8 thoughts on “Asal Sayı

  1. ıfademiz
    1.2003+2.2002+…+1002.1002+1003.1004+…+2002.2+2003.1

    =(1002-1001)*(1002+1001)+(1002-1000)*(1002+1000)+…+(1002-0)*(1002+0)+…(1002+1001)*(1002-1001)
    =1002^2-1001^2+1002^2-1000^2+…+1002^2-0^2+…+1002^2-1001^2
    =2003*1002^2-2*(1^2+2^2+…+1001^2)
    =2003*1002^2-2*1001*1002*2003/6
    =2003*1002*(3006-1001)/3
    =2003*334*2005
    =2*5*167*401*2003
    asal çarpanların toplamı da 2+5+167+401+2003=2578 bulunuyor.

  2. A=1
    b=2003

    ab+(a+1)(b-1) +(a-2)(b+2) …..(a+2002)(b-2002)

    ab +
    ab-a +b -1 +
    ab-2a+2b-4 +
    ab-3a+3b-9 +
    .
    .
    ab-2002a+2002b-2002*2002

    =

    2003ab-(2002*2003)/2)a +(2002*2003)/2)b-2002*2003*4005/6

    =1 341 349 010

    asal çarpanları 2 5 167 401 2003

    ==>2578

  3. P=1.2003+2.2002+3.2001+…+2001.3+2002.2+2003.1
    P=(1002+1001)*(1002-1001)+(1002+1000)*(1002-1000)+(1002+999)*(1002-999)+…+(1002+999)*(1002-999)+(1002+1000)*(1002-1000)+(1002+1001)*(1002-1001)
    P=(1002^2-1001^2)+(1002^2-1000^2)+(1002^2-999^2)+…+(1002^2-999^2)+(1002^2-1000^2)+(1002^2-1001^2)
    P=2003*1002^2-2*(1^2+2^2+3^2+…+999^2+1000^2+1001^2)
    P=2003*1002^2-2*1001*1002*2003*/6
    P=2003*(1002^2-2*1001*1002/6)
    P=2003*669670
    669670=2*5*167*401
    2+5+167+401+2003=2578

  4. P=1.2003+2.2002+3.2001+…+2001.3+2002.2+2003.1
    P=(1002+1001)*(1002-1001)+(1002+1000)*(1002-1000)+(1002+999)*(1002-999)+…+(1002+999)*(1002-999)+(1002+1000)*(1002-1000)+(1002+1001)*(1002-1001)
    P=(1002^2-1001^2)+(1002^2-1000^2)+(1002^2-999^2)+…+(1002^2-999^2)+(1002^2-1000^2)+(1002^2-1001^2)
    P=2003*1002^2-2*(1^2+2^2+3^2+…+999^2+1000^2+1001^2)
    P=2003*1002^2-2*1001*1002*2003*/6
    P=2003*(1002^2-2*1001*1002/6)
    P=2003*669670
    669670=2*5*167*401
    2+5+167+401+2003=2578

    Matematiksel çözümü budur arkadaşlar.

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