The first buyer bought ½ of the contents of the box with one free. Then the second buyer bought also ½ of the contents of the remaining and got one free. The same for the third buyer who bought ½ of the remaining with one free. The same for the fourth buyer who bought ½ of the remaining with one free and nothing left. How many dragon fruits were in the box?

[tex] \begin{array}{l} \textsf{Let x be the initial number of dragon fruits.} \\ \\ \textsf{Remaining dragon fruits after the first buyer:} \\ \\ \implies \sf \dfrac{1}{2}x - 1 \\ \\ \textsf{Remaining dragon fruits after the second buyer:} \\ \\ \implies \sf \dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1 \\ \\ \textsf{Remaining dragon fruits after the third buyer:} \\ \\ \implies \sf \dfrac{1}{2}\left[\dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1\right] \\ \\ \textsf{Remaining dragon fruits after the fourth buyer:} \\ \\ \implies \sf \dfrac{1}{2}\left\{\dfrac{1}{2}\left[\dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1\right]-1\right\}-1 = 0 \\ \\ \textsf{Solving for x,} \\ \\ \sf \dfrac{1}{2}\left\{\dfrac{1}{2}\left[\dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1\right]-1\right\}= 1 \\ \\ \sf \dfrac{1}{2}\left[\dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1\right]-1 = 2 \\ \\ \sf \dfrac{1}{2}\left[\dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1\right] = 3 \\ \\ \sf \dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right)-1 = 6 \\ \\ \sf \dfrac{1}{2}\left(\dfrac{1}{2}x - 1\right) = 7 \\ \\ \sf \dfrac{1}{2}x - 1 = 14 \\ \\ \sf \dfrac{1}{2}x = 15 \\ \\ \boxed{\sf x = 30} \end{array} [/tex]

#CarryOnLearning