Question Number 120469 by mathocean1 last updated on 31/Oct/20
$${c}\:{alculate}\:{the}\:{rest}\:{of}\:{the}\:{division}\:{of} \\ $$$$\mathrm{2}^{{n}} \:{by}\:\mathrm{3}\:;\:{n}\:\in\:\mathbb{N} \\ $$
Answered by mathmax by abdo last updated on 31/Oct/20
![n=2k ⇒2^n =2^(2k) =4^k we hsve 4≈1[3] ⇒4^n ≈1[3] ⇒r=1 n=2k+1 ⇒2^n =2^(2k+1) =2. 4^k we have 4^k ≈1[3] ⇒2.4^k ≈2[3] ⇒ ⇒r=2](https://www.tinkutara.com/question/Q120506.png)
$$\mathrm{n}=\mathrm{2k}\:\Rightarrow\mathrm{2}^{\mathrm{n}} \:=\mathrm{2}^{\mathrm{2k}} \:=\mathrm{4}^{\mathrm{k}} \:\:\mathrm{we}\:\mathrm{hsve}\:\mathrm{4}\approx\mathrm{1}\left[\mathrm{3}\right]\:\Rightarrow\mathrm{4}^{\mathrm{n}} \approx\mathrm{1}\left[\mathrm{3}\right]\:\Rightarrow\mathrm{r}=\mathrm{1} \\ $$$$\mathrm{n}=\mathrm{2k}+\mathrm{1}\:\Rightarrow\mathrm{2}^{\mathrm{n}} \:=\mathrm{2}^{\mathrm{2k}+\mathrm{1}} \:=\mathrm{2}.\:\mathrm{4}^{\mathrm{k}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{4}^{\mathrm{k}} \:\approx\mathrm{1}\left[\mathrm{3}\right]\:\Rightarrow\mathrm{2}.\mathrm{4}^{\mathrm{k}} \:\approx\mathrm{2}\left[\mathrm{3}\right]\:\Rightarrow \\ $$$$\Rightarrow\mathrm{r}=\mathrm{2} \\ $$