Question

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

Answer 1

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

Answer 2

n:9=1 rest 2
n=1x9+2
n=11
n:9=2 rest 3
n=2x9+3
n=21
n:9=3 rest 4
n=3x9+4
n=31
n:9=4 rest 5
n=4x9+5
n=41
n:9=5 rest 6
n=5x9+6
n=51
n:9=6 rest 7
n=6x9+7
n=61
n:9=7x9+8
n=71
R: nr. sunt : 11 ; 21 ; 31 ; 41 ; 51 ; 61 ; 71