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QUESTION: 1

On what dates of April 2001 did Wednesday fall?

Solution:

We shall find the day on 1st April, 2001.

1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April

(31 + 28 + 31 + 1) = 91 days ≡ 0 odd days.

Total number of odd days = (0 + 0 + 0) = 0

On 1st April, 2001 it was Sunday.

In April, 2001 Wednesday falls on 4^{th}, 11^{th}, 18^{th} and 25^{th}

QUESTION: 2

What was the day of the Week on 17th June 1998?

Solution:

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0

Odd days in 300 years = (5 x 3) ≡ 1

97 years has 24 leap years + 73 ordinary years.

Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 17) = 168 days

Therefore 168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.

Given day is Wednesday.

QUESTION: 3

How many days are there in x weeks x days

Solution:

x weeks x days = (7x + x) days

= 8x days.

QUESTION: 4

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

Solution:

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day

100 years have 5 odd days

75 years = (18 leap years + 57 ordinary years)

= [(18 x 2) + (57 x 1)]

= 93 (13 weeks + 2 days)

= 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day

Jan Feb Mar Apr May Jun Jul

31 + 29 + 31 + 30 + 31 + 30 + 16

= 198 days

= (28 weeks + 2 days)

Total number of odd days = (0 + 2) = 2

Required day was 'Tuesday'.

QUESTION: 5

What will be the day of the week 15^{th }August 2010?

Solution:

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.

Jan. Feb. March April Mayb June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.

Given day is Sunday

QUESTION: 6

It was Sunday on Jan 1, 2006. What was the day of the Week Jan 1, 2010

Solution:

On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.

∴ On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday.

QUESTION: 7

The last day of a Century cannot be

Solution:

100 years contain 5 odd days.

∴ Last day of 1st century is Friday.

200 years contain (5 x 2) ≡ 3 odd days.

∴ Last day of 2nd century is Wednesday.

300 years contain (5 x 3) = 15 ≡ 1 odd day.

∴ Last day of 3rd century is Monday.

400 years contain 0 odd day.

∴ Last day of 4th century is Sunday.

This cycle is repeated.

∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

QUESTION: 8

Today is Monday. After 61 days, it will be:

Solution:

Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.

After 61 days, it will be Saturday.

QUESTION: 9

On 8^{th} Feb,2005 it was Tuesday.What was the day of the Week on 8th Feb,2004?

Solution:

The year 2004 is a leap year. It has 2 odd days.

∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.

Hence, this day is Sunday.

QUESTION: 10

What was the day of the week on 28^{th }May 2006?

Solution:

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days

Jan. Feb. March April May

(31 + 28 + 31 + 30 + 28 ) = 148 days

∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day.

Given day is Sunday.

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