Question

Numerele naturale a,b si c indeplinesc conditiile:3a+2b+3c=35 si 2a+b+2c=18. Sa se determine: a] 8a+5b+8c b] a+b+c-a*b*c

Answer 1

3a+2b+3c=35
2a+b+2c=18
a. 8a+5b+8c=2(3a+2b+3c)+2a+b+2c
8a+5b+8c=2·35+18
8a+5b+8c=88
b. a+b+c=3a+2b+3c-2a-b-2c=35-18=17
3a+2b+3c=35 ⇔3(a+c)+2b=35
2a+b+2c=18 ⇔2(a+c)+b=18
Deoarece 2/2 ⇒2/2(a+c) pentru orice a;c∈N si 2/18 ⇒2/b ⇔b numar par cu conditia b<18.
Daca b=2 ⇒a+c=8 ⇒fals;
Daca b=4 ⇒a+c=7 ⇒fals;
Daca b=6 ⇒a+c=6 ⇒fals;
Daca b=8 ⇒a+c=5 ⇒fals;
Daca b=10 ⇒a+c=4 ⇒fals;
Daca b=12 ⇒a+c=3 ⇒fals;
Daca b=14 ⇒a+c=2 ⇒fals;
Daca b=16 ⇒a+c=1 ⇒convine deoarece 3·1+2·16=3+32=35 ⇒adevarat;
Asadar avem
a+b+c-a·b·c=17-0=17 .