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Determinate-m-such-that-m-is-written-abcca-in-base-5-and-is-written-bbab-in-base-8-




Question Number 124694 by mathocean1 last updated on 05/Dec/20
Determinate m such that  m is written abcca in base 5 and is  written bbab in base 8.
$${Determinate}\:{m}\:{such}\:{that} \\ $$$${m}\:{is}\:{written}\:{abcca}\:{in}\:{base}\:\mathrm{5}\:{and}\:{is} \\ $$$${written}\:{bbab}\:{in}\:{base}\:\mathrm{8}. \\ $$
Answered by floor(10²Eta[1]) last updated on 05/Dec/20
M=abcca_5 =bbab_8 ; 1≤a≤4, 0≤b, c≤4  a.5^4 +b.5^3 +c.5^2 +c.5+a=b.8^3 +b.8^2 +a.8+b  ⇒309a−226b+15c=0  309a−226b=−15c<0  309a<226b⇒a<b  309a−226b+15c≡4a−b≡0(mod5)  4a≡b(mod5)⇒(a, b)=(2, 3), (1, 4)  checking you see that (a,b)=(2,3)⇒c=4  M=23442_5 =3323_8
$$\mathrm{M}=\mathrm{abcca}_{\mathrm{5}} =\mathrm{bbab}_{\mathrm{8}} ;\:\mathrm{1}\leqslant\mathrm{a}\leqslant\mathrm{4},\:\mathrm{0}\leqslant\mathrm{b},\:\mathrm{c}\leqslant\mathrm{4} \\ $$$$\mathrm{a}.\mathrm{5}^{\mathrm{4}} +\mathrm{b}.\mathrm{5}^{\mathrm{3}} +\mathrm{c}.\mathrm{5}^{\mathrm{2}} +\mathrm{c}.\mathrm{5}+\mathrm{a}=\mathrm{b}.\mathrm{8}^{\mathrm{3}} +\mathrm{b}.\mathrm{8}^{\mathrm{2}} +\mathrm{a}.\mathrm{8}+\mathrm{b} \\ $$$$\Rightarrow\mathrm{309a}−\mathrm{226b}+\mathrm{15c}=\mathrm{0} \\ $$$$\mathrm{309a}−\mathrm{226b}=−\mathrm{15c}<\mathrm{0} \\ $$$$\mathrm{309a}<\mathrm{226b}\Rightarrow\mathrm{a}<\mathrm{b} \\ $$$$\mathrm{309a}−\mathrm{226b}+\mathrm{15c}\equiv\mathrm{4a}−\mathrm{b}\equiv\mathrm{0}\left(\mathrm{mod5}\right) \\ $$$$\mathrm{4a}\equiv\mathrm{b}\left(\mathrm{mod5}\right)\Rightarrow\left(\mathrm{a},\:\mathrm{b}\right)=\left(\mathrm{2},\:\mathrm{3}\right),\:\left(\mathrm{1},\:\mathrm{4}\right) \\ $$$$\mathrm{checking}\:\mathrm{you}\:\mathrm{see}\:\mathrm{that}\:\left(\mathrm{a},\mathrm{b}\right)=\left(\mathrm{2},\mathrm{3}\right)\Rightarrow\mathrm{c}=\mathrm{4} \\ $$$$\mathrm{M}=\mathrm{23442}_{\mathrm{5}} =\mathrm{3323}_{\mathrm{8}} \\ $$$$ \\ $$

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