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Question Number 127532 by MathSh last updated on 30/Dec/20

$$Solvetheequation:$$$$71a+3b=2021$$

Commented by mr W last updated on 30/Dec/20

$$seeQ44819$$

Answered by Ar Brandon last updated on 30/Dec/20

$$2021=28\left(71\right)+33$$$$=71\left(28\right)+3\left(11\right)$$$$(\mathrm{a},\mathrm{b})=(28,11)\mathrm{is}\mathrm{a}\mathrm{solution}$$

Commented by MathSh last updated on 30/Dec/20

$$Thankssir$$

Answered by floor(10²Eta[1]) last updated on 30/Dec/20

$$\mathrm{General}\mathrm{solution}\mathrm{to}\mathrm{a}\mathrm{Diophantine}\mathrm{equation}.$$$$\mathrm{ax}+\mathrm{by}=\mathrm{c}$$$$\mathrm{let}\mathrm{d}=\gcd (\mathrm{x},\mathrm{y})\Rightarrow \mathrm{d}\mid \mathrm{x},\mathrm{d}\mid \mathrm{y}\therefore \mathrm{d}\mid \mathrm{ax}+\mathrm{by}=\mathrm{c}$$$$\mathrm{ax}+\mathrm{by}\mathrm{has}\mathrm{a}\mathrm{solution}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{d}\mid \mathrm{c}.$$$$$$$$\mathrm{now}\mathrm{suppose}({\mathrm{x}}_0,{\mathrm{y}}_0)\mathrm{is}\mathrm{a}\mathrm{particular}\mathrm{solution},$$$$\mathrm{then},$$$${\mathrm{ax}}_0+{\mathrm{by}}_0=\mathrm{ax}+\mathrm{by}=\mathrm{c}$$$$\mathrm{a}({\mathrm{x}}_0-\mathrm{x})=\mathrm{b}(\mathrm{y}-{\mathrm{y}}_0)$$$$\mathrm{since}\mathrm{d}=\gcd (\mathrm{a},\mathrm{b})\Rightarrow \mathrm{a}={\mathrm{a}}^{\prime }\mathrm{d},\mathrm{b}={\mathrm{b}}^{\prime }\mathrm{d}$$$$\mathrm{where}\gcd ({\mathrm{a}}^{\prime },{\mathrm{b}}^{\prime })=1$$$${\mathrm{da}}^{\prime }({\mathrm{x}}_0-\mathrm{x})={\mathrm{db}}^{\prime }(\mathrm{y}-{\mathrm{y}}_0)$$$${\mathrm{a}}^{\prime }({\mathrm{x}}_0-\mathrm{x})={\mathrm{b}}^{\prime }(\mathrm{y}-{\mathrm{y}}_0)$$$$\Rightarrow \left(\mathrm{I}\right):{\mathrm{a}}^{\prime }\mid {\mathrm{b}}^{\prime }(\mathrm{y}-{\mathrm{y}}_0)\Rightarrow {\mathrm{a}}^{\prime }\mid (\mathrm{y}-{\mathrm{y}}_0)$$$$[\mathrm{because}\gcd ({\mathrm{a}}^{\prime },{\mathrm{b}}^{\prime })=1\Rightarrow {\mathrm{a}}^{\prime }\nmid {\mathrm{b}}^{\prime }]$$$$\Rightarrow \mathrm{y}-{\mathrm{y}}_0={\mathrm{a}}^{\prime }\mathrm{k}\Rightarrow \mathrm{y}={\mathrm{y}}_0+{\mathrm{a}}^{\prime }\mathrm{k}$$$$\Rightarrow \left(\mathrm{II}\right):{\mathrm{b}}^{\prime }\mid {\mathrm{a}}^{\prime }({\mathrm{x}}_0-\mathrm{x})\Rightarrow {\mathrm{b}}^{\prime }\mid ({\mathrm{x}}_0-\mathrm{x})$$$$\Rightarrow {\mathrm{x}}_0-\mathrm{x}={\mathrm{b}}^{\prime }\mathrm{k}\Rightarrow \mathrm{x}={\mathrm{x}}_0-{\mathrm{b}}^{\prime }\mathrm{k}$$$$\mathrm{so}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{to}\mathrm{a}\mathrm{diophantine}$$$$\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form}\mathrm{ax}+\mathrm{by}=\mathrm{c}\mathrm{is}$$$$(\mathrm{x},\mathrm{y})=({\mathrm{x}}_0-\frac{\mathrm{bk}}{\mathrm{d}},{\mathrm{y}}_0+\frac{\mathrm{ak}}{\mathrm{d}}),\mathrm{k}\in \mathbb{Z}$$$$\mathrm{where}({\mathrm{x}}_0,{\mathrm{y}}_0)\mathrm{is}\mathrm{a}\mathrm{particular}\mathrm{solution}$$$$$$$$\mathrm{now}\mathrm{that}\mathrm{you}\mathrm{know}\mathrm{this}\mathrm{try}\mathrm{to}\mathrm{solve}$$$$\mathrm{your}\mathrm{question}.$$$$$$

Commented by MathSh last updated on 31/Dec/20

$$Thankssir$$

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