22-22-what-is-hte-last-digit-

Question Number 176058 by Shrinava last updated on 11/Sep/22

$$\mathrm{22}^{\mathrm{22}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{hte}\:\mathrm{last}\:\mathrm{digit}? \\ $$

Answered by BaliramKumar last updated on 11/Sep/22

$$\mathrm{4} \\ $$

Commented by BaliramKumar last updated on 11/Sep/22

$$\mathrm{22}^{\mathrm{22}} \:=\:\mathrm{22}^{\mathrm{4}\left(\mathrm{5}\right)+\mathrm{2}} \:\:\:\:\:\:\:\left[{x}^{\mathrm{4}\left({n}\right)\:+\:{r}} \:=\:{x}^{{r}} \right]\:\:\mathrm{0}<{r}\leq\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{4}\:{Answer} \\ $$$$ \\ $$$${example}\:\mathrm{39857}^{\mathrm{2576}} \\ $$$$\mathrm{39857}^{\mathrm{4}\left({n}\right)+\mathrm{4}} \\ $$$$\mathrm{7}^{\mathrm{4}} \:=\:\mathrm{1}\:{Answer} \\ $$$$ \\ $$$${example}\:\:\:\:\:\:\:\mathrm{2345678}^{\mathrm{3566765}!} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}^{\mathrm{4}\left({n}\right)+\mathrm{4}} \:=\:\mathrm{8}^{\mathrm{4}} \:=\:\mathrm{6} \\ $$$$ \\ $$$${If}\:\:\:\:{x}!\:\:\:\:{then}\:\:\:\:{r}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({x}\:\geq\:\mathrm{4}\right) \\ $$

Commented by Shrinava last updated on 11/Sep/22

$$\mathrm{solution}\:\mathrm{please} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Sep/22

$$\mathrm{Your}\:\mathrm{rule}\:\mathrm{is}\:\boldsymbol{\mathrm{successful}}!\: \\ $$
$$\mathrm{Actually}\mathrm{I}\mathrm{think}:$$$$\mathrm{In}\mathrm{general}$$$$\mathrm{a}\equiv \mathrm{b}\left[10\right]\wedge \mathrm{m}\equiv \mathrm{n}\left[4\right]\iff {\mathrm{a}}^{\mathrm{m}}\equiv {\mathrm{b}}^{\mathrm{n}}\left[10\right]$$

Commented by Shrinava last updated on 13/Sep/22

$$\mathrm{cool}\mathrm{dear}\mathrm{sir}\mathrm{thankyou}$$

Answered by Rasheed.Sindhi last updated on 11/Sep/22

$$\mathrm{last}\mathrm{digit}={22}^{22}\mathrm{mod}10$$$$22\equiv 2\left(\mathrm{mod}10\right)$$$${22}^2\equiv 2^2\equiv 4\left(\mathrm{mod}10\right)\dots \dots \dots \dots .\left(\mathrm{i}\right)$$$${22}^5\equiv 2^5=32\equiv 2\left(\mathrm{mod}10\right)$$$${\left({22}^5\right)}^4\equiv 2^4=16\equiv 6\left(\mathrm{mod}10\right)$$$${22}^{20}\equiv 6\left(\mathrm{mod}10\right)\dots \dots \dots \dots \dots \dots \left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\times \left(\mathrm{ii}\right):$$$${22}^{22}\equiv 24\equiv 4\left(\mathrm{mod}10\right)$$

Answered by LordKazuma last updated on 11/Sep/22

$${22}^{22}\left(mod10\right)=2^{22}\left(mod10\right)$$$$={\left(2^3\right)}^7\cdot 2^1\left(mod10\right)$$$$={(-2)}^7\cdot 2\left(mod10\right)$$$$={\left(2^3\right)}^2\cdot (-2)\cdot 2\left(mod10\right)$$$$={(-2)}^2\cdot (-4)\left(mod10\right)$$$$=4\cdot (-4)\left(mod10\right)$$$$=(-16)\left(mod10\right)$$$$=(4-2\cdot 10)\left(mod10\right)=4$$$$sothelastdigitof{22}^{22}is4$$

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