0!=1(mod 5)
1!=1(mod 5)
2!=2(mod 5)
3!=1(mod 5)
4!=-1(mod 5)
for n>=5, n!=0(mod 5)
Therefore
(0!+...+64!)^2
=(1+1+2+1-1)^2(mod 5)
=4^2(mod 5)
=1(mod 5)
Therefore the remainder is 1
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