Question

1.aflati cel mai mic numar natural care impartit la 24,30 sau la 36 da de fiecare data restul 15<br />
2.impartiti numarul 158nin 3 parti invers proportionale cu numerele 3,5 si 8<br />

Answer 1

1) 24 =2^3 x 3

30 = 2 x 3 x 5

36 = 2^2 x 3^2

c.m.m.m.c (24,30,36) = 2^3 x 3^2 x 5 = 8 x 9 x 5 = 360

Numarul cautat = 360 + 15 = 375

2) 3a = 5b = 8c = k

a = k/3; b = k/5; c = k/8

a+b+c = 158 => k/3 + k/5 + k/8 = 158 => 40k/120 + 24k/120 + 15k/120 = 158

79k = 158 x 120 => k = 158x120/79 = 2x120 = 240

Atunci a = k/3 = 240/3 = 80

b = k/5 = 240/5 = 48

c = k/8 = 240/8 = 30