Iranian Mathematical Olympiad

Let $a_1,a_2,\dots ,a_n$ denote the lamps, and $b_1,b_2,\dots ,b_5$ denote the keys. First, it is easy to see that $n\ge 3$, because otherwise there are at most 4 subsets of lamps and it's in contradiction with the assumption of the problem. Construct a $10\times n$ table that each row corresponds to a triple set of keys and each column corresponds to a lamp.



In intersection of row $i$ and column $j$ assign number 1 if after switching 3 keys corresponding to row $i$, lamp $a_j$ turns on, and else assign 0. Denote the sum of row $i$'s numbers by $z_i$. Suppose lamp $a_i$ is connected to key $X_j$. Now sum of numbers in column $j$ is $\left(\genfrac{}{}{0ex}{}{X_j}{3}\right)+\left(\genfrac{}{}{0ex}{}{X_j}{1}\right)\left(\genfrac{}{}{0ex}{}{5-X_j}{2}\right)$, so

$$\sum _{i=1}^{10}{z}_i=\sum _{j=1}^n\left(\left(\genfrac{}{}{0ex}{}{X_j}{3}\right)+\left(\genfrac{}{}{0ex}{}{X_j}{1}\right)\left(\genfrac{}{}{0ex}{}{5-X_j}{2}\right)\right)(1)$$

$X_j$ is positive integer between 1 and 5. It's easy to check that $\left(\genfrac{}{}{0ex}{}{X_j}{3}\right)+\left(\genfrac{}{}{0ex}{}{X_j}{1}\right)\left(\genfrac{}{}{0ex}{}{5-X_j}{2}\right)$ phrase is always greater than 3. Because of equation 1, it is deduced that

$$\sum _{i=1}^{10}{z}_i\ge 4n.$$

Because of pigeonhole principle there exist $i$ so that $z_i\ge \lfloor \frac{4n}{10}\rfloor \ge 2$ and we're done. ■