$-3=(-1)\cdot1\cdot3$ is the product of three consecutive odd integers, so any integer that divides all such products must divide $-3$. The divisors of $-3$ are $-1,1,-3$, and $3$, so these are the only candidates. To finish the proof, we need only show that $3$ always does divide the product of three consecutive odd integers.
Let $n$ be any odd integer, and consider the product $m=n(n+2)(n+4)$. There are three possibilities.
- If $n$ is a multiple of $3$, then certainly so is $m$.
- If $n=3k+1$ for some integer $k$, then $n+2=3k+3=3(k+1)$ is a multiple of $3$, and therefore so is $m$.
- If $n=3k+2$ for some integer $k$, then $n+4=3k+6=3(k+2)$ is a multiple of $3$, and therefore so is $m$.
In every case, therefore, $m$ is a multiple of $3$, and $3$ is therefore the largest integer that divides every product of three consecutive odd integers.
Slightly more generally, note that multiplying $\rm\:m\, =\, k+an\:$ by any integer of the form $\rm\:1+bn\:$ doesn't change the remainder that $\rm\,m\,$ leaves when divided by $\rm\,n,\,$ i.e. the remainder stays = $\rm k,\,$ by
$$\rm (k+an)(1+bn)\, =\, k+n(a+b(k+an))$$
This is a special case $\rm\,j=1\,$ of $\rm\ mod\ n\!:\ \ \begin{eqnarray} x &\equiv&\,\rm k\\ \rm y &\equiv&\,\rm j\end{eqnarray}\ \Rightarrow\ xy\equiv\, k\, j,\ \ $ the Congruence Product Rule
That $\rm\: a+b(k+an)\in \Bbb Z\:$ follows from the fact that $\rm\:a,b,c\in \Bbb Z\:\Rightarrow\: a + b\,c\in \Bbb Z,\:$ since integers are closed under multiplication, thus $\rm\:bc\in \Bbb Z,\:$ and also under addition, hence $\rm\:a + bc\in\Bbb Z.\:$ Finally, that $\rm\,\Bbb Z\,$ is closed under the operations of addition and multiplication follows from the recursive definitions of addition and multiplication in Peano arithmetic. Your proof is correct.
Best Answer
You may notice that $$P=\frac{200!}{2^{100}\cdot 100!} $$ Now use that a prime $p$ occurs in $n!$ with multiplicity exactly $$\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\rfloor +\ldots $$