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author committer david 2009-01-12 18:52:53 (GMT) david 2009-01-12 18:52:53 (GMT) 61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233 (patch) 68d3699cc0c88bfefb11370070e399600e56d16a /controle-20081202.tex 3e3e73fa205b15950394f9feb8f2ed9254e3760b (diff) infmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.zipinfmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.tar.gzinfmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.tar.bz2
Make this a little more detailed.
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 diff --git a/controle-20081202.tex b/controle-20081202.texindex cc6b8bb..cd52eaf 100644--- a/controle-20081202.tex+++ b/controle-20081202.tex@@ -92,7 +92,7 @@ Les deux questions suivantes sont indépendantes. (note : $128 = 2^7$) ? \begin{corrige}-(1) Comme $11$ et $31$ sont premiers entre eux,+(1) Comme $11$ et $31$ sont premiers entre eux, le théorème chinois affirme $\mathbb{Z}/341\mathbb{Z} \cong (\mathbb{Z}/11\mathbb{Z}) \times (\mathbb{Z}/31\mathbb{Z})$, donc il suffit de prouver $a^{31} \equiv a \pmod{31}$ et $a^{31} \equiv a \pmod{11}$ pour tout \$a