We can rewrite the given congruences as follows:
abc \equiv 1 \pmod{11} \\ abd \equiv 2 \pmod{11} \\ acd \equiv 3 \pmod{11} \\ bcd \equiv 4 \pmod{11}
Multiplying the first three congruences together, we get:
a^2 bcd \equiv 1 \cdot 2 \cdot 3 \equiv 6 \pmod{11}
Multiplying the last three congruences together, we get:
bcd^2 \equiv 2 \cdot 3 \cdot 4 \equiv 24 \equiv 3 \pmod{11}
Therefore, a2bcd≡3(mod11). Dividing both sides by bcd, we get a2≡1(mod11). Since 1=12, a=1 (mod 11).
Now, we can substitute a=1 into the congruences abd≡2(mod11) and acd≡3(mod11) to get:
bd \equiv 2 \pmod{11} \\ cd \equiv 3 \pmod{11}
Multiplying these two congruences together, we get:
bcd^2 \equiv 2 \cdot 3 \equiv 6 \pmod{11}
Since bcd2≡3(mod11), we must have bcd≡3(mod11).
Finally, we substitute a=1 and bcd=3 into the congruence abc≡1(mod11) to get:
1 \cdot b \cdot 3 \equiv 1 \pmod{11}
Therefore, b≡1⋅1/3≡4(mod11).
Now, we can substitute a=1, b=4, and bcd=3 into the congruence abd≡2(mod11) to get:
1 \cdot 4 \cdot d \equiv 2 \pmod{11}
Therefore, d≡2/4≡5(mod11).
In conclusion, the solution to the system of congruences is a=1, b=4, c=1, and d=5.
