Question

scrieti numerele de forma 78ab care se împart exact la 5 și la 3

Answer 1

78ab divizibil cu 3 si cu 5

78ab divizibil cu 3 => 7+8+a+b ={15,18,21,24,27}

=> a+b={0,3,6,9,12}

78ab divizibil cu 5 =>b={0,5}

pentru b=0 =>a={0,3,6,9}

78ab={7800, 7830, 7860, 8790}

pentru b=5 =>a={1,4,7}

78ab={7815, 7845, 7875}

Answer 2

$\overline{78ab} \: \: \vdots \: \: cu \: 3 = > (7 + 8 + a + b) \: \: \vdots \: \: 3$

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

$pt. \: \: a = 1 = > b \: \in \: \left \{ 2,5,8 \right \}$

$pt. \: a = 2 = > b \: \in \: \left \{ 1,4,7 \right \}$

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

$pt. \: a = 4 = > b \: \in \: \left \{ 2,5,8\right \}$

$pt. \: a = 5 = > b \: \in \: \left \{ 1,4,7\right \}$

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

$pt. \: a = 7 = > b \: \in \: \left \{ 2,5,8 \right \}$

$pt. \: a = 8 = > b \: \in \: \left \{ 1,4,7 \right \}$

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

$= > numerele \: sunt\::$

$7800,7803,7806,7809,7812,7815 \\ 7818,7821,7824,7827,7830,7833 \\ 7836,7839,7842,7845,7848,7851 \\ 7854,7857,7860,7863,7866,7869 \\ 7872,7875,7878,7881,7884,7887 \\ 7890,7893,7896,7899.$

$\overline{78ab} \: \: \vdots \: \: 5 = > b \: \in \: \left \{ 0,5 \right \}$

$a \: \in \: \left \{ 0,1,2,3,4,5,6,7,8,9 \right \}$

$=>numerele \: sunt\::$

$7800,7805,7810,7815,7820,7825 \\ 7830,7835,7840,7845,7850,7855 \\ 7860,7865,7870,7875,7880,7885 \\ 7890,7895.$