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. \(S\) is non empty and \(|S|\) is a multiple of \(3\).
• 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. \(S=ϕ\)

Answer 1

Answer: (E)

Given: Let \( S = \{ n \in \mathbb{N} \mid n^3 + 3n^2 + 5n + 3 \text{ is not divisible by 3} \} \).

We analyze the polynomial modulo 3: \[ f(n) = n^3 + 3n^2 + 5n + 3 \] \[ f(n) \equiv n^3 + 2n \mod 3 \]

Using Fermat's Little Theorem (\( n^3 \equiv n \mod 3 \)): \[ f(n) \equiv n + 2n \equiv 3n \equiv 0 \mod 3 \]

This shows: \[ f(n) \text{ is divisible by 3 for all } n \in \mathbb{N} \]

Therefore, the set \( S \) contains no natural numbers: \[ S = \emptyset \]

Answer 2

Let \( P(n) = n^3 + 3n^2 + 5n + 3 \). We want to find the values of \( n \in \mathbb{N} \) such that \( P(n) \) is not divisible by 3.

We can rewrite \( P(n) \) as:

\[ P(n) = n^3 + 3n^2 + 5n + 3 = n^3 + 2n + 3 \pmod{3} \]

since \( 3n^2 \) and 3 are divisible by 3.

Now we analyze the values of \( n^3 + 2n \pmod{3} \) for \( n = 0, 1, 2 \):

• If \( n \equiv 0 \pmod{3} \), then \( n^3 + 2n \equiv 0 \pmod{3} \).
• If \( n \equiv 1 \pmod{3} \), then \( n^3 + 2n \equiv 1 + 2 \equiv 3 \equiv 0 \pmod{3} \).
• If \( n \equiv 2 \pmod{3} \), then \( n^3 + 2n \equiv 8 + 4 \equiv 12 \equiv 0 \pmod{3} \).

In all cases, \( n^3 + 2n \equiv 0 \pmod{3} \). Therefore, \( P(n) \equiv 0 \pmod{3} \) for all integers \( n \).

This means that \( n^3 + 3n^2 + 5n + 3 \) is always divisible by 3 for any integer \( n \). Consequently, \( S = \emptyset \), since there are no natural numbers \( n \) for which \( P(n) \) is not divisible by 3.

Therefore, the correct statement is (E).