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Question Number 77965 by mr W last updated on 12/Jan/20

$$solveforx,y,z\in \mathbb{N}$$$$35x+21y+60z=665$$

Commented by john santu last updated on 12/Jan/20

$$diopthantineequationsir?$$

Commented by mr W last updated on 12/Jan/20

$$yes$$

Commented by TawaTawa last updated on 12/Jan/20

$$35\mathrm{x}+3(7\mathrm{y}+20\mathrm{z})=665$$$$\mathrm{Let}\mathrm{u}=7\mathrm{y}+20\mathrm{z}$$$$35\mathrm{x}+3\mathrm{u}=665......\left(\mathrm{ii}\right)$$$$\mathrm{Particular}\mathrm{solution}.{\mathrm{x}}_1=1,{\mathrm{u}}_1=210$$$$\mathrm{General}\mathrm{solution}\mathrm{for}\left(\mathrm{ii}\right)$$$$\therefore \mathrm{x}=1+3\mathrm{m}$$$$35(1+3\mathrm{m})+3\mathrm{u}=665.....\left(\mathrm{ii}\right)$$$$\therefore 35+105\mathrm{m}+3\mathrm{u}=665$$$$\therefore 3\mathrm{u}=665-35-105\mathrm{m}$$$$\therefore 3\mathrm{u}=630-105\mathrm{m}$$$$\therefore \mathrm{u}=210-35\mathrm{m}\left(\mathrm{with}\mathrm{m}\mathrm{be}\mathrm{any}\mathrm{integer}\right)$$$$\mathrm{Particular}\mathrm{solution}\mathrm{for}$$$$7\mathrm{y}+20\mathrm{z}=1({\mathrm{y}}_1=3,{\mathrm{z}}_1=-1)$$$$\therefore \mathrm{y}=3+20\mathrm{k}$$$$\therefore 7(3+20\mathrm{k})+20\mathrm{z}=1,21+140\mathrm{k}+20\mathrm{z}=1$$$$\therefore 20\mathrm{z}=1-21-140\mathrm{k}$$$$\therefore 20\mathrm{z}=-20-140\mathrm{k},\mathrm{z}=-1-7\mathrm{k}$$$$$$$$\mathrm{General}\mathrm{solution}\mathrm{of}$$$$35\mathrm{x}+21\mathrm{y}+60\mathrm{z}=665$$$$\mathrm{y}=(3+20\mathrm{k})\mathrm{u}=3\mathrm{u}+20\mathrm{uk}=3\mathrm{u}+20\mathrm{n}$$$$\mathrm{y}=3(210-35\mathrm{m})+20\mathrm{n}=630-105\mathrm{m}+20\mathrm{n}$$$$$$$$\mathrm{z}=(-1-7\mathrm{k})\mathrm{u}=-\mathrm{u}-7\mathrm{uk}=-\mathrm{u}-7\mathrm{n}$$$$\mathrm{z}=-(210-35\mathrm{m})-7\mathrm{n}=-210+35\mathrm{m}-7\mathrm{n}$$$$\mathrm{Therefore},$$$$\mathrm{x}=1+3\mathrm{m}$$$$\mathrm{y}=630-105\mathrm{m}+20\mathrm{n}$$$$\mathrm{z}=-210+35\mathrm{m}-7\mathrm{n}$$$$\mathrm{m},\mathrm{n}\mathrm{be}\mathrm{any}\mathrm{integer}$$

Commented by TawaTawa last updated on 12/Jan/20

$$\mathrm{Sir}\mathrm{mrW},\mathrm{i}\mathrm{used}\mathrm{the}\mathrm{method}\mathrm{you}\mathrm{thought}\mathrm{me}\mathrm{sometime}\mathrm{ago}.$$$$\mathrm{Help}\mathrm{me}\mathrm{check}.$$

Commented by mr W last updated on 12/Jan/20

$$butincurrentcasewearesearching$$$$onlypositiveintergers.$$

Commented by TawaTawa last updated on 12/Jan/20

$$\mathrm{Alright}\mathrm{sir},\mathrm{i}\mathrm{will}\mathrm{learn}\mathrm{that}\mathrm{too}.$$

Commented by mr W last updated on 12/Jan/20

$$suchthatx,y,z>0yougetonlytwo$$$$solutions:$$$$m=0,n=-31or$$$$m=1,n=-26$$$$i.e.x=1,y=10,z=7$$$$orx=4,y=5,z=7$$

Commented by TawaTawa last updated on 12/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Commented by TawaTawa last updated on 12/Jan/20

$$\mathrm{Sir},\mathrm{from}\mathrm{bere},\mathrm{i}\mathrm{will}\mathrm{assume}\mathrm{the}\mathrm{values}\mathrm{for}\mathrm{m}\mathrm{and}\mathrm{n}?$$$$\mathrm{such}\mathrm{that}\mathrm{x},\mathrm{y},\mathrm{z}>0$$

Answered by key of knowledge last updated on 12/Jan/20

$$35\mathrm{x}\wedge 60\mathrm{z}\wedge 665\mid 5\Rightarrow 21\mathrm{y}\mid 5\Rightarrow \mathrm{y}={5\mathrm{y}}^{{}^{\prime }}$$$$35\mathrm{x}\wedge 21\mathrm{y}\wedge 665\mid 7\Rightarrow 60\mathrm{z}\mid 5\Rightarrow \mathrm{z}={7\mathrm{z}}^{{}^{\prime }}$$$$35\mathrm{x}+{105\mathrm{y}}^{{}^{\prime }}+{420\mathrm{z}}^{{}^{\prime }}=665\Rightarrow \mathrm{x}+{3\mathrm{y}}^{{}^{\prime }}+{12\mathrm{z}}^{{}^{\prime }}=19$$$$\left(i\right){\mathrm{z}}^{{}^{\prime }}=1,{\mathrm{y}}^{{}^{\prime }}=2,\mathrm{x}=1\Rightarrow \mathrm{z}=7,\mathrm{y}=10,\mathrm{x}=1$$$$\left(ii\right){\mathrm{z}}^{{}^{\prime }}=1,{\mathrm{y}}^{{}^{\prime }}=1,\mathrm{x}=4\Rightarrow \mathrm{z}=7,\mathrm{y}=5,\mathrm{x}=4$$

Commented by mr W last updated on 12/Jan/20

$$verynice!thanks!$$

Commented by jagoll last updated on 13/Jan/20

$$whatthemeaning60z\mid 5?$$$$dividedby5?ormod5?$$

Commented by mr W last updated on 13/Jan/20

$$60z\mid 5istypo.hemeans60z\mid 7,i.e.60z$$$$mustalsobedivisibleby7.$$

Commented by jagoll last updated on 13/Jan/20

$$ootyposir.allrightthankssir$$

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