Bakayev E.V.

Cheburashka has a set of 36 stones with masses of 1 g, 2 g, ..., 36 g, and Shapoklyak has super glue, one drop of which can glue two stones together (thus, three stones can be glued with two drops, and so on). Shapoklyak wants to glue the stones in such a way that Cheburashka cannot select one or several stones from the resulting set with a total mass of 37 g. What is the minimum number of drops of glue she needs to achieve her goal?
Example. By gluing stones with masses 1 and 18, 2 and 17, ..., 9 and 10, respectively, Cheburashka will get a set where each stone weighs from 19 to 36 grams, so one stone will be too little, and two will be too many.

Another way. By gluing all the stones with odd masses in pairs, Cheburashka will get a set where all stones have even masses, and Cheburashka will not be able to form an odd mass.

Estimate. If Cheburashka uses only 8 drops, then no more than 16 of the original stones will be involved in the gluing. Therefore, at least one of the 18 pairs

$\{1,36\},\{2,35\}, \ldots,\{18,19\}$ will remain "untouched," and Cheburashka will be able to choose it.

## Answer

9 drops.