Question

Aflati nr naturale x, y si z care verifica egalitatea :2 la puterea x +3 la puterea y +5 la puterea z =68

Answer 1

2^x +3^y +5^z =68

z={0;1;2}

z=0=>2^x +3^y +1 =68

2^x+3^y=67

x=6 si y=1

z=1 =>2^x +3^y +5 =68

2^x+3^y=63

3^y={1;3;9;27}si 2^x={1;2;4;8;16;32} => 2^x+3^y≠63

z=2 =>2^x +3^y +25 =68

2^x+3^y=43

=> x=4 si y=3

(x;y;z)={(4;3;2); (6;1;0)}

Answer 2

Răspuns:

descompunem pe 68 ca avand baza 2,3,5  si un exponent

68=16+27+25

68=2⁴+3³+5²=>x=4 , y=3 ,z=2