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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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