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Question Number 126400 by F_Nongue last updated on 20/Dec/20

If a_n =6^n +8^n find (a_(1991) /(49)).

$${If}\:{a}_{{n}} =\mathrm{6}^{{n}} +\mathrm{8}^{{n}} \:{find}\:\frac{{a}_{\mathrm{1991}} }{\mathrm{49}}. \\ $$

Commented by talminator2856791 last updated on 20/Dec/20

are you wanting remainder?

$$\:\mathrm{are}\:\mathrm{you}\:\mathrm{wanting}\:\mathrm{remainder}? \\ $$

Commented by MJS_new last updated on 20/Dec/20

a_(1991) >10^(1798) ⇒ (a_(1991) /(49))>10^(1796)

$${a}_{\mathrm{1991}} >\mathrm{10}^{\mathrm{1798}} \:\Rightarrow\:\frac{{a}_{\mathrm{1991}} }{\mathrm{49}}>\mathrm{10}^{\mathrm{1796}} \\ $$

Commented by MJS_new last updated on 20/Dec/20

the remainder is 42

$$\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{42} \\ $$

Commented by F_Nongue last updated on 20/Dec/20

yez I want the remainder

$${yez}\:{I}\:{want}\:{the}\:{remainder} \\ $$

Commented by MJS_new last updated on 20/Dec/20

the remainder you get by finding the periodical remainders of (a_n /(49)). start with n=1, 2, 3, ... until you have one full period. then find the position of 1991 within this period

$$\mathrm{the}\:\mathrm{remainder}\:\mathrm{you}\:\mathrm{get}\:\mathrm{by}\:\mathrm{finding}\:\mathrm{the}\:\mathrm{periodical} \\ $$$$\mathrm{remainders}\:\mathrm{of}\:\frac{{a}_{{n}} }{\mathrm{49}}.\:\mathrm{start}\:\mathrm{with}\:{n}=\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:... \\ $$$$\mathrm{until}\:\mathrm{you}\:\mathrm{have}\:\mathrm{one}\:\mathrm{full}\:\mathrm{period}.\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{1991}\:\mathrm{within}\:\mathrm{this}\:\mathrm{period} \\ $$

Answered by floor(10²Eta[1]) last updated on 20/Dec/20

6^(1991) +8^(1991) (mod 49) φ(49)=7^2 −7=42 6^(1991) =(6^(47) )^(42+17) ≡6^(17) 8^(1991) ≡8^(17) a_(1991) ≡6^(17) +8^(17) (mod 49) 6^2 =36≡−13(mod 49) 6^(17) =(6^2 )^8 .6≡13^8 .6=(13^2 )^4 .6≡22^4 .6 ≡(22^2 )^2 .6≡(−6)^2 .6≡6^2 .6≡−13.6≡−78≡20(mod 49) 8^2 =64≡15(mod 49) 8^(17) ≡(8^2 )^8 .8≡15^8 .8=(15^2 )^4 .8≡(−20)^4 .8=20^4 .8 =(20^2 )^2 .8≡8^2 .8≡15.8≡22(mod 49) a_(1991) ≡20+22≡42(mod49)

$$\mathrm{6}^{\mathrm{1991}} +\mathrm{8}^{\mathrm{1991}} \left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\phi\left(\mathrm{49}\right)=\mathrm{7}^{\mathrm{2}} −\mathrm{7}=\mathrm{42} \\ $$$$\mathrm{6}^{\mathrm{1991}} =\left(\mathrm{6}^{\mathrm{47}} \right)^{\mathrm{42}+\mathrm{17}} \equiv\mathrm{6}^{\mathrm{17}} \\ $$$$\mathrm{8}^{\mathrm{1991}} \equiv\mathrm{8}^{\mathrm{17}} \\ $$$$\mathrm{a}_{\mathrm{1991}} \equiv\mathrm{6}^{\mathrm{17}} +\mathrm{8}^{\mathrm{17}} \left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{36}\equiv−\mathrm{13}\left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\mathrm{6}^{\mathrm{17}} =\left(\mathrm{6}^{\mathrm{2}} \right)^{\mathrm{8}} .\mathrm{6}\equiv\mathrm{13}^{\mathrm{8}} .\mathrm{6}=\left(\mathrm{13}^{\mathrm{2}} \right)^{\mathrm{4}} .\mathrm{6}\equiv\mathrm{22}^{\mathrm{4}} .\mathrm{6} \\ $$$$\equiv\left(\mathrm{22}^{\mathrm{2}} \right)^{\mathrm{2}} .\mathrm{6}\equiv\left(−\mathrm{6}\right)^{\mathrm{2}} .\mathrm{6}\equiv\mathrm{6}^{\mathrm{2}} .\mathrm{6}\equiv−\mathrm{13}.\mathrm{6}\equiv−\mathrm{78}\equiv\mathrm{20}\left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\mathrm{8}^{\mathrm{2}} =\mathrm{64}\equiv\mathrm{15}\left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\mathrm{8}^{\mathrm{17}} \equiv\left(\mathrm{8}^{\mathrm{2}} \right)^{\mathrm{8}} .\mathrm{8}\equiv\mathrm{15}^{\mathrm{8}} .\mathrm{8}=\left(\mathrm{15}^{\mathrm{2}} \right)^{\mathrm{4}} .\mathrm{8}\equiv\left(−\mathrm{20}\right)^{\mathrm{4}} .\mathrm{8}=\mathrm{20}^{\mathrm{4}} .\mathrm{8} \\ $$$$=\left(\mathrm{20}^{\mathrm{2}} \right)^{\mathrm{2}} .\mathrm{8}\equiv\mathrm{8}^{\mathrm{2}} .\mathrm{8}\equiv\mathrm{15}.\mathrm{8}\equiv\mathrm{22}\left(\mathrm{mod}\:\mathrm{49}\right) \\ $$$$\mathrm{a}_{\mathrm{1991}} \equiv\mathrm{20}+\mathrm{22}\equiv\mathrm{42}\left(\mathrm{mod49}\right) \\ $$$$ \\ $$

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