Question

Aflati numerele naturale care impartite la9 dau catul si restul doua numere consecutivec( catul mai mic decat restul)

Answer 1

x:9=c+r
x=9c+r
r<9
c<r  numere consecutive 

c={1,2,3,4,5,6,7}

r={2,3,4,5,6,7,8}

nr sunt:
11,21,31,41,51,61,71

Answer 2

$x:9=cat+rest \\ cat\ \textless \ rest \\ x=cat*9+rest \\ cat=\{1,2,3,4,5,6,7\} \\ rest=\{2,3,4,5,6,7,8\} \\ Numerele~cautate~sunt:11,22,31,42,51,61,71. \\ Toate~numerele~din~multimea~"rest"~sunt~mai~mari~decat~numerele~ \\ ~din~multimea"cat".$