Question

Suma a 5 numere naturale consecutive se imparte la 21 si se obtine catul 9 si restul 6.Determinati cele 5 numere

Answer 1

n
n+1
n+2
n+3
n+4
(n+n+1+n+2+n+3+n+4):21=9 rest 6
(5n+10):21=9 rest 6
5n+10=21x9+6
5n+10=189+6
5n=195-10
5n=185 =>n=37
Numerele sunt 37, 38, 39, 40 si 41.

Answer 2

(a+a+1+a+2+a+3+a+4):21=9 r6
(5a+10):21=9 r6
5a+10=21×9+6
5a=189+6-10
5a=195-10
5a=185 /:5
a=37
b=38
c=39
d=40
e=41