Kaminin D.A.

In a row, all possible irreducible fractions with denominators not exceeding one hundred are written. Masha and Svetlana place "+" or "-" signs in front of any fraction that does not yet have a sign. They do this in turns, but it is known that Masha will have to make the last move and calculate the result of the operations. If the result turns out to be an integer, Svetlana will give her a chocolate bar. Can Masha get the chocolate bar regardless of Svetlana's actions?
Note that in the given sequence, there is an odd number of fractions. Indeed, if the proper fraction $a / b$ is irreducible, then the fraction $1-a / b=(b-a) / b$ is also a proper and irreducible fraction. These fractions are different, except for one case: $a=1, b=2$. Therefore, Masha should make the first move.

Then she can act as follows: on the first move, place any sign in front of the fraction $1 / 2$, for example, the sign “+”. The remaining fractions can be paired so that the sum of the fractions in each pair is

| equal to 1. Therefore, with all fractions except $1 / 4$ and $3 / 4$, the following strategy can be adhered to: if Svetlana places some sign in front of the fraction $a / b$, then Masha places the same sign in front of the fraction $1-a / b$. Thus, the sum of all fractions in such pairs will be an integer.

For the fractions $1 / 4$ and $3 / 4$, the strategy changes: in response to the sign placed by Svetlana in front of one of them, Masha should place the opposite sign in front of the other. Then $1 / 2+1 / 4-3 / 4=0$ or $1 / 2-1 / 4+3 / 4=1$, which means the result obtained by Masha will be an integer.

## Answer

She can.