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Re: [obm-l] Problema



1002 ~ -1001 (mod 2003)
1003 ~ -1000 (mod 2003)
...
2002 ~ -1 (mod 2003)
 
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1002*1003*...*2002 = (-1)(-2)...(-1001) = -(1*2*...*1001) mod (2003)
dessa forma temos
1*2*3*...*1001 + 1002*1003*...*2002 = 1*2*3*...*1001 - 1*2*3*...*1001 = 0 (mod 2003)
 
[ ]'s
----- Original Message -----
From: Benedito
Sent: Monday, August 04, 2003 10:56 AM
Subject: [obm-l] Problema

Problema
Mostre que o abaixo é divisível por 2003:
1*2*3*...*1001 + 1002*1003*...*2002.
 
Benedito Freire