If n is an integer, then (n3 - n) is always divisible by :
Options:
A . 4
B . 5
C . 6
D . 7
Answer: Option C $$\eqalign{ & \left( {{n^3} - n} \right){\text{ and n is any integer}} \cr & {\text{put n = 2 so, }}{{\text{2}}^3} - 2 = 6 \cr & {\text{It will be always divisible by 6}} \cr & {\text{(put n = 2, 3, 4}}....{\text{)}} \cr & {\text{(n = 2, 3, 4}}.....{\text{)}} \cr} $$
Submit Comment/FeedBack