n = d1 d2 $B!D(B dt$B$rK!$H$7$F!"$?$@0l$D$N2r$,$"$k!#(B
ni xi $B-q(B 1 (mod di )$B$N2r(B xi $B$r5a$a$k$3$H$,$G$-$k!#$=$3$G!"(B
x ≡ b1 n1 x1 + b2 n2 x2 + $B!D(B + bt nt xt (mod n)$B$H$9$l$P!"$3$N(B x $B$OL@$i$+$K$b$H$N9gF1<0$r$9$Y$FK~B-$9$k!#(B
$B$3$N$h$&$JO"N)9gF1<0$N@-$BCf9q?M>jM>DjM}(B (Chinese remainder theorem) $B$H8F$V!#(B