JEE Novice - (10)

Algebra Level 3

$\Large n^5-5n^3+4n$

If $n$ is a natural number and the above expression can always be expressed as the product of $k$ consecutive positive integers, then find the value of $k$ .

If you think that there is no largest value of $k$ , input 999 as your answer.

This question is a part of JEE Novices .

The answer is 5.

1 solution

Brian Charlesworth
May 26, 2015

We have $n^{5} - 5n^{3} + 4n = n(n^{4} - 5n^{2} + 4) = n(n^{2} - 4)(n^{2} - 1) =$

$n(n - 2)(n + 2)(n - 1)(n + 1) = (n - 2)(n - 1)(n)(n + 1)(n + 2).$

Thus the given expression can be expressed as the product of $\boxed{5}$ consecutive integers.

Did it the same way. But one thing to be noticed. it has been written that this expression can always be expressed as a product of K consecutive positive integers. But for value 1 and 2 of n only positive integers won't do.

Ansh Bhatt - 6 years ago

The given condition that the expression can always be expressed as the product of $k$ consecutive positive integers already implies that you're to consider $n\in\Bbb{Z_{\geq 3}}$ after getting the final factored form.

You should see that all of the presumptions of the problem are met before giving a counter-example. Else, the counter-example isn't valid. As the case here, the problem wording itself hints the solver to make that assumption about the restriction of domain of $n$ .

Prasun Biswas - 6 years ago

JEE Novice - (10)

This question is a part of JEE Novices .

The answer is 5.

1 solution

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