Question

The statement " $n^5-5n^3+4n$ is divisible by $120^{{}^{\prime \prime }}$ is true for

• A. n=1 only
• B. n=10 only
• C. n=100 only
• D. All positive integer values of n

Answer 1

Answer: (D)

We have, $120=3\times 5\times 8$
Therefore to show that given expression is divisible by $120$ , it is sufficient to show that $3,5$ and $8$ are the factors of $n^5-5n^3+4n$
Let us factorise the given expression
$n^5-5n^3+4n=n\left[n^4-5n^2+4\right]$
$=n\left[n^4-4n^2-n^2+4\right]$
$=n\left[n^2\left(n^2-4\right)-1\left(n^2-4\right)\right]$
$=n\left(n^2-4\right)\left(n^2-1\right)$
$=n\left(n^2-2^2\right)\left(n^2-1^2\right)$
$=n(n-2)(n+2)(n-1)(n+1)$
$=(n-2)(n-1)n(n+1)(n+2)$
Since, $(n-2),(n-1),n,(n+1)$ are $4$ consecutive integers. One of it must be divisible by $4.$ And at least two out of $5$ consecutive integers must be even. Thus, it must be divisible by $8$.
Now, again out of three consecutive integers $1$ must be divisible by $3$, and out of $5$ consecutive integers l must be divisible by $5.$
Thus, the given expression is divisible by $3,5,8$. Hence, it clear that $n^5-5n^3+4n$ is divisible by $120$ for all positive integer values of $n$.