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diamondminer
Pe: 2012-10-17T05:31:51+03:00 In: In:

13^n=6k + 7^n

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2 raspunsuri

  1. Anonim99

    ti-am raspuns la celalalt post
    faceti cumva si scrieti doar intr-un loc

  2. diamondminer

    Anonim99 wrote: ti-am raspuns la celalalt post
    faceti cumva si scrieti doar intr-un loc

    Am intrat din curiozitate la clasa a VIII-a (eu sunt intr-a VII-a) si am dat de postare. Cred ca mi-am insusit indicatiile!

        \[\begin{array}{l} \\ {\rm{Aratati ca }}p = {13^n} + {7^n} - 2 \vdots 9\\ - - - - - - - - - - - - - - - - - - - \\ {\rm{ }}p = {13^n} + {7^n} - 2 \Leftrightarrow {\rm{ }}p = \underbrace {{{13}^n} - 1}_{} + \underbrace {{7^n} - 1}_{}\\ {\rm{Din}}\left\{ {{x^0} + {x^1} + {x^2} + ... + {x^{n - 1}} = \frac{{{x^n} - 1}}{{x - 1}}} \right. \Rightarrow \left\{ \begin{array}{l} {13^n} - 1 = (13 - 1)({13^{n - 1}} + {13^{n - 2}} + ... + 13 + 1)\\ {7^n} - 1 = (7 - 1)({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1) \end{array} \right\} \Rightarrow p = 12({13^{n - 1}} + {13^{n - 2}} + ... + 13 + 1) + 6({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1) \Rightarrow \\ \Rightarrow p = \left\{ \begin{array}{l} 2\left[ {6({{13}^{n - 1}} + {{13}^{n - 2}} + ... + 13 + 1) + 3({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1)} \right]\\ 3\left[ {4({{13}^{n - 1}} + {{13}^{n - 2}} + ... + 13 + 1) + 2({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1)} \right]\\ 6\left[ {2({{13}^{n - 1}} + {{13}^{n - 2}} + ... + 13 + 1) + ({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1)} \right] \end{array} \right\} \Rightarrow \left\{ \begin{array}{l} p = M2 \Rightarrow p \vdots 2\\ p = M3 \Rightarrow p \vdots 3\\ p = M6 = 2*M3 \Rightarrow p \vdots 3 \end{array} \right\}\\ {\rm{Ne convin ultimele doua cazuri}}{\rm{.}}\\ {\rm{Trebuie sa demonstram ca si ce se gaseste in paranteza mare }}\left[ {...} \right]{\rm{ este multiplu de 3}}{\rm{.}}\\ \\ \\ {\rm{Daca }}{S_1} \vdots 3{\rm{ si }}S{}_2 \vdots 3{\rm{ atunci }}({S_1} + S{}_2) \vdots 3 \Rightarrow \left\{ \begin{array}{l} \left[ {{S_{d3}} = \underbrace {4({{13}^{n - 1}} + {{13}^{n - 2}} + ... + 13 + 1)}_{{S_1}} + \underbrace {2({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1}_{{S_2}})} \right]\\ \left[ {{S_{d6}} = \underbrace {2({{13}^{n - 1}} + {{13}^{n - 2}} + ... + 13 + 1)}_{{S_1}} + \underbrace {({7^{n - 1}} + {7^{n - 2}} + ... + 7 + 1}_{{S_2}})} \right] \end{array} \right\}\\ {\rm{Grupam convenabil cele doua sume }} \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} {S_1} = 4 + 4*13 + 4*{13^2} + ... + 4*{13^{n - 3}} + 4*{13^{n - 2}} + 4*{13^{n - 1}} \Leftrightarrow {S_1} = \underbrace {(4 + 4*13 + 4*{{13}^2})}_{3*244}\underbrace {(1 + ... + {{13}^{n - 3}} + {{13}^{n - 3}} + {{13}^{n - 3}})}_a\\ {S_2} = 2 + 2*7 + 2*{7^2} + ... + 2*{7^{n - 3}} + 2*{7^{n - 2}} + 2*{7^{n - 1}} \Leftrightarrow {S_2} = \underbrace {(2 + 2*7 + 2*{7^2})}_{3*38}\underbrace {(1 + ... + {7^{n - 3}} + {7^{n - 3}} + {7^{n - 3}})}_b \end{array} \right\}\\ \left\{ \begin{array}{l} {S_1} = 2 + 2*13 + 2*{13^2} + ... + 2*{13^{n - 3}} + 2*{13^{n - 2}} + 2*{13^{n - 1}} \Leftrightarrow {S_1} = \underbrace {(2 + 2*13 + 2*{{13}^2})}_{3*122}\underbrace {(1 + ... + {{13}^{n - 3}} + {{13}^{n - 3}} + {{13}^{n - 3}})}_a\\ {S_2} = 1 + 7 + {7^2} + ... + {7^{n - 3}} + {7^{n - 2}} + {7^{n - 1}} \Leftrightarrow {S_2} = \underbrace {(1 + 7 + {7^2})}_{3*19}\underbrace {(1 + ... + {7^{n - 3}} + {7^{n - 3}} + {7^{n - 3}})}_b \end{array} \right\} \end{array} \right]\\ \\ \Rightarrow \left\{ \begin{array}{l} {S_{d3}} = {s_1} + {s_2} = (3*244*a) + (3*38*b) = 3\left[ {(244*a) + (38*b)} \right]\\ {S_{d6}} = {s_1} + {s_2} = (3*122*a) + (3*19*b) = 3\left[ {(122*a) + (38*b)} \right] \end{array} \right\} \Rightarrow p = {13^n} + {7^n} - 2 \vdots 9 \end{array}\]

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