Question

n=1+3+3²+3³+...+3²⁰⁰⁸+3²⁰⁰⁹
a) Aflati restul impartirii lui n la 11
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Answer 1

 

$\displaystyle\bf\\n=1+3+3^2+3^3+...+3^{2008}+3^{2009}\\\\n=3^0+3^1+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9+...+\\\\+3^{2005}+3^{2006}+3^{2007}+3^{2008}+3^{2009}\\\\Observam~ca~suma~primilor~5~termeni=121\\3^0+3^1+3^2+3^3+3^4=1+3+9+27+81=121~si~121~\vdots~11\\Exponentii~sunt~de~la~0~la~2009.\\\\\implies~suma~are~2010~termeni\\2010~\vdots~5\\\\\implies~putem~grupa~cei~2010~termenii~in~grupe~de~cate~5~termeni.$

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$\displaystyle\bf\\n=\Big(3^0+3^1+3^2+3^3+3^4\Big)+\Big(3^5+3^6+3^7+3^8+3^9\Big)+...+\\\\+\Big(3^{2005}+3^{2006}+3^{2007}+3^{2008}+3^{2009}\Big)\\\\\\n=\Big(3^0+3^1+3^2+3^3+3^4\Big)+3^5\Big(3^0+3^1+3^2+3^3+3^4\Big)+...+\\\\+3^{2005}\Big(3^0+3^1+3^2+3^3+3^4}\Big)\\\\\\n=\Big(121\Big)+3^5\times\Big(121\Big)+...+3^{2005}\times\Big(121}\Big)\\\\\\n=121\Big(3^0+3^5+...+3^{2005}\Big)\\\\\\n~\vdots~11~deoarece~are~un~factor=121=11^2\\\\\boxed{\bf\implies~~n:11~ne~da~restul~=0}$