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Question Number 127940 by bramlexs22 last updated on 03/Jan/21

$$\{\begin{array}{l}3\mathrm{x}=1\left(\mathrm{mod}4\right)\\ 4\mathrm{x}=3\left(\mathrm{mod}5\right)\\ 5\mathrm{x}=7\left(\mathrm{mod}11\right)\end{array}$$

Answered by floor(10²Eta[1]) last updated on 03/Jan/21

$$3\mathrm{x}\equiv 1\left(\mathrm{mod}4\right)\Rightarrow \mathrm{x}\equiv 3\left(\mathrm{mod}4\right)$$$$\mathrm{x}=4\mathrm{a}+3$$$$4(4\mathrm{a}+3)\equiv 3\left(\mathrm{mod}5\right)$$$$\Rightarrow \mathrm{a}\equiv 1\left(\mathrm{mod}5\right)\Rightarrow \mathrm{a}=5\mathrm{b}+1$$$$\Rightarrow \mathrm{x}=4(5\mathrm{b}+1)+3=20\mathrm{b}+7$$$$5(20\mathrm{b}+7)\equiv 7\left(\mathrm{mod}11\right)$$$$\Rightarrow \mathrm{b}\equiv 5\left(\mathrm{mod}11\right)$$$$\mathrm{b}=11\mathrm{c}+5\Rightarrow \mathrm{x}=20(11\mathrm{c}+5)+7$$$$\mathrm{x}=220\mathrm{c}+107,\mathrm{c}\in \mathbb{Z}$$

Answered by liberty last updated on 04/Jan/21

$$\{\begin{array}{l}\mathrm{x}=3\left(\mathrm{mod}4\right)...\left(\mathrm{i}\right)\\ \mathrm{x}=2\left(\mathrm{mod}5\right)...\left(\mathrm{ii}\right)\\ \mathrm{x}=8\left(\mathrm{mod}11\right)...\left(\mathrm{iii}\right)\end{array}$$$$\mathrm{for}\left(\mathrm{i}\right)\Rightarrow 55\mathrm{a}=3\left(\mathrm{mod}4\right)$$$$-\mathrm{a}=3\left(\mathrm{mod}4\right);\mathrm{a}=-3\left(\mathrm{mod}4\right)$$$$\mathrm{for}\left(\mathrm{ii}\right)\Rightarrow 44\mathrm{b}=2\left(\mathrm{mod}5\right)$$$$4\mathrm{b}=2\left(\mathrm{mod}5\right);\mathrm{b}=3\left(\mathrm{mod}5\right)$$$$\mathrm{for}\left(\mathrm{iii}\right)\Rightarrow 20\mathrm{c}=8\left(\mathrm{mod}11\right)$$$$-2\mathrm{c}=8\left(\mathrm{mod}11\right);\mathrm{c}=-4\left(\mathrm{mod}11\right)$$$$\mathrm{general}\mathrm{solution}$$$$\therefore 55(-3)+44\left(3\right)+20(-4)+220\mathrm{k};\mathrm{k}\in \mathbb{Z}$$$$\mathrm{i}.\mathrm{e}220\mathrm{k}-113\mathrm{or}220\mathrm{k}+107;\mathrm{k}\in \mathbb{Z}$$

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