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Find-the-three-last-digits-of-5-9999-




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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