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|---|---|---|
| Calendar: Concepts of Odd Days and Leap Years | Finding the Day of the Week for a Given Date | Calendar Repetition Problems |
| Solving Calendar Problems | ||
Calendars
Calendar: Concepts of Odd Days and Leap Years
Calendar problems involve determining the day of the week for a given date, calculating the number of days between dates, or understanding patterns in the calendar. These problems rely on the concepts of leap years and 'odd days'.
Odd Days:
The number of days in a standard week is 7. When we divide the total number of days in a given period by 7, the remainder represents the number of Odd Days. These odd days indicate how many days the day of the week will shift forward from a starting day.
For example, if a period has 8 days, $8 \div 7$ gives a remainder of 1. So, there is 1 odd day. If the period started on a Monday, 8 days later it would be Monday + 1 day = Tuesday.
- A period of 7 days (a week) has $7 \div 7$, remainder 0 $\implies 0$ odd days.
- A period of 10 days has $10 \div 7$, remainder 3 $\implies 3$ odd days.
- A period of 30 days (e.g., April) has $30 \div 7$, remainder 2 $\implies 2$ odd days.
- A period of 31 days (e.g., January) has $31 \div 7$, remainder 3 $\implies 3$ odd days.
Number of Odd Days in Months:
The number of odd days in a month depends on the number of days in that month:
| Month | Number of Days | Odd Days ($Days \div 7$ Remainder) |
|---|---|---|
| January | 31 | 3 |
| February (Normal Year) | 28 | 0 |
| February (Leap Year) | 29 | 1 |
| March | 31 | 3 |
| April | 30 | 2 |
| May | 31 | 3 |
| June | 30 | 2 |
| July | 31 | 3 |
| August | 31 | 3 |
| September | 30 | 2 |
| October | 31 | 3 |
| November | 30 | 2 |
| December | 31 | 3 |
Leap Year:
A Leap Year is a calendar year that contains an extra day, making it 366 days long, as opposed to a Normal Year (or Common Year) which has 365 days. The extra day is added to the month of February, which has 29 days in a leap year and 28 days in a normal year.
Leap years occur to keep the calendar year synchronized with the astronomical year or seasonal year, which is slightly longer than 365 days.
The rules for determining if a year is a leap year are:
1. A year is a leap year if it is exactly divisible by 4.
2. However, if a year is exactly divisible by 100, it is NOT a leap year, UNLESS it is also exactly divisible by 400.
Examples:
- 1996: Divisible by 4, not by 100 $\implies$ Leap Year.
- 2000: Divisible by 4, by 100, AND by 400 $\implies$ Leap Year.
- 2004: Divisible by 4, not by 100 $\implies$ Leap Year.
- 1900: Divisible by 4, by 100, BUT NOT by 400 $\implies$ NOT a Leap Year (Normal Year).
- 2024: Divisible by 4, not by 100 $\implies$ Leap Year.
- 2100: Divisible by 4, by 100, BUT NOT by 400 $\implies$ NOT a Leap Year (Normal Year).
Number of Odd Days in a Year:
- Normal Year (365 days): $365 \div 7 = 52$ weeks and 1 day. Remainder = 1 odd day.
- Leap Year (366 days): $366 \div 7 = 52$ weeks and 2 days. Remainder = 2 odd days.
The extra day in a leap year contributes one extra odd day compared to a normal year.
Number of Odd Days in a Century:
To find the number of odd days in a period of 100, 200, 300, or 400 years, we count the number of leap years and normal years within that period.
Consider 100 consecutive years (e.g., from year 1 to 100, or 1901 to 2000). According to the rules:
- Number of years divisible by 4: 25 (4, 8, ..., 100)
- Number of years divisible by 100: 1 (100)
- Number of years divisible by 400: 0 (if the 100-year period does not contain a multiple of 400, e.g., 1901-2000)
Number of leap years in 100 years (excluding the case where the century ends with a multiple of 400) $= 25 - 1 = 24$.
Number of normal years in 100 years $= 100 - 24 = 76$.
Total odd days in 100 years $= (\text{Odd days in 76 normal years}) + (\text{Odd days in 24 leap years}) = (76 \times 1) + (24 \times 2) = 76 + 48 = 124$ odd days.
Number of odd days in 124 days $= 124 \div 7$. $124 = 17 \times 7 + 5$. Remainder $= 5$.
So, 100 years have 5 odd days.
Now consider longer century periods:
- 100 years: 5 odd days
- 200 years: Total odd days = $2 \times 5 = 10$. $10 \div 7$, remainder $= 3$ odd days.
- 300 years: Total odd days = $3 \times 5 = 15$. $15 \div 7$, remainder $= 1$ odd day.
- 400 years: Total odd days = $4 \times 5 + 1$ (The extra '1' is for the year 400, which is a leap year according to the rule. If we consider 1 to 400, years 100, 200, 300 are not leap years, but 400 is. Number of leap years in 400 years = $100 - 3 = 97$. Total odd days = $97 \times 2 + (400-97) \times 1 = 194 + 303 = 497$. $497 \div 7 = 71$ with remainder 0. OR $4 \times 5$ odd days from 4x100 year blocks $+1$ for the extra leap year in 400 years (relative to 100, 200, 300 calculation) $= 20+1 = 21$. $21 \div 7$, remainder $= 0$.
The number of odd days in a century:
| Period | Odd Days |
|---|---|
| 100 years | 5 |
| 200 years | 3 |
| 300 years | 1 |
| 400 years | 0 |
The cycle of odd days repeats every 400 years.
Competitive Exam Notes:
Understanding Odd Days and Leap Years is fundamental to solving calendar problems.
- Odd Days: Remainder when total days are divided by 7. Determines the shift in the day of the week.
- Leap Year Rule: Divisible by 4, but century years (divisible by 100) only if divisible by 400.
- Odd Days per Year: Normal year = 1 odd day, Leap year = 2 odd days.
- Odd Days per Century: Memorize or quickly derive: 100 yrs = 5, 200 yrs = 3, 300 yrs = 1, 400 yrs = 0. This pattern repeats.
- Day Codes: A standard mapping (often 0=Sunday, 1=Monday, etc.) is used for the final number of odd days.
Finding the Day of the Week for a Given Date
A common type of calendar problem involves determining the day of the week corresponding to a specific date (past or future). This is done by calculating the total number of 'odd days' from a known reference point to the given date. A universally accepted reference point is the beginning of the Christian Era, i.e., 1st January of the year 0001.
Reference Days and Codes:
The total number of odd days calculated from the reference point (1st Jan 0001) is mapped to a specific day of the week. A standard mapping assigns codes to the days based on the remainder obtained when the total odd days are divided by 7:
| Number of Odd Days (Remainder when divided by 7) | Day of the Week |
|---|---|
| 0 | Sunday |
| 1 | Monday |
| 2 | Tuesday |
| 3 | Wednesday |
| 4 | Thursday |
| 5 | Friday |
| 6 | Saturday |
Note that 1st January 0001 is considered a Monday (1 odd day from an imaginary 0th day). Sometimes, problems might use a different reference point or mapping, but this is the standard convention for calculating historical/future dates.
Steps to Find the Day for a Given Date:
Let the given date be D day of M month of Year YYYY.
1. Calculate odd days in the completed centuries before Year YYYY: Find the number of odd days in (YYYY - 1) years. Break down (YYYY - 1) into centuries (multiples of 400, then 100, 200, 300) and the remaining years. Use the odd days per century table (400 yrs -> 0, 300 yrs -> 1, 200 yrs -> 3, 100 yrs -> 5).
2. Calculate odd days in the completed years of the current century: Find the number of odd days in the years from the start of the current century up to (YYYY - 1). This is the number of years remaining after accounting for full centuries in step 1. Calculate the number of leap years in this period and the number of normal years. Sum their odd days (Leap years $\times 2 +$ Normal years $\times 1$).
3. Calculate odd days in the months and days of the given year: Find the number of odd days from the 1st of January of Year YYYY up to the given date (D day of M month). Sum the odd days for each completed month before month M, and add the number of days D in month M. Determine if Year YYYY is a leap year to correctly count odd days for February.
4. Sum Total Odd Days: Add the total number of odd days from steps 1, 2, and 3.
5. Find the Remainder: Divide the total number of odd days by 7 and find the remainder.
6. Map to Day of the Week: Use the remainder (day code) and the reference table to find the corresponding day of the week.
Example Calculation: Finding the Day for 26th January 1950 (Indian Republic Day)
We need to find the day on 26th January 1950.
Period elapsed before 26th Jan 1950 = 1949 full years + days from 1st Jan 1950 to 26th Jan 1950.
1. Odd days in 1949 years:
1949 years = 1600 years + 300 years + 49 years.
Odd days in 1600 years ($4 \times 400$) $= 0$ odd days.
Odd days in 300 years $= 1$ odd day.
Now, odd days in the remaining 49 years (from 1901 to 1949).
Number of leap years from 1901 to 1949 = Number of years divisible by 4 in this period (1904, 1908, ..., 1948).
Number of leap years $=\lfloor\frac{1949}{4}\rfloor - \lfloor\frac{1900}{4}\rfloor = 487 - 475 = 12$. (Years are 1904, 1908, ..., 1944, 1948 - 12 leap years).
Number of normal years in 49 years $= 49 - 12 = 37$.
Odd days in 37 normal years $= 37 \times 1 = 37$.
Odd days in 12 leap years $= 12 \times 2 = 24$.
Total odd days in 49 years $= 37 + 24 = 61$.
Number of odd days in 61 days $= 61 \div 7$. $61 = 8 \times 7 + 5$. Remainder $= 5$ odd days.
Total odd days in 1949 years = Odd days in 1600 yrs + Odd days in 300 yrs + Odd days in 49 yrs
Odd days in 1949 years $= 0 + 1 + 5 = 6$ odd days.
2. Odd days in the year 1950 up to 26th January:
1950 is a normal year (not divisible by 4).
We need odd days from Jan 1st to Jan 26th, 1950.
Number of days in January up to 26th $= 26$ days.
Number of odd days in 26 days $= 26 \div 7$. $26 = 3 \times 7 + 5$. Remainder $= 5$ odd days.
Odd days in 1950 up to Jan 26 $= 5$ odd days.
3. Total Odd Days:
Total odd days from 1st Jan 0001 to 26th Jan 1950 = Odd days in 1949 years + Odd days in 1950 up to Jan 26
Total odd days $= 6 + 5 = 11$ odd days.
Find the final remainder when total odd days are divided by 7:
$11 \div 7$. $11 = 1 \times 7 + 4$. Remainder $= 4$ odd days.
4. Map to Day of the Week:
Using the day code table (0=Sunday, 1=Monday, ...):
4 odd days correspond to Thursday.
So, 26th January 1950 was a $\boldsymbol{Thursday}$.
Competitive Exam Notes:
Finding the day of the week is a common calendar problem. Practice the systematic approach of calculating odd days.
- Reference: Start from 1st Jan 0001 (Monday = 1 odd day).
- Odd Days Calculation: Sum odd days from centuries, completed years in the current century, and months/days in the current year.
- Odd Days per Century: Remember 400->0, 300->1, 200->3, 100->5.
- Odd Days per Year: Normal year->1, Leap year->2.
- Odd Days per Month: Memorize (31 days->3, 30 days->2, Feb Normal->0, Feb Leap->1).
- Leap Year Check: Apply the rules correctly (Divisible by 4, century exception unless divisible by 400).
- Final Remainder: Divide total odd days by 7. The remainder gives the day code (0=Sun, 1=Mon, ...).
- Practice: Practice with various dates to become efficient in calculating odd days.
Calendar Repetition Problems
Calendar repetition problems involve finding the next year (or a subsequent year) that will have the exact same calendar as a given year. This means that every date in the repeating year falls on the same day of the week as it did in the original year. The key condition for a calendar to repeat is that the total number of odd days between the end of the original year and the end of the year preceding the repeating year must be a multiple of 7 (i.e., the cumulative sum of odd days is $0 \pmod 7$).
Condition for Calendar Repetition:
The calendar of Year A will be identical to the calendar of Year B (where B > A) if:
1. Both Year A and Year B are of the same type (both are Normal Years or both are Leap Years).
2. The total number of odd days accumulated from the end of Year A up to the end of Year B-1 is a multiple of 7 (sum $\equiv 0 \pmod 7$).
To find the repeating year, we calculate the cumulative number of odd days for consecutive years starting from the year immediately following the given year, until the cumulative sum becomes a multiple of 7. Let the given year be $Y$. We calculate odd days for $Y+1, Y+2, \dots$, summing them up. When the cumulative sum is a multiple of 7, say after $n$ years (from $Y+1$ up to $Y+n$), then the day of the week on January 1st of year $Y+n+1$ will be the same as January 1st of year $Y+1$. If $Y+n+1$ is the same type of year as $Y+1$, the calendar repeats.
However, working with the given year $Y$ itself is simpler. The calendar of year Y repeats with the calendar of year Y+n if the sum of odd days from the end of year Y to the end of year Y+n-1 is a multiple of 7. Or, if the sum of odd days of the years Y, Y+1, ..., Y+n-1 is a multiple of 7. This is incorrect. The sum of odd days of the years from Year A to Year B-1 determines the day of the week of Jan 1st of Year B relative to Jan 1st of Year A.
Let's use the cumulative odd days from the end of the original year.
Calculating Repetition Intervals:
We list the number of odd days for consecutive years starting from the year *after* the given year and find the cumulative sum modulo 7.
Example: When does a Normal Year (N) repeat? Let's say we start counting from a Normal year (1 odd day).
| Year (relative to start) | Type | Odd Days | Cumulative Odd Days (from end of start year) | Cumulative $\pmod 7$ |
|---|---|---|---|---|
| +1 | N | 1 | 1 | 1 |
| +2 | N | 1 | $1+1=2$ | 2 |
| +3 | N | 1 | $2+1=3$ | 3 |
| +4 | L | 2 | $3+2=5$ | 5 |
| +5 | N | 1 | $5+1=6$ | 6 |
| +6 | N | 1 | $6+1=7$ | 0 |
The cumulative odd days become $0 \pmod 7$ after 6 years. This means the day of the week on January 1st, Year (Start + 7) is the same as January 1st, Year (Start + 1).
Let's check specific years:
- Consider a Normal year immediately after a Leap Year (e.g., 2001). Years from 2002: 2002(1), 2003(1), 2004(2), 2005(1), 2006(1), 2007(1). Sum = 7. This means Jan 1, 2008 is same day as Jan 1, 2002. Let's check the pattern relative to the *given* year. Calendar of Year Y repeats in Year Y+n if sum of odd days from Y to Y+n-1 is 0 mod 7. No, this is not the correct logic.
The calendar of year X repeats in year Y if Jan 1st of X and Jan 1st of Y are the same day AND both X and Y are the same type (Normal/Leap).
Let's use the cumulative sum of odd days starting from Jan 1st of the given year relative to Jan 1st of subsequent years.
Jan 1 Year Y vs Jan 1 Year Y+n. We need the number of odd days from Jan 1 Year Y to Dec 31 Year Y+n-1 to be $0 \pmod 7$. Summing the odd days of years Y, Y+1, ..., Y+n-1.
Example: Calendar of 2001 (Normal). Sum of odd days from 2001 to 2001+n-1.
2001(1), 2002(1), 2003(1), 2004(2), 2005(1), 2006(1). Sum = 7. So Dec 31, 2006 is same day as Dec 31, 2001 + 7 odd days = Same day. So Jan 1, 2007 is same day as Jan 1, 2002 + 1 day. This means Jan 1, 2007 is the same day as Jan 1, 2001 + sum of odd days of 2001, 2002, 2003, 2004, 2005, 2006 = 7 odd days = same day. So Jan 1, 2007 is the same day as Jan 1, 2001. Both are normal years. So 2001 calendar repeats in 2007. This is a 6-year interval. This happens for a Normal year following a Leap Year.
Consider a Normal year two years after a Leap Year (e.g., 2003). Sum odd days from 2003:
2003(1), 2004(2), 2005(1), 2006(1), 2007(1), 2008(2), 2009(1), 2010(1), 2011(1), 2012(2). Sum = 13. No. 2003(1), 2004(2), 2005(1), 2006(1), 2007(1), 2008(2), 2009(1), 2010(1), 2011(1), 2012(2), 2013(1). Sum = 14. Sum of odd days from 2003 to 2013 = 14. So Jan 1, 2014 is same day as Jan 1, 2003. Both are normal years. Calendar repeats in 2014. This is an 11-year interval (2014 - 2003 = 11). This happens for a Normal year two years after a Leap Year.
Consider a Normal year three years after a Leap Year (e.g., 2004+3=2007). Sum odd days from 2007:
2007(1), 2008(2), 2009(1), 2010(1), 2011(1), 2012(2), 2013(1), 2014(1), 2015(1), 2016(2). Sum = 13. No. 2007(1), 2008(2), ..., 2017(1). Sum of odd days from 2007 to 2017 (11 years): Normal years (2007,09,10,11,13,14,15,17) = 8*1 = 8. Leap years (2008,12,16) = 3*2 = 6. Total = 14. So Jan 1, 2018 is same day as Jan 1, 2007. Both are normal years. Calendar repeats in 2018. This is an 11-year interval (2018 - 2007 = 11). This happens for a Normal year three years after a Leap Year.
Consider a Leap Year (e.g., 2000). Sum odd days from 2000:
2000(2), 2001(1), 2002(1), 2003(1), 2004(2), 2005(1), ..., 2027(1). Sum of odd days from 2000 to 2027 (28 years). Number of leap years = 7 (2000, 04, ..., 24, 28 - NO, 2000,04,...,24 is 7 leap years). Yes, 7 leap years. Normal years = 21. Total odd days = $7 \times 2 + 21 \times 1 = 14 + 21 = 35$. $35 \div 7$, remainder 0. So Jan 1, 2028 is same day as Jan 1, 2000. Both are leap years. Calendar repeats in 2028. This is a 28-year interval.
Summary of Repetition Intervals (ignoring century exceptions for now):
- Leap Year: Repeats after 28 years.
- Normal Year immediately after a Leap Year (e.g., Year 4k+1): Repeats after 6 years.
- Normal Year 2 years after a Leap Year (e.g., Year 4k+2): Repeats after 11 years.
- Normal Year 3 years after a Leap Year (e.g., Year 4k+3): Repeats after 11 years.
Exceptions occur when crossing a century year that is not a leap century (e.g., 1900, 2100). These years break the 28-year leap cycle. For instance, the calendar of 1897 (Normal, 4k+1) repeats after 6 years in 1903. The calendar of 1900 (Normal, but breaks 28yr cycle) does not repeat after 6 years. The calendar of 1901 (Normal, 4k+1) repeats after 12 years (in 1913, not 6 years) because 1900 was not a leap year. Similarly for 1902, 1903 etc.
For competitive exams, the simple 6, 11, 11, 28 year pattern usually holds unless a non-leap century is involved in the interval.
Example 1. The calendar for the year 2017 will be the same as which year?
Answer:
The year 2017 is a Normal Year (2017 is not divisible by 4).
The preceding year is 2016, which is a Leap Year ($2016 \div 4 = 504$).
So, 2017 is a normal year immediately following a leap year (form 4k+1).
According to the pattern, a normal year immediately following a leap year repeats after 6 years.
The calendar for 2017 will repeat in the year $2017 + 6 = 2023$.
Let's verify by checking the cumulative odd days from 2017 to 2022:
- 2017 (N): 1
- 2018 (N): 1 (Cumulative 2)
- 2019 (N): 1 (Cumulative 3)
- 2020 (L): 2 (Cumulative 5)
- 2021 (N): 1 (Cumulative 6)
- 2022 (N): 1 (Cumulative 7 $\equiv 0 \pmod 7$)
The cumulative sum of odd days from the end of 2017 to the end of 2022 is 7 (a multiple of 7). This means the day of the week on 1st January 2023 is the same as 1st January 2017.
Also, 2023 is a Normal Year (2023 is not divisible by 4).
Since both 2017 and 2023 are normal years and start on the same day, their calendars are identical.
The calendar for the year 2017 will be the same as the year $\boldsymbol{2023}$.
Example 2. The calendar for the year 2025 will be the same as which year?
Answer:
The year 2025 is a Normal Year (2025 is not divisible by 4).
The preceding year is 2024, which is a Leap Year ($2024 \div 4 = 506$).
So, 2025 is a normal year immediately following a leap year (form 4k+1).
According to the pattern, a normal year immediately following a leap year repeats after 6 years.
The calendar for 2025 will repeat in the year $2025 + 6 = 2031$.
Let's verify by checking the cumulative odd days from 2025 to 2030:
- 2025 (N): 1
- 2026 (N): 1 (Cumulative 2)
- 2027 (N): 1 (Cumulative 3)
- 2028 (L): 2 (Cumulative 5)
- 2029 (N): 1 (Cumulative 6)
- 2030 (N): 1 (Cumulative 7 $\equiv 0 \pmod 7$)
The cumulative sum of odd days from the end of 2025 to the end of 2030 is 7. This means Jan 1st, 2031 is the same day as Jan 1st, 2025.
Also, 2031 is a Normal Year (2031 is not divisible by 4).
Since both 2025 and 2031 are normal years and start on the same day, their calendars are identical.
The calendar for the year 2025 will be the same as the year $\boldsymbol{2031}$.
Competitive Exam Notes:
Calendar repetition problems can often be solved by knowing the standard intervals for repetition based on the type of year.
- Leap Year (LY): Repeats after 28 years (e.g., 2008 $\to$ 2036).
- Year after Leap Year (LY+1): Repeats after 6 years (e.g., 2001 $\to$ 2007, 2005 $\to$ 2011, 2017 $\to$ 2023, 2021 $\to$ 2027, 2025 $\to$ 2031).
- Year 2 years after Leap Year (LY+2): Repeats after 11 years (e.g., 2002 $\to$ 2013, 2006 $\to$ 2017).
- Year 3 years after Leap Year (LY+3): Repeats after 11 years (e.g., 2003 $\to$ 2014, 2007 $\to$ 2018).
- Century Exceptions: Be aware that these patterns can be altered when crossing a century year that is not a leap century (e.g., 1900, 2100). For example, 1901 (LY+1) repeats in $1901+12 = 1913$ (not 6 years). However, for typical exam questions within the 20th and 21st centuries (not crossing 1900 or 2100), the standard 6, 11, 11, 28 pattern is usually applicable.
- Verification: If unsure, calculate the cumulative odd days from the end of the original year until the end of the year preceding the suspected repeating year. If the sum is a multiple of 7 AND the suspected repeating year is the same type (N or L), the calendar repeats.
Solving Calendar Problems
Solving calendar problems involves applying the concepts of leap years and odd days to determine the day of the week for specific dates, calculate the day after a certain number of days, or understand patterns in the calendar. These problems require careful calculation of odd days and logical reasoning.
Example 1. If today is Wednesday, what will be the day after 50 days?
Answer:
We need to find the number of odd days in a period of 50 days.
Number of odd days = Remainder when 50 is divided by 7.
$50 \div 7 = 7$ weeks with a remainder.
$50 = 7 \times 7 + 1$.
Remainder $= 1$.
So, there is 1 odd day in a period of 50 days.
The starting day is Wednesday. We need to count 1 day forward from Wednesday.
Wednesday + 1 day = Thursday.
The day after 50 days will be $\boldsymbol{Thursday}$.
Example 2. If 1st January 2022 was a Saturday, what was the day of the week on 1st January 2023?
Answer:
We need to determine the number of odd days between 1st January 2022 and 1st January 2023.
This period is exactly one year, from Jan 1st to Jan 1st of the next year.
We need to determine if the year 2022 is a leap year to know the number of days in this period.
2022 is not divisible by 4 ($2022 \div 4 = 505.5$).
So, 2022 is a Normal Year (365 days).
The period from 1st January 2022 to 1st January 2023 spans exactly 365 days.
Number of odd days in 365 days = Remainder when 365 is divided by 7.
$365 \div 7 = 52$ weeks with a remainder.
$365 = 52 \times 7 + 1$.
Remainder $= 1$.
So, there is 1 odd day between 1st January 2022 and 1st January 2023.
Given that 1st January 2022 was a Saturday. We add 1 odd day to Saturday.
Saturday + 1 day = Sunday.
The day of the week on 1st January 2023 was $\boldsymbol{Sunday}$.
Example 3. If 20th July 2024 was a Saturday, what was the day of the week on 20th July 2025?
Answer:
We need to determine the number of odd days between 20th July 2024 and 20th July 2025.
This period is exactly one year, from July 20th to July 20th of the next year.
We need to determine if the year 2024 is a leap year because the period includes February 2025.
2024 is divisible by 4 ($2024 \div 4 = 506$). It is not a century year divisible by 100.
So, 2024 is a Leap Year (366 days).
The period from 20th July 2024 to 20th July 2025 spans exactly 365 days (it does not include the leap day of 2024, which is Feb 29, 2024. It does not include Feb 29, 2025 as 2025 is normal year).
The year from 20th July 2024 to 20th July 2025 includes Feb 2025 (28 days) but not Feb 2024 (29 days). This is a normal 365-day period.
Number of odd days in 365 days = Remainder when 365 is divided by 7.
$365 = 52 \times 7 + 1$.
Remainder $= 1$.
So, there is 1 odd day between 20th July 2024 and 20th July 2025.
Given that 20th July 2024 was a Saturday. We add 1 odd day to Saturday.
Saturday + 1 day = Sunday.
The day of the week on 20th July 2025 was $\boldsymbol{Sunday}$.
Competitive Exam Notes:
Problems asking for the day of the week for a date within a few years of a known date are common. The key is to count the number of odd days in the intervening period.
- Period between two dates: Count the number of days between the two dates. This might span full years and parts of years.
- Full Year Period: If the period is exactly one year (e.g., from Date X Year Y to Date X Year Y+1), the number of odd days is 1 if the period includes exactly one February of 28 days (Normal Year) and 2 if the period includes exactly one February of 29 days (Leap Year). Check if the leap day (Feb 29) falls within the period.
- Multiple Years: For a period of several years, sum the odd days contributed by each year (1 for normal, 2 for leap) in that period.
- Within the Same Year: Count the number of days or sum the odd days of the months between the two dates in the same year.
- Counting Forward/Backward: Add odd days to the starting day for future dates, subtract for past dates. Remember to count cyclically through the days of the week.