# 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.

• Comments are not for extended discussion; this conversation has been moved to chat.
– Pedro
Apr 10 at 14:59

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.

• Thanks a lot, but is my approach workable by showing the infeasibility of $8n^2=5k−9 \cup (2n^2−5)=5j−9.$ Apr 10 at 3:45
• You're welcome. However, it seems your approach is not quite correct, if I understand correctly. If you want to show that $8n^2(2n^2 - 5) + 9$ is a multiple of $5$, then $8n^2(2n^2 - 5) = 5m - 9$ for some integer $m$. This doesn't necessarily require that $8n^2 = 5k - 9$ or $2n^2 - 5 = 5j - 9$. For example, using $n = 3$, we get $8(9)(18-5)+9 = 945$, but $8(9) = 72 \neq 5k - 9$ and $2(9) - 5 = 13 \neq 5j - 9$. Apr 10 at 3:58
• Kindly elaborate as you have shown that $5| 8n^2 +9$. Apr 10 at 4:17
• With $n = 3$, then $8n^2 + 9 = 8(9) + 9 = 81$, but $5 \nmid 81$, so I've not shown $5 \mid 8n^2 + 9$. Thus, I'm not quite sure what you wish for me to elaborate on. Apr 10 at 4:22
• Sorry, for confusion caused by my last comment. Kindly ignore that. Apr 10 at 4:58

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 minor issue with your end statement of "... noting any three consecutive members of an arithmetic sequence must have at least one member divisible by $3$" is with examples like $1, 4, 7, \ldots$ . It would be true if the common difference is restricted to being non-divisible by $3$. Apr 10 at 22:33

We use the fact that every number $$n\neq 3$$. can be written as $$3k+1$$ or $$3k+2$$. We chek this in:

$$A=(2n-1)(2n+1)(2n+3)(2n+5)$$

1. $$n=3k+1\Rightarrow 2n+1=6k+3\Rightarrow 3|A$$

$$2n+3=6k+5 \Rightarrow 5 \nmid A$$

$$2n+5=6k+7 \Rightarrow 7 \nmid A$$

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.

• Can there be approach using the fact the all divisors being tested are primes, or a product of. Apr 10 at 9:40