Question

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

Answer 1

d/9=c si r,r<9;
⇒r=[0,1,2,3,4,5,6,7,8];
⇒r=c+1⇒c<r;
⇒r≠0;
⇒r=[1,2,3,4,5,6,7,8];
⇒c=[0,1,2,3,4,5,6,7];
⇒d=9·0+1=0+1=1;
⇒d=9·1+2=9+2=11;
⇒d=9·2+3=18+3=21;
⇒d=9·3+4=27+4=31;
⇒d=9·4+5=36+5=41;
⇒d=9·5+6=45+6=51;
⇒d=9·6+7=54+7=61;
⇒d=9·7+8=63+8=71;
⇒d=[1,11,21,31,41,51,61,71];

Answer 2

n = 9·c + c + 1 = 10c + 1    r ∈ {1,2,3,4,5,6,7,8} ⇒ c ∈ {0,1,2,3,4,5,6,7}
n = 10c +1  ⇒ n ∈ {1,11,21,31,41,51,61,71}