Question

aflati numerele de forma:
123a divizibile cu trei
63bc divizibile cu 9 si 10
25ab divizibile cu 9
URGENT!!

Answer 1

$\star)\overline{123a} \: \: \vdots \: \: 3 = > (1 + 2 + 3 + a) \: \: \vdots \: \: 3$

$1 + 2 + 3 + 0 = 6 \: \: \vdots \: \: 3$

$1 + 2 + 3 + 3 = 9 \: \: \vdots \: \: 3$

$1 + 2 + 3 + 6 = 12 \: \: \vdots \: \: 3$

$1 + 2 + 3 + 9 = 15 \: \: \vdots \: \: 3$

$= > a \: \in \: \left \{0,3,6,9 \right \}$

$= > numerele \: sunt \: :\:1230,1233,1236,1239.$

$\star)\overline{63bc} \: \: \vdots \: \: 9 \: si \: 10$

$\overline{63bc} \: \: \vdots \: \: 9 = > (6 + 3 + b + c) \: \: \vdots \: \: 9$

$\overline{63bc} \: \: \vdots \: \: 10 = > c = 0$

$6 + 3 + 0 + 0 = 9 \: \: \vdots \: \: 9$

$6 + 3 + 9 + 0 = 18 \: \: \vdots \: \: 9$

$= > numerele \: sunt \: :\:6300,6390.$

$\star)\overline{25ab} \: \: \vdots \: \: 9 = > (2 + 5 + a + b) \: \: \vdots \: \: 9$

$2 + 5 + 0 + 2 = 9 \: \: \vdots \: \: 9$

$2 + 5 + 1 + 1 = 9 \: \: \vdots \: \: 9$

$2 + 5 + 2 + 0 = 9 \: \: \vdots \: \: 9$

$2 + 5 + 2 + 9 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 3 + 8 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 4 + 7 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 5 + 6 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 6 + 5 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 7 + 4 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 8 + 3 = 18 \: \: \vdots \: \: 9$

$2 + 5 + 9 + 2 = 18 \: \: \vdots \: \: 9$

$= > numerele \: sunt \: :\:2502,2511,2520,2529,2538,2547,2556 ,2565,2574,2583,2592.$