MHB Prove Multiples of 2003 are Divisible by 2003

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The discussion focuses on proving that the expression formed by the product of odd numbers up to 2001 added to the product of even numbers up to 2002 is divisible by 2003. Utilizing Wilson's theorem, it establishes that for a prime number p, the factorial of (p-1) is congruent to (p-1) modulo p. The products of odd and even numbers are denoted as P1 and P2, respectively, leading to the conclusion that their product modulo 2003 is equivalent to 2002. The analysis shows that x, representing P1 modulo 2003, must be a divisor of 2002, resulting in specific values for x and y. Ultimately, it is concluded that the sum of P1 and P2 is divisible by 2003.
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prove:

$1\times 3\times 5\times --------\times 2001+2\times 4\times 6\times --------\times 2002$

is a multiple of 2003
 
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Let:
$$P_1 = 1 \times 3 \times \cdots \times 2001$$
$$P_2 = 2 \times 4 \times \cdots \times 2002$$
Then:
$$P_1 = 1 \times 3 \times \cdots \times 2001 \equiv (-2002) \times (-2000) \times \cdots \times (-2) \equiv (-1)^{1001} P_2 \equiv -P_2 \pmod{2003}$$
Hence:
$$P_1 + P_2 \equiv 0 \pmod{2003}$$
$\blacksquare$
 
[sp]By Wilson's theorem,

$$(p-1)!\equiv p-1\pmod{p}$$

for any prime $$p$$.

Letting the product of odd numbers be $$P_1$$ and the product of even numbers be $$P_2$$,

$$P_1\equiv x\pmod{2003}\quad[1]$$

and

$$P_2\equiv y\pmod{2003}\quad[2]$$

So,

$$P_1\cdot P_2\equiv xy\pmod{2003}$$

$$\Rightarrow2002!\equiv xy\pmod{2003}\implies xy=2002$$ and, from $$[1]+[2]$$, $$\dfrac{x^2+2002}{2003x}$$ must be an integer:

$$x$$ must be a divisor of $$2002$$: $$\dfrac{x\cdot x+n\cdot x}{2003x}=\dfrac{x(x+n)}{2003x}\Rightarrow 2003\geq x+n$$

$$\Rightarrow x\in\{1,2002\}\Rightarrow y=\dfrac{2002}{x}$$ and $$P_1+P_2$$ is divisible by $$2003$$.[/sp]
 
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