Question

determ toate nr.nat.cuprinse intre 50 si 250 care la impartirea cu 3,8 si 12 dau de fiecare data restul1

Answer 1

n:3=c1r1
n:8=c2r1
n:12=c3 r1
scriu teorema impartirii cu rest
n=3c1+1=8c2+1=12c3+1
scad 1
n-1=3c1=8c2=12c3
n-1 e un multiplu comun al nr 3,8,12  cu n care se afla in (50,250)
aflu [3,8,12]
3=3
8=2^3
12=2^2*3
[3,8,12]=2^3*3=24
M24∈{24,48,72,96, 120, 144,168,192, 216,240.....}
daca n-1 apartine multimii de mai sus ⇒ n-1 ∈{72,96,120,144,192,216,240}
n∈{73,97,121,145,193,217,241}