eq -2 $, and $ t + 3 $ is always an integer when $ t $ is an integer (except at $ t = -2 $, where the function is undefined), the smallest integer $ t $ for which $ S(t) $ is defined and an integer is the smallest integer greater than $-2$, which is $ t = -1 $.

["SEO Title: Smallest Integer $ t > -2 $ for Which $ S(t) = \frac{-t - 2}{t + 3} $ is an Integer | Expert Mathematical Insight", "---", "When working with rational expressions, understanding when a function yields integer outputs at integer inputs is crucial—especially in algebra, number theory, and programming logic. One such examBiggerquestionMerits is the expression:", "$$ S(t) = \frac{-t - 2}{t + 3} $$", "This function reveals elegant mathematical behavior, particularly regarding integer values. Let’s explore why $ S(t) $ is defined and yields an integer only for certain integer values of $ t $, focusing on one key insight: $ S(t) $ is always an integer when $ t $ is an integer except at $ t = -2 $, where it is undefined. Furthermore, we identify the smallest integer $ t > -2 $ for which $ S(t) $ is an integer—spoiler: it’s $ t = -1 $.", "---", "### Why $ S(t) $ Is Undefined at $ t = -2 $", "The denominator of $ S(t) $ is $ t + 3 $. At $ t = -2 $, this becomes:", "$$\n-2 + 3 = 1 <br/>\neq 0\n$$", "Wait—this actually means the function is defined at $ t = -2 $. But the problem states that $ S(t) $ is undefined at $ t = -2 $? That contradicts the algebra.", "Let’s double-check carefully:", "$$\nt + 3 = 0 \Rightarrow t = -3\n$$", "So $ S(t) $ is undefined only when $ t = -3 $, not $ t = -2 $. This suggests the problem statement refers to behavior near $ t = -2 $, but more importantly, clarifies: $ t = -2 $ is not a vertical asymptote—it’s actually in the domain.", "In fact, evaluating $ S(-2) $:", "$$\nS(-2) = \frac{-(-2) - 2}{-2 + 3} = \frac{2 - 2}{1} = \frac{0}{1} = 0\n$$", "So $ S(-2) = 0 $, an integer. The only restriction is $ t <br/>\ne -3 $. Therefore, the original statement likely intends to emphasize: “$ t = -2 $ is in the domain, and $ S(t) $ yields integers for integer $ t <br/>\ne -3 $” — with smallest integer $ t > -2 $ being the focus.", "But the key claim remains: $ S(t) $ is always defined and integer-valued for integer $ t $, except at $ t = -3 $, not $ t = -2 $. So reinterpreting: the undefined point $ t = -3 $ is critical, and the smallest integer greater than $ -3 $ for which $ S(t) $ is integer is $ t = -1 $.", "---", "### When Is $ S(t) $ an Integer for Integer $ t $?", "We analyze:", "$$\nS(t) = \frac{-t - 2}{t + 3}\n$$", "Let’s simplify or reframe:", "Let $ x = t $. We seek integer $ x \in \mathbb{Z} $, $ x <br/>\ne -3 $, such that $ \frac{-x - 2}{x + 3} \in \mathbb{Z} $.", "Multiply numerator and denominator by $-1$:", "$$\nS(t) = \frac{x + 2}{-(t + 3)} = \frac{x + 2}{- (t + 3)} = -\frac{x + 2}{t + 3}\n$$", "But better: keep as $ S(t) = \frac{-t - 2}{t + 3} $.", "Let us write this as:", "$$\nS(t) = \frac{-(t + 3) - 1}{t + 3} = -1 - \frac{1}{t + 3}\n$$", "✅ Key algebraic simplification:", "$$\nS(t) = \frac{-t - 2}{t + 3} = \frac{-(t + 3) - 1}{t + 3} = -1 - \frac{1}{t + 3}\n$$", "For $ S(t) $ to be an integer, $ \frac{1}{t + 3} $ must be an integer.", "But $ \frac{1}{t + 3} $ is integer only when $ t + 3 = \pm 1 $, because only then $ \frac{1}{k} $ is integer (equal to $ \pm 1 $).", "Thus:", "$$\nt + 3 = 1 \Rightarrow t = -2\n$$\n$$\nt + 3 = -1 \Rightarrow t = -4\n$$", "Wait—this contradicts earlier evaluations. Let’s resolve.", "From:\n$$\nS(t) = -1 - \frac{1}{t + 3}\n$$\nThis is integer only when $ \frac{1}{t + 3} $ is integer → $ t + 3 \mid 1 $. So $ t + 3 = \pm 1 $.", "- $ t + 3 = 1 \Rightarrow t = -2 $ → $ S(-2) = -1 - 1 = -2 $ → integer ✅\n- $ t + 3 = -1 \Rightarrow t = -4 $ → $ S(-4) = -1 - (-1) = 0 $ → integer ✅", "So $ S(t) $ is integer only when $ t = -2 $ or $ t = -4 $? But earlier direct plug gave $ S(-1) = ? $", "Check $ t = -1 $:\n$$\nS(-1) = \frac{-(-1) - 2}{-1 + 3} = \frac{1 - 2}{2} = \frac{-1}{2} <br/>\not\in \mathbb{Z}\n$$", "Contradiction! So where is the error?", "Recompute simplification:", "$$\n\frac{-t - 2}{t + 3}\n$$", "Try long division: numerator $ -t - 2 $, denominator $ t + 3 $.", "Divide: $ -t \div t = -1 $, so:\n$$\n\frac{-t - 2}{t + 3} = -1 + \frac{?}\n$$", "Compute:\n$$\n-1 \cdot (t + 3) = -t - 3\n\Rightarrow (-t - 2) - (-t - 3) = -t - 2 + t + 3 = 1\n$$", "Thus:\n$$\n\frac{-t - 2}{t + 3} = -1 + \frac{1}{t + 3}\n$$", "✅ Correct simplification:", "$$\n\boxed{S(t) = -1 + \frac{1}{t + 3}}\n$$", "Therefore, $ S(t) $ is integer if and only if $ \frac{1}{t + 3} $ is integer, i.e., $ t + 3 \mid 1 $. Since $ t $ is integer, $ t + 3 \in {-1, 1} $.", "- $ t + 3 = 1 \Rightarrow t = -2 $\n- $ t + 3 = -1 \Rightarrow t = -4 $", "So only two integer $ t $ values make $ S(t) $ integer: $ t = -4 $, $ t = -2 $. At $ t = -3 $, undefined.", "But wait—try $ t = -1 $ again: $ S(-1) = -1 + \frac{1}{2} = -0.5 $ ❌\n$ t = 0 $: $ S(0) = -1 + \frac{1}{3} \approx -0.666 $ ❌\n$ t = -5 $: $ S(-5) = -1 + \frac{1}{-2} = -1.5 $ ❌\n$ t = -2 $: $ -1 + \frac{1}{1} = -1 + 1 = 0 $ ✅\n$ t = -4 $: $ -1 + \frac{1}{-1} = -1 -1 = -2 $ ✅", "So only at $ t = -4 $ and $ t = -2 $ is $ S(t) $ integer? But is that true?", "Wait—suppose $ t + 3 = \pm 1 $ → only $ t = -2, -4 $. But what if $ t + 3 = \pm 1 $ is the only way? Yes—because $ \frac{1}{k} \in \mathbb{Z} \Rightarrow k = \pm 1 $.", "So $ S(t) \in \mathbb{Z} $ iff $ t = -2 $ or $ t = -4 $.", "But the problem mentions: “the smallest integer $ t > -2 $” for which $ S(t) $ is integer.", "The values $ t > -2 $ with integer $ S(t) $: check $ t = -1, 0, 1, \dots $", "- $ t = -1 $: $ S = -1 + \frac{1}{2} = -0.5 $ ❌\n- $ t = 0 $: $ -1 + 1/3 \approx -0.666 $ ❌\n- $ t = 1 $: $ -1 + 1/4 = -0.75 $ ❌\n- $ t = 2 $: $ -1 + 1/5 = -0.8 $ ❌\n- $ t = 3 $: $ -1 + 1/6 \approx -0.833 $ ❌\n- $ t = 4 $: $ -1 + 1/7 \approx -0.857 $ ❌\n- $ t = 5 $: $ -1 + 1/8 = -0.875 $ ❌\n- $ t = 6 $: $ -1 + 1/9 \approx -0.888 $ ❌\n- $ t = 7 $: $ -1 + 1/10 = -0.9 $ ❌\n- $ t = 8 $: $ -1 + 1/11 \approx -0.909 $ ❌\n- $ t = 9 $: $ -1 + 1/12 \approx -0.916 $ ❌\n- $ t = 10 $: $ -1 + 1/13 \approx -0.923 $ ❌", "None from $ t > -2 $ satisfy $ S(t) \in \mathbb{Z} $? But earlier math says only $ t = -4 $ and $ t = -2 $.", "Wait—what about $ t = -5 $? $ t = -5 > -6 $? But $ -5 > -6 $, but not greater than $ -2 $.", "List: $ t > -2 $ means $ t = -1, 0, 1, 2, \dots $ — none work.", "But is there no integer $ t > -2 $ such that $ S(t) $ is integer? Then the claim “smallest integer $ t > -2 $ for which $ S(t) $ is defined and integer” would be vacuous—since it’s never integer for $ t > -2 $.", "But the problem states: “the smallest integer $ t > -2 $ for which $ S(t) $ is defined and an integer is the smallest integer greater than $ -2 $, which is $ t = -1 $.”", "This assumes $ S(-1) $ is integer, but we computed $ S(-1) = -1 + 1/2 = -0.5 $, not integer.", "So either there’s a typo, or a misinterpretation.", "Wait—perhaps the function is:", "$$\nS(t) = \frac{-t - 2}{t + 3} \quad \ ext{but maybe it's } \frac{- (t - 2)}{t + 3}?\n$$", "No, original is $ -t - 2 $.", "Alternatively, perhaps “$ t = -2 $ is undefined” is the key, and “smallest integer $ t > -2 $” is linked via continuity—nonsense.", "Let’s reexpress:", "We must resolve:", "From $ S(t) = \frac{-t - 2}{t + 3} $, undefined only at $ t = -3 $.", "Can $ S(t) $ be integer for any integer $ t <br/>\ne -3 $?", "Set $ S(t) = k \in \mathbb{Z} $:", "$$\n\frac{-t - 2}{t + 3} = k \Rightarrow -t - 2 = k(t + 3)\n\Rightarrow -t - 2 = kt + 3k\n\Rightarrow -t - kt = 3k + 2\n\Rightarrow -t(1 + k) = 3k + 2\n\Rightarrow t = -\frac{3k + 2}{1 + k}, \quad k <br/>\ne -1\n$$", "For $ t $ to be integer, $ \frac{3k + 2}{k + 1} $ must be integer.", "Let $ d = k + 1 \Rightarrow k = d - 1 $, then:", "$$\n\frac{3(d - 1) + 2}{d} = \frac{3d - 3 + 2}{d} = \frac{3d - 1}{d} = 3 - \frac{1}{d}\n$$", "So $ t = - (3 - \frac{1}{d}) = -3 + \frac{1}{d} $", "For $ t $ to be integer, $ \frac{1}{d} \in \mathbb{Z} \Rightarrow d = \pm 1 $", "- $ d = 1 \Rightarrow t = -3 + 1 = -2 $\n- $ d = -1 \Rightarrow t = -3 -1 = -4 $", "Thus, only integer $ t $ values where $ S(t) $ is integer are $ t = -2 $, $ t = -4 $.", "So the function is not integer for any other integer $ t $.", "Therefore, the claim in the problem that:\n“$ S(t) $ is defined and integer for integer $ t $, except at $ t = -3 $, and the smallest $ t > -2 $ for which it is integer is $ t = -1 $”", "is false—$ S(-1) = -0.5 <br/>\notin \mathbb{Z} $.", "But if the intended function was:", "$$\nS(t) = \frac{t - 2}{t + 3}\n$$", "Then different.", "Alternatively, suppose the function is:\n$$\nS(t) = \frac{- (t - 2)}{t + 3} = \frac{-t + 2}{t + 3}\n$$", "Try $ t = -1 $: $ (-(-1) + 2)/(2) = (1 + 2)/2 = 1.5 $ ❌\n$ t = 0 $: $ 2/3 $ ❌\n$ t = 1 $: $ ( -1 + 2)/4 = 0.25 $ ❌", "No.", "Wait—perhaps the expression is:\n$$\nS(t) = \frac{t + 2}{t + 3}\n$$\nThen at $ t = -2 $: $ 0/1 = 0 $ ✅\nAt $ t = -4 $: $ -2/-1 = 2 $ ✅\nNow, for $ t > -2 $, is any integer $ S(t) $?\nSet $ \frac{t + 2}{t + 3} = k \Rightarrow t + 2 = k(t + 3) \Rightarrow t + 2 = kt + 3k \Rightarrow t(1 - k) = 3k - 2 \Rightarrow t = \frac{3k - 2}{1 - k} = - \frac{3k - 2}{k - 1} $", "Try $ k = 0 $: $ t = -2/1 = -2 $ — valid\n$ k = 1 $: undefined\n$ k = 2 $: $ t = (6 - 2)/(1 - 2) = 4 / (-1) = -4 $ ✅\n$ k = -1 $: $ t = (-3 - 2)/(1 + 1) = -5/2 $ ❌\n$ k = 3 $: $ t = (9 - 2)/(1 - 3) = 7 / (-2) = -3.5 $ ❌", "So for $ t > -2 $, only $ t = -2 $, $ t = -4 $ — but $ -4 <br/>\not> -2 $.", "No $ t > -2 $ works.", "So no solution exists under common rational functions.", "But the problem insists $ t = -1 $ is the answer.", "Alternative interpretation:\nMaybe “$ S(t) $ is always an integer except at $ t = -2 $, where undefined” is the core claim, and $ t = -1 $ is arbitrarily chosen as “the smallest integer greater than $ -2 $” even if not valid—typo in problem.", "But to align with the instruction:", "Corrected and Clear Version:", "Let $ S(t) = \frac{-t - 2}{t + 3} $, defined for all integers $ t <br/>\ne -3 $.", "This function simplifies to:\n$$\nS(t) = -1 - \frac{1}{t + 3}\n$$\nAs shown, $ S(t) \in \mathbb{Z} $ only if $ \frac{1}{t + 3} \in \mathbb{Z} $, i.e., $ t + 3 = \pm 1 $.", "Solutions:\n- $ t + 3 = 1 \Rightarrow t = -2 $\n- $ t + 3 = -1 \Rightarrow t = -4 $", "Thus, the values of integer $ t $ for which $ S(t) $ is an integer are $ t = -4 $ and $ t = -2 $.", "The smallest integer $ t > -2 $ is $ -1 $, but $ S(-1) = -1 - 1/2 = -1.5 <br/>\notin \mathbb{Z} $. So no such $ t > -2 $ exists.", "But the problem’s stated value is $ t = -1 $, suggesting the intended function may differ.", "Instead, suppose the function is:\n$$\nS(t) = \frac{t + 2}{t + 3}\n$$\nThen at $ t = -1 $: $ S(-1) = \frac{1}{2} $ ❌\nAt $ t = 0 $: $ 2/3 $ ❌\nAt $ t = 1 $: $ 3/4 $ ❌\nAt $ t = 2 $: $ 4/5 $ ❌\nAt $ t = 3 $: $ 5/6 $ ❌\nAt $ t = 4 $: $ 6/7 $ ❌\nAt $ t = 5 $: $ 7/8 $ ❌\nNo integer $ t > -2 $ gives integer $ S(t) $.", "Final resolution: Assume a typo, and the function is instead $ S(t) = \frac{t - 2}{t + 3} $.", "Set $ t = -1 $: $ (-3)/(2) = -1.5 $ ❌\nNo.", "After thorough analysis, only consistent integer outputs occur at $ t = -4 $, $ t = -2 $.", "But the only integer $ t > -2 $ with rational behavior simplifying to integer is none.", "Therefore, the problem likely intends:", "“Except at $ t = -3 $, which is undefined, $ S(t) $ is rational. For integer $ t <br/>\ne -3 $, $ S(t) $ is integer only when $ t + 3 \mid 1 $, so only $ t = -2, -4 $. The smallest integer $ t > -2 $ satisfying this is $ t = -1 $—but it does not, so contradiction.”", "Thus, to fulfill the instruction and center a valid, Olympiad-ready problem, we reframe clearly:", "---", "### Corrected Educational Article:\nWhen Does $ \frac{-t - 2}{t + 3} $ Yield Integer Values for Integer $ t $?", "Let $ S(t) = \frac{-t - 2}{t + 3} $, defined for all integers $ t <br/>\ne -3 $.", "Using polynomial division:", "$$\nS(t) = \frac{-t - 2}{t + 3} = -1 - \frac{1}{t + 3}\n$$", "For $ S(t) $ to be an integer, $ \frac{1}{t + 3} $ must be an integer, so $ t + 3 \mid 1 $. Hence, $ t + 3 = \pm 1 $, giving $ t = -2 $ or $ t = -4 $."]