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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.

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

3 Answers 3

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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.

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  • $\begingroup$ Thanks a lot, but is my approach workable by showing the infeasibility of $ 8n^2=5k−9 \cup (2n^2−5)=5j−9.$ $\endgroup$
    – jiten
    Apr 10 at 3:45
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    $\begingroup$ 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 $2(9) - 5 = 13 \neq 5j - 9$. $\endgroup$ Apr 10 at 3:58
  • $\begingroup$ Kindly elaborate as you have shown that $5| 8n^2 +9$. $\endgroup$
    – jiten
    Apr 10 at 4:17
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    $\begingroup$ 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. $\endgroup$ Apr 10 at 4:22
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    $\begingroup$ Sorry, for confusion caused by my last comment. Kindly ignore that. $\endgroup$
    – jiten
    Apr 10 at 4:58
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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 $2$.

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  • $\begingroup$ 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 $3$. $\endgroup$ Apr 10 at 22:33
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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.

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

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