Question

The largest possible number by which the expression \[{{n}^{3}}-n\] is divisible for all the possible integral values of n is
A. 2
B. 3
C. 4
D. 5
E. 6

Answer 1

Hint: For this problem to solve we have to split the given expression to three consecutive numbers, here the logic is if there are three consecutive numbers if the taken number is odd then the consecutive numbers will be even.

Complete step-by-step solution -
Given expression is \[{{n}^{3}}-n\]
By further simplifying we get, \[{{n}^{3}}-n=n\left( {{n}^{2}}-1 \right)\]
\[{{n}^{3}}-n=\left( n-1 \right)\left( n \right)\left( n+1 \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Above there are three consecutive numbers
If there are three consecutive numbers if the taken number is odd then the consecutive numbers will be even.
One of the values among them will be an even number. so it will be divisible by 2.
If there are three consecutive numbers if the taken number is even then the consecutive numbers will be odd.
Also, as there are three consecutive numbers, one of the values will be odd this means it will be divisible by 3.
Since, \[{{n}^{3}}-n\] is divisible by 2 and 3 both , therefore, \[{{n}^{3}}-n\] is divisible by 6.
Therefore the answer is option E.

Note: For this type of problem there will be a logic like if an odd number is there then its consecutive numbers will be even. By following this we have found out the largest possible number. Take care while doing calculations.