Răspuns:
5.
abc + bca + cab
= a·10²+b·10+c+ b·10²+c·10+a+ c·10²+a·10+b =
= a(10²+10+1)+b(10²+10+1)+c(10²+10+1) =
= (10²+10+1)(a+b+c) =
= 111·(a+b+c) =
= 37·3·(a+b+c) - divizibil cu 37
6.
x- numar truse continand cate 4 creioane
y - numar truse continand cate 2 stilouri
⇒ numar total creioane = 4·x
numar total stilouri = 2·y
32 - numar total creioane+ stilouri
⇒ 32=4·x+2·y
10 numar total truse ⇒ x+y=10
⇒[tex]\left \{ {{4x+2y=32} \atop {x+y=10}} \right. [/tex]
x+y=10 ⇒ y=10-x
4·x+2·y=32 ⇔4·x+2·(10-x)=32 ⇔ 4x+20-2x=32 ⇔ 2x = 12 ⇒ x=6
y=10-x=10-6=4
⇒ 6 truse creioane
4 truse stilouri
