[tex]\overline{abc}=100a+10b+c[/tex]
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⇒ [tex]\overline{abc}[/tex] este divizibil cu 7
[tex]\overline{abcd}=1000a+100b+10c+d[/tex]
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⇒[tex]\overline{abcd}[/tex] este divizibil cu 7
Un alt exercitiu gasesti aici: https://brainly.ro/tema/1047790
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