Question

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

Answer 1

cat 0 rest 1............9*0+1=1
cat1 rest2................9*1+2=11
cat2 rest3.................9*2+3=21
cat3 rest4.................9*3+4=31
 cat4 rest5..............................41
cat5 rest6...............................51
cat6 rest 7..............................61
cat7 rest8...............................71

deci numerele sunt 1;11;21;.....;71

Answer 2

Câtul=C
Restul=R

X÷9=C rest R. R<9
=> R={ 8,7,6,...,0}
C=>C={7,6,5,...,0}
Cazul 1.
X÷9=7 rest 8
X=9×7+8
X=71

Cazul 2.
X÷9=6 rest 7
X=9×6+7
X=61

Cazul 3.
X÷9=5 rest 6
X=9×5+6
X=51
.........

Cazul 8.
X÷9=0 rest 1
X=9×0+1
X=1

=>X={71;61;51;...;1}
Sper că te-am ajutat :-)