Calendar Questions and Answers

Answer: B

31st Dec, 2011 = (2010 years + Period from 1.1.2011 to 31.12.2011)

Odd days in 2000 years = 0

10 years has 2 leap years + 8 ordinary years.

Number of odd days in 10 years ( 2 x 2 + 8) = 5 odd days.

31st,Dec => complete year of 2011 (non-leap year) = 1 odd day.

Total number of odd days = (0 + 5 + 1) = 6.

Given day is saturday.

Answer: C

Correct answer: Saturday

Only the remainder after dividing by 7 matters, since each complete week returns to the same day.

\[61 = 8 \times 7 + 5\]

Counting five days on from Monday gives Saturday.

Answer: C

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Answer: D

A year is a leap year if it is divisible by 4, except that a century year must also be divisible by 400.

All four are century years, so each must be divisible by 400.

\[ 2500 \div 400 = 6.25 \ \times, \quad 1700 \div 400 = 4.25 \ \times, \quad 1900 \div 400 = 4.75 \ \times \]

\[ 2400 \div 400 = 6 \ \checkmark \]

So 2400 is the leap year.

Answer: B

Count odd days backwards from 2006 until they reach a multiple of seven.

\[ 2005:1,\ 2004:2,\ 2003:1,\ 2002:1,\ 2001:1,\ 2000:2,\ 1999:1,\ 1998:1,\ 1997:1,\ 1996:2,\ 1995:1 \]

\[ \text{total} = 14 \quad\Longrightarrow\quad 0 \text{ odd days} \]

So the 2006 calendar had been used in 1995.

Answer: D

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 days.
Calendar for the year 2018 will be the same as for the year 2007.

Answer: C

A year is a leap year if it is divisible by 4, except that a century year must also be divisible by 400.

\[ 2000 \div 400 = 5 \ \checkmark, \qquad 2008 \div 4 = 502 \ \checkmark, \qquad 2012 \div 4 = 503 \ \checkmark \]

\[ 2009 \div 4 = 502.25 \ \times \]

So 2009 is not a leap year.

Answer: C

Count forward from the given day and reduce modulo 7:

\[ 25 - 3 = 22 \text{ days} \]

\[ 22 = 3 \times 7 + 1 \quad\Longrightarrow\quad 22 \equiv 1 \pmod 7 \]

So the 25th is one day past the day of the 3rd:

\[ \text{Tuesday} + 1 = \textbf{Wednesday} \]

Listing them confirms it: the Tuesdays are the 3rd, 10th, 17th and 24th, so the 25th is the Wednesday that follows.

Answer: C

Only the remainder on division by 7 matters - the odd days.

Track the weekday of 5 January year by year. An ordinary year adds one day; a leap year adds two, but only if 29 February falls within the interval - which it does for the year following a 5 January in a leap year.

\[ 2018 \text{ Fri} \to 2019 \text{ Sat} \to 2020 \text{ Sun} \to 2021 \text{ Tue} \]

\[ \to 2022 \text{ Wed} \to 2023 \text{ Thu} \to 2024 \ \textbf{Fri} \]

The jump from 2020 to 2021 skips Monday because 29 February 2020 lies between the two dates.