Question

5. Determinaţi cifrele x şi y în fiecare dintre cazurile: a) 2 | 35x; b) 3 | 15x; e) 6| 2xy; f) 15 | x32y; c) 5| 43x; g) 45 | 1x7y; d) 9 | 42x; h) 18 | 4x3y.​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Explicație pas cu pas:

a)

$2 \ \Big| \ \overline {35x}$

$x \in \Big\{ 0; 2; 4; 6; 8\Big\}$

b)

$3 \ \Big| \ \overline {15x}$

$3 \ \Big| \ (1+5+x) \iff 3 \ \Big| \ (6+x) \iff 3 \ \Big| \ x$

$x \in \Big\{ 0; 3; 6; 9\Big\}$

c)

$5 \ \Big| \ \overline {43x}$

$x \in \Big\{ 0; 5\Big\}$

d)

$9 \ \Big| \ \overline {42x}$

$9 \ \Big| \ (4+2+x) \iff 9 \ \Big| \ (6+x)$

$x = 3$

e)

$6 \ \Big| \ \overline {2xy}$

$2 \ \Big| \ \overline {2xy} \implies y \in \Big\{ 0; 2; 4; 6; 8\Big\}$

$3 \ \Big| \ \overline {2xy} \implies 3 \ \Big| \ (2+x+y)$

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

$y = 2 \implies 3 \ \Big| \ (4+x) \implies x \in \Big\{2; 5; 8\Big\}$

$y = 4 \implies 3 \ \Big| \ (6+x) \implies x \in \Big\{0; 3; 6; 9\Big\}$

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

$y = 8 \implies 3 \ \Big| \ (10+x) \implies x \in \Big\{2; 5; 8\Big\}$

f)

$15 \ \Big| \ \overline {x32y}$

$5 \ \Big| \ \overline {x32y} \implies y \in \Big\{ 0; 5\Big\}$

$3 \ \Big| \ \overline {x32y} \implies 3 \ \Big| \ (x+3+2+y) \iff 3 \ \Big| \ (x+5+y)$

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

$y = 5 \implies 3 \ \Big| \ (x+10) \implies x \in \Big\{2; 5; 8\Big\}$

g)

$45 \ \Big| \ \overline {1x7y}$

$5 \ \Big| \ \overline {1x7y} \implies y \in \Big\{ 0; 5\Big\}$

$9 \ \Big| \ \overline {1x7y} \implies 9 \ \Big| \ (1+x+7+y) \iff 9 \ \Big| \ (x+8+y)$

$y = 0 \implies 9 \ \Big| \ (x+8) \implies x = 1$

$y = 5 \implies 9 \ \Big| \ (x+13) \implies x = 5$

h)

$18 \ \Big| \ \overline {4x3y}$

$2 \ \Big| \ \overline {4x3y} \implies y \in \Big\{ 0; 2; 4; 6; 8\Big\}$

$9 \ \Big| \ \overline {4x3y} \implies 9 \ \Big| \ (4+x+3+y) \iff 9 \ \Big| \ (x+7+y)$

$y = 0 \implies 9 \ \Big| \ (x+7) \implies x = 2$

$y = 2 \implies 9 \ \Big| \ (x+9) \implies x \in \Big\{0; 9\Big\}$

$y = 4 \implies 9 \ \Big| \ (x+11) \implies x = 7$

$y = 6 \implies 9 \ \Big| \ (x+13) \implies x = 5$

$y = 8 \implies 9 \ \Big| \ (x+15) \implies x = 3$