In a year N, the 259th day of the year is a Saturday. In the year N+l, the…

Key insight: the difference in day numbers across consecutive years determines whether year N is a leap year.

• Compute days from the 259th day of year N to the 222nd day of year N+1: (days in year N − 259) + 222 = days in year N − 37.

• Both given days are Saturdays, so the total difference must be a multiple of 7: days in year N − 37 ≡ 0 (mod 7). Thus days in year N ≡ 37 ≡ 2 (mod 7).

• The only reasonable calendar value with remainder 2 mod 7 is 366, so year N is a leap year (366 days).

• Now compute days from the 119th day of year N−1 to the 259th day of year N: (365 − 119) + 259 = 246 + 259 = 505 days.

• 505 ≡ 1 (mod 7), so moving from the 119th day of year N−1 to the 259th day of year N advances the weekday by 1.

• Therefore the 119th day of year N−1 is one weekday earlier than the 259th day of year N. Since the 259th day of year N is Saturday, the 119th day of year N−1 is Friday.

Answer: Friday