31st July, 1993 = (1992 years + period from 1st January, 1993 to 31st July, 1993)
1600 years have 0 odd days and 300 years have 1 odd day.
Now, the period from 1900 to 1992 have 69 ordinary years and 23 leap years
= [latex](69 \times 1 + 23 \times 2)[/latex] = 115 odd days = (16 weeks + 3 days) = 3 odd days.
Januart |
February |
March |
April |
May |
June |
July |
31 |
28 |
31 |
30 |
31 |
30 |
31 |
= 212 days = (30 weeks + 2 days) = 2 odd days
Therefore, total number of odd days = 1 + 3 + 2 = 6 odd days.
Therefore, the required day was Saturday.