Find value of n so 120 ∣ 5n(n^2 −1)


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Question Number 54897 by gunawan last updated on 14/Feb/19

$Find value of n so 120 ∣ 5 n (n^{2} - 1)$
Commented by jahanara@gmail.com last updated on 02/May/19

$s o e a s y$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/Feb/19

	$n (n^{2} - 1) = 24$ $(n - 1) n (n + 1) = 24 = 2 \times 3 \times 4 \to (3 - 1) \times 3 \times (3 + 1)$ $n = 3$ $f r o m r e a s o n i n g i t i s s h o w n . . .$
	Answered by kaivan.ahmadi last updated on 14/Feb/19

	$24 ∣ n (n^{2} - 1)$ $(n - 1) n (n + 1) = 24 k = 2 \times 3 \times 4 \times k$ $the multiple of three sequential nuber$ $must divide 24$ $n = 3$ $n = 7$ $n = 8$ $n = 9$ $n = 12$ $n = 13$ $n = 15$ $n = 16$ $⋮$ $excusme, \overset{}{} i cant find general result$
	Commented by MJS last updated on 15/Feb/19

	$case 1$ $2 ∣ (n - 1) \land 4 ∣ (n + 1)$ $1.1 3 ∣ (n - 1) \Rightarrow n = 12 k + 7$ $1.2 3 ∣ n \Rightarrow n = 12 k + 3$ $1.3 3 ∣ (n + 1) \Rightarrow n = 12 k + 11$ $case 2$ $4 ∣ (n - 1) \land 2 ∣ (n + 1)$ $2.1 3 ∣ (n - 1) \Rightarrow n = 12 k + 1$ $2.2 3 ∣ n \Rightarrow n = 12 k + 9$ $2.3 3 ∣ (n + 1) \Rightarrow n = 12 k + 5$ $case 3$ $8 ∣ n$ $3.1 3 ∣ (n - 1) \Rightarrow n = 24 k + 16$ $3.2 3 ∣ n \Rightarrow n = 24 k$ $3.3 3 ∣ (n + 1) \Rightarrow n = 24 k + 8$ $\Rightarrow n = 2 k + 1 \lor n = 8 k$
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