Question

suma a patru numere naturale consecutive se inparte la 23si se obtine catul 3 si restul 5 ajutatima

Answer 1

[a+(a+1)+(a+2)+(a+3)]:23= 3 rest 5
(4a+6):23= 3 rest 5 => 4a+6= 23×3+5 => 4a+6= 69+5 => 4a+6= 74 => 4a= 74-6 =>
=> 4a= 68 => a= 68:4 => a= 17
a+1= 18
a+2= 19
a+3= 20

(17+18+19+20):23= 74:23= 3 rest 5

Answer 2

(x + x+1 + x+2 + x+3) : 23 = 3  (rest 5)

              (4x + 6) ; 23 = 3 (rest 5)  ⇔  4x+6=23*3+5

                                                            4x = 68 ⇒  x = 17 

R:     17;18;19;20