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Prove-that-3-3n-26n-1-is-divisible-by-676-where-n-Z-

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Question Number 2545 by Rasheed Soomro last updated on 22/Nov/15
Prove that 3^(3n) −26n−1 is divisible by 676, where n∈Z^+
Provethat33n−26n−1isdivisibleby676,wheren∈Z+
Commented by Yozzi last updated on 22/Nov/15
Use induction.
Useinduction.
Answered by Yozzi last updated on 22/Nov/15
Let P(n)=3^(3n) −26n−1=(27)^n −26n−1 P(1)=27−26−1=0=676×0 P(2)=27^2 −52−1=676=1×676 P(3)=27^3 −26×3−1=29×676 Let P(n) be the proposition that, ∀n∈Z^+ , P(n)=676m where P(n)=27^n −26n−1 and m∈Z^≥ . For n=1, P(1)=27−26−1=0=676×0 0∈Z^≥ ⇒m=0 ∴ P(n) is true for n=1. Assume P(n) is true for n=k, i.e P(k)=676m m∈Z^≥ . For n=k+1, observe the difference Δ=P(k+1)−P(k). P(k+1)=27^(k+1) −26(k+1)−1 Δ=27×27^k −26k−26−1−27^k +26k+1 Δ=26×27^k −26 Δ=26(27^k −1) ⇒P(k+1)=P(k)+26(27^k −1) P(k+1)=676m+26(27^k −1) Now, 27^k −1=(27−1)(Σ_(r=0) ^(k−1) 27^r )=26(Σ_(r=0) ^(k−1) 27^r ) ∴P(k+1)=676m+676Σ_(r=0) ^(k−1) 27^r P(k+1)=676(m+Σ_(r=0) ^(k−1) 27^r )=676q where q∈Z^+ since {m+Σ_(r=0) ^(k−1) 27^r }∈Z^+ . ⇒ P(k+1) is true if P(k) is true. Since P(1) is true then, by P.M.I, P(n) is true ∀n∈Z^+ .
LetP(n)=33n−26n−1=(27)n−26n−1P(1)=27−26−1=0=676×0P(2)=272−52−1=676=1×676P(3)=273−26×3−1=29×676LetP(n)bethepropositionthat,∀n∈Z+,P(n)=676mwhereP(n)=27n−26n−1andm∈Z⩾.Forn=1,P(1)=27−26−1=0=676×00∈Z⩾⇒m=0∴P(n)istrueforn=1.AssumeP(n)istrueforn=k,i.eP(k)=676mm∈Z⩾.Forn=k+1,observethedifferenceΔ=P(k+1)−P(k).P(k+1)=27k+1−26(k+1)−1Δ=27×27k−26k−26−1−27k+26k+1Δ=26×27k−26Δ=26(27k−1)⇒P(k+1)=P(k)+26(27k−1)P(k+1)=676m+26(27k−1)Now,27k−1=(27−1)(∑k−1r=027r)=26(∑k−1r=027r)∴P(k+1)=676m+676∑k−1r=027rP(k+1)=676(m+∑k−1r=027r)=676qwhereq∈Z+since{m+∑k−1r=027r}∈Z+.⇒P(k+1)istrueifP(k)istrue.SinceP(1)istruethen,byP.M.I,P(n)istrue∀n∈Z+.
Commented by Rasheed Soomro last updated on 22/Nov/15
NICE^(Very) !
NICEVery!
Answered by Rasheed Soomro last updated on 22/Nov/15
AN EASY Approach p(n): 3^(3n) −26n−1 is divisible by 676 Case−1_(−) :For n=1 3^(3n) −26n−1=3^3 −26−1=0 and 676∣0 Hence p(n) is true for n=1 Case−2 : Assume that for n=k , p(n) is true I−e 676 ∣ (3^(3k) −26k−1) Now for n=k+1, we have 3^(3(k+1)) −26(k+1)−1 =3^(3k+3) −26k−26−1 =27.3^(3k) −26k−27 =27.3^(3k) −27(26k)−27+26(26k) =27(3^(3k) −26k−1)+676k ∵ 676 ∣ (3^(3k) −26k−1) [By hypothesis]] and 676 ∣ 676k ∴ 676 ∣ 27(3^(3k) −26k−1)+676k ∴ p(n) is true for n=k+1,if it is true for n=k QED
ANEASYApproachp(n):33n−26n−1isdivisibleby676Case−1−:Forn=133n−26n−1=33−26−1=0and676∣0Hencep(n)istrueforn=1Case−2:Assumethatforn=k,p(n)istrueI−e676∣(33k−26k−1)Nowforn=k+1,wehave33(k+1)−26(k+1)−1=33k+3−26k−26−1=27.33k−26k−27=27.33k−27(26k)−27+26(26k)=27(33k−26k−1)+676k∵676∣(33k−26k−1)[Byhypothesis]]and676∣676k∴676∣27(33k−26k−1)+676k∴p(n)istrueforn=k+1,ifitistrueforn=kQED
Commented by Yozzi last updated on 22/Nov/15
Nicely □
Nicely◻
Commented by Rasheed Soomro last updated on 23/Nov/15
□ THANKS! Actually I saw this problem and its solution in a book and wanted to share with friends on this platform. Therefore in order to get oppertunity I posted the question. I like your answer because its your own whereas my answer is bookish and not my own.Anyway it is such that it be shared.
◻THANKS!ActuallyIsawthisproblemanditssolutioninabookandwantedtosharewithfriendsonthisplatform.ThereforeinordertogetoppertunityIpostedthequestion.Ilikeyouranswerbecauseitsyourownwhereasmyanswerisbookishandnotmyown.Anywayitissuchthatitbeshared.

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