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Question Number 48675 by rahul 19 last updated on 27/Nov/18

Find remainder when 27^(40) is divided by 12 ?

$${Find}\:{remainder}\:{when}\:\mathrm{27}^{\mathrm{40}} \:{is}\:{divided} \\ $$$${by}\:\mathrm{12}\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

(24+3)^(40) =f(24)+3^(40) 3^(40) =3^(39) ×3 (27)^(13) ×3 (24+3)^(13) ×3 ={g(24)+3^(13) }×3 =3×g(24)+3^(14) (3^3 )^3 ×3^5 (24+3)^3 ×3^5 {h(24)+3^3 }×3^5 (3^3 )^2 ×3^2 (24+3)^2 ×3^2 {24^2 +2×24×3+3^2 }×3^2 3^4 =(24+3)×3=72+9 so remainder is 9 pls check...

$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{40}} ={f}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{40}} \\ $$$$\mathrm{3}^{\mathrm{40}} =\mathrm{3}^{\mathrm{39}} ×\mathrm{3} \\ $$$$\left(\mathrm{27}\right)^{\mathrm{13}} ×\mathrm{3} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{13}} ×\mathrm{3} \\ $$$$=\left\{{g}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{13}} \right\}×\mathrm{3} \\ $$$$=\mathrm{3}×{g}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{14}} \\ $$$$\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{3}} ×\mathrm{3}^{\mathrm{5}} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{3}} ×\mathrm{3}^{\mathrm{5}} \\ $$$$\left\{{h}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{3}} \right\}×\mathrm{3}^{\mathrm{5}} \\ $$$$\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\left\{\mathrm{24}^{\mathrm{2}} +\mathrm{2}×\mathrm{24}×\mathrm{3}+\mathrm{3}^{\mathrm{2}} \right\}×\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{4}} =\left(\mathrm{24}+\mathrm{3}\right)×\mathrm{3}=\mathrm{72}+\mathrm{9} \\ $$$${so}\:{remainder}\:{is}\:\mathrm{9}\:{pls}\:{check}... \\ $$$$ \\ $$

Commented by rahul 19 last updated on 27/Nov/18

wow ���� thanks sir ����

Answered by MJS last updated on 27/Nov/18

n∈N^★ : 3^(2n−1) ≡3mod12; 3^(2n) ≡9mod12 27^(2n−1) =3^(3(2n−1)) ⇒ 27^(2n−1) ≡3mod12 27^(2n) =3^(3(2n)) ⇒ 27^(2n) ≡9mod12 27^(40) =27^(2×20) ⇒ 27^(40) ≡9mod12

$${n}\in\mathbb{N}^{\bigstar} :\:\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} \equiv\mathrm{3mod12};\:\mathrm{3}^{\mathrm{2}{n}} \equiv\mathrm{9mod12} \\ $$$$\mathrm{27}^{\mathrm{2}{n}−\mathrm{1}} =\mathrm{3}^{\mathrm{3}\left(\mathrm{2}{n}−\mathrm{1}\right)} \:\Rightarrow\:\mathrm{27}^{\mathrm{2}{n}−\mathrm{1}} \equiv\mathrm{3mod12} \\ $$$$\mathrm{27}^{\mathrm{2}{n}} =\mathrm{3}^{\mathrm{3}\left(\mathrm{2}{n}\right)} \:\Rightarrow\:\mathrm{27}^{\mathrm{2}{n}} \equiv\mathrm{9mod12} \\ $$$$\mathrm{27}^{\mathrm{40}} =\mathrm{27}^{\mathrm{2}×\mathrm{20}} \:\Rightarrow\:\mathrm{27}^{\mathrm{40}} \equiv\mathrm{9mod12} \\ $$

Commented by rahul 19 last updated on 27/Nov/18

what is ≡mod ...? can′t understand anything...

$${what}\:{is}\:\equiv{mod}\:...? \\ $$$${can}'{t}\:{understand}\:{anything}... \\ $$

Commented by MJS last updated on 27/Nov/18

a≡bmodc means a has the same remainder as b when divided by c.

$${a}\equiv{b}\mathrm{mod}{c}\:\mathrm{means}\:{a}\:\mathrm{has}\:\mathrm{the}\:\mathrm{same}\:\mathrm{remainder} \\ $$$$\mathrm{as}\:{b}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:{c}. \\ $$

Commented by rahul 19 last updated on 27/Nov/18

thanks sir ����

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