so basically we need to avoid one years each 4 years and count more 7 to get a year with same identical calendar date ... that would be each two 11 years and one 6 years
OpenStudy (ikram002p):
so,
1999-1915=0 mod 7 xD
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OpenStudy (ikram002p):
bhahaha xD
OpenStudy (ikram002p):
this is nice to show what day next year would be
OpenStudy (rsadhvika):
I see... if you have time could you help me with solution
OpenStudy (ikram002p):
solution for what :O
OpenStudy (rsadhvika):
i mean proof
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OpenStudy (ikram002p):
like proper proof ?
ok , main i dea is to show that 1999-1915 =0 mod 7
OpenStudy (rsadhvika):
\[Y_1 \equiv Y2 \pmod {7}\] doesnt necessarily mean they will have exact same "day name" on each day of the entire year rigt ?
OpenStudy (rsadhvika):
If thats the case we will be having that pattern every 7 years i guess
OpenStudy (ikram002p):
nope , count 7+4=11
when we have leap in the start
and 7-1 when we dont have
OpenStudy (rsadhvika):
thats for advancing a year ?
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