Calendar deals with odd days, leap year, ordinary year, counting of odd days, and day of the week related to odd days. To find the day of the week on a given date concept of odd days is used. The article **Calendar Quiz 4** lists important **Calendar** practice questions for competitive exams like **RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO Exams** and etc.

Examples:

- Years like 1988, 2008 are leap year (divisible by 4).

Centuries like 2000, 2400 are leap year ( divisible by 400).

Years like 1999, 2003 are not leap year (not divisible by 4).

Centuries like 1700, 1800 are not leap year ( not divisible by 400).

In a century, there is 76 ordinary year and 24 leap year.

An ordinary year has one odd day.

Now, as an ordinary year adds 1 odd day and a leap year adds 2 odd days

we have:

2001, 2002, 2003, 2005, 2006 – 1 odd day each

2004 – 2 odd days

Hence, at the end of 2006 total number of odd days = 7 or 0

Therefore, the calendar for the year 2001 is repeated in the year 2007.

1600 years have 0 odd days and 300 years have 1 odd day.

Now, the period from 1900 to 1992 have 69 ordinary years and 23 leap years

= [latex](69 \times 1 + 23 \times 2)[/latex] = 115 odd days = (16 weeks + 3 days) = 3 odd days.

Januart | February | March | April | May | June | July |
---|---|---|---|---|---|---|

31 | 28 | 31 | 30 | 31 | 30 | 31 |

= 212 days = (30 weeks + 2 days) = 2 odd days

Therefore, total number of odd days = 1 + 3 + 2 = 6 odd days.

Therefore, the required day was Saturday.

200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.

300 years have 15 odd days or 2 week + 1 odd day. Hence, the last day of the 3rd century is a Monday.

400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.

• For a year to be a leap year, it should be divisible by 4.

• No century is a leap year unless it is divisible by 400.

Hence, the year 2100 is not a leap year as it is not divisible by 400.

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