MHB Proving: 2003 Is a Product of Natural Numbers

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The discussion focuses on proving that the product of the odd numbers up to 2001 and the even numbers up to 2002 is a multiple of 2003. The second part of the product can be expressed as (2003-1)(2003-3)...(2003-2001), indicating that every term in this expansion contains 2003, except for the last term. This last term, which is negative, cancels out the first part of the product. Consequently, it is established that 2003 divides every term in the expression. The proof is concluded with a sense of accomplishment.
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$prove :1\times 3\times 5\times---\times 1999\times 2001
+2\times 4\times 6\times---\times 2000\times 2002$
is a multiple of 2003
 
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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:D
 
perfect (Clapping)
 
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