There are 360 stickers in a box, divided equally among a group of children. If there are six fewer children in the group, each of the remaining children would receive two additional stickers. How many children are there initially?

Let \( n \) be the initial number of children. Each child would receive \( \frac{360}{n} \) stickers. If there are six fewer children, the number of children becomes \( n - 6 \). Each of these remaining children would then receive \( \frac{360}{n - 6} \) stickers. According to the problem, if there are six fewer children, each remaining child receives two additional stickers compared to the original distribution. This gives us the equation: \[ \frac{360}{n - 6} = \frac{360}{n} + 2 \] To eliminate the fractions, we can multiply through by \( n(n - 6) \): \[ 360n = 360(n - 6) + 2n(n - 6) \] Expanding both sides: \[ 360n = 360n - 2160 + 2n^2 - 12n \] Now, we can simplify this equation. Subtract \( 360n \) from both sides: \[ 0 = -2160 + 2n^2 - 12n \] Rearranging gives: \[ 2n^2 - 12n + 2160 = 0 \] Dividing the entire equation by 2 simplifies it: \[ n^2 - 6n + 1080 = 0 \] Next, we can use the quadratic formula to solve for \( n \): \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -6 \), and \( c = 1080 \): \[ n = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1080}}{2 \cdot 1} \] \[ n = \frac{6 \pm \sqrt{36 - 4320}}{2} \] \[ n = \frac{6 \pm \sqrt{-4284}}{2} \] Since the discriminant \( -4284 \) is negative, it indicates that there are no real solutions for \( n \) in this context. Let's check our calculations again. Revisiting the equation: \[ \frac{360}{n - 6} - \frac{360}{n} = 2 \] Finding a common denominator: \[ \frac{360n - 360(n - 6)}{n(n - 6)} = 2 \] \[ \frac{360n - 360n + 2160}{n(n - 6)} = 2 \] \[ \frac{2160}{n(n - 6)} = 2 \] Cross-multiplying gives: \[ 2160 = 2n(n - 6) \] \[ 2160 = 2n^2 - 12n \] \[ 2n^2 - 12n - 2160 = 0 \] \[ n^2 - 6n - 1080 = 0 \] Now we can use the quadratic formula again: \[ n = \frac{6 \pm \sqrt{36 + 4320}}{2} \] \[ n = \frac{6 \pm \sqrt{4356}}{2} \] \[ n = \frac{6 \pm 66}{2} \] Calculating the two possible values for \( n \): 1. \( n = \frac{72}{2} = 36 \) 2. \( n = \frac{-60}{2} = -30 \) (not valid) Thus, the initial number of children is: \[ \boxed{36} \]