📘 Quiz

Solution: Step 1: Calculate the total number of years between 1971 and 2001 (excluding the current day of 2001 for odd day calculation and focusing on the full years between the two dates). Number of years from end of 1971 to end of 2000 = 2000 - 1971 = 29 years. Adding 2001 makes it 30 years. Step 2: Identify the leap years within this period (from 1972 to 2000, inclusive). Leap years are: 1972, 1976, 1980, 1984, 1988, 1992, 1996, 2000. Total leap years = 8. Step 3: Calculate the number of ordinary years. Ordinary years = Total years - Leap years = 30 - 8 = 22 years. Step 4: Calculate the total odd days from Dec 9, 1971 to Dec 9, 2001. Each ordinary year has 1 odd day. Each leap year has 2 odd days. Total odd days = (22 * 1) + (8 * 2) = 22 + 16 = 38 odd days. Step 5: Find the remainder when total odd days are divided by 7: 38 ÷ 7 = 5 remainder 3. So, there are 3 odd days. Step 6: Since we are going backward from 2001 to 1971, subtract the odd days from Sunday. Sunday - 3 days = Thursday. Therefore, December 9, 1971, was a Thursday.

Solution: Step 1: Decompose the date for odd day calculation: July 16, 1776 is considered as (1775 full years + the period from Jan 1 to Jul 16, 1776). Step 2: Calculate odd days for the 1775 full years. We break it down as: 1775 = 1600 + 100 + 75. - Odd days in 1600 years = 0 (as 1600 is a multiple of 400). - Odd days in next 100 years = 5. - For the remaining 75 years (1701 to 1775): Number of leap years = 75 ÷ 4 = 18. Ordinary years = 75 - 18 = 57. Total odd days = (18 × 2) + (57 × 1) = 36 + 57 = 93. 93 ÷ 7 = 13 weeks and 2 odd days ⇒ So, 2 odd days. Total odd days from 1775 years = 0 (1600) + 5 (100) + 2 (75) = 7 ≡ 0 (mod 7). Step 3: Calculate odd days from Jan 1 to July 16, 1776. Since 1776 is a leap year, February has 29 days. Days in months from Jan to June: Jan (31), Feb (29), Mar (31), Apr (30), May (31), Jun (30) = total 182 days. Add 16 days of July ⇒ 182 + 16 = 198 days. 198 ÷ 7 = 28 weeks and 2 odd days ⇒ 2 odd days. Step 4: Add to previous total. Odd days from 1775 years = 0. Odd days in 1776 up to July 16 = 2. Total odd days = 0 + 2 = 2. Therefore, the day on July 16, 1776 was **Tuesday**.

Solution: Step 1: Understand the condition for a calendar to repeat: The total number of odd days between the initial year and the repeating year must be a multiple of 7 (i.e., 0 odd days), and both years must be of the same type (both ordinary or both leap years). Step 2: The year 1990 is an ordinary year (1 odd day). Step 3: Calculate cumulative odd days starting from the year after 1990: - 1991 (Ordinary): 1 odd day. Cumulative = 1. - 1992 (Leap): 2 odd days. Cumulative = 1 + 2 = 3. - 1993 (Ordinary): 1 odd day. Cumulative = 3 + 1 = 4. - 1994 (Ordinary): 1 odd day. Cumulative = 4 + 1 = 5. - 1995 (Ordinary): 1 odd day. Cumulative = 5 + 1 = 6. - 1996 (Leap): 2 odd days. Cumulative = 6 + 2 = 8 (i.e., 1 mod 7). (Cannot be 1996 as 1990 is ordinary, 1996 is leap). - 1997 (Ordinary): 1 odd day. Cumulative = 1 + 1 = 2. - 1998 (Ordinary): 1 odd day. Cumulative = 2 + 1 = 3. - 1999 (Ordinary): 1 odd day. Cumulative = 3 + 1 = 4. - 2000 (Leap): 2 odd days. Cumulative = 4 + 2 = 6. - 2001 (Ordinary): 1 odd day. Cumulative = 6 + 1 = 7 (i.e., 0 mod 7). Step 4: Analyze the result for 2001. The cumulative odd days from 1990 to the end of 2000 is 14, which is a multiple of 7 (14 mod 7 = 0). The year 2001 starts after 0 cumulative odd days from 1990. Both 1990 and 2001 are ordinary years. Therefore, the calendar of 2001 will be the same as that of 1990.

Solution: Step 1: Identify the year and determine if it's a leap year. The year is 2012. Since 2012 is divisible by 4, it is a leap year. This means February has 29 days. Step 2: Calculate the number of days for each month within the specified range (both inclusive). Days in February (from 3rd to 29th) = 29 - 3 + 1 = 27 days. Days in March = 31 days. Days in April (from 1st to 18th) = 18 days. Step 3: Sum the total number of days. Total days = 27 (February) + 31 (March) + 18 (April) = 76 days. Therefore, there are 76 days from February 3, 2012, to April 18, 2012 (both inclusive).

Solution: Step 1: Determine yesterday's day. If the day before yesterday was Thursday, then yesterday was Friday. Step 2: Determine today's day. If yesterday was Friday, then today is Saturday. Step 3: Determine when Sunday will be. If today is Saturday, then tomorrow will be Sunday. Therefore, Sunday will be tomorrow.

Solution: Step 1: Identify the given date and day: January 1, 2004, was a Thursday. Step 2: Determine if the year 2004 is a leap year. 2004 is divisible by 4, so it is a leap year. Step 3: Calculate the number of odd days in the year 2004. A leap year has 366 days, which is 52 weeks and 2 odd days. Since we are going from Jan 1, 2004, to Jan 1, 2005, we consider the entire year 2004. So, there are 2 odd days. Step 4: Add the odd days to the day of the week for January 1, 2004. Thursday + 2 odd days = Saturday. Therefore, January 1, 2005, will be a Saturday.

Solution: Step 1: Decompose the date for odd day calculation: August 15, 1947 is considered as (1946 full years + period from Jan 1 to Aug 15, 1947). Step 2: Calculate odd days for the 1900 full years. 1900 years = (1600 years + 300 years). Odd days in 1600 years = 0. Odd days in 300 years = 1. Total odd days for 1900 years = 1 odd day. Step 3: Calculate odd days for the remaining 46 years (1901-1946). Number of leap years in 46 years = 46 ÷ 4 = 11 leap years. Ordinary years = 46 - 11 = 35 ordinary years. Odd days from 46 years = (11 * 2) + (35 * 1) = 22 + 35 = 57 odd days. Convert to modulo 7: 57 ÷ 7 = 8 weeks and 1 odd day. Total odd days up to end of 1946 = 0 + 1 + 1 = 2 odd days. Step 4: Calculate odd days for the period from Jan 1, 1947, to Aug 15, 1947. Year 1947 is an ordinary year. Days in Jan=31 (3 odd), Feb=28 (0 odd), Mar=31 (3 odd), Apr=30 (2 odd), May=31 (3 odd), Jun=30 (2 odd), Jul=31 (3 odd), Aug=15 (1 odd). Total odd days for these months = 3+0+3+2+3+2+3+1 = 17 odd days. Convert to modulo 7: 17 ÷ 7 = 2 weeks and 3 odd days. Step 5: Sum all the odd days. Total odd days = (Odd days from 1946 years) + (Odd days in 1947 up to Aug 15) = 2 + 3 = 5 odd days. Step 6: Map the total odd days to the day of the week. 0 = Sunday, 1 = Monday, ..., 5 = Friday. 5 odd days correspond to Friday. Therefore, August 15, 1947, was a Friday.

Solution: Step 1: Calculate the number of odd days in 30 days. Divide 30 by 7 (the number of days in a week): 30 ÷ 7 = 4 with a remainder of 2. So, there are 2 odd days. Step 2: Since we are looking for a day in the past, subtract the odd days from today's day. Today is Thursday. Thursday - 2 days = Tuesday. Therefore, 30 days ago it was Tuesday.

Solution: Step 1: Decompose the date for odd day calculation: January 26, 1950 is considered as (1949 full years + period from Jan 1 to Jan 26, 1950). Step 2: Calculate odd days for the 1900 full years. 1900 years = 1600 years + 300 years. Odd days in 1600 years = 0. Odd days in 300 years = 1. Total odd days for 1900 years = 1 odd day. Step 3: Calculate odd days for the remaining 49 years (1901-1949). Number of leap years in 49 years = 49 ÷ 4 = 12 leap years (1904, 1908, ..., 1948). Number of ordinary years = 49 - 12 = 37 ordinary years. Odd days from 49 years = (12 * 2) + (37 * 1) = 24 + 37 = 61 odd days. Convert to modulo 7: 61 ÷ 7 = 8 weeks and 5 odd days. Step 4: Calculate odd days for the period from Jan 1 to Jan 26, 1950. Days in January 1950 = 26 days. Convert to modulo 7: 26 ÷ 7 = 3 weeks and 5 odd days. Step 5: Sum all the odd days. Total odd days = (Odd days from 1900 years) + (Odd days from 49 years) + (Odd days from Jan 1950) = 1 + 5 + 5 = 11 odd days. Step 6: Find the remainder when total odd days are divided by 7: 11 ÷ 7 = 1 remainder 4. So, there are 4 odd days. Step 7: Map the total odd days to the day of the week. 0 = Sunday, 1 = Monday, ..., 4 = Thursday. 4 odd days correspond to Thursday. Therefore, January 26, 1950, was a Thursday.

Solution: Step 1: Decompose the date for odd day calculation: January 1, 1901 is considered as (1900 full years + 1st day of 1901). Step 2: Calculate odd days for the 1900 full years. 1900 years = 1600 years + 300 years. Number of odd days in 1600 years = 0 (since 1600 is a multiple of 400). Number of odd days in 300 years = 1 (Standard rule: 100 years = 5 odd days, 200 years = 3 odd days, 300 years = 1 odd day, 400 years = 0 odd days). Total odd days for 1900 years = 0 + 1 = 1 odd day. Step 3: Calculate odd days for the period from Jan 1, 1901, to Jan 1, 1901. For 1st January, we count 1 day. So, 1 odd day. Step 4: Sum all the odd days. Total odd days = (Odd days from 1900 years) + (Odd days in 1901 up to Jan 1) = 1 + 1 = 2 odd days. Step 5: Map the total odd days to the day of the week. 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday. 2 odd days correspond to Tuesday. Therefore, January 1, 1901, was a Tuesday.

Solution: Step 1: Calculate the number of odd days in 59 days. Divide 59 by 7 (the number of days in a week): 59 ÷ 7 = 8 with a remainder of 3. So, there are 3 odd days. Step 2: Since we are looking for a day in the future, add the odd days to today's day. Today is Thursday. Thursday + 3 days = Sunday. Therefore, the day after 59 days will be Sunday.

Solution: Step 1: Calculate the number of days in each month within the given range (2005 is an ordinary year). Days in May (from 13th to 31st) = 31 - 13 + 1 = 19 days. Days in June = 30 days. Days in July = 31 days. Days in August (from 1st to 19th) = 19 days. Step 2: Sum the total number of days. Total days = 19 (May) + 30 (June) + 31 (July) + 19 (August) = 99 days. Step 3: Calculate the number of odd days from the total days. Divide the total days by 7: 99 ÷ 7 = 14 weeks with a remainder of 1. Therefore, there is 1 odd day in the given period.

Solution: Step 1: Decompose the date for odd day calculation: September 21, 1987 is considered as (1986 full years + period from Jan 1 to Sep 21, 1987). Step 2: Calculate odd days for the 1986 full years. 1986 years = (1600 years + 300 years + 86 years). Odd days in 1600 years = 0. Odd days in 300 years = 1. For 86 years (1901-1986): Leap years in 86 years = 86 ÷ 4 = 21 leap years. Ordinary years = 86 - 21 = 65 ordinary years. Odd days from 86 years = (21 * 2) + (65 * 1) = 42 + 65 = 107 odd days. Convert to modulo 7: 107 ÷ 7 = 15 weeks and 2 odd days. Total odd days up to end of 1986 = 0 + 1 + 2 = 3 odd days. Step 3: Calculate odd days for the period from Jan 1, 1987, to Sep 21, 1987. Year 1987 is an ordinary year. Days in Jan=31 (3 odd), Feb=28 (0 odd), Mar=31 (3 odd), Apr=30 (2 odd), May=31 (3 odd), Jun=30 (2 odd), Jul=31 (3 odd), Aug=31 (3 odd), Sep=21 (0 odd). Total odd days for these months = 3+0+3+2+3+2+3+3+0 = 19 odd days. Convert to modulo 7: 19 ÷ 7 = 2 weeks and 5 odd days. Step 4: Sum all the odd days. Total odd days = (Odd days from 1986 years) + (Odd days in 1987 up to Sep 21) = 3 + 5 = 8 odd days. Step 5: Find the remainder when total odd days are divided by 7: 8 ÷ 7 = 1 remainder 1. So, there is 1 odd day. Step 6: Map the total odd days to the day of the week. 1 odd day corresponds to Monday. Therefore, September 21, 1987, was a Monday.

Solution: Step 1: Identify the given information: December 1, 1991, is a Sunday. Step 2: Determine the date of the first Tuesday in December 1991. If Dec 1 is Sunday, then Dec 2 is Monday, and Dec 3 is Tuesday. So, the first Tuesday is December 3, 1991. Step 3: Calculate the date of the fourth Tuesday. To find successive Tuesdays, add 7 days for each week. First Tuesday: Dec 3. Second Tuesday: Dec 3 + 7 = Dec 10. Third Tuesday: Dec 10 + 7 = Dec 17. Fourth Tuesday: Dec 17 + 7 = Dec 24. Therefore, the fourth Tuesday of December 1991 is December 24, 1991.

Solution: Step 1: Identify the current date and its day: August 5th, Wednesday, in a leap year. Step 2: Determine the nature of the next three years. Since the current year is a leap year (e.g., 20XX), the next three consecutive years will be ordinary years (e.g., 20XX+1, 20XX+2, 20XX+3). Step 3: Calculate odd days for each of the next three years. An ordinary year has 1 odd day. Therefore, 3 ordinary years will have 3 * 1 = 3 odd days. Step 4: Add the total odd days to the current day of the week. Starting from Wednesday, add 3 odd days: Wednesday + 1 day = Thursday, Thursday + 1 day = Friday, Friday + 1 day = Saturday. Thus, August 5th after 3 years will be a Saturday.

Solution: Step 1: Break down 126 years into standard periods for odd day calculation. 126 years = 100 years + 26 years. Step 2: Calculate odd days for the first 100 years. A period of 100 years has 5 odd days (76 ordinary years * 1 odd day + 24 leap years * 2 odd days = 76 + 48 = 124 days = 17 weeks + 5 days). Step 3: Calculate odd days for the remaining 26 years. Identify leap years within these 26 years: 26 ÷ 4 = 6 leap years. Ordinary years = 26 - 6 = 20 ordinary years. Odd days from 26 years = (6 * 2) + (20 * 1) = 12 + 20 = 32 odd days. Convert to modulo 7: 32 ÷ 7 = 4 weeks with a remainder of 4 odd days. Step 4: Sum the odd days from both periods. Total odd days = (Odd days from 100 years) + (Odd days from 26 years) = 5 + 4 = 9 odd days. Step 5: Convert the total odd days to a value less than 7. 9 ÷ 7 = 1 week with a remainder of 2 odd days. Therefore, there are 2 odd days in 126 years.

Solution: Step 1: Determine the number of leap years in 400 consecutive years. A standard cycle of 400 years contains 97 leap years (e.g., 1600, 2000, 2400 are leap years, but 1700, 1800, 1900 are not). Step 2: In a leap year, February has 29 days. So, the 29th day in February occurs 97 times. Step 3: For the other 11 months (January, March, April, ..., December), the 29th day occurs in every year, regardless of whether it's a leap year or not. Step 4: Calculate total occurrences for these 11 months: 400 years * 11 months/year = 4400 times. Step 5: Sum the occurrences for February and the other 11 months: 97 + 4400 = 4497 times. Therefore, the 29th day of the month occurs 4497 times in 400 consecutive years.