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7n-1-mod-5-What-is-the-general-form-of-n-




Question Number 75186 by Hassen_Timol last updated on 08/Dec/19
7n ≡ 1 (mod 5)  What is the general form of n ?
$$\mathrm{7}{n}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{form}\:\mathrm{of}\:{n}\:? \\ $$
Commented by mr W last updated on 08/Dec/19
7n=5m+1  ⇒n=5k+3
$$\mathrm{7}{n}=\mathrm{5}{m}+\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{5}{k}+\mathrm{3} \\ $$
Commented by Hassen_Timol last updated on 08/Dec/19
How do you get that from first to second  line, please Sir ?
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\mathrm{that}\:\mathrm{from}\:\mathrm{first}\:\mathrm{to}\:\mathrm{second} \\ $$$$\mathrm{line},\:\mathrm{please}\:\mathrm{Sir}\:? \\ $$
Commented by mr W last updated on 08/Dec/19
see my comment inQ19198 or here:
$${see}\:{my}\:{comment}\:{inQ}\mathrm{19198}\:{or}\:{here}: \\ $$
Commented by mr W last updated on 08/Dec/19
Commented by Hassen_Timol last updated on 08/Dec/19
Thank you sir, may god bless you
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{may}\:\mathrm{god}\:\mathrm{bless}\:\mathrm{you} \\ $$
Commented by mr W last updated on 08/Dec/19
see also Q66832 for other methods
$${see}\:{also}\:{Q}\mathrm{66832}\:{for}\:{other}\:{methods} \\ $$
Commented by jagannath.02 last updated on 08/Dec/19
sir please solve the thermodynamics question thag i have posted... please sir..
$${sir}\:{please}\:{solve}\:{the}\:{thermodynamics}\:{question}\:{thag}\:{i}\:{have}\:{posted}…\:{please}\:{sir}.. \\ $$
Answered by Rio Michael last updated on 08/Dec/19
7n ≡ 1(mod 5)  by trial and error we look for   a number n which when multiplied to   7 gives a remainder of 1 when divided by 5.   so   n=1 does not work  n = 2 does not work  n = 3 ⇒ 7(3) = 21    21 = 4(5) + 1 remainder of 1 hence it works  ∴  n = 3 and we write  n ≡ 3(mod 5)  hence n = 5k + 3,  k∈Z
$$\mathrm{7}{n}\:\equiv\:\mathrm{1}\left({mod}\:\mathrm{5}\right) \\ $$$${by}\:{trial}\:{and}\:{error}\:{we}\:{look}\:{for}\: \\ $$$${a}\:{number}\:{n}\:{which}\:{when}\:{multiplied}\:{to}\: \\ $$$$\mathrm{7}\:{gives}\:{a}\:{remainder}\:{of}\:\mathrm{1}\:{when}\:{divided}\:{by}\:\mathrm{5}. \\ $$$$\:{so}\: \\ $$$${n}=\mathrm{1}\:{does}\:{not}\:{work} \\ $$$${n}\:=\:\mathrm{2}\:{does}\:{not}\:{work} \\ $$$${n}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{7}\left(\mathrm{3}\right)\:=\:\mathrm{21}\: \\ $$$$\:\mathrm{21}\:=\:\mathrm{4}\left(\mathrm{5}\right)\:+\:\mathrm{1}\:{remainder}\:{of}\:\mathrm{1}\:{hence}\:{it}\:{works} \\ $$$$\therefore\:\:{n}\:=\:\mathrm{3}\:{and}\:{we}\:{write}\:\:{n}\:\equiv\:\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$$${hence}\:{n}\:=\:\mathrm{5}{k}\:+\:\mathrm{3},\:\:{k}\in\mathbb{Z}\: \\ $$
Commented by Hassen_Timol last updated on 09/Dec/19
Thank you Sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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