The number of positive integers $n$ for which $n^{3} - 8 n^{2} + 20 n - 13$ is a prime number is

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3$
$n^{3} - 8 n^{2} + 20 n - 13$
$= (n - 1) (n^{2} - 7 n + 13)$
This means $n - 1 = \pm 1$
or $n^{2} - 7 n + 13 = \pm 1$

Case $1 :$
$n - 1 = \pm 1$
$\Rightarrow n = 0, 2$
For $n = 0, n^{3} - 8 n^{2} + 20 n - 13 = - 13$
For $n = 2, n^{3} - 8 n^{2} + 20 n - 13 = 3$

Case $2 :$
$n^{2} - 7 n + 13 = \pm 1$
$\Rightarrow n = 3, 4$
For $n = 3, n^{3} - 8 n^{2} + 20 n - 13 = 2$
For $n = 4, n^{3} - 8 n^{2} + 20 n - 13 = 3$

Hence, there are $3$ values of $n$ for which $n^{3} - 8 n^{2} + 20 n - 13$ is prime.

Suggest Corrections

Similar questions

n

n^{3} - 8 n^{2} + 20 n - 13

n^{4} - 20 n^{2} + 4

Write the negation of the statement, for all positive integers n, $n^{2}$ + 41n + 41 is a prime number.

n

2^{n} 1

n^{2} = n^{3} = n

Join BYJU'S Learning Program