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| author | david <david> | 2009-01-12 18:52:53 +0000 |
|---|---|---|
| committer | david <david> | 2009-01-12 18:52:53 +0000 |
| commit | 61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233 (patch) | |
| tree | 68d3699cc0c88bfefb11370070e399600e56d16a | |
| parent | 3e3e73fa205b15950394f9feb8f2ed9254e3760b (diff) | |
| download | infmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.tar.gz infmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.tar.bz2 infmdi720-61797dc2666eec5f4f7cfeb4ae70c1e8d9f53233.zip | |
Make this a little more detailed.
| -rw-r--r-- | controle-20081202.tex | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/controle-20081202.tex b/controle-20081202.tex index 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 |