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Question Number 135692 by liberty last updated on 15/Mar/21

$$$$Solve the system of congruences 2x≡1(mod5) 3x≡2(mod7) 4x≡1(mod11)

Answered by floor(10²Eta[1]) last updated on 15/Mar/21

$$2\mathrm{x}\equiv 1\left(\mathrm{mod}5\right)\Rightarrow \mathrm{x}\equiv 3\left(\mathrm{mod}5\right)\Rightarrow \mathrm{x}=5\mathrm{a}+3$$$$3(5\mathrm{a}+3)=15\mathrm{a}+9\equiv \mathrm{a}+2\equiv 2\left(\mathrm{mod}7\right)$$$$\Rightarrow \mathrm{a}\equiv 0\left(\mathrm{mod}7\right)\Rightarrow \mathrm{a}=7\mathrm{b}\Rightarrow \mathrm{x}=35\mathrm{b}+3$$$$4(35\mathrm{b}+3)=140\mathrm{b}+12\equiv 8\mathrm{b}+1\equiv 1\left(\mathrm{mod}11\right)$$$$8\mathrm{b}\equiv 0\left(\mathrm{mod}11\right)\Rightarrow \mathrm{b}\equiv 0\left(\mathrm{mod}11\right)\Rightarrow \mathrm{b}=11\mathrm{c}$$$$\Rightarrow \mathrm{x}=385\mathrm{c}+3,\mathrm{c}\in \mathbb{Z}$$

Commented by liberty last updated on 15/Mar/21

$$thankyou$$

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