Solve all solutions to the following system of linear congruences:&nbsp;x≡1mod2,x≡2mod3,x≡3mod5,x≡4mod7,x≡5mod11.

Step 1 The system of linear congruence’s are given by, x≡1mod2 x≡2mod3 x≡3mod5 x≡4mod7 x≡5mod11 Step 2 According to the Chinese Remainder Theorem, M=2⋅3⋅5⋅7⋅11=2310 M1=M2=23102=1155 M2=M3=23103=770 M3=M5=23105=462 M4=M7=23107=330 M5=M11=231011=210 Step 3 Find the value of y1,y2,y3,y4,y5. 1y1≡1mod2⇒y1=1 2y2≡1mod3⇒y2=2 2y3≡1mod5⇒y3=3 1y4≡1mod7⇒y4=1 1y5≡1mod11⇒y5=1 Step 4 Solution is, x=(1M1y1+2M2y2+3M3y3+4M4y4+5M5y5)modM x=(1×1155×1+2×770×2+3×462×3+4×330×1+5×210×1)mod2310 x=10763mod2310 x=1403mod2310

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Ramsey

Answered question

2021-02-04

Solve all solutions to the following system of linear congruences: $x \equiv 1 mod 2, x \equiv 2 mod 3, x \equiv 3 mod 5, x \equiv 4 mod 7, x \equiv 5 mod 11$ .

Answer & Explanation

oppturf

Skilled2021-02-05Added 94 answers

Step 1
The system of linear congruence’s are given by,
$x \equiv 1 mod 2$
$x \equiv 2 mod 3$
$x \equiv 3 mod 5$
$x \equiv 4 mod 7$
$x \equiv 5 mod 11$
Step 2
According to the Chinese Remainder Theorem,
$M = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$
$M_{1} = \frac{M}{2} = \frac{2310}{2} = 1155$
$M_{2} = \frac{M}{3} = \frac{2310}{3} = 770$
$M_{3} = \frac{M}{5} = \frac{2310}{5} = 462$
$M_{4} = \frac{M}{7} = \frac{2310}{7} = 330$
$M_{5} = \frac{M}{11} = \frac{2310}{11} = 210$
Step 3
Find the value of $y_{1}, y_{2}, y_{3}, y_{4}, y_{5}$ .
$1 y_{1} \equiv 1 mod 2 \Rightarrow y_{1} = 1$
$2 y_{2} \equiv 1 mod 3 \Rightarrow y_{2} = 2$
$2 y_{3} \equiv 1 mod 5 \Rightarrow y_{3} = 3$
$1 y_{4} \equiv 1 mod 7 \Rightarrow y_{4} = 1$
$1 y_{5} \equiv 1 mod 11 \Rightarrow y_{5} = 1$
Step 4
Solution is,
$x = (1 M_{1} y_{1} + 2 M_{2} y_{2} + 3 M_{3} y_{3} + 4 M_{4} y_{4} + 5 M_{5} y_{5}) mod M$
$x = (1 \times 1155 \times 1 + 2 \times 770 \times 2 + 3 \times 462 \times 3 + 4 \times 330 \times 1 + 5 \times 210 \times 1) mod 2310$
$x = 10763 \mod 2310$
$x = 1403 \mod 2310$

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