Question

aflati numerele naturale care împărțite la 9 dau Catul si restul doua nr consecutive

Answer 1

n : 9 = c rest c-1
I<R ===> R={0;1;2;3;4;5;6;7;8}

n= 9 * c + c-1
n= 9c + c-1
n=10c -1

1) R=0 <=> c-1=0 --->c=1
n=10*1-1
n=10-1
n=9

2) R=1 <=> c-1=1 --->c=2
n=10*2-1
n=20-1
n=19

3) R=2 <=> c-1=2 --->c=3
n=10*3-1
n=30-1
n=29

4) R=3 <=> c-1=3 --->c=4
n=10*4-1
n=40-1
n=39

5) R=4 <=> c-1=4 --->c=5
n=10*5-1
n=50-1
n=49

6) R=5 <=> c-1=5 --->c=6
n=10*6-1
n=60-1
n=59

7) R=6 <=> c-1=6 --->c=7
n=10*7-1
n=70-1
n=69

8) R=7 <=> c-1=7 --->c=8
n=10*8-1
n=80-1
n=79

9) R=8 <=> c-1=8 --->c=9
n=10*9-1
n=90-1
n=89

Sn={9;19;29;39;49;59;69;79;89}

Answer 2

$x:9= a\ rest\ a+1 \Leftrightarrow x=9a+a+1 \Leftrightarrow x=10a+1\ \ \ (*)$

$restul\ a+1 \ \textless \ 9 \Leftrightarrow a\ \textless \ 8\ \ \ (**)$

$(*),\ (**) \Longrightarrow x \in\{1, 11, 21, 31, 41, 51, 61, 71\}$