Question

calculati suma tuturor numerelor naturale de trei cifre care împartite la31 dau restul 13

Answer 1

Fie a un numar natural de 3 cifre astfel incat
$a:31=c,rest\ 13$
$a=31c+13$

a fiind de 3 cifre inseamna ca
$100 \leq a \leq 999$
[tex]100 \leq 31c+13 \leq 999 \\ 83 \leq 31c \leq 986 \\ \frac{83}{31} \leq c \leq \frac{986}{31} \\ 2,8... \leq c \leq 31,8...[/tex]
Dar c este numar natural deci c={3,4,5,...,30}

a={106,137,168,199,230,261,292,323,354,385,416,447,478,509,540,571,602,633,664,695,726,757,788,819,850,881,912,943}

106+137+168+199+230+261+292+323+354+385+416+447+478+509+540+571+602+633+664+695+726+757+788+819+850+881+912+943=14686