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Question Number 103209 by nimnim last updated on 13/Jul/20
Answered by bramlex last updated on 13/Jul/20
n = 7k+1 ...(1)×6 n= 6p+3 ...(2)×7 n = 5q +2...(3) ⇒ (2)−(1) ⇒n = 42(p−k)+15 ...(4) set p−k = l consider 5q+2 = 42l+15 ⇒2 (mod 5) = 42l +15 ⇒2l = 2 (mod 5), l = 1 (mod 5) ⇔ l = 5t+1 n = 42l +15 = 42(5t+1)+15 n = 210t + 57
$$\mathrm{n}\:=\:\mathrm{7k}+\mathrm{1}\:…\left(\mathrm{1}\right)×\mathrm{6} \\ $$$$\mathrm{n}=\:\mathrm{6p}+\mathrm{3}\:…\left(\mathrm{2}\right)×\mathrm{7} \\ $$$$\mathrm{n}\:=\:\mathrm{5q}\:+\mathrm{2}…\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)−\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{n}\:=\:\mathrm{42}\left(\mathrm{p}−\mathrm{k}\right)+\mathrm{15}\:…\left(\mathrm{4}\right) \\ $$$$\mathrm{set}\:\mathrm{p}−\mathrm{k}\:=\:{l} \\ $$$$\mathrm{consider}\: \\ $$$$\mathrm{5q}+\mathrm{2}\:=\:\mathrm{42}{l}+\mathrm{15}\: \\ $$$$\Rightarrow\mathrm{2}\:\left({mod}\:\mathrm{5}\right)\:=\:\mathrm{42}{l}\:+\mathrm{15} \\ $$$$\Rightarrow\mathrm{2}{l}\:=\:\mathrm{2}\:\left({mod}\:\mathrm{5}\right),\:{l}\:=\:\mathrm{1}\:\left({mod}\:\mathrm{5}\right) \\ $$$$\Leftrightarrow\:{l}\:=\:\mathrm{5}{t}+\mathrm{1} \\ $$$${n}\:=\:\mathrm{42}{l}\:+\mathrm{15}\:=\:\mathrm{42}\left(\mathrm{5}{t}+\mathrm{1}\right)+\mathrm{15} \\ $$$${n}\:=\:\mathrm{210}{t}\:+\:\mathrm{57}\: \\ $$
Commented by nimnim last updated on 13/Jul/20
thanks
$${thanks} \\ $$
Answered by floor(10²Eta[1]) last updated on 13/Jul/20
(1)x≡1(mod 7)⇒x=7a+1 (2)x≡3(mod 6) (3)x≡2(mod 5) (2):7a+1≡3(mod 6)⇒a≡2(mod 6) ⇒a=6b+2 ⇒x=7(6b+2)+1=42b+15 (3):42b+15≡2(mod 5)⇒2b≡2(mod 5) ⇒b≡1(mod 5)⇒b=5c+1 ⇒x=42(5c+1)+15 ⇒x=210c+57, c∈Z
$$\left(\mathrm{1}\right)\mathrm{x}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{x}=\mathrm{7a}+\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\mathrm{x}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{6}\right) \\ $$$$\left(\mathrm{3}\right)\mathrm{x}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\left(\mathrm{2}\right):\mathrm{7a}+\mathrm{1}\equiv\mathrm{3}\left(\mathrm{mod}\:\mathrm{6}\right)\Rightarrow\mathrm{a}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{a}=\mathrm{6b}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{7}\left(\mathrm{6b}+\mathrm{2}\right)+\mathrm{1}=\mathrm{42b}+\mathrm{15} \\ $$$$\left(\mathrm{3}\right):\mathrm{42b}+\mathrm{15}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{2b}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{b}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{5}\right)\Rightarrow\mathrm{b}=\mathrm{5c}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{42}\left(\mathrm{5c}+\mathrm{1}\right)+\mathrm{15} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{210c}+\mathrm{57},\:\mathrm{c}\in\mathbb{Z} \\ $$
Commented by nimnim last updated on 13/Jul/20
Thanks
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Answered by 1549442205 last updated on 13/Jul/20
From the condition that number when divided by 6 gives remainder 3 and when divided by 5 gives remainder 2 it follows that if adding that number to 3 we get a multiple of 30.Hence Denoting by n the number we need find then n=30k−3.On the other hands,n=7m+1.We infer 30k+3=7m+1⇒m=((30k−4)/7)=4k+((2k−4)/7) Put ((2k−4)/7)=p⇒k=((7p+4)/2)=3p+2+(p/2) Put (p/2)=q⇒p=2q⇒k=7q+2 m=30q+8⇒n=210q+57.Thus,the number we need find is n=210q+57(q∈N)
$$\mathrm{From}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{that}\:\mathrm{number}\:\mathrm{when} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{6}\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{3}\:\mathrm{and} \\ $$$$\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5}\:\mathrm{gives}\:\mathrm{remainder}\:\mathrm{2} \\ $$$$\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{if}\:\mathrm{adding}\:\mathrm{that}\:\mathrm{number}\: \\ $$$$\mathrm{to}\:\mathrm{3}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{30}.\mathrm{Hence} \\ $$$$\mathrm{Denoting}\:\mathrm{by}\:\mathrm{n}\:\mathrm{the}\:\:\mathrm{number}\:\mathrm{we}\:\mathrm{need}\:\mathrm{find} \\ $$$$\mathrm{then}\:\mathrm{n}=\mathrm{30k}−\mathrm{3}.\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hands},\mathrm{n}=\mathrm{7m}+\mathrm{1}.\mathrm{We} \\ $$$$\mathrm{infer}\:\mathrm{30k}+\mathrm{3}=\mathrm{7m}+\mathrm{1}\Rightarrow\mathrm{m}=\frac{\mathrm{30k}−\mathrm{4}}{\mathrm{7}}=\mathrm{4k}+\frac{\mathrm{2k}−\mathrm{4}}{\mathrm{7}} \\ $$$$\mathrm{Put}\:\frac{\mathrm{2k}−\mathrm{4}}{\mathrm{7}}=\mathrm{p}\Rightarrow\mathrm{k}=\frac{\mathrm{7p}+\mathrm{4}}{\mathrm{2}}=\mathrm{3p}+\mathrm{2}+\frac{\mathrm{p}}{\mathrm{2}} \\ $$$$\mathrm{Put}\:\frac{\mathrm{p}}{\mathrm{2}}=\mathrm{q}\Rightarrow\mathrm{p}=\mathrm{2q}\Rightarrow\mathrm{k}=\mathrm{7q}+\mathrm{2} \\ $$$$\mathrm{m}=\mathrm{30q}+\mathrm{8}\Rightarrow\mathrm{n}=\mathrm{210q}+\mathrm{57}.\mathrm{Thus},\mathrm{the}\: \\ $$$$\mathrm{number}\:\mathrm{we}\:\mathrm{need}\:\mathrm{find}\:\mathrm{is} \\ $$$$\boldsymbol{\mathrm{n}}=\mathrm{210}\boldsymbol{\mathrm{q}}+\mathrm{57}\left(\boldsymbol{\mathrm{q}}\in\mathbb{N}\right) \\ $$$$ \\ $$
Commented by nimnim last updated on 13/Jul/20
Thanks
$${Thanks} \\ $$

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