+0

# system of congruences pls help

0
17
2
+59

abc = 1 (mod 11),

abd = 2 (mod 11),

acd = 3 (mod 11),

bcd =4 (mod 11).

solve for a, b, c, and d in mod 11.

Oct 7, 2023

#1
+671
-1

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.

Oct 7, 2023
#2
+118627
+1

a=10, b=6, c=9, d=7