A market vendor's basket contains eggs. The first buyer buys half of the eggs and one more. The second buyer buys half of the remaining eggs and one more. The third buyer also gets exactly half of the eggs left in the basket and one more. How many eggs were in the basket?
I. solution (with equation): If we denote the number of eggs by $x$, then according to the problem, the first buyer takes $\frac{x}{2}+1$ eggs and leaves $\frac{x}{2}-1$. The second buyer takes $\frac{1}{2}\left(\frac{x}{2}-1\right)+1$ eggs and leaves $\frac{1}{2}\left(\frac{x}{2}-1\right)-1$. The third buyer takes $\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]^{2}+1$ eggs and leaves $\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]-1$, which, according to the problem, equals zero. Therefore,

$$
\begin{array}{r}
\frac{1}{2}\left[\frac{1}{2}\left(\frac{x}{2}-1\right)-1\right]-1=0 \\
\frac{x}{4}-\frac{1}{2}-1=2
\end{array}
$$

from which

$$
x=14
$$

Orlik Péter (Bp. V., Eötvös g. I. o. t.)

II. solution (by reasoning): Let's put the eggs back into the basket. The third buyer returns the last 1 egg they took, which was half of the eggs they found, so they return 1 more egg. The second buyer returns the extra egg they took, and since the 3 eggs in the basket represent half of what the second buyer found, the second buyer adds 3 more eggs to make the total 6. The first buyer returns the extra 1 egg they took, and the 7 eggs in the basket represent half of what the first buyer found, so there were 14 eggs in the basket.

Szabó Anna (Jászberény, Tanítóképző II. o. t.)

III. solution: Generally,

the number of eggs before the last purchase is 2,

the number of eggs before the second-to-last purchase is $2(2+1)=2^{2}+2=6$,

the number of eggs before the purchase before that is $2\left(2^{2}+2+1\right)=2^{3}+2^{2}+2=14$, and so on

$\vdots$

for $n$ buyers, the number of eggs before the first purchase is

$$
2\left(2^{n-1}+2^{n-2}+\ldots+2+1\right)=2\left(2^{n}-1\right)
$$

according to the formula for the sum of a geometric series.

Biszterszky Sára (Miskolc, 13. sz. gépip. techn. II. o. t.)

Note: This problem is a typical example (especially when considering more than 3 buyers) that equations are not always the shortest path. Sometimes simple arithmetic can lead to the solution more quickly. (L. Perelman: "Entertaining Algebra" p. 108.)