Question

problema e asa:daca a, b, c sunt nr. naturale si a+b+c =18 iar 2a+b+2c=25, atunci calculeaza:

(a+b)x(3a+2b+3c) 

multumesc!

Answer 1

a+b+c=18    (1) ⇒ a+c=18-b ⇒ 2(18-b) + b = 25 ⇒ 36 - b = 25 ⇒b=11  ⇒ a+c = 7   (a=0,c=7), (a=1,c=6), (a=2,c=5), (a=3,c=4),(a=4,c=3), (a=5,c=2), (a=6,c=1), (a=7,c =0)    ⇒     ( a+b) ∈{11,12,13,14,15,16,17,18}

2a+b+2c=25  (2)      din  (1)+(2) ⇒3a+2b+3c=43

(a+b)×(3a+2b+3c) =11×43 sau 12×43 sau.........18×43 ....dar, cred ca se cerea (a+b+c)×(3a+2b+3c) = 18×43 = 774