If-n-is-any-even-number-then-n-n-2-20-is-always-divisible-by-

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 .