Question 1

For a certain month, the dates of three of the Sundays are even numbers. Then, the 15th of the that month falls on a

Accepted Answer

Concept & Weekly Date Patterns: Days of the week are spaced exactly 7 days apart. Consequently, the parity (even/odd) of the dates for a specific weekday alternates. If the 1st is even, the next is odd, the next is even, and so on. Step-by-Step Solution: - We are told that three Sundays in a single month have even dates. - Since a month has a maximum of 31 days, a day of the week can appear at most 5 times. - Let's map the sequence of even (E) and odd (O) dates for consecutive weeks: E, O, E, O, E. - To have three even-numbered Sundays, the month MUST have exactly 5 Sundays, and they must follow the E-O-E-O-E pattern. - The lowest possible even date for the first Sunday is the 2nd. - Let's test if the 1st Sunday is on the 2nd: - 1st Sunday: 2nd (Even) - 2nd Sunday: 2 + 7 = 9th (Odd) - 3rd Sunday: 9 + 7 = 16th (Even) - 4th Sunday: 16 + 7 = 23rd (Odd) - 5th Sunday: 23 + 7 = 30th (Even) - This perfectly satisfies the condition of having three even-dated Sundays (2nd, 16th, 30th) within a valid month length (30 or 31 days). - If the 16th is a Sunday, we can find the day for the 15th by stepping backward by one day. - Sunday - 1 day = Saturday. Exam Strategy & Shortcut: Memorize that if a month has 3 even occurrences of a weekday, those dates MUST be the 2nd, 16th, and 30th. Once you know the 16th is that specific weekday (Sunday), the 15th is always the day before (Saturday). Common Pitfall: Trying to start the first Sunday on an odd date (like the 1st or 3rd), which only yields two even Sundays (8, 22 or 10, 24). Final Answer: Therefore, the correct answer is Saturday.

Question 2

The calendar for the year 2009 will be the same as that of the year

Accepted Answer

Concept & Calendar Repetition Cycle: A calendar year repeats when the cumulative sum of odd days from the starting year becomes exactly divisible by 7, and the resulting matching year is of the exact same type (ordinary or leap). Step-by-Step Solution: - Let's count the odd days starting from 2009. - 2009 (ordinary) = 1 odd day - 2010 (ordinary) = 1 odd day - 2011 (ordinary) = 1 odd day - 2012 (leap) = 2 odd days - 2013 (ordinary) = 1 odd day - 2014 (ordinary) = 1 odd day - Sum of odd days from 2009 to 2014 = 1 + 1 + 1 + 2 + 1 + 1 = 7. - Since 7 \pmod 7 = 0, the year immediately following 2014 will have the identical calendar. - The next year is 2015. - Verify year type: Both 2009 and 2015 are ordinary years, so the entire calendar matches perfectly. Exam Strategy & Shortcut: Determine how many years the target year is past the previous leap year. 2009 is 1 year after a leap year (2008). Ordinary years that immediately follow a leap year ALWAYS repeat after exactly 6 years. 2009 + 6 = 2015. Common Pitfall: Assuming all calendars repeat every 11 years. Only ordinary years that are 2 or 3 years past a leap year repeat in 11 years. Ordinary years 1 year past a leap year repeat in 6 years. Final Answer: Therefore, the correct answer is 2015.

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