\( 120 = 3n(n+1) → n(n+1) = 40 → n ≈ 6.3 → n=6: 6×7=42>40, n=5:5×6=30<40 — no.

["Solving the Equation ( 120 = 3n(n+1) ): A Step-by-Step Exploration", "Finding integer solutions to equations involving quadratic expressions is a common challenge in algebra and math problem-solving. One such problem is solving:", "[\n120 = 3n(n+1)\n]", "At first glance, this equation might seem straightforward, but determining the exact integer value of ( n ) requires careful step-by-step analysis.", "---", "### Step 1: Simplify the Equation", "Start by dividing both sides by 3 to simplify:", "[\n\frac{120}{3} = n(n+1) \implies 40 = n(n+1)\n]", "Now we face the key equation:", "[\nn(n+1) = 40\n]", "This means we’re looking for two consecutive integers whose product is 40.", "---", "### Step 2: Estimate ( n ) Using Approximation", "Since ( n(n+1) ) lies between two perfect squares, approximate ( n ) by solving the quadratic equation:", "[\nn^2 + n - 40 = 0\n]", "Using the quadratic formula ( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), with ( a=1, b=1, c=-40 ):", "[\nn = \frac{-1 \pm \sqrt{1 + 160}}{2} = \frac{-1 \pm \sqrt{161}}{2}\n]", "Estimate ( \sqrt{161} \approx 12.7 ), so:", "[\nn \approx \frac{-1 + 12.7}{2} = \frac{11.7}{2} \approx 6.3\n]", "This tells us ( n ) is close to 6 or 7.", "---", "### Step 3: Test Integer Values Near the Approximation", "Test neighboring integers to find which satisfies ( n(n+1) = 40 ):", "- For ( n = 6 ):\n ( 6 \ imes 7 = 42 ) (too large)\n- For ( n = 5 ):\n ( 5 \ imes 6 = 30 ) (too small)", "Neither gives exactly 40, meaning no integer solution satisfies the equation exactly.", "---", "### Step 4: Interpret the Result", "Even though ( n \approx 6.3 ) gives ( n(n+1) \approx 42 > 40 ), and ( n = 5 ) yields ( 30 < 40 ), we conclude:", "- There is no integer value of ( n ) such that ( 3n(n+1) = 120 )\n- The closest values are 30 and 42, suggesting a potential rounding or estimation error if trying to solve approximately", "---", "### Alternative Insight: Solve for ( n ) Exactly", "Rewriting the original:", "[\n3n(n+1) = 120 \implies n(n+1) = 40\n]", "This quadratic has solutions:", "[\nn = \frac{-1 \pm \sqrt{161}}{2}\n]", "Since ( \sqrt{161} ) is irrational, ( n ) is irrational, confirming no exact integer solution exists.", "---", "### Conclusion", "While algebraic manipulation leads us to a quadratic form with no integer roots, testing nearby integers confirms ( n = 6 ) yields ( 3n(n+1) = 126 ), and ( n = 5 ) gives 120 < 120 — wait, correction:", "Wait:\n- ( n = 5 \Rightarrow 5 \ imes 6 = 30 \Rightarrow 3 \ imes 30 = 90 <br/>\ne 120 )\n- ( n = 6 \Rightarrow 6 \ imes 7 = 42 \Rightarrow 3 \ imes 42 = 126 <br/>\ne 120 )", "So actually, neither integer gives exactly 120.", "Thus, the equation ( 3n(n+1) = 120 ) has no integer solution. The problem demonstrates how real-world equations sometimes resist simple integer answers, encouraging deeper inspection beyond guesswork.", "---", "### Key Takeaways", "- Simplify equations carefully before solving\n- Approximate roots help guide exact testing\n- Not all quadratic equations yield integer solutions – accept irrational answers when needed\n- Understanding algebra supports better estimation and validation", "---", "For further exploration, try similar equations like ( 3n(n+1) = 126 ), where ( n = 6 ) works perfectly:\n( 3 \ imes 6 \ imes 7 = 126 )", "---", "Keywords: solve equation ( 120 = 3n(n+1) ), integer solutions, quadratic approximation, ( n(n+1) = 40 ), algebra problem solving", "---", "If you're interested in finding approximate values or rounding in math contests and real-world applications, they’re valuable skills beyond just “getting the right answer.”"]