Question

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

Answer 1

[x + (x+1) + (x+2) + (x+3) + (x+4)] : 21 = 9 r 6
=> (5x + 10) : 21 = 9 r 6
Conform T. I. R: (5x + 10) = 9 * 21 + 6
=> (5x + 10) = 195
=> 5x = 185
=> x = 185/5
=> x = 37

Deci cele 5 numere sunt 37, 38, 39, 40 și 41.

Answer 2

Fie n, n+1, n+2, n+3 si n+4-cele 5 nr. naturale consecutive.
$\frac{n+n+1+n+2+n+3+n+4}{21} =9 (rest 6)$
$\frac{5n+10}{21} =9 (rest 6)$
$D=I*C+R, R\ \textless \ I$
[tex] \frac{5n+10}{21} =9 (rest 6)=\ \textgreater \ 5n+10=21*9+6, 6\ \textless \ 21 5n+10=189+6 5n+10=195 5n=195-10 5n=185 n=185:5 n=37[/tex]
Numere sunt:37, 38, 39, 40 si 41.