6. (3 points) Given the cryptarithm: ЖАЛО + ЛОЖА = ОСЕНЬ. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter А.
Answer: 8

Solution: The rebus can be rewritten as ОСЕНЬ $=($ ЖА + ЛО $) \cdot 101$. First, this means that the last digit of ЖА + ЛО is Ь. Second, if ЖА + ЛО $<100$, the result will be a four-digit number. Let $Ж А+Л О=1 Х Ь$, where $X$ is some digit.

Then ОСЕНЬ $=1 Х Ь 00+1 Х Ь$. If $\mathrm{b}<9$, then the second and fourth digits of this number should match, but they do not. Therefore, $\mathrm{b}=9$. Consequently, $\mathrm{A}+\mathrm{O}=9$. But О is 1, so А is 8.

There are many examples, for instance, $7861+6178=14039$.

7. (4 points) Alice and Bob each have three equal segments. First, Alice breaks one of the segments into two unequal parts. Then, Bob breaks another of the original segments into any two parts. As a result, there are five segments, from which ten sets of three segments can be chosen. Alice wins if at least 4 of these ten sets form a triangle. Otherwise, Bob wins. Who will win if both players play optimally?

Answer: Bob wins.

Solution: Let the original segments have a length of 1, and Alice breaks one of the segments into parts $x$ and $1-x$, with $x<1-x$ for definiteness. Then Bob needs to break another segment into parts $1-y$ and $y$ such that $y$ is very small, specifically, the inequalities $y<x$, $y<(1-x)-x$, and $y<(1-y)-(1-x)$, or $2 y<x$, are satisfied. Clearly, he can choose such a $y$.

Using the triangle inequality: to form a triangle, it is necessary and sufficient that the largest side of the triangle is less than the sum of the other two. With such a choice of $y$, it is not greater than the difference between any two other segments, and therefore cannot form a triangle with any of them. There are exactly 6 sets of three segments that include the segment $y$. Additionally, the remaining segment of length 1 and the segments Alice obtained also do not form a triangle. This is the seventh set of segments that does not form a triangle. Thus, Alice is left with only 3 sets of segments that can form a triangle, and she loses.