Calendar rules to solve questions on calendar. Comprehensive calendar concepts for solving any calendar problem. Exercise on Calendar problems with answers.

Exercise problems solved with detailed concepts.

**Sections** are,

**Odd days technique**in solving Calendar problems.- Calendar units.
**Specific properties of Calendar elements**of century, year, month and week in Calendar problems.**Working with weekdays and dates**in Calendar problems using odd days.- Odd days in 12 months of a year.
- Odd days in a normal year and a leap year.
- Odd days in a normal century and leap year century.
- Odd days in 400 years.
**Starting day of the Calendar**in general use.**Net odd days by modulo 7 operation.**

**Examples**on How to solve Calendar problems.**Exercise**on Calendar problems**with answers.****Solution to Calendar problem exercise.**

### Odd days technique in solving Calendar problems

Odd days technique is an ingenious way to **squeeze the whole period** between two dates **into a single number ranging from 0 to 6** and this *represents the gap in weekdays between the two dates.*

This technique is *based on Euclid’s division lemma*.

Essentially **it drops all completed weeks** in a period and **gets the remainder** by dividing a number of days spanning a period of month, year or a century by 7.

This **remainder is the gap in weekdays ranging from 0 to 6 between the two dates,** making it very easy to **find the weekday of the latter date.**

To speed up the process,

Instead of time-consuming calculation of total number of days and divide it by 7 to get the remainder,

it is quickly obtainedjust by adding odd day for months, years or centuries step by step doing modulo 7 operation on the result of addition at each step.

*This simulates finding the remainder by dividing the total number of days by 7.*

We will use here the calendar that is in general use, the calendar that starts with 1 AD and says, this year is 2021 AD.

### Calendar units

**Day:** The smallest interval or unit in a calendar is a day. *The link with a clock and calendar is in the fact that 24 hours make a day.*

**Week:** Seven days make a week, the next larger period of time. Weekdays are, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday **in sequence** that repeats. These names are easy to remember compared to numbers.

**Month:** The next larger interval or unit in a calendar is a month and finally 12 months make a year. Months are, January, February, March, April, May, June, July, August, September, October, November and December **in sequence** and repeating over years.

**General Concept:** No fixed number of weeks or days make a month or year, **though generally,** if nothing else is specified, we take,

- a month has 4 weeks
- a month has 30 days
- a year has 52 weeks
- a year has 365 days

**Printed Calendars** that we see are yearly calendars, but actually,

A calendar is a series of years starting in the past at some defined point of zero and proceeding into the future.

We divide continuing time into years for convenience of leading our lives.

### Specific properties of Calendar elements in Calendar problems

#### Properties of a year

- There are two kinds of years:
**Ordinary year**and**Leap year**. - An ordinary year has 365 days and a leap year has 366 days, one day extra. This extra day appears in February which in a leap year has 29 days instead of 28.

#### Leap year rule

A year to be a leap year, it must satisfy two conditions.

1. It must be divisible by 4, such as 2016, 5324 and so on. By this rule every century year should be a leap year.

2. But

for a century year to be a leap year, there is asecond rule, a century year must be divisible by 400 to be a leap year. 1600, 2000, 2400 are leap years but 1900, 2100, 2200 are not.

### Working with weekdays and dates in Calendar problems using Odd days

Most times we work with weekdays and the dates. **To go forward or backward in terms of number of days, months or years** and still be able to know the weekday and date, a special concept of **odd days** has been created.

Number of odd days in a given number of days of a period is

the remainder when dividing the period in weeks, that is, dividing by 7.

For example, *January always has 31 days, so its number of odd days is 3* which is the remainder after dividing 31 by 7.

**Odd days in 12 months in a year**

An year has 12 months, each with a fixed number of days and and so with fixed number of odd days.

The following is a month-wise chart of number of odd days. Number of odd days for each month is fixed for any year, except for February that has NO odd days for a normal year and 1 odd day for a leap year.

#### Odd days in a year

An

ordinary year with 365 dayshas 52 weeks and1 odd day, while aleap year has 2 odd daysas it has1 day extra in February.

**That’s why,** the *same date after one year is the next day of the week if it is an ordinary year.*

*14-11-2014 was a Friday, so 14-11-2015 must have been a Saturday.*

#### Odd days in a century

First century starts with year 01 and ended with year 100. Similarly, 21st century started with 1st January, year 2001 and will end on 31st December, year 2100. The year next, year 2101 will usher in a new 22nd century.

In an

ordinary century starting from year 01, there will be24 leap yearsand76 ordinary years.

The **24 numbers of leap years** will start with year 04 and end with year 96. This is because the ending year of the century, say, **2100 won’t be a leap year** *as it is not divisible by 400.*

So total **number of odd days** in an **ordinary century** is,

24×2+76=12424×2+76=124.

Dividing by 7 again, the remainder odd days in an ordinary century is, (124−119)=5(124−119)=5. Thus we have the fixed rule,

In an

ordinary century, resultantnet odd dayswill be5.

In a century with century year as a leap year, say in 2400 which is divisible by 400, there will be one odd day extra, **that is, 6**. So the fixed rule for a century for which end year is divisible by 400 is,

For a

leap year century, number ofodd days will be 6.

#### Odd days in 400 years

This gives us the measure of number of odd days *in a larger span of 400 years as*, **remainder of dividing (5+5+5+6)=21 by 7 giving 0.** Thus we have the fixed rule,

For

400 yearsnumber of resultantodd days is 0.

#### Starting day of the calendar in general use

There must be a start days of the calendar we use. It is defined as,

1st day of 1 AD is a Monday, and so

0th day is a Sunday.

*All other weekdays follow from this definition.*

#### Net odd days by Modulo 7 operation

Modulo operation is part of extended arithmetic. It simply is the remainder after dividing an integer by a number, in our case by 7.

For example,

modulo 7(5+5+4+3)=modulo 7(17)=3modulo 7(5+5+4+3)=modulo 7(17)=3, the remainder after dividing 17 by 7.

In odd day accumulation over months of a year, or years of a century, after add up the number of odd days are added up for the individual periods, the sum is divided by 7 and to get the remainder as **number of net odd days in the whole period**.

In this way, instead of time-consuming calculation of total number of days and divide it by 7 to get the remainder, it is quickly obtained just

by adding odd days for months, years or centuries step by step doing modulo 7 operation on the result of addition at each step.

*This simulates finding the remainder by dividing the total number of days by 7 following Euclid’s division lemma.*

**For example,** to calculate the net odd days **between 17th April 2015** and **6th March 2021**, an extended period spanning years, months and days, the period is divided into five parts,

**Part 1:****Remaining days**of the**starting month April**and equivalent odd days – a=13a=13, (April is 30 day month, START COUNTING FROM DAY AFTER START DATE).**Part 2:****Remaining months**in the starting year,**May to December 2015**, and equivalent odd days – b=21b=21, (add up the month-wise odd days from the table above).**Part 3:****Intervening complete years****between starting year 2015 and ending year 2021, that is 2016 to 2020,**and equivalent odd days – c=7c=7, (2016 and 2020 are leap years with 2 odd days each, rest three years 1 odd day each).**Part 4:****Number of complete months**in the ending year 2021 up to the**ending month February**and equivalent odd says – d=3d=3, (January and February).**Part 5:****Number of days**up to the ending date and equivalent odd days – e=6e=6, (6 days up to 6th March).

Total number of odd days between the two dates is,

a+b+c+d+e=13+21+7+3+6=50a+b+c+d+e=13+21+7+3+6=50, adding up the odd days in each period.

Net odd days obtained by dividing this total odd days by 7 and get it as the remainder: 1.

This is modulo 7 operation at each step.

As it was **Friday on 17th April 2015**, *with Net odd days 1 up to 6th March 2021*, this **end date of 6th March 2021 has to be Friday plus 1 day, that is, Saturday.**

Now we will consolidate these concepts by solving a few examples of calendar math problems.

### Examples of Calendar problems

#### Calendar math problem 1

If it is a Monday on 1st October 2012, what day will it be on 21st December, 2013?

#### Calendar math problem solution 1

Till 1st October 2013 it is intervening 1 ordinary year with 365 days and 1 odd day. Rest number of days up to 21st December is 81 and odd days 4.

Total odd days = 4 + 1 = 5.

So the weekday on 21st December 2013 will be 5 days after Monday which is a **Saturday**.

#### Calendar math problem example 2

If it is a Monday on 1st October 2012, what day will it be on 21st March, 2411?

#### Calendar math problem solution 2

Given a weekday and date, finding out a future or past weekday when date is given requires finding out **number of odd days in the number of intervening days**.

**Number of odd days in 2012:**

Total odd days, (30+30+31)=91. So **odd days in 2012 is 0** (as 7 divides 91 fully with 0 remainder).

**Number of odd days in 2013-2100:**

88 years left have 21 leap years each with 2 odd days and 67 ordinary years each with 1 odd day. So total odd days is,

21×2+67=10921×2+67=109, dividing by 7 gives **remainder odd digits as 4**.

Note that year **2100 is not a leap year** (year 400, 800, 1200, 1600, 2000, 2400, and so on are leap years).

**Number of odd days in 22nd century, 23rd century and 24th century:**

Each ordinary century has 24 leap years and 76 ordinary years, whereas in a century with century year as a leap year, we get 25 leap years and 75 ordinary years. So in an* ordinary century we have the number of odd days* as,

Number of odd days in an ordinary centuryNumber of odd days in an ordinary century

=modulo 7(76×1+24×2)=modulo 7(124)=5=modulo 7(76×1+24×2)=modulo 7(124)=5

If the **century is a leap year century,** its number of odd days turns to 6.

Here we have three intervening centuries 22nd, 23rd and 24th, *last being a leap year century*.

So number of odd days in these three centuries is,

modulo 7(5+5+6)=2modulo 7(5+5+6)=2.

**Number of odd days in first 10 years of 25th century:**

In 25th century in the first 10 years the number of odd days will be,

modulo 7(8×1+2×2)=5modulo 7(8×1+2×2)=5, 2404 and 2408 will be leap years.

**Number of odd days in 25th century up to 21st March, 2411:**

Lastly in 2411 up to 21st March 2411, we have (31+28+21)=80 days and so **3 odd days**.

**Total number of odd days:**

Finally, Total intervening odd number of days is,

modulo 7(0+4+2+5+3)=0modulo 7(0+4+2+5+3)=0.

Thus 21st March 2411 would also be a **Monday**.

#### Calendar math problem 3

What was the weekday on 20 June 1354?

#### Calendar math problem Solution 3

The usual calendar starts with 1st January, 1 AD as a Monday, that is, **0th day of January, 1 AD on a Sunday.**

This mechanism enables you to get intervening odd days up to say, 31st December 100 AD as exactly 5 (full 100 years difference as starting point is 0) and so the weekday will be 5 days after Sunday, that is, Friday.

Thus, in the first **1200 years number of odd days is 0** as each 400 years number of odd days is 0 (5+5+5+6=21=>0).

In **13th century odd days is 5**.

In **first 53 years of 14th century**, 13 leap years and 40 ordinary years contribute to odd days as,

modulo 7(13×2+40)=modulo 7(66)=3modulo 7(13×2+40)=modulo 7(66)=3.

**In 1354** (a normal year) up to 20th June, number of days is = (31 + 28 + 31 + 30 + 31 + 20) = 171 giving a **number of odd days 3**.

Thus,

Total odd days=modulo 7(0+5+3+3)=modulo 7(11)=4Total odd days=modulo 7(0+5+3+3)=modulo 7(11)=4.

So the required weekday is 4 days after Sunday, that is, **Thursday**.

### Exercise on Calendar problems

**Q1.** The calendar for the year 2007 will be the same for the year,

- 2014
- 2016
- 2017
- 2018

**Q2.** If today 26th January, 2021 is Tuesday, what will be the day 351 days from now?

- Tuesday
- Monday
- Wednesday
- Thursday

**Q3.** If Feb 12th, 1986 falls on Wednesday then Jan 1st, 3086 falls on which day?

- Wednesday
- Friday
- Thursday
- Tuesday

**Q4.** On what dates of July, 2004 did Monday fall?

- 6th, 20th, 27th, 13th
- 12th, 7th, 19th, 28th
- 3rd, 10th, 24th, 17th
- 5th, 12th, 19th, 26th

**Q5.** It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

- Friday
- Saturday
- Monday
- Wednesday

**Q6.** The second day of a month is Sunday, What will be the last day of the next month which has 31 days?

- Friday
- Saturday
- Sunday
- Monday

**Q7.** The last day of a century cannot be a

- Sunday
- Monday
- Thursday
- Wednesday

**Q8.** If 3rd February, 1999 was Wednesday then what day was 25th January, 2001?

- Thursday
- Sunday
- Wednesday
- Tuesday

**Q9.** What will be the day on 26th January four years later if today is Tuesday, 26th January, 2021?

- Friday
- Sunday
- Monday
- Wednesday

**10.** Any date in March of a year and the day of the week on corresponding date in November that year will be.

- Same day
- Next to next day
- Next day
- Previous day

### Answers to the Exercise on Calendar problems

**Q1. Answer:** Option d: 2018.

**Q2. Answer:** Option c: Wednesday.

**Q3. Answer:** Option b: Friday.

**Q4. Answer:** Option d: 5th, 12th, 19th, 26th.

**Q5. Answer:** Option a. Friday.

**Q6. Answer:** Option d. Monday.

**Q7. Answer:** Option c. Thursday.

**Q8. Answer:** Option a. Thursday.

**Q9. Answer:** Option b. Sunday.

**Q10. Answer:** Option a. Same day.

### Solution to exercise on Calendar problems

#### Calendar exercise question 1.

The calendar for the year 2007 will be the same for the year,

- 2014
- 2016
- 2017
- 2018

#### Solution to Calendar exercise question 1.

Calendar for a year will be same with calendar of 2007,

- if the year is not a leap year, as 2007 is not a leap year, and
- intervening years starting from 2007 together has odd days 0.

Let’s jot down the intervening years with odd days:

2007 – odd days 1, 2008 – leap year – odd days 2, 2009 – odd days 1, 2010 – odd days 1, 2011 – odd days -1, 2012 – leap year odd days 2, 2013 – odd days 1.

Up to 2013 odd days is 9 – **net 2.** So 2014 1st January won’t have same day as 1st January 2007.

Continuing again,

2014 – odd days 1, 2015 – odd days 1, 2016 – leap year cannot have same calendar with 2007 and has odd days 2.

Total odd days between 1st January 2007 and 1st January 2017 is 6. Add 1 more odd day for 2017 to odd day 7, net 0, to have 1st January 2018 same day as 1st January 2007.

**Answer:** Option d: 2018.

#### Calendar exercise question 2.

If today 26th January, 2021 is Tuesday, what will be the day 351 days from now?

- Tuesday
- Monday
- Wednesday
- Thursday

#### Solution to Calendar exercise question 2.

350 days from today is 50 complete weeks with 0 odd days, and 351 days from today results in 1 net odd day.

Today being Tuesday, 351 days from today it will be Wednesday.

**Answer:** Option c: Wednesday.

#### Calendar exercise question 3.

If Feb 12th, 1986 falls on Wednesday then Jan 1st, 3086 falls on which day?

- Wednesday
- Friday
- Thursday
- Tuesday

#### Solution to Calendar exercise question 3.

Backtrack from February 12 by (31+11)=42(31+11)=42 days with 0 net odd days to 1st January 1986 as also a Wednesday.

Secondly, 1901 to 1986 having same number of odd days as 3001 to 3086, between 1st January 1901 to 1st January 3001 there will be 11 number of complete centuries, Out of which three centuries, 2000, 2400 and 2800 will be leap year centuries with 6 odd days each.

Total odd days in these intervening 11 centuries are then (5×11+3)=58(5×11+3)=58, that is net odd days 58−56=258−56=2.

1st January 1986 being a Wednesday, 1st January 3086 will be 2 more days ahead, that is, a Friday.

**Answer:** Option b: Friday.

#### Calendar exercise question 4.

On what dates of July, 2004 did Monday fall?

- 6th, 20th, 27th, 13th
- 12th, 7th, 19th, 28th
- 3rd, 10th, 24th, 17th
- 5th, 12th, 19th, 26th

#### Solution to Calendar exercise question 4.

As no reference date and day are given, we have to use the starting day and date of the Calendar used.

Starting day and date of Calendar in use: 1st January, 1 AD was a Monday.

Going ahead 5 numbers of 400 years, that is 2000 years, 1st January 2001 must also be Monday. This is because number of odd days in every 400 years starting at the beginning is 0.

Going ahead three years 2001, 2002 and 2003 each with odd days of 1, total odd days up to 1st January 2004 from 1st January 2001 is 3. 1st January 2004 is then a Thursday.

Going ahead six months of January (odd days 3), February(odd days 1 as it is a leap year), March (odd days 3), April (odd days 2), May (odd days 3) and June (odd days 2) add up to total odd days from 1st January, 2004 to 1st July 2004 to 14, net odd days 0. So 1st July 2004 is a Thursday.

Add 4 more days to reach the 1st Monday of July 2004. It is 5th July, 2004.

The Mondays in July, 2004 are then on 5th, 12th, 19th and 26th.

**Answer:** Option d: 5th, 12th, 19th, 26th.

#### Calendar exercise question 5.

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

- Friday
- Saturday
- Monday
- Wednesday

#### Solution to Calendar exercise question 5.

Covering three intervening years 2006 (odd days 1), 2007 (odd days 1), 2008 (odd days 2) and 2009 (odd days 1) with total odd days 5, five days ahead of Sunday will be Friday on 1st January, 2010.

**Answer:** Option a: Friday.

#### Calendar exercise question 6.

The second day of a month is Sunday, What will be the last day of the next month which has 31 days?

- Friday
- Saturday
- Sunday
- Monday

#### Solution to Calendar exercise question 6.

Second day of present month being a Sunday, last day of the previous month must be a Friday.

Odd days in the present month may have four possible months,

- If the present month is February, and it is a leap year, odd days is 1. With 3 more odd days in next month of 31 days, total odd days is 4. 31st of next month Tuesday. As there is no Tuesday among the options, this possibility of present month is invalid.
- If the present month is February, and it is a normal year, odd days is 0. With 3 more odd days in next month of 31 days, total odd days is 3. 31st of next month is Monday. It is option d.
**This should be the answer**if other possibilities are invalid. Otherwise question is wrong. - Present month is any month with number of days 30 and odd days 2, total odd days up to 31st of next month 5 and day on 31st Wednesday. And as this day is not among the options, this possibility is also invalid.
- Lastly, present month may be July with odd days 3, total odd days up to 31st of next month 6, and day on 31st Thursday. As expected, this day is also not there among the options. So this possibility is invalid.

Finally then, 31st of next month must be a Monday.

**Answer:** Option d: Monday.

#### Calendar exercise question 7.

The last day of a century cannot be a

- Sunday
- Monday
- Thursday
- Wednesday

#### Solution to Calendar exercise question 7.

Starting day of 1st century 1st January 1 AD was a Monday. As 1st century was not a leap year century, it had 5 odd days so that 1st January 101 was Saturday and 31st December 100 AD a Friday.

The second century of the calendar also had 5 odd days so that 31st December 200 AD was Wednesday.

The third century similarly had 5 odd days so that 31st December 300 AD was a Monday.

But the 4th century had 6 odd days so that 31st December 400 AD was a Sunday.

Three of these four days are in the choices and Friday doesn’t appear in the choices.

Then it can be be concluded that last day of the first four centuries cannot be a Thursday with certainty.

But what about other centuries?

Answer is simple. As odd days in 400 years beginning from the start of the calendar is 0, the same pattern and calendar days will be repeated every 400 years.

In general then, last day of no century can be a Thursday.

**Answer:** Option c: Thursday.

#### Calendar exercise question 8.

If 3rd February, 1999 was Wednesday then what day was 25th January, 2001?

- Thursday
- Sunday
- Wednesday
- Tuesday

#### Solution to Calendar exercise question 8.

3rd February, 1999 was a Wednesday. Instead of traversing 1999 month-wise, backtrack to 1st January 1999 by 33 day or odd days 5. Going back by 5 days from Wednesday, it was a Friday on 1st January 1999.

The normal year 1999 had 1 odd day and leap year 2000 with 2 odd days, total 3 odd days. So 1st January 2001 was a Monday.

Up to 25th January is 24 more days, that is 3 odd days. so 25th January, 2001 was Thursday.

**Answer:** Option a: Thursday.

#### Calendar exercise question 9.

What will be the day on 26th January four years later if today is Tuesday, 26th January, 2021?

- Friday
- Sunday
- Monday
- Wednesday

#### Solution to Calendar exercise question 9.

26th January 2021 to 26th January 2025 will be equivalent to 1st January 2021 to 1st January 2025. Intervening years – 2021 (odd days 1), 2022 (odd days 1), 2023 (odd days 1) and 2024 (odd days 2) – total odd days 5.

So 26th January 2021 being a Tuesday, 26th January, 2025 will be going ahead by 5 days, a Sunday.

**Answer:** Option b: Sunday.

#### Calendar exercise question 10.

Any date in March of a year and the day of the week on corresponding date in November that year will be.

- Same day
- Next to next day
- Next day
- Previous day

#### Solution to Calendar exercise question 10.

March has 31 days and November has 30 days. So first conclusion:

March 31st won’t have any corresponding date in November.

We have to find correspondence between first 30 days of the two months.

Second conclusion,

As these two years do not have February between them, number of odd days between 1st of March and 1st of November will be fixed and same for the first 30 days.

March – 3 odd days, April – 2 odd days, May 3 odd days, June – 2 odd days, July and August both with 3 odd days, September – 2 odd days and October with 3 odd days make a total of 21 odd days, that is 0 net odd days.

So 1st March and 1st November will have same day like every other 29 days of the two months.