Calendar Questions and Answers

Answer: A

Look backwards from 1989 until the odd days total a multiple of seven.

\[ 1988:2,\ 1987:1,\ 1986:1,\ 1985:1,\ 1984:2,\ 1983:1,\ 1982:1,\ 1981:1,\ 1980:2,\ 1979:1,\ 1978:1 \]

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

So 1989 repeats the calendar of 1978.

Answer: B

Given that 2.12.91 is the first Tuesday
Hence we can assume that 4.12.91 is the first Thursday

If we add 7 days to 4.12.91, we will get second Thursday
If we add 14 days to 4.12.91, we will get third Thursday
If we add 21 days to 4.12.91, we will get fourth Thursday

=> fourth Thursday = (4.12.91 + 21 days) = 25.12.91.

Answer: D

Find the day for 1 April 2001, then step in sevens.

Odd days to the end of 2000: a multiple of 400 years leaves 0. For 1 January to 1 April 2001, an ordinary year:

\[ 31 + 28 + 31 + 1 = 91 \equiv 0 \pmod 7 \]

So 1 April 2001 was a Sunday, which puts the first Wednesday on the 4th. Adding weeks gives

\[ 4,\; 11,\; 18,\; 25 \]

April has 30 days, so a fifth Wednesday would need the 32nd and does not exist - which is why this list has four dates where the other options have five.

Answer: C

\[ k \text{ weeks } + k \text{ days} = 7k + k = 8k \text{ days} \]

The week count is multiplied by 7 and the day count added, giving a single term in \( k \).

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.

• \( 1900 \div 400 \) is not whole ✗
• \( 1700 \div 400 \) is not whole ✗
• \( 1100 \div 400 \) is not whole ✗
• \( 1600 \div 400 = 4 \) ✓

So 1600 is the leap century year here.

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: 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: D

Going back one year shifts the day by the number of odd days in the year crossed.

The span from 8 December 2006 to 8 December 2007 lies wholly inside ordinary territory - 2008 is the leap year, and its 29 February is not in this range - so the span is 365 days:

\[ 365 \equiv 1 \pmod 7 \]

The later date is therefore one day further on, so the earlier one is one day back:

\[ \text{Saturday} - 1 = \textbf{Friday} \]

Answer: C

Correct answer: 2018

A calendar repeats when the odd days since the starting year sum to a multiple of 7. Counting from 2007 onward, ordinary years contribute 1 each and the leap years 2008, 2012 and 2016 contribute 2 each.

\[1+2+1+1+1+2+1+1+1+2+1 = 14\]

Fourteen is a multiple of 7, and that total is reached at the end of 2017, so 2018 repeats the calendar of 2007.

Answer: C

Let today be \( D \). Yesterday is \( D - 1 \).

Suppose that day were tomorrow instead. If \( D - 1 \) is tomorrow, then "today" in that shifted week is \( D - 2 \), and we are told it would be Friday:

\[ D - 2 = \text{Friday} \quad\Longrightarrow\quad D = \textbf{Sunday} \]

Checking: if today were Sunday, yesterday was Saturday; treating Saturday as tomorrow puts today at Friday, as required.