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2014-11-03T11:14:21+02:00
[(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