Question

Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row 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 did. Determine the initial numbers of balls.

Answer 1

Let the number of rows in which the balls are arranged to form an equilateral triangle be $n$.
According to the given condition the total number of balls
$S=1+2+3+...+n=\frac{n(n+1)}{2}$
(since number of balls in the kth row of equilateral triangle is k)
$\therefore S+669={(n-8)}^2$
$\Rightarrow \frac{n(n+1)}{2}+669=n^2-16n+64$
$\Rightarrow n^2+n+1338=2n^2-32n+128$
$\Rightarrow n^2-33n-1210=0$
$\Rightarrow n=\frac{33\pm \sqrt{1089+4840}}{2}$
$\Rightarrow n=\frac{33\pm 77}{2}$
$\Rightarrow n=55$
Therefore initial number of balls
$=\frac{n(n+1)}{2}=\frac{55\times 56}{2}=1540$