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Question-121834




Question Number 121834 by bemath last updated on 12/Nov/20
Answered by liberty last updated on 12/Nov/20
we have gcd(93,27) = 3 and 6 divides by 3  therefore there exist solutions.   we have 93 = 3×27+12 ∧ 27 = 2×12+3   12 = 4×3 + 0    3 = 1×27−2×12       = 1×27−2(1×93−3×27)      = 7×27−2×93  thus −2×93 + 7×27 = 3  so −4×93−(−14)×27 = 6  we get general solutions is  { ((x=−4−9k)),((y=−14−31k)) :}  for all k∈Z
$$\mathrm{we}\:\mathrm{have}\:\mathrm{gcd}\left(\mathrm{93},\mathrm{27}\right)\:=\:\mathrm{3}\:\mathrm{and}\:\mathrm{6}\:\mathrm{divides}\:\mathrm{by}\:\mathrm{3} \\ $$$$\mathrm{therefore}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{solutions}.\: \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{93}\:=\:\mathrm{3}×\mathrm{27}+\mathrm{12}\:\wedge\:\mathrm{27}\:=\:\mathrm{2}×\mathrm{12}+\mathrm{3} \\ $$$$\:\mathrm{12}\:=\:\mathrm{4}×\mathrm{3}\:+\:\mathrm{0}\: \\ $$$$\:\mathrm{3}\:=\:\mathrm{1}×\mathrm{27}−\mathrm{2}×\mathrm{12}\: \\ $$$$\:\:\:\:=\:\mathrm{1}×\mathrm{27}−\mathrm{2}\left(\mathrm{1}×\mathrm{93}−\mathrm{3}×\mathrm{27}\right) \\ $$$$\:\:\:\:=\:\mathrm{7}×\mathrm{27}−\mathrm{2}×\mathrm{93} \\ $$$$\mathrm{thus}\:−\mathrm{2}×\mathrm{93}\:+\:\mathrm{7}×\mathrm{27}\:=\:\mathrm{3} \\ $$$$\mathrm{so}\:−\mathrm{4}×\mathrm{93}−\left(−\mathrm{14}\right)×\mathrm{27}\:=\:\mathrm{6} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{is}\:\begin{cases}{\mathrm{x}=−\mathrm{4}−\mathrm{9k}}\\{\mathrm{y}=−\mathrm{14}−\mathrm{31k}}\end{cases} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{k}\in\mathbb{Z}\: \\ $$$$ \\ $$

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