On what date of August, 1988 did Friday fall ?

Concept: the odd-days method

Every date maps to a weekday using odd days worked out modulo 7. A complete 400-year cycle contributes exactly 0 odd days (it is an exact multiple of 7 weeks); the running totals for 100, 200, and 300 years are 5, 3, and 1 odd day respectively. Within any span of ordinary years, a leap year contributes 2 odd days and a non-leap year contributes 1 odd day, since 366 mod 7 = 2 and 365 mod 7 = 1. Adding the odd days from complete years to the odd days from the elapsed part of the current year, then reducing the sum modulo 7, gives one number from 0 to 6 that is read against a fixed reference: 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday.

Application: finding the weekday of 1 August 1988

1. Split the 1987 complete years before 1988 as 1600 + 300 + 87.

2. 1600 complete years contribute 0 odd days (four full 400-year cycles).

3. The next 300 years (1601-1900) contribute 1 odd day.

4. In the remaining 87 years (1901-1987), the leap years are 1904, 1908, ..., 1984 - that is 21 leap years, leaving 66 ordinary years.

5. Odd days from these 87 years = (21 x 2) + (66 x 1) = 42 + 66 = 108 days = 15 weeks + 3 odd days, so this block contributes 3.

6. Odd days for all 1987 complete years = 0 + 1 + 3 = 4.

7. 1988 is itself a leap year, so days elapsed from 1 January to 1 August 1988 = 31 (Jan) + 29 (Feb) + 31 (Mar) + 30 (Apr) + 31 (May) + 30 (Jun) + 31 (Jul) + 1 (reaching 1 Aug) = 214 days.

8. 214 days = 30 weeks + 4 odd days, so this span contributes 4.

9. Total odd days = 4 + 4 = 8 = 1 week + 1 odd day, so the net odd day is 1, which the reference table reads as Monday.

10. So 1 August 1988 was a Monday. Counting forward, the first Friday of the month is 4 days later, on the 5th, and every 7 days after that - the 12th, 19th, and 26th - is also a Friday.

Cross-check: an independent formula

Zeller's congruence gives the weekday of 5 August 1988 directly: with month m = 8, day q = 5, and year split as century K = 19, year-in-century J = 88, h = (q + floor(13(m+1)/5) + J + floor(J/4) + floor(K/4) - 2K) mod 7 = (5 + 23 + 88 + 22 + 4 - 38) mod 7 = 104 mod 7 = 6, which on Zeller's own reference (0 = Saturday, 1 = Sunday, ..., 6 = Friday) is Friday - the same result reached by the odd-days count above, confirming the two methods agree.

So the date on which Friday fell in August 1988 is the 5th (and again on the 12th, 19th, and 26th).