Question Number 48675 by rahul 19 last updated on 27/Nov/18
$${Find}\:{remainder}\:{when}\:\mathrm{27}^{\mathrm{40}} \:{is}\:{divided} \\ $$$${by}\:\mathrm{12}\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18
$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{40}} ={f}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{40}} \\ $$$$\mathrm{3}^{\mathrm{40}} =\mathrm{3}^{\mathrm{39}} ×\mathrm{3} \\ $$$$\left(\mathrm{27}\right)^{\mathrm{13}} ×\mathrm{3} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{13}} ×\mathrm{3} \\ $$$$=\left\{{g}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{13}} \right\}×\mathrm{3} \\ $$$$=\mathrm{3}×{g}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{14}} \\ $$$$\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{3}} ×\mathrm{3}^{\mathrm{5}} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{3}} ×\mathrm{3}^{\mathrm{5}} \\ $$$$\left\{{h}\left(\mathrm{24}\right)+\mathrm{3}^{\mathrm{3}} \right\}×\mathrm{3}^{\mathrm{5}} \\ $$$$\left(\mathrm{3}^{\mathrm{3}} \right)^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\left(\mathrm{24}+\mathrm{3}\right)^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} \\ $$$$\left\{\mathrm{24}^{\mathrm{2}} +\mathrm{2}×\mathrm{24}×\mathrm{3}+\mathrm{3}^{\mathrm{2}} \right\}×\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{3}^{\mathrm{4}} =\left(\mathrm{24}+\mathrm{3}\right)×\mathrm{3}=\mathrm{72}+\mathrm{9} \\ $$$${so}\:{remainder}\:{is}\:\mathrm{9}\:{pls}\:{check}… \\ $$$$ \\ $$
Commented by rahul 19 last updated on 27/Nov/18
wow ����
thanks sir ����
Answered by MJS last updated on 27/Nov/18
$${n}\in\mathbb{N}^{\bigstar} :\:\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} \equiv\mathrm{3mod12};\:\mathrm{3}^{\mathrm{2}{n}} \equiv\mathrm{9mod12} \\ $$$$\mathrm{27}^{\mathrm{2}{n}−\mathrm{1}} =\mathrm{3}^{\mathrm{3}\left(\mathrm{2}{n}−\mathrm{1}\right)} \:\Rightarrow\:\mathrm{27}^{\mathrm{2}{n}−\mathrm{1}} \equiv\mathrm{3mod12} \\ $$$$\mathrm{27}^{\mathrm{2}{n}} =\mathrm{3}^{\mathrm{3}\left(\mathrm{2}{n}\right)} \:\Rightarrow\:\mathrm{27}^{\mathrm{2}{n}} \equiv\mathrm{9mod12} \\ $$$$\mathrm{27}^{\mathrm{40}} =\mathrm{27}^{\mathrm{2}×\mathrm{20}} \:\Rightarrow\:\mathrm{27}^{\mathrm{40}} \equiv\mathrm{9mod12} \\ $$
Commented by rahul 19 last updated on 27/Nov/18
$${what}\:{is}\:\equiv{mod}\:…? \\ $$$${can}'{t}\:{understand}\:{anything}… \\ $$
Commented by MJS last updated on 27/Nov/18
$${a}\equiv{b}\mathrm{mod}{c}\:\mathrm{means}\:{a}\:\mathrm{has}\:\mathrm{the}\:\mathrm{same}\:\mathrm{remainder} \\ $$$$\mathrm{as}\:{b}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:{c}. \\ $$
Commented by rahul 19 last updated on 27/Nov/18
thanks sir ����