On which day of the week did 15 August 1947 fall?

Concept: To find the day of the week for any date, add up the total number of “odd days” — days left over after removing complete weeks — contributed by (a) every year before the date's year, and (b) the months and days already elapsed within that year. A block of 100 years contributes 5 odd days, 200 years contribute 3, 300 years contribute 1, and a full 400-year block contributes 0 (the leap-year correction resets the cycle). The final odd-day count is then read off the cycle 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday.

Applying this to 15 August 1947:

1. Odd days for all years up to 1600 = 0, by the 400-year cycle rule.

2. For 1601–1900 (three complete centuries: 1601–1700, 1701–1800, 1801–1900), each century contributes 5 odd days, so 5 + 5 + 5 = 15, which reduces to 1 odd day (15 − 14 = 1, removing two full weeks).

3. For the 46 elapsed years from 1901 to 1946, the leap years are 1904, 1908, …, 1944 — 11 of them — leaving 35 non-leap years. Odd days = (11 × 2) + (35 × 1) = 22 + 35 = 57, which reduces to 1 odd day (57 − 56 = 1, removing eight full weeks).

4. For the elapsed part of 1947 itself: January (31 days) = 3, February 1947 (not a leap year, 28 days) = 0, March (31) = 3, April (30) = 2, May (31) = 3, June (30) = 2, July (31) = 3, and 1–15 August = 15. Their sum is 3 + 0 + 3 + 2 + 3 + 2 + 3 + 15 = 31, which reduces to 3 odd days (31 − 28 = 3, removing four full weeks).

5. Adding every contribution: 0 (up to 1600) + 1 (1601–1900) + 1 (1901–1946) + 3 (Jan–15 Aug 1947) = 5 total odd days.

6. Reading 5 off the cycle 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday gives Friday.

Cross-check: This matches the historical record independently: 15 August 1947, India's Independence Day, is documented across standard calendar references as having fallen on a Friday.

Result: Friday is the correct answer.