Problem: What was the day of the week on 16th July, 1776?Solution:
16th July, 1776 = (1775 years + Period from 1.1.1776 to 16.7.1776)
Counting of odd days:
Number of odd days in 1600 years = 0
Number of odd days in 100 years = 5
75 years = 18 leap years + 57 ordinary years
= (18 × 2 + 57 × 1) odd days = 93 odd days
= (13 weeks + 2 days) ≡ 2 odd days
∴ 1775 years have
= (0 + 5 + 2) odd days = 7 odd days ≡ 0 odd day.
Jan. Feb. March April May June July (31 + 29 + 31 + 30 + 31 + 30 + 16) = 198 days
198 days = (28 weeks + 2 days) ≡ 2 odd days.
∴ Total number of odd days = (0 + 2) = 2.
Hence, the required day is Tuesday.
Problem: What was the day of the week on 15th August, 1947?Solution:
15th August, 1947 = (1946 years + Period from 1.1.1947 to 15.8.1947)
Odd days in 1600 years = 0
Odd days in 300 years = (5 × 3) = 15 ≡ 1
46 years = (11 leap years + 35 ordinary years)
= (11 × 2 + 35 × 1) odd days = 57 odd days
= (8 weeks + 1 day) ≡ 1 odd day.
∴ Odd days in 1946 years = (0 + 1+ 1) = 2.
Jan. Feb. March April May 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 = (2 + 3) = 5.
Hence, the required day is Friday.
Problem: What was the day of the week on 4th June, 2002?Solution:
4th June, 2002 = (2001 years + Period from 1.1.2002 to 4.6.2002)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Odd days in 1 ordinary year = 1
Odd days in 2001 years = (0 + 0 + 1) = 1
Jan. Feb. March April May June
= 155 days
(31 + 28 + 31 + 30 + 31 + 4)
= 22 weeks + 1 day ≡ 1 odd day
Total number of odd days = (1 + 1) = 2 ∴ Required day is Tuesday.
Problem: On what dates of March 2005 did Friday fall?Solution:
First, we find the day on 1.3.2005
1.3.2005 = (2004 years + Period from 1.1.2005 to 1.3.2005)
Odd days in 1600 years = 0 Odd days in 400 years = 0
4 years = (1 leap year + 3 ordinary years)
= (1 × 2 + 3 × 1) odd days = 5 odd days
Jan. Feb. March
(31 + 28 + 11)
= 60 days = (8 weeks + 4 days) ≡ 4 odd days.
Total number of odd days = (0 + 0 + 5 + 4) = 9 ≡ 2 odd days
∴ 1.3.2005 was Tuesday. So, Friday lies on 4.3.2005 Hence, Friday lies on 4th, 11th, 18th and 25th of March, 2005
Problem: Prove that the calendar for the year 2003 will serve for the year 2014.Solution:
We must have the same day on 1.1.2003 and 1.1.2014.
So, the number of odd days between 31.12.2002 and 31.12.2013 must be 0.
This period has 3 leap years and 8 ordinary years.
Number of odd days = (3 × 2 + 8 × 1) = 14 ≡ 0 odd day
∴ Calendar for the year 2003 will serve for the year 2014.
Problem: January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008?Solution:
The year 2007 is an ordinary year. So, it has 1 odd day.
1st day of the year 2007 was Monday.
1st day of the year 2008 will be 1 day beyond Monday. Hence, it will be on Tuesday.
Problem: January 1, 2008 is Tuesday. What day of the week lies on Jan. 1, 2009?Solution:
The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.
Problem: On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec. 2006?Solution:
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 8th Dec, 2007 will be 1 day beyond the
day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday.
Problem: On 6th March, 2005 Monday falls. What was the day of the week on 6th March, 2004?Solution:
The year 2004 is a leap year. So, it has 2 odd days.
∴ The day on 6th March, 2005 will be 2 days beyond the day on 6th March, 2004.
But, 6th March, 2005 is Monday.
∴ 6th March, 2004 is Saturday.
Problem: The calendar for the year 2007 will be the same for the year: (a) 2014 (b) 2016 (c) 2017 (d) 2018
Count the number of odd days from the year 2007 onwards to
get the sum equal to 0 odd day.
| Year | 2007 | 2008 | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017 |
| Odd day | 1 | 2 | 1 | 1 | 1 | 2 | 1 | 1 | 1 | 2 | 1 |
Sum = 14 odd days ≡ 0 odd day.
∴ The calendar for the year 2018 will be the same as for the year 2007.
Problem: On what dates of April, 2001 did Wednesday fall?
(a) 1st, 8th, 15th, 22nd, 29th
(b) 2nd, 9th, 16th, 23rd, 30th
(c) 3rd, 10th, 17th, 24th
(d) 4th, 11th, 18th, 25th
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
= 91 days ≡ 0
odd days. (31 + 28 + 31 + 1)
Problem: What was the day of the week on 17th June, 1998?
(a) Monday (b) Tuesday
(c) Wednesday (d) ThursdaySolution:
Wednesday
Problem: What was the day of the week on 28th May, 2006?
(a) Thursday (b) Friday
(c) Saturday (d) Sunday
Solution:
Sunday
Problem: What will be the day of the week on 15th August, 2010?
(a) Sunday (b) Monday
(c) Tuesday (d) Friday
Solution:
Sunday
Problem: Today is Monday. After 61 days, it will be
(a) Wednesday (b) Saturday
(c) Tuesday (d) Thursday
Solution:
Saturday
Problem: The last day of a century cannot be
(a) Monday (b) Wednesday
(c) Tuesday (d) Friday
Solution:
Tuesday
Problem: Which of the following is not a leap year?
(a) 700 (b) 800
(c) 1200 (d) 2000
Solution:
700
Problem: How many days are there in x weeks x days?
(a) 7x2 (b) 8x
(c) 14x (d) 7
Solution:
8x
Problem: It was Sunday on Jan 1, 2006. What was the day of the week on Jan 1, 2010?
(a) Sunday (b) Saturday
(c) Friday (d) Wednesday
Solution:
Friday
Problem:On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
(a) Tuesday (b) Monday
(c) Sunday (d) Wednesday
Solution:
Sunday
Problem: For a certain month, the dates of three of the Sundays are even numbers. Then, the 15th of the that month falls on a [SSC—CGL (Tier I) Exam, 2012]
(a) Thursday (b) Friday
(c) Saturday (d) Sunday
Solution:
The dates of three of the Sundays are even number is
2, 9, 16, 23, 30.
So, on 16th of that month = Sunday, 15th of that month falls on a Saturday
Problem: What was the day of the week on 15 August, 1947? [DMRC— Customer Relationship Assistant (CRA) Exam, 2016]
(a) Saturday (b) Friday
(c) Thursday (d) Wednesday
Solution:
15 August 1947 means 1946 complete years + first 7
months up to July 1947 + 15 days of August 1947
1600 years have 0 odd days.
300 years have 1 odd day
46 years have 11 leap years and 35 ordinary years
= (11 × 2) + (35 × 1)
= 22 + 35 = 57 odd days
= 8 × 7 + 1 odd days
= 8 weeks + 1 odd day
Up to 1946 there are 1 + 1 = 2 odd days
January 1947 ⇒ 3 odd days
February 1947 ⇒ 0 odd days
(1947 is a normal year)
March 1947 ⇒ 3 odd days
April 1947 ⇒ 2 odd days
May 1947 ⇒ 3 odd days
June 1947 ⇒ 2 odd days
July 1947 ⇒ 3 odd days
Up to 15 August ⇒ 15 odd days
Total number of odd days up to 15 August 1947
= 2 + 3 + 0 + 3 + 2 + 3 + 2 + 3 + 15 = 33 odd days.
Hence, 15th August 1947 was Friday.
Problem: The calendar for the year 2009 will be the same as that of the year [DMRC— Customer Relationship Assistant (CRA) Exam, 2016]
(a) 2013 (b) 2014
(c) 2015 (d) 2014
Solution:
2008 was a leap year.
A leap year has two odd days.
Suppose the year 2008 starts with a Monday.
Then the first day of 2009 was Wednesday.
Now,
First day of 2010 ⇒ Thursday
First day of 2011 ⇒ Friday
First day of 2012 ⇒ Saturday
Because 2012 was a leap year
First day of 2013 ⇒ Monday
First day of 2014 ⇒ Tuesday
First day of 2015 ⇒ Wednesday
Thus, the calendar for the year 2009 was the same as that of the year 2015.