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Afyonlu03
@Afyonlu03
June 2021
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merci bien de repondre
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winner123
Verified answer
Bonsoir
n + n +1 + n +2 = 3 n +3 donc divisible par 3
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Isomega
1+2+3=6 6/3 = 2
3+4+5 = 12 => 12/3 =4
sinon la démonstration,
a+b+c
b=a+1
c=b+1 =a+1+1
a+b+c=a+a+a+1+1+1=3a+3=3(a+1)
si on prend 3+4+5=12
3(3+1)=3*4=12 => c'est bien un nombre divisible par 3
par conséquent 3 nombre consécutif, leur somme est toujours divisible par 3
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Afyonlu03
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thannnnnks!!!!!!!!!!!!!!!!!!!
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merci bien de m'aider
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Faites moi l'exercice 5 merci les gens
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Faites l'ex 23, que le a et b merci :))
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Verified answer
Bonsoirn + n +1 + n +2 = 3 n +3 donc divisible par 3
1+2+3=6 6/3 = 2
3+4+5 = 12 => 12/3 =4
sinon la démonstration,
a+b+c
b=a+1
c=b+1 =a+1+1
a+b+c=a+a+a+1+1+1=3a+3=3(a+1)
si on prend 3+4+5=12
3(3+1)=3*4=12 => c'est bien un nombre divisible par 3
par conséquent 3 nombre consécutif, leur somme est toujours divisible par 3