Home/ Intrebari/Q 73252
Urmator
In Process
monicasteluta
user (0)
Pe: 2011-09-18T12:25:42+03:00 In: In:

Fie A

Fie A = 4 la puterea 2011
a) Det. restul imp. lui A la 3
b) Demonstrati ca A are cel putin 1207 cifre

Similare

3 raspunsuri

  1. maestru (V)

    Ai dreptate, red_dog! Multumesc!

    • 0
  2. veteran (III)

    4^{2011}=4\cdot 4^{2010}=(3+1)4^{2010}=3\cdot 4^{2010}+4^{2010}=\\ =3\cdot 4^{2010}+4\cdot 4^{2009}=3\cdot 4^{2010}+3\cdot 4^{2009}+4^{2009}=3\left(4^{2010}+4^{2009}\right)+4^{2009}

    Continuand in acest mod se obtine

    4^{2011}=3\left(4^{2010}+4^{2009}+\ldots+4^2+4+1\right)+1

    deci restul impartirii la 3 este 1.

    • 0
  4. user (0)

    un ajutor? :-ss si la b) ?

    • 0
Raspunde

Raspunde