Let there be NNN binary relations from {A,B,C,D}\{A, B, C, D\}{A,B,C,D} to {1,2,3,4,5}\{1,2,3,4,5\}{1,2,3,4,5}. Calculate N mod 13N \bmod 13Nmod13.
參考答案與解析
N mod 13=9N \bmod 13 = 9Nmod13=9。從{A,B,C,D}\{A,B,C,D\}{A,B,C,D}到{1,...,5}\{1,...,5\}{1,...,5}的二元關係即{A,B,C,D}×{1,...,5}\{A,B,C,D\}\times\{1,...,5\}{A,B,C,D}×{1,...,5}（20個元素）的任意子集，故N=220=1048576N=2^{20}=1048576N=220=1048576，1048576 mod 13=91048576 \bmod 13 = 91048576mod13=9。
