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6-273-8-273-49-prove-the-division-

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Question Number 98448 by Quvonchbek last updated on 14/Jun/20
6^(273) +8^(273) :49 prove the division
$$\:\:\:\:\:\:\:\mathrm{6}^{\mathrm{273}} +\mathrm{8}^{\mathrm{273}} \:\::\mathrm{49}\:\:\:\boldsymbol{{prove}}\:\:\boldsymbol{{the}}\:\:\boldsymbol{{divi}\mathrm{s}{ion}} \\ $$
Commented by Rasheed.Sindhi last updated on 14/Jun/20
49 ∣ 2^(273) (3^(273) +4^(273) ) ∵ 49 ∤ 2^(273) ∴ 49 ∣ (3^(273) +4^(273) ) Or 3^(273) +4^(273) ≡0(mod 49)
$$\:\:\:\:\:\:\mathrm{49}\:\mid\:\mathrm{2}^{\mathrm{273}} \left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right) \\ $$$$\because\:\mathrm{49}\:\nmid\:\mathrm{2}^{\mathrm{273}} \:\:\therefore\:\mathrm{49}\:\mid\:\left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right) \\ $$$$\:\:\:{Or}\:\:\:\:\:\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \equiv\mathrm{0}\left({mod}\:\mathrm{49}\right) \\ $$
Commented by mr W last updated on 14/Jun/20
do we need to prove that 3^(273) +4^(273) ≡0 (mod 49) here sir?
$${do}\:{we}\:{need}\:{to}\:{prove}\:{that} \\ $$$$\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \equiv\mathrm{0}\:\left({mod}\:\mathrm{49}\right)\: \\ $$$${here}\:{sir}? \\ $$
Commented by Rasheed.Sindhi last updated on 14/Jun/20
Yes sir, If I understood the question.
$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{If}\:\mathrm{I}\:\mathrm{understood}\:\mathrm{the}\:\mathrm{question}. \\ $$
Commented by 1549442205 last updated on 14/Jun/20
3^(273) +4^(273) =27^(91) +64^(91) =91.(27^(90) −27^(89) .64+27^(88) .64^2 −...−27.64^(89) +64^(90) ) Since 91=13.7,it is enough to prove that A=(27^(90) −27^(89) .64+27^(88) .64^2 −...−27.64^(89) +64^(90) )⋮7 Since 64^k =(63+1)^k =1(mod7)and 27^(90−k) =(28−1)^(90−k) =(−1)^(90−k) (mod7), so 27^(90−k) .64^k =(−1)^(90−k) .=(−1)^(90−k) (mod7)(k=0,1...90) it follows that A=1−1+1−1+...−1+1=1(mod7) because A has 91 terms Therefore A isn′t divisible by 7 and so 3^(273) +4^(273) isn′t divisible by 49 it is only divisible by 7 I answered the question of Mr.W
$$ \\ $$$$\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} =\mathrm{27}^{\mathrm{91}} +\mathrm{64}^{\mathrm{91}} =\mathrm{91}.\left(\mathrm{27}^{\mathrm{90}} −\mathrm{27}^{\mathrm{89}} .\mathrm{64}+\mathrm{27}^{\mathrm{88}} .\mathrm{64}^{\mathrm{2}} −…−\mathrm{27}.\mathrm{64}^{\mathrm{89}} +\mathrm{64}^{\mathrm{90}} \right) \\ $$$$\mathrm{Since}\:\mathrm{91}=\mathrm{13}.\mathrm{7},\mathrm{it}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\mathrm{A}=\left(\mathrm{27}^{\mathrm{90}} −\mathrm{27}^{\mathrm{89}} .\mathrm{64}+\mathrm{27}^{\mathrm{88}} .\mathrm{64}^{\mathrm{2}} −…−\mathrm{27}.\mathrm{64}^{\mathrm{89}} +\mathrm{64}^{\mathrm{90}} \right)\vdots\mathrm{7} \\ $$$$\mathrm{Since}\:\mathrm{64}^{\mathrm{k}} =\left(\mathrm{63}+\mathrm{1}\right)^{\mathrm{k}} =\mathrm{1}\left(\mathrm{mod7}\right)\mathrm{and} \\ $$$$\mathrm{27}^{\mathrm{90}−\mathrm{k}} =\left(\mathrm{28}−\mathrm{1}\right)^{\mathrm{90}−\mathrm{k}} =\left(−\mathrm{1}\right)^{\mathrm{90}−\mathrm{k}} \left(\mathrm{mod7}\right), \\ $$$$\mathrm{so}\:\mathrm{27}^{\mathrm{90}−\mathrm{k}} .\mathrm{64}^{\mathrm{k}} =\left(−\mathrm{1}\right)^{\mathrm{90}−\mathrm{k}} .=\left(−\mathrm{1}\right)^{\mathrm{90}−\mathrm{k}} \left(\mathrm{mod7}\right)\left(\mathrm{k}=\mathrm{0},\mathrm{1}…\mathrm{90}\right) \\ $$$$\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{A}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+…−\mathrm{1}+\mathrm{1}=\mathrm{1}\left(\mathrm{mod7}\right) \\ $$$$\mathrm{because}\:\mathrm{A}\:\mathrm{has}\:\mathrm{91}\:\mathrm{terms} \\ $$$$\mathrm{Therefore}\:\mathrm{A}\:\mathrm{isn}'\mathrm{t}\:\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\:\mathrm{and}\:\mathrm{so}\: \\ $$$$\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{49} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{only}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\mathrm{I}\:\mathrm{answered}\:\mathrm{the}\:\mathrm{question}\:\mathrm{of}\:\mathrm{Mr}.\mathrm{W} \\ $$$$ \\ $$
Commented by mr W last updated on 14/Jun/20
thanks sir! but 3^(273) +4^(273) is divisible by 49.
$${thanks}\:{sir}! \\ $$$${but}\:\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \:{is}\:{divisible}\:{by}\:\mathrm{49}. \\ $$
Answered by Rasheed.Sindhi last updated on 14/Jun/20
6^(273) +8^(273) ≡0(mod 49) ⇒2^(273) (3^(273) +4^(273) )≡0(mod 49) ⇒(3^(273) +4^(273) )≡0(mod 49) [∵ 49∤2^(273) ] ^• 4^(21) ≡1(mod 49) (4^(21) )^(13) ≡(1)^(13) (mod 49) 4^(273) ≡1(mod 49)............(i) ^• 3^(42) ≡1(mod 49) (3^(42) )^6 ≡(1)^6 (mod 49) 3^(252) ≡1(mod 49) 3^(252) .3^(21) ≡1.3^(21) ≡48(mod 49) 3^(273) ≡48(mod 49)...........(ii) (i)+(ii): 3^(273) +4^(273) ≡49≡0(mod49) 3^(273) +4^(273) ≡0(mod49) Proved
$$\mathrm{6}^{\mathrm{273}} +\mathrm{8}^{\mathrm{273}} \equiv\mathrm{0}\left({mod}\:\mathrm{49}\right) \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{273}} \left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right)\equiv\mathrm{0}\left({mod}\:\mathrm{49}\right) \\ $$$$\Rightarrow\left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right)\equiv\mathrm{0}\left({mod}\:\mathrm{49}\right)\:\left[\because\:\mathrm{49}\nmid\mathrm{2}^{\mathrm{273}} \right] \\ $$$$\:\:^{\bullet} \:\:\mathrm{4}^{\mathrm{21}} \equiv\mathrm{1}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\left(\mathrm{4}^{\mathrm{21}} \right)^{\mathrm{13}} \equiv\left(\mathrm{1}\right)^{\mathrm{13}} \left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}^{\mathrm{273}} \equiv\mathrm{1}\left({mod}\:\mathrm{49}\right)…………\left({i}\right) \\ $$$$\:^{\bullet} \:\:\:\mathrm{3}^{\mathrm{42}} \equiv\mathrm{1}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{3}^{\mathrm{42}} \right)^{\mathrm{6}} \equiv\left(\mathrm{1}\right)^{\mathrm{6}} \left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{252}} \equiv\mathrm{1}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{252}} .\mathrm{3}^{\mathrm{21}} \equiv\mathrm{1}.\mathrm{3}^{\mathrm{21}} \equiv\mathrm{48}\left({mod}\:\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{273}} \equiv\mathrm{48}\left({mod}\:\mathrm{49}\right)………..\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right):\:\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \equiv\mathrm{49}\equiv\mathrm{0}\left({mod}\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \equiv\mathrm{0}\left({mod}\mathrm{49}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{Proved} \\ $$
Commented by mr W last updated on 14/Jun/20
great!
$${great}! \\ $$

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