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Lin
Pe: 2011-06-07T14:30:29+03:00 In: In:

Problema

Demonstrati ca daca a+c = b+d, atunci numarul abcd barat este divizibil cu 11.

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3 raspunsuri

  1. dan
    expert (VI)

    Lin wrote: Demonstrati ca daca a+c = b+d, atunci numarul abcd barat este divizibil cu 11.

    Multumesc!

    \rm{\overline{abcd}=1000a+100b+10c+d}=10(100a+c)+(100b+d) \\ dar \;a+c=b+d \;\Rightarrow \;pentru\;a=b\;si\; c=d\;atunci \;10*(100a+c)+(100a+c)=(100a+c)(10+1)\,\,\vdots\,\,11

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  2. bedrix
    expert (VI)

    Mai general:
    abcd barat = 1000*a+100*b+10*c+d = 1000*a+10*c+99*b+b+d =
    =1000*a+10*c+99*b+a+c =
    1001*a+99*b+11*c = 11*(91*a+9*b+c) = M11

  4. ambri
    veteran (III)

    \rm{Notez a+c=b+d=n (1)\\ \bar {abcd}=1000a+100b+10c+d=990a+10a+99b+b+10c+d= (990a+99b)+(10a+10c)+(b+d) = \\ =(M_{11}+M_{11}) +10(a+c)+(b+d) (2)\\ (1), (2) \Rightarrow \bar { abcd }= M_{11}+10n+n = \\ =M_{11}+11n = M_{11}+M_{11} = M_{11}.\bl}

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