Question

4. Demonstrați că:
a) dacă 3/(a+6)→ 3|a
c) dacă 5 (a+20) ⇒ 5|a
b) dacă 2/(a+4) ⇒ 2/a
d) 7/(a+14b+21) ⇒7|ae) 9/(a+18b+9c) ⇒
9|a

Answer 1

Explicație pas cu pas:

a)

$\dfrac{a + 6}{3} = \dfrac{a}{3} + \dfrac{6}{3} = \dfrac{a}{3} + 2 \in \mathbb{Z} \\ \implies \dfrac{a}{3} \in \mathbb{Z} \implies 3 \ \Big| \ a$

c)

$\dfrac{a + 20}{5} = \dfrac{a}{5} + \dfrac{20}{5} = \dfrac{a}{5} + 4 \in \mathbb{Z} \\ \implies \dfrac{a}{5} \in \mathbb{Z} \implies 5 \ \Big| \ a$

b)

$\dfrac{a + 4}{2} = \dfrac{a}{2} + \dfrac{4}{2} = \dfrac{a}{2} + 2 \in \mathbb{Z} \\ \implies \dfrac{a}{2} \in \mathbb{Z} \implies 2 \ \Big| \ a$

d)

$\dfrac{a + 14b + 21}{7} = \dfrac{a}{7} + \dfrac{7(2b + 3)}{7} = \dfrac{a}{7} + (2b + 3) \in \mathbb{Z} \\ \implies \dfrac{a}{7} \in \mathbb{Z} \implies 7 \ \Big| \ a$

e)

$\dfrac{a + 18b + 9c}{9} = \dfrac{a}{9} + \dfrac{9(2b + 1)}{9} = \dfrac{a}{9} + (2b + 1) \in \mathbb{Z} \\ \implies \dfrac{a}{9} \in \mathbb{Z} \implies 9 \ \Big| \ a$