Question

Aflati numerele  naturale care impartite la 9 dau catul si restul doua numere naturale consecutive ( catul mai mic decat restul )

Answer 1

D:9=C+R
D=9C+R
R<9
C<R

R-nr natural
R={1,2,3,4,5,6,7,8}

C-nr natural 
C={2,3,4,5,6,7}

1)R=1 =>C=0   nu este solutie

2)R=2  =>C=1
D=9*1+2=9+2=11

3)R=3  =>C=2
D=9*2+3=18+3=21

4)R=4  =>C=3
D=9*3+4=27+4=31

5)R=5  =>C=4
D=9*4+5=36+5=41

6)R=6  =>C=5
D=9*5+6=45+6=51

7)R=7  =>C=6
D=9*6+7=54+7=61

8)R=8  =>C=7
D=9*7+8=63+8=71