# Find all solutions to the following system of linear congruences: x-= 1 mod 2, x -= 2 mod 3, x-= 3 mod 5, x -= 4 mod 7, x -= 5 mod 11.

Find all solutions to the following system of linear congruences: $x\equiv 1\text{mod}2,x\equiv 2\text{mod}3,x\equiv 3\text{mod}5,x\equiv 4\text{mod}7,x\equiv 5\text{mod}11$.
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Step 1
The system of linear congruence’s are given by,
$x\equiv 1\text{mod}2$
$x\equiv 2\text{mod}3$
$x\equiv 3\text{mod}5$
$x\equiv 4\text{mod}7$
$x\equiv 5\text{mod}11$
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 ${y}_{1},{y}_{2},{y}_{3},{y}_{4},{y}_{5}$.
$1{y}_{1}\equiv 1\text{mod}2⇒{y}_{1}=1$
$2{y}_{2}\equiv 1\text{mod}3⇒{y}_{2}=2$
$2{y}_{3}\equiv 1\text{mod}5⇒{y}_{3}=3$
$1{y}_{4}\equiv 1\text{mod}7⇒{y}_{4}=1$
$1{y}_{5}\equiv 1\text{mod}11⇒{y}_{5}=1$
Step 4
Solution is,
$x=\left(1{M}_{1}{y}_{1}+2{M}_{2}{y}_{2}+3{M}_{3}{y}_{3}+4{M}_{4}{y}_{4}+5{M}_{5}{y}_{5}\right)\text{mod}M$
$x=\left(1×1155×1+2×770×2+3×462×3+4×330×1+5×210×1\right)\text{mod}2310$
$x=10763\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}2310$
$x=1403\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}2310$