Răspuns:

Explicație pas cu pas:

Explicație pas cu pas:

a)

[tex]2 \ \Big| \ \overline {35x}[/tex]

[tex]x \in \Big\{ 0; 2; 4; 6; 8\Big\}[/tex]

b)

[tex]3 \ \Big| \ \overline {15x}[/tex]

[tex]3 \ \Big| \ (1+5+x) \iff 3 \ \Big| \ (6+x) \iff 3 \ \Big| \ x \\ [/tex]

[tex]x \in \Big\{ 0; 3; 6; 9\Big\}[/tex]

c)

[tex]5 \ \Big| \ \overline {43x}[/tex]

[tex]x \in \Big\{ 0; 5\Big\}[/tex]

d)

[tex]9 \ \Big| \ \overline {42x}[/tex]

[tex]9 \ \Big| \ (4+2+x) \iff 9 \ \Big| \ (6+x)[/tex]

[tex]x = 3[/tex]

e)

[tex]6 \ \Big| \ \overline {2xy}[/tex]

[tex]2 \ \Big| \ \overline {2xy} \implies y \in \Big\{ 0; 2; 4; 6; 8\Big\}\\[/tex]

[tex]3 \ \Big| \ \overline {2xy} \implies 3 \ \Big| \ (2+x+y)[/tex]

[tex]y = 0 \implies 3 \ \Big| \ (2+x) \implies x \in \Big\{1; 4; 7\Big\} \\ [/tex]

[tex]y = 2 \implies 3 \ \Big| \ (4+x) \implies x \in \Big\{2; 5; 8\Big\}\\[/tex]

[tex]y = 4 \implies 3 \ \Big| \ (6+x) \implies x \in \Big\{0; 3; 6; 9\Big\}\\[/tex]

[tex]y = 6 \implies 3 \ \Big| \ (8+x) \implies x \in \Big\{1; 4; 7\Big\}\\[/tex]

[tex]y = 8 \implies 3 \ \Big| \ (10+x) \implies x \in \Big\{2; 5; 8\Big\}\\[/tex]

f)

[tex]15 \ \Big| \ \overline {x32y}[/tex]

[tex]5 \ \Big| \ \overline {x32y} \implies y \in \Big\{ 0; 5\Big\}[/tex]

[tex]3 \ \Big| \ \overline {x32y} \implies 3 \ \Big| \ (x+3+2+y) \iff 3 \ \Big| \ (x+5+y) \\ [/tex]

[tex]y = 0 \implies 3 \ \Big| \ (x+5) \implies x \in \Big\{1; 4; 7\Big\}\\[/tex]

[tex]y = 5 \implies 3 \ \Big| \ (x+10) \implies x \in \Big\{2; 5; 8\Big\}\\[/tex]

g)

[tex]45 \ \Big| \ \overline {1x7y}[/tex]

[tex]5 \ \Big| \ \overline {1x7y} \implies y \in \Big\{ 0; 5\Big\}[/tex]

[tex]9 \ \Big| \ \overline {1x7y} \implies 9 \ \Big| \ (1+x+7+y) \iff 9 \ \Big| \ (x+8+y) \\ [/tex]

[tex]y = 0 \implies 9 \ \Big| \ (x+8) \implies x = 1\\[/tex]

[tex]y = 5 \implies 9 \ \Big| \ (x+13) \implies x = 5\\[/tex]

h)

[tex]18 \ \Big| \ \overline {4x3y}[/tex]

[tex]2 \ \Big| \ \overline {4x3y} \implies y \in \Big\{ 0; 2; 4; 6; 8\Big\}\\[/tex]

[tex]9 \ \Big| \ \overline {4x3y} \implies 9 \ \Big| \ (4+x+3+y) \iff 9 \ \Big| \ (x+7+y)\\[/tex]

[tex]y = 0 \implies 9 \ \Big| \ (x+7) \implies x = 2\\[/tex]

[tex]y = 2 \implies 9 \ \Big| \ (x+9) \implies x \in \Big\{0; 9\Big\}\\[/tex]

[tex]y = 4 \implies 9 \ \Big| \ (x+11) \implies x = 7\\[/tex]

[tex]y = 6 \implies 9 \ \Big| \ (x+13) \implies x = 5\\[/tex]

[tex]y = 8 \implies 9 \ \Big| \ (x+15) \implies x = 3\\[/tex]