Question

a)Aratati ca daca (2a+3b+c) este divizibil cu 7 , atunci 7 divide abc in baza zece
b)Aratati ca daca 7 divide (6a+2b+3c+d) , atunci abcd in baza zece divizibil cu 7.

Answer 1

$\overline{abc}=100a+10b+c$

100a+10b+c=98a+2a+7b+3b+c

100a+10b+c=98a+7b+2a+3b+c

100a+10b+c=7(14a+b)+2a+3b+c

7 | 2a+3b+c

7 | 7(14a+b) ⇒ 7 | 7(14a+b)+2a+3b+c⇒ $\overline{abc}$ este divizibil cu 7

$\overline{abcd}=1000a+100b+10c+d$

1000a+100b+10c+d=994a+6a+98b+2b+7c+3c+d

1000a+100b+10c+d=994a+98b+7c+6a+2b+3c+d

1000a+100b+10c+d=7(142a+14b+c)+6a+2b+3c+d

7 | 6a+2b+3c+d

7 | 7(142a+14b+c)⇒ 7 | 7(142a+14b+c)+6a+2b+3c+d⇒$\overline{abcd}$ este divizibil cu 7

Un alt exercitiu gasesti aici: https://brainly.ro/tema/1047790

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