Question 44

For how many positive integers ‘n’, $$n^3 - 8n^2 + 20n - 13$$ is a prime?

Solution

$$n^3- 8n^2+20n -13$$

= $$(n-1)\times (n^2 - 7n + 13 )$$

In order to make (n-1) prime it has to be +/-1 or $$(n^2 - 7n + 13 )$$ has to be +/-1

n-1 = +/-1

= n = 0 or 2

$$n^3- 8n^2+20n -13$$ = -13 or 3

$$(n^2 - 7n + 13 ) = +/-1$$

= n is 3 or 4.

For this we will get $$n^3- 8n^2+20n -13$$ = 2 or 3

We know that prime numbers are positive so the prime numbers will be 2,3 and 4

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