Răspuns :
restul trebuie sa fie mai mic decat impartitorul
r∈{0, 1, 2, 3, 4, 5, 6, 7, 8}
n:9=c rest r c=r n=9c+r
n=9r+r=10r
suma = 10·0+10·1+10·2+.....+10·8=0+10+20+...+80=360
r∈{0, 1, 2, 3, 4, 5, 6, 7, 8}
n:9=c rest r c=r n=9c+r
n=9r+r=10r
suma = 10·0+10·1+10·2+.....+10·8=0+10+20+...+80=360