Calendar Questions and Answers

Answer: A

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 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 = (0 + 0 + 4 + 3) = 7  0 odd days.

Given day is Sunday.

Answer: D

Let us find the day on 1st July, 2004.

2000 years have 0 odd day. 3 ordinary years have 3 odd days.

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 1

= 183 days = (26 weeks + 1 day) = 1 t .

Total number of odd days = (0 + 3 + 1) odd days = 4 odd days. '

:. 1st July 2004 was 'Thursday',-,-

Thus, 1st Monday in July 2004 _as on 5th July.

Hence, during July 2004, Monday fell on 5th, 12th, 19th and 26th. .

Answer: A

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

• \( 800 \div 400 = 2 \) ✓
• \( 1200 \div 400 = 3 \) ✓
• \( 2000 \div 400 = 5 \) ✓
• \( 700 \div 400 = 1.75 \) ✗

So 700 is not a leap year, even though it is divisible by 4. The 400-year rule exists because the tropical year is a little under \( 365.25 \) days.

Answer: C

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 = 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
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

Answer: D

Count odd days in stages.

2000 years leave 0. 2001 to 2005 holds one leap year, 2004:

\[ 4(1) + 1(2) = 6 \]

1 January to 28 May 2006, an ordinary year:

\[ 31 + 28 + 31 + 30 + 28 = 148 \equiv 148 - 147 = 1 \pmod 7 \]

\[ \text{total} = 0 + 6 + 1 = 7 \equiv 0 \pmod 7 \]

Zero odd days is Sunday.

Answer: B

The year 2004 is a jump year.
In this way, it has 2 odd days.
∴ The day on sixth March, 2005 will be 2 days.

past the day on sixth Walk, 2004.
Yet, sixth March, 2005 is Monday.
∴ sixth March, 2004 is Saturday.

Answer: C

16 June, 1993 = (1992 years + Period from 1.1.1993 to 16.6.1993)

Odd days in 1600 years = 0

Odd days in 300 years = 1

92 years = (69 ordinary years + 23 leap year) = (69 x 1 + 23 x 2)= 3 odd days

Jan. Feb. March April May June

(31 + 28 + 31 + 30 + 31 + 16 ) = 167 days

167 days = (23 weeks + 6 days) =6 odd days.

Total number of odd days = (0 + 1 + 3 + 6) = 3 odd days.

Given day is Wednesday.

Answer: A

Count the odd days up to the date.

2000 years. Complete centuries in multiples of 400 leave no odd days, so 2000 years contribute 0.

2001 to 2009. Nine years of which 2004 and 2008 are leap:

\[ 7(1) + 2(2) = 11 \equiv 4 \pmod 7 \]

1 January to 15 August 2010. 2010 is ordinary:

\[ 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 \equiv 227 - 224 = 3 \pmod 7 \]

\[ \text{total odd days} = 0 + 4 + 3 = 7 \equiv 0 \pmod 7 \]

Zero odd days corresponds to Sunday.

Answer: A

9th May, 2001 = (2000 years + Period from 1.1.2001 to 9.5.2001)

Odd days in 1600 years = 0

Odd days in 400 years = 0

Jan. Feb. March April May 

(31 + 28 + 31 + 30 + 9) = 129 days

129 days = (18 weeks + 3 day)= 3 odd day.

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

Given day is Wednesday.

Answer: D

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

Step back one year from 8 February 2005 to 8 February 2004. That interval contains 29 February 2004, so it is 366 days long.

\[ 366 \rightarrow 2 \text{ odd days} \]

\[ \text{Tuesday} - 2 = \textbf{Sunday} \]

Going backwards the odd days are subtracted, and the leap day must be counted because 29 February 2004 lies inside the interval.