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Wenice
Pe: 2011-09-30T22:12:40+03:00 In: In:

Suma lui Gauss?

Sa se arate ca:
a) 1+2^1+2^2+2^3+…+2^197 se divide cu 7
b) 7^1+7^2+7^3+…+7^150 se divide cu 19

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4 raspunsuri

  1. edy8
    maestru (V)

    \rm{S= 1+2^1+2^2+2^3+...+2^{197} =2^0+2^1+2^2+2^3+...+2^{197}\\ Se observa ca suma contine 198 de termeni. Dar, 198 =3\cdot 66 \Rightarrow suma S poate fi scrisa astfel : \\S=(1+2^1+2^2)+2^3(1+2^1+2^2)+2^6(1+2^1+2^2)+... +2^{195}(1+2^1+2^2)=7+2^3\cdot 7 +2^6 \cdot 7 +... +2^{195} \cdot 7\\\Rightarrow S=7\cdot (1+2^3+2^6+... + 2^{195}) \Rightarrow S\in M_7\Rightarrow S \vdots 7.\bl}

    \rm{b) 7+7^2+7^3=7+49+343 =399=19\cdot 21\\Termenii sumei pot fi asociati cate trei, astfel :\\S=(7+7^2+7^3)+7^3\cdot (7+7^2+7^3)+7^6\cdot (7+7^2+7^3)+ ... +7^{147}\cdot (7+7^2+7^3) =399+7^3\cdot 399+7^6\cdot 399 + ... +7^{147}\cdot 399\\\Rightarrow S= 399\cdot (1+7^3+7^6+ ... +7^{147}) = 19\cdot 21 \cdot (1+7^3+7^6+ ... +7^{147})\Rightarrow S\in M_{19}\Rightarrow S\vdots {19}.\bl}

  2. Wenice

    Excelenta prezentare,complimente!
    Multumesc

  4. Wenice

    Mai am una interesanta:
    abcd+cdab se divide cu 101

    🙄

  5. J3anina
    maestru (V)

    Scrii numerele in baza zece si vei obtine: 1010a+101b+1010c+101d

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