Question

(27 la puterea 9 +3*2 la puterea 36+15*3 la puterea 24+6 la puterea 26)impartit la 3 LA PUTEREA 25
Determinati restul impartirii

Answer 1

[(3^3)^9+3*2^36+5*3*3^24+(2*3)^26]:3^25=
[3^27+3*2^36+5*3^25+2^26*3^26]:3^25=

[3^25*3^2+5*3^25+3^25*3*2^26+3*2^36]:3^25=
{3^25 (3^2+5+3*2^26) + 3*2^26 } : 3^25=
Conform teoremei impartirii cu rest  D=C*I+R , unde
D=3^25*3^2+5*3^25+3^25*3*2^26+3*2^36 , I= 3^25 ,  C= 3^2+5+3*2^26 ⇒ R= 3*2^26