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3-444-4-333-Find-the-remainder-when-dividing-the-number-by-7-




Question Number 227371 by Math1 last updated on 18/Jan/26
3^(444)   +  4^(333)   Find the remainder when dividing the  number by 7
$$\mathrm{3}^{\mathrm{444}} \:\:+\:\:\mathrm{4}^{\mathrm{333}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{dividing}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{by}\:\mathrm{7} \\ $$
Answered by A5T last updated on 19/Jan/26
φ(7)=6  444=6q and 333=6k+3  ⇒3^(444) +4^(333) =(3^6 )^q +(4^6 )^k 4^3 ≡1+1∙4^3 ≡65≡2(mod 7)
$$\phi\left(\mathrm{7}\right)=\mathrm{6} \\ $$$$\mathrm{444}=\mathrm{6q}\:\mathrm{and}\:\mathrm{333}=\mathrm{6k}+\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{444}} +\mathrm{4}^{\mathrm{333}} =\left(\mathrm{3}^{\mathrm{6}} \right)^{\mathrm{q}} +\left(\mathrm{4}^{\mathrm{6}} \right)^{\mathrm{k}} \mathrm{4}^{\mathrm{3}} \equiv\mathrm{1}+\mathrm{1}\centerdot\mathrm{4}^{\mathrm{3}} \equiv\mathrm{65}\equiv\mathrm{2}\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$
Answered by Kassista last updated on 24/Jan/26
  Observe that 4 and 3 are coprimes of 7, therefore,  by Fermat′s Little Theorem:  a^(p−1) ≡1 (mod p), gcd(a, p)=1  here, p=7⇒p−1=6  3^(444) +4^(333) ≡(3^6 )^(74) +(4^6 )^(55) ∙4^3 ≡1^(74) +1^(55) ∙64 ≡ 65 ≡ 2 (mod7)
$$ \\ $$$${Observe}\:{that}\:\mathrm{4}\:{and}\:\mathrm{3}\:{are}\:{coprimes}\:{of}\:\mathrm{7},\:{therefore}, \\ $$$${by}\:{Fermat}'{s}\:{Little}\:{Theorem}: \\ $$$${a}^{{p}−\mathrm{1}} \equiv\mathrm{1}\:\left({mod}\:{p}\right),\:{gcd}\left({a},\:{p}\right)=\mathrm{1} \\ $$$${here},\:{p}=\mathrm{7}\Rightarrow{p}−\mathrm{1}=\mathrm{6} \\ $$$$\mathrm{3}^{\mathrm{444}} +\mathrm{4}^{\mathrm{333}} \equiv\left(\mathrm{3}^{\mathrm{6}} \right)^{\mathrm{74}} +\left(\mathrm{4}^{\mathrm{6}} \right)^{\mathrm{55}} \centerdot\mathrm{4}^{\mathrm{3}} \equiv\mathrm{1}^{\mathrm{74}} +\mathrm{1}^{\mathrm{55}} \centerdot\mathrm{64}\:\equiv\:\mathrm{65}\:\equiv\:\mathrm{2}\:\left({mod}\mathrm{7}\right) \\ $$

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