Question

Balls are arranged in rows to form an equilateral triangle (1 in the first row, 2 in the second, and so on). If 669 more balls are added, all the balls can be arranged in a square and each side of the square contains 8 balls fewer than the number of balls on a side of the triangle. What was the initial number of balls?

• A. 1600
• B. 1500
• C. 1540
• D. 1690

Hint:

Convert shape-wording to number formulas: triangles \(\to T_n=\frac{n(n+1)}{2}\), squares \(\to s^2\). Then use the side-length relation to link both counts and solve.

Answer 1

The Correct Option is C

Step 1: Translate to algebra. 
Let the triangular arrangement have \(n\) balls on a side. 
Initial balls \(= T_n = \dfrac{n(n+1)}{2}\) (triangular number). 
After adding \(669\) balls, we get a square with side \(s\). 
Given: the square's side is 8 less than the triangle's side \(\Rightarrow s = n - 8\). 

Step 2: Form the equation. 
Square count \(= s^2 = (n-8)^2\). 
But \(s^2 = T_n + 669 = \dfrac{n(n+1)}{2} + 669\). 
Thus \((n-8)^2 = \dfrac{n(n+1)}{2} + 669\). 

Step 3: Solve for \(n\). 
\( (n^2 - 16n + 64) = \dfrac{n^2 + n}{2} + 669 \). 
Multiply by 2: \(2n^2 - 32n + 128 = n^2 + n + 1338\). 
\(\Rightarrow n^2 - 33n - 1210 = 0\). 
Discriminant: \(D = (-33)^2 + 4\cdot1210 = 1089 + 4840 = 5929 = 77^2\). 
\(\Rightarrow n = \dfrac{33 + 77}{2} = 55\) (positive root). 

Step 4: Compute the initial triangular number and verify. 
Initial balls \(= T_{55} = \dfrac{55 \cdot 56}{2} = 55 \cdot 28 = \boxed{1540}\). 
Check: square side \(s = 55 - 8 = 47\) \(\Rightarrow s^2 = 2209\). 
\(T_{55} + 669 = 1540 + 669 = 2209 = 47^2\) (verified).