Prove Multiples of 2003 are Divisible by 2003

[Albert1]

prove:

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

is a multiple of 2003

 

[Nono713]

Spoiler

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:
$$P_1 + P_2 \equiv 0 \pmod{2003}$$
$\blacksquare$

 

[Greg]

[sp]By Wilson's theorem,

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

for any prime \(\displaystyle p\).

Letting the product of odd numbers be \(\displaystyle P_1\) and the product of even numbers be \(\displaystyle P_2\),

\(\displaystyle P_1\equiv x\pmod{2003}\quad[1]\)

and

\(\displaystyle P_2\equiv y\pmod{2003}\quad[2]\)

So,

\(\displaystyle P_1\cdot P_2\equiv xy\pmod{2003}\)

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

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

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