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*2+1=18+1=19

3)R=3  =>C=2
D=9*3+2=27+2=29

4)R=4  =>C=3
D=9*4+3=36+3=39

5)R=5  =>C=4
D=9*5+4=45+4=49

6)R=6  =>C=5
D=9*6+5=54+5=59

7)R=7  =>C=6
D=9*7+6=63+6=69

8)R=8  =>C=7
D=9*8+7=72+7=79