3. (1994 China National Training Team Exam Question) Find all arrays composed of four positive integers $a, b, c, d$, such that the product of any three numbers in the array, when divided by the remaining one, leaves a remainder of 1.
3. From the condition $b c d=k a+1$, it is evident that $a \geqslant 2$, and $a$ is coprime with $b, c, d$, otherwise the remainder of $b c d$ divided by $a$ would be 0, contradicting the condition. Similarly, $b, c, d$ are all no less than 2, and $a, b, c, d$ are pairwise coprime. Without loss of generality, assume $2 \leqslant a5, d \geqslant 7, \frac{1}{6}+\frac{1}{6 c d}=\frac{1}{c}+\frac{1}{d}<\frac{2}{c} \quad$ (1), thus $c<12$. Since $(a, c)=1$, hence $c \leqslant 11$. From (1), we can derive $d=6+\frac{35}{c-6}$. Given that $(c-6) \mid 35$ and $c \leqslant 11$, we get $c=7$ or $c=11$. When $c=7$, $d=41$; when $c=11$, $d=13$. Therefore, the required quadruples $(a, b, c, d)=(2,3,7,41)$ or $(2,3,11,13)$.