A Multiple of 2003

[Albert]

$prove :1\times 3\times 5\times---\times 1999\times 2001
+2\times 4\times 6\times---\times 2000\times 2002$
is a multiple of 2003

 

[mathworker]

the second part can be written as
(2003-1)(2003-3)......(2003-2001) hence every term in the expansion contains 2003 exept the last term that is (1*3*5*....2001)(-1)^1001 hence it is negative and it cancel's out the first part of the question and hence 2003 divides every term ..
hence proved

 

[Albert]

perfect