1. **State the problem:** We need to prove that 6 is a factor of the expression $n(n^2 + 5)$ for all integers $n \geq 1$. 2. **Rewrite the expression:** The expression is $n(n^2 + 5) = n^3 + 5n$. 3. **Analyze divisibility by 2:** - If $n$ is even, then $n$ is divisible by 2, so $n^3 + 5n$ is divisible by 2. - If $n$ is odd, then $n^3$ is odd and $5n$ is odd, so $n^3 + 5n$ is even because odd + odd = even. Thus, $n^3 + 5n$ is divisible by 2 for all integers $n$. 4. **Analyze divisibility by 3:** We check $n^3 + 5n$ modulo 3. - Note that $5n \equiv 2n \pmod{3}$ since $5 \equiv 2 \pmod{3}$. - So, $n^3 + 5n \equiv n^3 + 2n \pmod{3}$. - Factor out $n$: $n^3 + 2n = n(n^2 + 2)$. - Consider $n \pmod{3}$: - If $n \equiv 0 \pmod{3}$, then $n(n^2 + 2) \equiv 0$. - If $n \equiv 1 \pmod{3}$, then $n^2 + 2 \equiv 1 + 2 = 3 \equiv 0$, so product is $1 \times 0 = 0$. - If $n \equiv 2 \pmod{3}$, then $n^2 + 2 \equiv 4 + 2 = 6 \equiv 0$, so product is $2 \times 0 = 0$. In all cases, $n^3 + 5n \equiv 0 \pmod{3}$. 5. **Conclusion:** Since $n^3 + 5n$ is divisible by both 2 and 3 for all $n \geq 1$, it is divisible by their product 6. **Final answer:** $6$ is a factor of $n(n^2 + 5)$ for all integers $n \geq 1$.