Question

Determinati numerele naturale cuprinse intr 1200 si 1300 care impartite la 9,6,3 dau restul 8,5,2.

Answer 1

n=9a+8
n=6b+5
n=3c+2

n+1=9(a+1)
n+1=6(b+1)
n+1=3(c+1)

n+1= multiplu comun al numerelor 9, 6 si 3
[9,6,3]=18
1200<18k<1300
67≤k≤72
n+1=67 x 18=1206, n=1205
n+1=68 x 18=1224, n=1223
n+1=69 x 18=1242, n=1241
n+1=70 x 18=1260, n=1259
n+1=71 x 18=1278, n=1277
n+1=72 x 18=1296, n=1295