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

Find the three last digits of 5^(9999) .

$${Find}\:{the}\:{three}\:{last}\:{digits}\:{of}\:\mathrm{5}^{\mathrm{9999}} . \\ $$

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=2×4999+1 ⇒ answer is 125

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

Answered by Serlea last updated on 26/Mar/20

5=5(mod1000) 5^2 =25(mod1000) 5^4 =625(mod1000) 5^6 =625(mod1000) So for all even number ′′x′ 5^x =625(mod1000) if x≥4 5^3 =125(mod1000) 5^5 =125(mod1000) So for all odd number ′′Y′′ 5^y =125(mld1000) if Y≥3 With that 5^(9999) =5^(9998+1) =(5^(9998) )(5^1 ) =625×5(mod1000) =5^4 (mod1000) =125

$$\mathrm{5}=\mathrm{5}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{5}^{\mathrm{2}} =\mathrm{25}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{5}^{\mathrm{4}} =\mathrm{625}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{5}^{\mathrm{6}} =\mathrm{625}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{all}\:\mathrm{even}\:\mathrm{number}\:''\mathrm{x}' \\ $$$$\mathrm{5}^{\mathrm{x}} =\mathrm{625}\left(\mathrm{mod1000}\right)\:\mathrm{if}\:\mathrm{x}\geqslant\mathrm{4} \\ $$$$\mathrm{5}^{\mathrm{3}} =\mathrm{125}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{5}^{\mathrm{5}} =\mathrm{125}\left(\mathrm{mod1000}\right) \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{all}\:\mathrm{odd}\:\mathrm{number}\:''\mathrm{Y}'' \\ $$$$\mathrm{5}^{\mathrm{y}} =\mathrm{125}\left(\mathrm{mld1000}\right)\:\mathrm{if}\:\mathrm{Y}\geqslant\mathrm{3} \\ $$$$\mathrm{With}\:\mathrm{that} \\ $$$$\mathrm{5}^{\mathrm{9999}} =\mathrm{5}^{\mathrm{9998}+\mathrm{1}} =\left(\mathrm{5}^{\mathrm{9998}} \right)\left(\mathrm{5}^{\mathrm{1}} \right) \\ $$$$=\mathrm{625}×\mathrm{5}\left(\mathrm{mod1000}\right) \\ $$$$=\mathrm{5}^{\mathrm{4}} \left(\mathrm{mod1000}\right) \\ $$$$=\mathrm{125} \\ $$$$ \\ $$$$ \\ $$

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