Question

for which natural numbers n is n^3-10n^2+28n-19 a prime number

Answer 1

Answer:

Let S = n^3-10n^2+28n-19

Start putting the values of n as natural numbers in above equation

n=0 , we get S = -19

n=1 , we get S = 0

n=2 , we get S = 5 -----> Prime number

n=3 , we get S = 2 -----> Prime number

n=4 , we get S = -3

n=5 , we get S = -4

n=6 , we get S = 5 -----> Prime number

n=7 , we get S = 30

n=8 , we get S = 77

n=9 , we get S = 152

n=10 , we get S = 261

So, at n = 2,3,6 we get S as a prime number

Answer 2

To find:

For which natural numbers $n$, $n^{3}-10n^{2}+28n-19$ is a prime number

Step-by-step explanation:

Let, $f(n)=n^{3}-10n^{2}+28n-19$

When $n=1,\:f(1)=0$

When $n=2,\:f(2)=5$

When $n=3,\:f(3)=2$

When $n=4,\:f(4)=-3$

When $n=5,\:f(5)=-4$

When $n=6,\:f(6)=5$

When $n=7,\:f(7)=30$

When $n=8,\:f(8)=77$

When $n=9,\:f(9)=152$

When $n=10,\:f(10)=261$

When $n=11,\:f(11)=410$ and so on.

Conclusion. So far, we have obtained only three cases where $f(n)$ is a prime number, when $n=2,3,6,...$.

Final answer: $n=2,3,6,...$

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