Let the day of the week to be found on a given date. To find that, we are going to use the concept of odd days

**Odd Days :**

In a given period, the number of days more than the complete weeks are called odd days.

**Leap Year :**

(i) Every year is divisible by 4 is a leap year, if it is not a century.

(ii) Every 4th century is a leap year and no other century is a leap year.

Examples :

Each of the years 1948, 2004, 1676 is a leap year

Each of the years 400, 1200, 1600, 2000 is a leap year

None of the years 100, 2001, 2005, 2100 is a leap year

**Note :**

A leap years has 366 days and the month February has 29 days

**Ordinary Year :**

The year which is not a leap year is called an ordinary year. An ordinary year has 365 days and the month February has 28 days.

**Shortcut to find the no. of leap years in the given no. of years :**

Divide the given number of years by 4 and take the integer part. That is the number of leap years available in the given number of years.

**Example :**

Let us find the number of leap years in 75 years. Divide 75 by 4.

That is,

75/4 = 18.75

In this, integer part is 18. There are 18 leap years in 75 years.

**Caution : **

When we find the number of leap years in 100 years, we will divide 100 by 4 as explained above.

Then,

100/4 = 25

Here, there are 25 numbers from 1 to 100 which are divisible by 25.

Even though 100 is divisible by 4, it is not leap year.

So, number of leap years in 100 years is

= 25 - 1

= 24

**Counting of Odd Days :**

(i) 1 leap year has 366 days = 52 weeks + 2 days.

So, 1 leap year has 2 odd days

(ii) 1 ordinary year = 365 days = 52 weeks + 1 day.

So, 1 ordinary year has 1 odd day

100 years = 76 ordinary years + 24 leap years

100 years = 76x1 + 24x2 = 76 + 48 = 124 days

100 years = 17 weeks + 5 days = 5 odd days

So, 100 years has 5 odd days

Number of odd days in 100 years = 5

Number of odd days in 200 years = 3

Number of odd days in 300 years = 1

Number of odd days in 400 years = 0

Similarly, each one of 800 years, 1200 years, 1600 years and 2000 years etc has 0 odd days.

**Days of the week related to odd days :**

0 odd days ---> Sunday

1 odd day ---> Monday

2 odd days ---> Tuesday

3 odd days ---> Wednesday

4 odd days ---> Thursday

5 odd days ---> Friday

6 odd days ---> Saturday

**Problem 1 : **

What was the day of the week on 16th July,1776 ?

**Solution : **

16th July, 1776 :

= 1775 years + Period from 01.01.1776 to 16.07.1776

1775 :

= 1600 + 100 + 75

No.of odd days in 1600 years = 0 ----(1)

No.of odd days in 100 years = 5 ----(2)

No. of odd days in 75 years :

= 18 leap years + 57 ordinary years

= 18x2 + 57x1

= 93 days

= 13 weeks + 2 days

= 2 odd days ----(3)

From 01.01.1776 to 16.07.1776, we have,

Jan - 31 days

Feb - 29 days

Mar - 31 days

Apr - 30 days

May - 31 days

Jun - 30 days

July - 16 days

Total = 198 days

= 28 weeks + 2 days

= 2 odd days ----(4)

Add (1), (2), (3) and (4).

= 0 + 5 + 2 + 2

= 9 days

= 1 week + 2 days

= 2 odd days

2 odd days is corresponding to Tuesday

So, the day of the week on 16th July,1776 is Tuesday.

**Problem 2 : **

On what dates of March 2005 did Friday fall ?

**Solution : **

First, we have to find the day on 01.03.2005.

01.03.2005 :

= 2004 years + Period from 01.01.2005 to 01.03.2005

2004 :

= 2000 + 4

Number of odd days in 2000 years = 0 ----(1)

Number of odd days in 4 years :

= 1 leap year + 3 ordinary year

= 1x2 + 3x1

= 5 days ----(2)

From 01.01.2005 to 01.03.2005, we have,

Jan - 31 days

Feb - 28 days

Mar - 1 day

Total = 60 days.

= 8 weeks + 4 days

= 4 odd days ----(3)

Add (1), (2) and (3).

= 0 + 5 + 4

= 9 days

= 1 week + 2 days

= 2 odd days

2 odd days is corresponding to Tuesday.

That is, 01.03.2005 was Tuesday.

Then, Friday lies on 04.03.2005.

So, the dates of March, 2015 on which Fridays fell are

04, 11, 18 and 25

**Problem 3 : **

For which of the following years, will the calendar for the year 2007 be the same ?

(a) 2014

(b) 2016

(c) 2017

(d) 2018

**Solution : **

For example, if our answer is 2018, we must have the same day on 01.01.07 & 01.01.18 and number of odd days from 01.01.07 to 31.12.17 must be zero.

Let us check the same thing for each of the given options.

**Option (a) 2015 : **

No. of odd days from 01.01.2007 to 31.12.2014 :

= 2 leap years + 6 ordinary years

= 2x2 + 6x1

= 10 days

= 1 week + 3 days

= 3 odd days [not correct]

**Option (b) 2016 : **

No. of odd days from 01.01.2007 to 31.12.2015 :

= 2 leap years + 7 ordinary years

= 2x2 + 7x1

= 11 days

= 1 week + 4 days

= 4 odd days [not correct]

**Option (c) 2017 :**

No. of odd days from 01.01.2007 to 31.12.2016 :

= 3 leap years + 7 ordinary years

= 3x2 + 7x1

= 13 days

= 1 weeks + 6 days

= 6 odd days [not correct]

**Option (d) 2018 : **

No.of odd days from 01.01.2007 to 31.12.2017 :

= 3 leap years + 8 ordinary years

= 3x2 + 8x1

= 14 days

= 2 weeks + 0 day

= 0 odd day [correct]

So, Calendar for the year 2018 will be the same as for the year 2007.

**Problem 4 : **

How many days are there in 'x' weeks 'x' days ?

**Solution : **

Number of days in 'x' weeks = 7x

Number of days in 'x' days = x

Number of days in 'x' weeks 'x'days :

= 7x + x

= 8x

So, the number of days in 'x' weeks x days is 8x.

**Problem 5 : **

The last day of a century can not be

(a) Monday

(b) Tuesday

(c) Wednesday

(d) Thursday

**Solution : **

From the notes given on this web page, we have

Number of odd days in 100 years (1st century) = 5

So, the last day of 1st century is Friday.

Number of odd days in 200 years (2nd century) = 3

So, the last day of 2nd century is Wednesday.

Number of odd days in 300 years (3rd century) = 1

So, the last day of 3rd century is Monday

Number of odd days in 400 years (4th century) = 0

So, the last day of 4th century is Sunday

This pattern is repeated

Therefore, the last day of a century can not be Tuesday or Thursday or Saturday.

So, option (c) Tuesday is the correct answer.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

If you have any feedback about our math content, please mail us :

**v4formath@gmail.com**

We always appreciate your feedback.

You can also visit the following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**