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Question Number 86018 by Maclaurin Stickker last updated on 26/Mar/20

$$Findthethreelastdigitsof{5}^{9999}.$$

Answered by MJS last updated on 26/Mar/20

$$5^{1}005$$$$5^{2}025$$$$5^{3}125$$$$5^{4}625$$$$5^{5}125$$$$5^{6}625$$$$...$$$$5^{2n+1}125$$$$5^{2n}625$$$$9999=2\times 4999+1\Rightarrow \mathrm{answer}\mathrm{is}125$$

Answered by Serlea last updated on 26/Mar/20

$$5=5\left(\mathrm{mod}1000\right)$$$$5^2=25\left(\mathrm{mod}1000\right)$$$$5^4=625\left(\mathrm{mod}1000\right)$$$$5^6=625\left(\mathrm{mod}1000\right)$$$$\mathrm{So}\mathrm{for}\mathrm{all}\mathrm{even}\mathrm{number}{}^{″}{\mathrm{x}}^{\prime }$$$$5^{\mathrm{x}}=625\left(\mathrm{mod}1000\right)\mathrm{if}\mathrm{x}⩾4$$$$5^3=125\left(\mathrm{mod}1000\right)$$$$5^5=125\left(\mathrm{mod}1000\right)$$$$\mathrm{So}\mathrm{for}\mathrm{all}\mathrm{odd}\mathrm{number}{}^{″}{\mathrm{Y}}^{″}$$$$5^{\mathrm{y}}=125\left(\mathrm{mld}1000\right)\mathrm{if}\mathrm{Y}⩾3$$$$\mathrm{With}\mathrm{that}$$$$5^{9999}=5^{9998+1}=\left(5^{9998}\right)\left(5^1\right)$$$$=625\times 5\left(\mathrm{mod}1000\right)$$$$=5^4\left(\mathrm{mod}1000\right)$$$$=125$$$$$$$$$$

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