Question

Aflati numarulr natural care impartit la 9 dau catul si restul doua numere naturale consecutive (catul mai mic ca restul )

Answer 1

N:9= c si r
r=(c+1) <9 => c<8
N=9c+c+1
N=10c+1
c=0 =>10*0+1=1   => 1:9= 0 rest 1
c=1 =>10*1+1=11 => 11:9= 1 rest 2
c=2 =>10*2+1=21 => 21:9= 2 rest 3
c=3 =>10*3+1=31 => 31:9= 3 rest 4
c=4 =>10*4+1=41 => 41:9= 4 rest 5
c=5 =>10*5+1=51 => 51:9= 5 rest 6
c=6 =>10*6+1=61 => 61:9= 6 rest 7
c=7 =>10*7+1=71 => 71:9= 7 rest 8

N∈{1,11,21,31,41,51,61,71}