Question

Scrieti toate numerele divizibile cu 3 de forma
a)7a5
b)98a
c)4a8a
d)a45a​

Answer 1

Răspuns:

$\bf \green{a)~ ~\overline{7a5}\in \{705,735,765,795\}~}$

$\bf \blue{b)~~\overline{98a}\in \{981,984,987\}~}$

$\bf \purple{c)~~\overline{4a8a}\in \{4080,4383,4686,4989\}~}$

$\bf\red{d)~~\overline{4a5a}\in \{4353,4656,4959\}~}$

Explicație pas cu pas:

Criteriul de divizibilitate cu 3: "Un număr de divide cu 3 dacă și numai dacă suma cifrelor numărului respectiv se divide cu 3"

a)

$\bf \overline{7a5} ~ \vdots~ 3\Leftrightarrow \text {\bf daca suma cifrelor se divide cu 3 }$

$\bf a - cifra$

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

$\bf 7+a+5 =( 12+a) \in M_{3}\Rightarrow( 12+a) \in \{12, 15, 18, 21 \}$

$\bf \Rightarrow a \in \{0, 3,6,9\}\Rightarrow\green{\boxed{\bf ~\overline{7a5}\in \{705,735,765,795\}~}}$

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b)

$\bf 9 + 8+a =( 17+a) \in M_{3}\Rightarrow( 17+a) \in \{18, 21,24 \}$

$\bf \Rightarrow a \in \{1,4,7\}\Rightarrow\blue{\boxed{\bf~\overline{98a}\in \{981,984,987\}~}}$

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c)

$\bf 4 + a+8+a =( 12+2a) \in M_{3}\Rightarrow( 12+2a) \in \{12,15,18,21,24,27,30 \}$

$\bf \Rightarrow a \in \{0,3,6,9\}\Rightarrow\purple{\boxed{\bf~\overline{4a8a}\in \{4080,4383,4686,4989\}~}}$

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d)

$\bf 4+a+5+a =( 9+2a) \in M_{3}\Rightarrow( 9+2a) \in \{9,12,15,18,21,24,27 \}$

$\bf \Rightarrow a \in \{3,6,9\}\Rightarrow\red{\boxed{\bf~\overline{4a5a}\in \{4353,4656,4959\}~}}$

==pav38==