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From: remy <remy@fctpas.fr>
Subject: =?UTF-8?Q?crypto_sym=c3=a9trique?=
Date: Fri, 20 Nov 2020 11:24:49 +0100
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bonjour

il y avait longtemps que je n'avais pas post=C3=A9 ici
donc un petit exercice pour le confinement



alice et bob partagent un secret ici 2,3,5
mais avant le pourquoi de comment

p=3D157 =3D 157
p%2 =3D 1
p%3 =3D 1
p%5 =3D 2
p%7 =3D 3
p%11 =3D 3
p%13 =3D 1
p%17 =3D 4

q=3D7 =3D 7
q%2 =3D 1
q%3 =3D 1
q%5 =3D 2
q%7 =3D 0
q%11 =3D 7
q%13 =3D 7
q%17 =3D 7

a=3D157-7 =3D 150
a%2 =3D 0
a%3 =3D 0
a%5 =3D 0
a%7 =3D 3
a%11 =3D 7
a%13 =3D 7
a%17 =3D 14

donc en tout logique

b=3D36231  =3D 36231
bob=3D(alice+b)+ 2*3^2*5^10*56987 + 2^11*3*5*123 =3D 10017249908698
bob%2 =3D 0
bob%3 =3D 1
bob%5 =3D 3
bob%7 =3D 4
bob%11 =3D 10
bob%13 =3D 6
bob%17 =3D 5

bob a envoyer
b%2 =3D 1
b%3 =3D 0
b%5 =3D 1

que alice retrouve assez facilement

(bob-7)%2 =3D 1
(bob-7)%3 =3D 0
(bob-7)%5 =3D 1


remy


--=20
http://remyaumeunier.chez-alice.fr/
toujours autant dyslexique