Zhendarov R.G.

On the board, the product $a_{1} a_{2} \ldots a_{100}$ is written, where $a_{1}, \ldots, a_{100}$ are natural numbers. Consider 99 expressions, each of which is obtained by replacing one of the multiplication signs with an addition sign. It is known that the values of exactly 32 of these expressions are even. What is the maximum number of even numbers among $a_{1}, a_{2}, \ldots, a_{100}$ that could have been?
Consider the leftmost even number $a_{i}$ and the rightmost even number $a_{k}$. Note that the sums with indices from $i$ to $k-1$ are even and only they are (in sums with smaller indices, the first addend is odd and the second is even; in sums with larger indices, it is the opposite).

Thus, $k-i=32$. The number of even numbers will be the greatest if all numbers between $a_{i}$ and $a_{k}$ are also even. In this case, the number of even numbers will be 33.

## Answer

33 numbers.