sept 2019 =

[Number of months with 28 days] x28 +

[Number of months with 29 days] x29 +

[Number of months with 30 days] x30 +

[Number of months with 31 days] x31 +

Remaining number of days

Months with 30 days = {April, June}

Months with 31 days = {March, May, July, August}

Remaining number of days = 27 days remaining in February + 19 days in September

Number of days = 2×30 + 4×31 + 27+19

=60+124+46

= 230

By using remainder property of numbers, the calculation can be simplified as:

Number of days = 2 x 2 + 4×3+(-1)+(-2) = 13

** **

**Reason:**

On dividing 30 by 7, remainder =2

On dividing 31 by 7, remainder =3

On dividing 27 by 7, remainder = 6 or (-1)

On dividing 19 by 7, remainder = 5 or (-2)

Odd days = 6

The 6th day from Friday is Thursday and thus that is the required day.

**Eg. Given that 31st March 2006 is a Tuesday, find the day on 31st March 2008.**

Soln: Total number of days between the mentioned dates =

[Number of months with 28 days] x 28 +

[Number of months with 29 days] x 29 +

[Number of months with 30 days] x 30 +

[Number of months with 31 days] x 31 + Remaining number of days

= 1×28 + 1×29 + 8×30 + 14×31+0

= 28+29+240+434+0

= 731

By using remainder property of numbers, the calculation can be simplified as:

Number of days = (1 x 0) + (1×1)+ (1 x 2)+0 x 3 = 3

Number of odd days = 3

Third Day from Tuesday is Friday.

Thus the day of given date is Friday.

**An easier way would be:**

31st March 2006 to 31st March 2007 will have 365 days in between. (Since it is one year and 2007 is a normal year and Feb 2007 will have only 28 days)

31st March 2007 to 31st March 2008 will have 366 days in between (Since it is one year and 2008 is a leap year with Feb 2008 having 29 days)

Total No. days = 365+366 = 731.

Odd days = remainder when dividing 731 by 7 = 3

No. of odd days in a non-leap year = odd days in 365 days = 1

No. of odd days in a leap year = odd days in 366 days = 2

**No. of odd days in 100 years**:

100 Years have 24 leap years and 76 non-leap years

[100/4 = 25.

But, the 100th year is a non-leap year.

So leap years = 25 – 1 = 24]

No. of odd days =Remainder on dividing [(24 x 2)+(76 x1)] by 7

= Remainder on dividing 124 by 7

= **5 **

**No. of odd days in 200 years:**

200 Years have 48 leap years and 152 non-leap years

[200/4 = 50.

But, the 100th year and 200th year are non-leap years.

So leap years = 50 – 2 = 48]

No. of odd days = Remainder on dividing [(48 x 2)+(152 x 1)] by 7

= Remainder on dividing 248 by 7

= **3**

**No. of odd days in 300 years:**

300 Years have 72 leap years and 228 non-leap years

[300/4 = 75.

But, the 100th year,200th year and 300th year are non-leap years.

So leap years = 75 – 3 = 72]

No. of odd days = Remainder on dividing [(72 x 2)+(228 x 1)] by 7

= Remainder on dividing 372 by 7

= **1**

**No. of odd days in 400 years:**

400 Years have 72 leap years and 228 non-leap years

[400/4 = 100.

But, the 100th year,200th year and 300th year are non-leap years. 400th year is a leap year.

So leap years = 100 – 3 = 97]

No. of odd days = Remainder on dividing [(97 x 2)+(303 x 1)] by 7

= Remainder on dividing 497 by 7

= **0**

This means that if 16th April 2020 is a Thursday, 16th April 2420 will also be a Thursday.

A significant fact this conveys is that : **Any calendar can be reused every 400 years**

The calendar used from 1 AD to 400 AD can be used for year 401 to 800, 801 to 1200 AD, 1201 to 1600 AD, 1601 to 2000 and so on.

1st January of every set of 400 years is a Monday.

1st January 1201, 1st January 1601, 1st Jan 2001, 1st Jan 2401 are all Mondays.

**Eg. What day was it on 5th August 1998?**

Soln:

5th August 1998 = 1997 years + (Period from 1st Jan 1998 to 5th Aug 1998)

= 1600 + 300 + 97 + (Period from 1st Jan 1998 to 5th Aug 1998)

**No. of odd days:**

**Odd days in 1600 years** = 0

**Odd days in 300 years** =1

**Odd days in 97 years** =Remainder on dividing (24 x 2 + 73 x 1) by 7 = 2

(Since there are 24 leap years and 73 non-leap years in 97 years)

**Odd days in Period from 1st Jan 1998 to 5th Aug 1998 **

= Remainder of (31 + 28+ 31+30+31+30+31+5) divided by 7

=Remainder of (3 + 0 + 3+2+3+2+3+ 5) divided by 7

= 0

Total no. of odd days = 0 + 1+ 2 + 0 = **3**

** ****Hence the given day was 3rd day from Sunday and hence a Wednesday**

**Note:**

**Property of Numbers:**

Let us divide 1000 by 7. We get **remainder as 6**.

Split 1000 into two parts – say 600 and 400.

Remainder when 600 is divided by 7 = **5 **

Remainder when 400 is divided by 7 = **1**

Adding up both remainders, **we get 6 again**.

This is valid for any divisor and any dividend