If n is any even number, then n(n^2 +20) is always divisible by?


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Question Number 112994 by Aina Samuel Temidayo last updated on 10/Sep/20

$If n is any even number, then$ $n (n^{2} + 20) is always divisible by ?$
	Answered by mr W last updated on 10/Sep/20

	$n = 2 k$ $2 k (4 k^{2} + 20) = 8 k (k^{2} + 5)$ $o n e f r o m k a n d k^{2} + 5 i s e v e n$ $\Rightarrow n (n^{2} + 20) i s a l w a y s d i v i s i b l e b y 16 .$
	Commented by Aina Samuel Temidayo last updated on 10/Sep/20

	$n = 2 k (k \in Z)$ $How is k^{2} + 5 even ?$
	Commented by mr W last updated on 10/Sep/20

	$i {d i d n}^{'} t s a y t h a t k^{2} + 5 i s e v e n . i s a i d$ $o n e o f k a n d k^{2} + 5 i s e v e n .$ $i f k i s e v e n t h e n k^{2} + 5 i s o d d .$ $i f k i s o d d t h e n k^{2} + 5 i s e v e n .$ $t h a t m e a n s$ $k (k^{2} + 5) i s a l w a y s e v e n .$
	Commented by Aina Samuel Temidayo last updated on 10/Sep/20

	$Oh . Thanks .$
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