Chapter 2 • Unit I • Section 2.2

Exercise 2.2: Calendar

Comprehensive theory, odd days calculations, century distributions, and authentic CBSE model solutions (2026-27).

Structure of the Calendar: Ordinary Year vs. Leap Year

A clock apprises us about smaller spans of time (seconds, minutes, hours), whereas a calendar depicts the days, weeks, and months of a specific year.

Parameter Ordinary Year Leap Year
Number of Days 365 Days 366 Days
Number of Weeks 52 Weeks and 1 Day 52 Weeks and 2 Days
Number of Odd Days 1 Odd Day ($365 = 52 \times 7 + 1$) 2 Odd Days ($366 = 52 \times 7 + 2$)
Definition of Odd Day:
The extra day(s) left after counting an exact number of complete 7-day weeks is considered an Odd Day. That is, for any given duration $N$ days:
$$\text{Odd Days} = N \pmod 7 = \text{Remainder when } N \text{ is divided by } 7.$$

Calculating Number of Odd Days in a Month (Juggle Your Brain)

Since every 7 consecutive days form a complete week, the odd days in each calendar month are determined by the remainder of its days divided by 7:

MonthTotal DaysOdd Days ($Days \pmod 7$)
January31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
February28 / 29 days$28 = 4\text{ wks} + 0 \implies \mathbf{0}$ (ordinary)
$29 = 4\text{ wks} + 1 \implies \mathbf{1}$ (leap)
March31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
April30 days$\frac{30}{7} = 4\text{ wks and } 2\text{ days} \implies \mathbf{2}$
May31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
June30 days$30 = 4\text{ wks} + 2 \implies \mathbf{2}$
MonthTotal DaysOdd Days ($Days \pmod 7$)
July31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
August31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
September30 days$30 = 4\text{ wks} + 2 \implies \mathbf{2}$
October31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$
November30 days$30 = 4\text{ wks} + 2 \implies \mathbf{2}$
December31 days$31 = 4\text{ wks} + 3 \implies \mathbf{3}$

Decoding the Day of the Week

Coding and decoding is foundational to modern computing and cryptography (e.g. phone passwords, bank debit cards, hashing). In calendar mathematics, decoding a date into a day of the week is performed via Modulo 7 arithmetic.

0 = Sunday 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday
Example 5 (CBSE Support Material): If $30^{\text{th}}$ March 2020 was Monday. What would be the day after 61 days?
Show Step-by-Step Solution
Reference Date: 30 March 2020
Reference Day: Monday
Duration: 61 days

Step 1: Divide duration by 7:
$\frac{61}{7} = 8\text{ weeks and } 5\text{ days}$ ($61 = 8 \times 7 + 5$).

Step 2: The day after 8 complete weeks ($56^{\text{th}}$ day) is Monday.
$57^{\text{th}}$ day = Tuesday
$58^{\text{th}}$ day = Wednesday
$59^{\text{th}}$ day = Thursday
$60^{\text{th}}$ day = Friday
$61^{\text{st}}$ day = Saturday.

Answer: The $61^{\text{st}}$ day will be a Saturday.

Leap Year Rules & Odd Days in a Century

A normal year is a leap year if it is divisible by 4. Fun Fact: A number is divisible by 4 if its last two digits are divisible by 4. However, for century years (100, 200, 300, 400, 1900, 2000), it must be divisible by 400 to be a leap year.

Illustration: What day of the week was on 31st December 100 A.D.?
  • Step 1: Total number of years $= 100$. Number of leap years $= 24$ (because $100 / 4 = 25$, but century year 100 is not divisible by 400, so $25 - 1 = 24$).
  • Step 2: Number of ordinary years $= 100 - 24 = 76\text{ ordinary years}$.
  • Step 3: Odd days in 100 years $= (76 \times 1) + (24 \times 2) = 76 + 48 = \mathbf{124\text{ days}}$.
  • Step 4: Divide by 7: $124 = (17 \times 7) + \mathbf{5\text{ odd days}}$.
  • Conclusion: 5 odd days corresponds to Friday. Hence, 31st December 100 A.D. was a Friday!
Stretch Your Mind: Century Odd Days & Final Century Days
Century Year Calculation of Odd Days Net Odd Days Last Day of Century
100100 years5Friday
200$5 + 5 = 10 = (7 \times 1) + 3$3Wednesday
300$5 \times 3 = 15 = (7 \times 2) + 1$1Monday
400$(5 \times 4) + 1 = 21$ (multiple of 7)0Sunday
500$400 + 100 \implies 0 + 5$5Friday
600$400 + 200 \implies 0 + 3$3Wednesday
700$400 + 300 \implies 0 + 1$1Monday
800$2 \times 400 \implies 0$0Sunday
1000$800 + 200 \implies 0 + 3$3Wednesday
1400$1200 + 200 \implies 0 + 3$3Wednesday
1600Multiple of $400 \implies 0$0Sunday
1700$1600 + 100 \implies 0 + 5$5Friday
2000Multiple of $400 \implies 0$0Sunday
2100$2000 + 100 \implies 0 + 5$5Friday
Theorem (Last Day of a Century):
Since centuries have odd days of 5, 3, 1, or 0, the last day of a century can only be Friday, Wednesday, Monday, or Sunday. It can never be Tuesday, Thursday, or Saturday.
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