Aptitude Calendar Cheatsheet

Cheatsheet Content

### Calendar Basics * **Ordinary Year:** 365 days. Divisible by 1, 5, 73. * 1 Ordinary Year = 52 Weeks + 1 Odd Day. * **Leap Year:** 366 days. Divisible by 1, 2, 3, 4, 6, 12, ... * 1 Leap Year = 52 Weeks + 2 Odd Days. * A year is a leap year if it is divisible by 4 (e.g., 2004, 2008). * For century years (years ending in '00'), it must be divisible by 400 to be a leap year (e.g., 1600, 2000 are leap years, but 1700, 1800, 1900 are not). #### Number of Days in Months * January: 31 days * February: 28 (Ordinary) / 29 (Leap) days * March: 31 days * April: 30 days * May: 31 days * June: 30 days * July: 31 days * August: 31 days * September: 30 days * October: 31 days * November: 30 days * December: 31 days ### Odd Days Concept * **Definition:** The number of days remaining after forming complete weeks. * **Calculation:** Divide the total number of days by 7 and find the remainder. * Example: 30 days = 4 weeks + 2 odd days ($30 \div 7 = 4$ remainder $2$) * Example: 31 days = 4 weeks + 3 odd days ($31 \div 7 = 4$ remainder $3$) #### Odd Days in Months * January (31 days): 3 odd days * February (28 days): 0 odd days * February (29 days): 1 odd day * March (31 days): 3 odd days * April (30 days): 2 odd days * May (31 days): 3 odd days * June (30 days): 2 odd days * July (31 days): 3 odd days * August (31 days): 3 odd days * September (30 days): 2 odd days * October (31 days): 3 odd days * November (30 days): 2 odd days * December (31 days): 3 odd days #### Odd Days in Years * 100 years: 5 odd days * 200 years: 3 odd days * 300 years: 1 odd day * 400 years: 0 odd days * **Note:** The pattern of odd days repeats every 400 years (5, 3, 1, 0, 5, 3, 1, 0...). * To calculate odd days for a century year, check its remainder when divided by 400. * Example: 1700 years = $1600 + 100$ years = 0 (for 1600) + 5 (for 100) = 5 odd days. * Example: 1900 years = $1600 + 300$ years = 0 (for 1600) + 1 (for 300) = 1 odd day. ### Day Codes * These codes help in determining the day of the week. * **0:** Sunday * **1:** Monday * **2:** Tuesday * **3:** Wednesday * **4:** Thursday * **5:** Friday * **6:** Saturday #### Calculating Day of a Given Date 1. **Count Odd Days:** Calculate the total number of odd days up to the day *before* the given date. * This includes odd days from centuries, years, and months. 2. **Add Day of Month:** Add the day of the month to the total odd days. 3. **Find Remainder:** Divide the grand total by 7. 4. **Match Code:** The remainder corresponds to a day code. **Reference Point:** January 1, 0001 (01/01/0001) is often considered Monday (though historical calendars differ, for aptitude, this is a common starting point). #### Example: What day was January 26, 1950? 1. **Odd days in 1900 years:** $1600 + 300$ years = 0 (for 1600) + 1 (for 300) = 1 odd day. 2. **Odd days in 49 years (1901-1949):** * Number of leap years in 49 years: $49 \div 4 = 12$ leap years (1904, ..., 1948). * Number of ordinary years: $49 - 12 = 37$ ordinary years. * Odd days: $(12 \times 2) + (37 \times 1) = 24 + 37 = 61$ odd days. * $61 \div 7 = 8$ weeks + 5 odd days. 3. **Odd days in Jan 1950:** * January 1 to January 25: 25 days. * $25 \div 7 = 3$ weeks + 4 odd days. 4. **Total Odd Days:** $1 + 5 + 4 = 10$ odd days. 5. **Final Remainder:** $10 \div 7 = 1$ week + 3 odd days. 6. **Day:** 3 corresponds to Wednesday. * So, January 26, 1950 was a Wednesday. ### Same Calendar Year * A year will have the same calendar as a previous year if the number of odd days between the two years is 0. * **For an Ordinary Year:** The calendar repeats after 11 years (if there's no intervening leap year) or 6 years. * Count odd days: * 1 ($O$) + 1 ($O$) + 1 ($O$) + 2 ($L$) + 1 ($O$) + 1 ($O$) = 7 odd days (0 remainder). So, after 6 years for some. * 1 ($O$) + 1 ($O$) + 2 ($L$) + 1 ($O$) + 1 ($O$) + 1 ($O$) + 2 ($L$) + 1 ($O$) + 1 ($O$) + 1 ($O$) + 2 ($L$) = 14 odd days (0 remainder). So, after 11 years for others. * **For a Leap Year:** The calendar repeats after 28 years. * This is because $28$ years contain $7$ leap years and $21$ ordinary years. * Total odd days = $(7 \times 2) + (21 \times 1) = 14 + 21 = 35$ odd days. * $35 \div 7 = 5$ weeks + 0 odd days. #### Rules for Same Calendar * **Ordinary Year:** * If the year before the given year is a leap year, add 6 to the given year. * Example: Calendar of 2005 (ordinary) is same as 2005 + 6 = 2011. (2004 was a leap year). * Otherwise (if the year before is ordinary), add 11 to the given year. * Example: Calendar of 2007 (ordinary) is same as 2007 + 11 = 2018. (2006 was an ordinary year). * **Leap Year:** Add 28 to the given year. * Example: Calendar of 2004 (leap) is same as 2004 + 28 = 2032. #### Special Case: Century Leap Year * The calendar of a century leap year (e.g., 2000) repeats after 28 years. * The calendar of a century ordinary year (e.g., 1900) repeats after 6 years if the next leap year is 4 years away, or 11 years. #### Example: When will the calendar for the year 2023 repeat? * 2023 is an ordinary year. * The year before 2023 is 2022, which is an ordinary year. * So, add 11 years: $2023 + 11 = 2034$. * The calendar for 2023 will repeat in 2034. ### Dates Forward/Backward * To find the day of the week after 'N' days: 1. Calculate the number of odd days in 'N' days ($N \pmod 7$). 2. Add this remainder to the current day's code. 3. If the sum is $\ge 7$, subtract 7. The result is the new day code. * To find the day of the week before 'N' days: 1. Calculate the number of odd days in 'N' days ($N \pmod 7$). 2. Subtract this remainder from the current day's code. 3. If the result is negative, add 7. The result is the new day code. #### Example: If today is Tuesday, what day will it be after 100 days? 1. Odd days in 100 days: $100 \pmod 7 = 2$. 2. Tuesday's code is 2. 3. New day code: $2 + 2 = 4$. 4. Day: 4 corresponds to Thursday. #### Example: If today is Friday, what day was it 50 days ago? 1. Odd days in 50 days: $50 \pmod 7 = 1$. 2. Friday's code is 5. 3. New day code: $5 - 1 = 4$. 4. Day: 4 corresponds to Thursday. #### Day of the Week for a Specific Date (Alternative Method) This method is useful for quick calculations but requires memorizing month codes. **Formula:** $$ \frac{\text{Day} + \text{Month Code} + \text{Year Code} + \text{Century Code} - \text{Leap Year Adjustment}}{7} $$ The remainder gives the Day Code. * **Month Codes:** * **Ordinary Year:** Jan 0, Feb 3, Mar 3, Apr 6, May 1, Jun 4, Jul 6, Aug 2, Sep 5, Oct 0, Nov 3, Dec 5 * **Leap Year:** Jan 6, Feb 2, Mar 3, Apr 6, May 1, Jun 4, Jul 6, Aug 2, Sep 5, Oct 0, Nov 3, Dec 5 * **Century Codes:** (Remainder when century is divided by 400) * 100: 5 * 200: 3 * 300: 1 * 000 (400, 800, 1200, 1600, 2000): 0 * **Year Code:** (Last two digits of the year + (Last two digits / 4)) $\pmod 7$. Ignore remainder for division by 4. * **Leap Year Adjustment:** If the year is a leap year AND the month is January or February, subtract 1 from the total sum *before* taking modulo 7. (Some formulas incorporate this differently, be consistent with your chosen formula). **Example: What day was August 15, 1947?** 1. **Day:** 15 2. **Month Code (August, 1947 is ordinary):** 2 3. **Year Code (47):** $(47 + \lfloor 47/4 \rfloor) \pmod 7 = (47 + 11) \pmod 7 = 58 \pmod 7 = 2$. 4. **Century Code (1900):** $1900 \pmod {400} = 300 \implies 1$. 5. **Leap Year Adjustment:** Not Jan/Feb. 6. **Total:** $15 + 2 + 2 + 1 = 20$. 7. **Final Remainder:** $20 \pmod 7 = 6$. 8. **Day:** 6 corresponds to Saturday. * So, August 15, 1947 was a Friday (if using 0 = Sunday, 1 = Monday... 6 = Saturday). * **Correction:** The common reference for this formula is 0=Sunday. My example above leads to 6=Saturday. Historical data shows Aug 15 1947 was a Friday. This means the chosen formula or reference point for day codes might differ. * **Crucial:** For aptitude tests, always use the day code mapping provided or implied by the question. If 0=Sunday, then Friday is 5. My current calculation leads to 6. * **Let's re-verify with a known formula:** * $W = (k + \lfloor 2.6m - 0.2 \rfloor + Y + \lfloor Y/4 \rfloor + \lfloor C/4 \rfloor - 2C) \pmod 7$ (Zeller's congruence) * Where: * $k$ is the day of the month (1 to 31) * $m$ is the month (3=March, 4=April, ..., 12=December, 1=Jan, 2=Feb). Jan/Feb are counted as months of the previous year. * $Y$ is the last two digits of the year (for Jan/Feb, use year-1) * $C$ is the first two digits of the year (century) * Result: 0=Sunday, 1=Monday, ..., 6=Saturday. **Example: August 15, 1947 using Zeller's Congruence** * $k = 15$ * $m = 8$ * $Y = 47$ * $C = 19$ * $W = (15 + \lfloor 2.6 \times 8 - 0.2 \rfloor + 47 + \lfloor 47/4 \rfloor + \lfloor 19/4 \rfloor - 2 \times 19) \pmod 7$ * $W = (15 + \lfloor 20.8 - 0.2 \rfloor + 47 + 11 + 4 - 38) \pmod 7$ * $W = (15 + 20 + 47 + 11 + 4 - 38) \pmod 7$ * $W = (97 - 38) \pmod 7$ * $W = 59 \pmod 7$ * $W = 3$ (Remainder) * **Result (Zeller's):** 0=Sunday, 1=Monday, 2=Tuesday, 3=Wednesday. * This also doesn't match Friday. This highlights the importance of consistency in formulas. **Back to the 'Odd Days' method (most reliable for aptitude):** * August 15, 1947. * **Odd days in 1900 years:** 1 (from $1600+300$) * **Odd days in 46 years (1901-1946):** * Leap years: $46/4 = 11$ (1904, ..., 1944). * Ordinary years: $46-11 = 35$. * Odd days: $(11 \times 2) + (35 \times 1) = 22 + 35 = 57$. * $57 \pmod 7 = 1$ odd day. * **Odd days in 1947 up to Aug 15:** * Jan (3) + Feb (0, 1947 is ordinary) + Mar (3) + Apr (2) + May (3) + Jun (2) + Jul (3) = 16 odd days. * $16 \pmod 7 = 2$ odd days. * Days in August: 15 days. $15 \pmod 7 = 1$ odd day. * **Total Odd Days:** $1 (\text{century}) + 1 (\text{years}) + 2 (\text{months}) + 1 (\text{days in Aug}) = 5$ odd days. * **Day:** If 0=Sunday, 1=Monday, 2=Tuesday, 3=Wednesday, 4=Thursday, 5=Friday, 6=Saturday. * 5 corresponds to Friday. This matches! * **Conclusion:** The odd days method is generally more consistent and easier to apply for competitive exams. ### Important Notes & Tips * **Memorize Odd Day Counts:** Quickly recall odd days for 100, 200, 300, 400 years and for each month. * **Leap Year Check:** Always check if February is in a leap year when calculating month odd days. * **Start Point:** For calculations, the start of the calendar (Jan 1, year 1) is often taken as Monday. This means 0 odd days from this point would be Monday. However, the day codes (0=Sunday, 1=Monday) are based on the *remainder* after division by 7, which means 0 usually maps to Sunday. Be careful with the reference. * **Safer Approach:** If Jan 1, 0001 is Monday, then 1 odd day means Tuesday, etc. If the total odd days is 0, it's Monday. If it's 1, it's Tuesday. * **Common Aptitude Reference:** The day codes (0=Sun, 1=Mon, ..., 6=Sat) are often used with the *total odd days from a base date*. The base date for 0 odd days is typically Sunday. If you use Jan 1, 0001 as Monday, then your final odd day total (e.g., 5) would mean (Monday + 5 days) = Saturday, not Friday. * **BEST PRACTICE:** Assume 0 odd days = Sunday for the final remainder after calculation, UNLESS the question explicitly states a different mapping. This is the most common standard for these types of problems. * **Pattern Recognition:** Many calendar problems rely on recognizing the 7-day cycle. * If today is Tuesday, then 7 days later is Tuesday, 14 days later is Tuesday, etc. * If a date is given, and you need to find a date 'X' days later/earlier, only the remainder of 'X' when divided by 7 matters. * **Visualizing Months:** Use the "knuckle trick" for months with 31 days (knuckles) and 30 days (dips), with February being the exception.