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walonbac
@walonbac
January 2021
1
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Bonjour à tous !!
Voici la question que je dois résoudre :
Démontrer que 1 exposant 2011 + 2 exposant 2011 + 3 exposant 2011 + 4 exposant 2011 est divisible par 5
(je trouve que ça ne l'est pas)
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scoladan
Verified answer
Bonjour,
1^2011 = 1 [5]
2^
4
= 16 = 3x5 + 1 = 1 [5]
==> 2^2011 = 2^(502x
4
+ 3) = (2^4)^502 x 2^3
==> 2^2011 = 1^502 + 8 [5] = 8 [5]
3^
4
= 81 = 16x5 + 1
==> 3^2011 = 3^(502x
4
+ 3) = (3^4)^502 x 3^3
==> 3^2011 = 1^502 + 9 [5] = 9 [5]
4^
2
= 16 = 3x5 + 1
==> 4^2011 = 4^(
2
x1005 + 1) = (4^2)^1005 x 4^1
==> 4^2011 = 1^1005 + 4 [5] = 4 [5)
Somme congrue à 1 + 8 + 9 + 4 = 22 [5]
Comme toi, je trouve non divisible par 5.
Erreur énoncé ??
2 votes
Thanks 1
walonbac
merci
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Lista de comentários
Verified answer
Bonjour,1^2011 = 1 [5]
2^4 = 16 = 3x5 + 1 = 1 [5]
==> 2^2011 = 2^(502x4 + 3) = (2^4)^502 x 2^3
==> 2^2011 = 1^502 + 8 [5] = 8 [5]
3^4 = 81 = 16x5 + 1
==> 3^2011 = 3^(502x4 + 3) = (3^4)^502 x 3^3
==> 3^2011 = 1^502 + 9 [5] = 9 [5]
4^2 = 16 = 3x5 + 1
==> 4^2011 = 4^(2x1005 + 1) = (4^2)^1005 x 4^1
==> 4^2011 = 1^1005 + 4 [5] = 4 [5)
Somme congrue à 1 + 8 + 9 + 4 = 22 [5]
Comme toi, je trouve non divisible par 5.
Erreur énoncé ??