What was the day on 4th January 2002?
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Given : 4th January 2002
To Find : Day
Solution:
ZEllER’S Rule Formula
F = K + [(13×M – 1)/5] + D + [D/4] + [C/4] -2C
K = 4 ( Date )
M = 11 ( as March is 1st month )
D = 01 ( as Jan & Feb are counted in previous year)
C = 20
F = 4 + [(13×11 – 1)/5] + 01 + [01/4] + [20/4] -2(20)
=> F = 4 + 28 + 01 + 0 + 5 - 40
=> F = -2
-2 = -7 + 5
5 = FRIDAY
4th January 2002 FRIDAY
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