1. For how many integers $n$ with $1 \leq n \leq 2024$ is the expression $\frac{n(n+3)}{27}$ equal to an integer?

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∻Solution by Steps∻ ‖ step 1 ⋮ We need to determine for how many integers $n$ with $1 \leq n \leq 2024$ the expression $\frac{n(n+3)}{27}$ is an integer. ‖ ‖ step 2 ⋮ For $\frac{n(n+3)}{27}$ to be an integer, $n(n+3)$ must be divisible by 27. ‖ ‖ step 3 ⋮ We need to find the values of $n$ such that $n(n+3) \equiv 0 \pmod{27}$. ‖ ‖ step 4 ⋮ This means $n$ or $n+3$ must be divisible by 27. ‖ ‖ step 5 ⋮ Let's consider the two cases separately: 1. $n \equiv 0 \pmod{27}$ 2. $n+3 \equiv 0 \pmod{27}$ ‖ ‖ step 6 ⋮ For the first case, $n = 27k$ where $k$ is an integer. The range $1 \leq n \leq 2024$ gives $1 \leq 27k \leq 2024$. Solving for $k$, we get $1 \leq k \leq \left\lfloor \frac{2024}{27} \right\rfloor = 74$. So, there are 74 values of $n$ in this case. ‖ ‖ step 7 ⋮ For the second case, $n+3 = 27m$ where $m$ is an integer. This gives $n = 27m - 3$. The range $1 \leq n \leq 2024$ gives $1 \leq 27m - 3 \leq 2024$. Solving for $m$, we get $4 \leq 27m \leq 2027$ or $4/27 \leq m \leq 2027/27$. This simplifies to $1 \leq m \leq 75$. So, there are 75 values of $n$ in this case. ‖ ‖ step 8 ⋮ However, we must exclude the overlap where $n$ is both $27k$ and $27m - 3$. This happens when $27k = 27m - 3$, or $k = m - \frac{1}{9}$, which is not possible for integer $k$ and $m$. Therefore, there is no overlap. ‖ ‖ step 9 ⋮ Adding the two cases, we get $74 + 75 = 149$ values of $n$ for which $\frac{n(n+3)}{27}$ is an integer. ‖ ∻Answer∻ ⚹ 149 ⚹ ∻Key Concept∻ ⚹Divisibility by 27⚹ ∻Explanation∻ ⚹To determine when $\frac{n(n+3)}{27}$ is an integer, we need $n(n+3)$ to be divisible by 27. This occurs when either $n$ or $n+3$ is a multiple of 27.⚹◊How can we determine the values of integer $n$ where the expression $\frac{n(n+3)}{27}$ results in an⍭ Generate me a similar question◊

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