Question

Fie S suma numerelor naturale abc cu abc=11 (a+b+c)+cba.Justificati daca S este patrat perfect.
Dau coroana.

Answer 1

Salut!

Folosim scrierea in baza 10!
Avem ca 100a + 10b + c =11a + 11b + 11c + 100c + 10b + a; unde a si c sunt nenule;
Atunci, 88a = 110c + 11b <=> 8a = 11c + b <=> 8( a - c ) = 3c + b => 8 divide pe 3c + b;
Fie b = 0 =>8 / 3c => c = 8 => 8a = 88 => a = 11, Fals;
Fie b = 1 => 8 / 3c + 1 => c = 5 => 8a = 56 => a = 7 => 715;
Fie b = 2 => 8 / 3c + 2 => c = 2 => 8a = 24 => a = 3 => 322;
Fie b = 3 => 8 / 3c + 3 =>  c = 5 => 8a = 58, Fals;
Fie b = 4 => 8 / 3c + 4 => c = 4 => 8a = 48 => a = 6 => 644;
Fie b = 5 => 8 / 3c + 5 => c = 1 => 8a = 16 => a = 2 => 251;
Fie b = 6 => 8 / 3c + 6 => c = 6 => 8a = 72 => a = 9 => 966;
Fie b = 7 => 8 / 3c + 7 => c = 3 => 8a = 40 => a = 5 > 573;
Fie b = 8 => 8 / 3c + 8 => c = 0, Fals;
Fie b = 9 => 8 / 3c + 9 => c = 5 => 8a = 64 => a = 8 => 895;

Suma este 4366.

Nu e patrat perfect pentru ca 66^2 = 4356 si 67^2 = 4489;

Bafta!