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Question Number 166392 by leicianocosta last updated on 19/Feb/22

Answered by som(math1967) last updated on 19/Feb/22

$$n=1$$$$3^4+2^7=81+128=209=11\times 19$$$$multipleof11$$$$3^{2m+2}+2^{6m+1}=11k\left(say\right)$$$$\therefore 2^{6m+1}=11k-3^{2m+2}$$$$nowform+1$$$$3^{2m+4}+2^{6m+7}$$$$3^{2m+4}+2^{6m+1}\times 2^6$$$$3^{2m+4}+(11k-3^{2m+2})\times 2^6$$$$=3^{2m+4}-3^{2m+2}\times 2^6+11k\times 2^6$$$$=11k\times 2^6-3^{2m+2}(2^6-3^2)$$$$=11k\times 2^6-3^{2m+2}\times 55$$$$=11(k\times 2^6-3^{2m+2}\times 5)$$$$multopleof11$$$$\therefore truefor(m+1)$$$$\therefore 3^{2n+2}+2^{6n+1}ismultipleof11$$

Answered by JDamian last updated on 19/Feb/22

$$q=3^{2n+2}+2^{6n+1}=3^{2(n+1)}+2^{5n+n+1}$$$$q=9^{n+1}+2^{n+1}\cdot {32}^n$$$$$$$$r=qmod11$$$$r=[{(11-2)}^{n+1}+2^{n+1}{(33-1)}^n]mod11=$$$$=[{(-2)}^{n+1}+2\cdot 2^n\cdot {(-1)}^n]mod11=$$$$=[(-2){(-2)}^{\mathit{n}}+2\cdot {(-2)}^{\mathit{n}}]mod11=$$$$=[(-2+2)\cdot {(-2)}^n]mod11=$$$$=[0\cdot {(-2)}^n]mod11=$$$$=0$$

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