Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 669 more balls are added, then all the balls can be arranged in the shape of a square and each of the sides then contains 8 balls less than each side of the triangle had. The initial number of balls is

### Concept & Formula This problem is built on the concept of **Arithmetic Progressions** and **Sum of N Natural Numbers**. An equilateral triangle arrangement of balls means the total number of balls is the sum of the first $n$ natural numbers. The formula for the sum of the first $n$ natural numbers is: $$Total = \frac{n(n+1)}{2}$$ The problem requires us to equate this initial sum plus the added balls to the area of a square arrangement, leading to a quadratic equation. ### Step-by-Step Solution **Given:** * Let the number of rows (or balls in the side) of the initial triangle be $n$. * Initial number of balls = $\frac{n(n+1)}{2}$. * Balls added = 669. * Side of the new square = $n - 8$. * Total balls in square = $(n - 8)^2$. **Calculation:** Step 1: Set up the core equation. $\frac{n(n+1)}{2} + 669 = (n - 8)^2$ Step 2: Expand and simplify. $n^2 + n + 1338 = 2(n^2 - 16n + 64)$ $n^2 + n + 1338 = 2n^2 - 32n + 128$ Step 3: Bring all terms to one side to form a standard quadratic equation. $n^2 - 33n - 1210 = 0$ Step 4: Factorize the quadratic equation. We need two numbers that multiply to -1210 and add to -33. The factors are -55 and 22. $(n - 55)(n + 22) = 0$ Since the number of balls cannot be negative, $n = 55$. Step 5: Find the initial number of balls. Initial balls = $\frac{55 \times (55+1)}{2} = \frac{55 \times 56}{2} = 55 \times 28 = 1540$. ### Exam Strategy & Shortcut **Reverse Engineering from Options:** Solving large quadratics like $n^2 - 33n - 1210 = 0$ can eat up time. Instead, use the options to test the final state. If initial balls = 1540 (Option B). New total after adding 669 = $1540 + 669 = 2209$. Check if 2209 is a perfect square. $\sqrt{2209} = 47$. So the square side is 47. If the square side is 47, the triangle side $n$ should be $47 + 8 = 55$. Check if $n=55$ yields 1540 balls in the triangle: $\frac{55 \times 56}{2} = 1540$. It matches perfectly! Option B is correct. ### Common Pitfall The most frequent error is misinterpreting "each side of the triangle". Students sometimes confuse this with the perimeter. In a triangular stacking arrangement (like bowling pins), the number of rows is exactly equal to the number of balls in the bottom-most side. ### Final Answer **Therefore, the correct answer is 1540.**