16th June, 1999 = (1900 years + 98 years + period from 1.1.1999 to 16.6.1999)
Now first deal with 1998 complete years; then how many odd days are there in 1998 years?
Odd days in 1600 years = 0
Odd days in 300 years = (5 × 3) = 1
98 years have 24 leap years + 74 ordinary years. Number of odd days in 98 years (24 × 2 + 74) = 122 = 3 odd days.
Now come to calculation of odd days in period from 1.1.1999 to 16.6.1999.
167 days = 23 complete weeks plus 6 odd days (this result is obtained by dividing 167 days by 7).
Total number of odd days = (0 + 1 + 3 + 6) = 10 ⇒ 3 odd days
Thus, the day asked in the question is Wednesday
Now first deal with 1998 complete years; then how many odd days are there in 1998 years?
Odd days in 1600 years = 0
Odd days in 300 years = (5 × 3) = 1
98 years have 24 leap years + 74 ordinary years. Number of odd days in 98 years (24 × 2 + 74) = 122 = 3 odd days.
Now come to calculation of odd days in period from 1.1.1999 to 16.6.1999.
| Months | Jan | Feb | March | April | May | June | Total |
| Odd Days | 31 | 28 | 31 | 30 | 31 | 16 | 167 |
Total number of odd days = (0 + 1 + 3 + 6) = 10 ⇒ 3 odd days
Thus, the day asked in the question is Wednesday
