MCQ about the product of $4$ consecutive odd numbers The product of four consecutive odd numbers must be …  
(A) A multiple of  3, but not necessarily of  9. 
(B) A multiple of  5 . 
(C) A multiple of  7.
(D) A multiple of  9. 
(E) A multiple of  3×5×7×9= 945.
Let $n\in \mathbb{Z}$, then
$(2n-1)(2n+1)(2n+3)(2n-3)$
$= (4n^2-1)(4n^2-9)$
$= 16n^4-40n^2+9$
$= 8n^2(2n^2-5)+9.$
For being a multiple of $9$, need $8n^2(2n^2-5)=9m$, for some suitable value of $m\in \mathbb{Z}$; or
$$2n^2-5=9m\cup n=3j, \text{for suitable} \,\,j\in \mathbb{Z}.$$
But, it gets more confusing to pursue further using my approach. Is my approach workable?
Want to add that if take case of showing not divisible by $5$, then : $x+9= 5k$, then $x= 5k-9$.
Hence, for $j, k\in \mathbb{Z}$, $$8n^2 = 5k-9\cup (2n^2-5)= 5j-9.$$ 
Both options are seemingly not possible, though not clear how to show theoretically the impossibility of both.
 A: Let $n$ be the first of the consecutive odd integers, with
$$f(n) = n(n+2)(n+4)(n+6)$$
If $n \equiv 0 \pmod{3}$, then $3 \mid f(n)$ (i.e., $f(n)$ is an integral multiple of $3$). If $n \equiv 1 \pmod{3}$, then $n + 2 \equiv 0 \pmod{3}$ so $3 \mid f(n)$. Finally, if $n \equiv 2 \pmod{3}$, then $n + 4 \equiv 0 \pmod{3}$ so $3 \mid f(n)$. Thus, in all cases, $f(n)$ is a multiple of $3$.
To show $f(n)$ is not necessarily a multiple of $9$, if $n \equiv 2 \pmod 9$ (e.g., $n = 11$), then $f(n) \equiv 2(4)(6)(8) \equiv 6 \pmod{9}$. Thus, option (A) is correct, plus options (D) and (E) are incorrect.
Note if $n \equiv 2 \pmod{5}$ (e.g., $n = 7$), then $f(n) \equiv 2(4)(6)(8) \equiv 4 \pmod{5}$, and if $n \equiv 2 \pmod{7}$ (e.g., $n = 9$), then $f(n) \equiv 2(4)(6)(8) \equiv 6 \pmod{7}$. Thus, these cases show that $f(n)$ is not necessarily a multiple of either $5$ or $7$, so both options (B) and (C) are not correct.
A: This query is subject to simple resolution by exemplary cases.
$1\times 3\times 5\times 7$ is not divisible by $9$
$7\times 9\times 11\times 13$ is not divisible by $5$
$9\times 11\times 13\times 15$ is not divisible by $7$
Since there are simple counterexamples to B, C, D, and E, the only possible answer is A.
You can satisfy yourself that A is true by noting any three consecutive members of an arithmetic sequence must have at least one member divisible by $3$, except (as pointed out in the comment by John Omielan) when the common difference in the arithmetic sequence is divisible by $3$, which is not the case for consecutive odd numbers. There the difference is $2$.
A: We use the fact that every number $n\neq 3$. can be written as $3k+1$ or $3k+2$. We chek this in:
$A=(2n-1)(2n+1)(2n+3)(2n+5)$

*

*$n=3k+1\Rightarrow 2n+1=6k+3\Rightarrow 3|A$
$2n+3=6k+5 \Rightarrow 5 \nmid A$
$2n+5=6k+7 \Rightarrow 7 \nmid A$
2): $n=3k+2$
$2n-1=6k \Rightarrow 3\mid A$
$2n+3=6k+7 \Rightarrow 7\nmid A$
$2n+1=6k+5\Rightarrow 5\nmid A$
The common point is that numbers 5 and 7 do not divide A except:
$n=3k, k=0 \Rightarrow 2n+5=5 \Rightarrow 5 \mid A$
$k=2, n=2k+3=7$
$k=4, 2k+5= 13$
$k=4, 2k+1=9$
So options  B, C , D and E  are exceptions and only option A can always be true.
Reply to comment: When we say every number it includes all primes and composites.
