What was the day of the week on 17th June, 1998?

### Concept & Total Odd Days from Base Year 0 To find the exact day of the week for any historical date, calculate the total number of odd days from the start of the Gregorian calendar (Year 1) to the given date. 0 = Sunday, 1 = Monday, ..., 6 = Saturday. ### Step-by-Step Solution - Calculate odd days up to the completed year 1997. - Split 1997 into $1600 + 300 + 97$. - 1600 years have 0 odd days. - 300 years have 1 odd day. - Evaluate the remaining 97 years: - Leap years = $97 \div 4 = 24$ leap years. - Ordinary years = $97 - 24 = 73$ ordinary years. - Odd days = $(24 \times 2) + (73 \times 1) = 48 + 73 = 121$. - $121 \pmod 7 = 2$ odd days. - Total odd days for completed years = $0 + 1 + 2 = 3$ odd days. - Now count odd days for 1998 up to June 17: - Jan(3) + Feb(0) + Mar(3) + Apr(2) + May(3) + Jun(17). - Sum = $3 + 0 + 3 + 2 + 3 + 17 = 28$. - $28 \pmod 7 = 0$ odd days. - Total odd days = $3 (\text{from years}) + 0 (\text{from months}) = 3$. - Day 3 corresponds to Wednesday. ### Exam Strategy & Shortcut Memorize century odd days: 1900 has 1 odd day. From 1900, evaluate 97 years directly: $(97 + (97 \div 4)) \pmod 7 = (97 + 24) \pmod 7 = 121 \pmod 7 = 2$. Total for 1997 = 1 + 2 = 3. ### Common Pitfall Forgetting to use the *completed* year (1997) for the annual calculation and mistakenly using 1998, effectively double-counting the months. ### Final Answer Therefore, the correct answer is **Wednesday**.