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I found this problem in a 2001 Math Contest RMO.

Here is what I have tried :

$\prod_{j=1}^{1000} 2j = 2^{1000} * 1000!$

and

$\prod_{k=0}^{1000} (2k+1) =(\prod_{k=1}^{1000} (2k+1))$

For the "product of odd numbers", there will be several terms, two of which will $(2^{1000})$ and $(1)$.

That is all i could do.

How to proceed ?

  • 0
    Similar question: [Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$](http://math.stackexchange.com/questions/1058993/prove-that-2-cdot-4-cdot-6-cdot-cdot-4000-1-cdot-3-cdot-5-cdot) – 2017-01-15

2 Answers 2

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$2001=3\cdot 23\cdot 29$, so the product of the odd integers is itself a multiple of $2001$.

Among the even integers, we find $2\cdot 3$, $2\cdot 23$, and $2\cdot 29$, so the product of the even integers is also a multiple of $2001$.

The difference between two multiples of $2001$ is a multiple of $2001$.

(Any halfway serious entrant in a math contest will know that contest problems often include the year of the contest as a magic constant, so they will have memorized its prime factorization and other elementary properties in advance. In any case $2001$ is obviously a multiple of $3$, and the factorization $3\cdot 667$ works equally well for this argument).

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    +2 for the second part of the answer ;-) – 2017-01-14
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Note that $\mod 2001$:

$$1\equiv -2000, 3\equiv -1998, \cdots,1999\equiv-2$$

Multiply these all together:

$$\text{product of the first 1000 odd numbers}\equiv \text{product of the first 1000 even numbers}\cdot(-1)^{1000}$$

$1000$ is even, so $(-1)^{1000}=1$, and we are done.

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    +1; this generalizes nicely to the case where 2001 is prime! – 2017-01-14