Question

Let $S=${$n∈NIn^3+3n^2+5n+3$ is not divisible by $3$}.Then, which of the following statements is true about $S$

• A. $S=ϕ$
• B. $|S|≥2$ and |S|$$$ is a multiple of 5$
• C. $|S|$ is infinite
• D. $S$ is non empty and $|S|$ is a multiple of $3$
• E. $S$ is non empty and $|S|$ is a multiple of $3$.

Answer 1

Answer: (E)

$S$ is non empty and $|S|$ is a multiple of $3$.

Answer 2

Given that:

$S=${$n∈NIn^3+3n^2+5n+3$ is not divisible by $3$}.

Here as $n∈N$

let us test:

take, $n=1$

$⇒1^3+3.1^2+5.1+3=12$

take $n=2$

$⇒2^3+3.2^2+5.2+3=33$

take, $n=3$

$⇒3^3+3.3^2+5.3+3=72$

All these above cases proves that S is divisible by $3$

Hence,$S={n∈N : n^3+3n^2+5n+3}$ is divisible by $3$

So, S is non empty and $|S|$is a multiple of $3$ (_Ans.)