6. (3 points) Given the cryptarithm: RIVER + SQUARE = ABVAD. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter B.
Answer: 2

Solution: The rebus can be rewritten as ABVAD $=(\mathrm{KA}+\mathrm{PE}) \cdot 101$. First, this means that the last digit of $\mathrm{KA}+\mathrm{PE}$ is D. Second, if $\mathrm{KA}+\mathrm{PE}<100$, the result will be a four-digit number. Let $\mathrm{KA}+\mathrm{PE}=1X$D, where $X$ is some digit.

Then ABVAD $=1X$D$00+1X$D. If D $<9$, then the second and fourth digits of this number should match, but they do not. Therefore, D $=9$ and the second digit of the number ABVAD is one more than the fourth. But A is 1, so B is 2.

There are many examples, for instance, $5861+6158=12019$.

7. (4 points) Alice and Bob have three equal segments. First, Alice breaks one of the segments into two parts. Then, Bob breaks another of the original segments into two arbitrary parts. As a result, five segments are obtained, from which ten ways to choose three segments are possible. Alice wins if at least 4 of these ten ways give triples of segments that form a triangle. Otherwise, Bob wins. Who will win with correct play by both opponents?

Answer: Alice wins.

Solution: We use 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.

Let the original segments have a length of 1. Alice breaks one of the segments into two equal parts. Let Bob break his segment into parts of length $x$ and $1-x$, and for definiteness, let $x<1-x$. $1-x<\frac{1}{2}+\frac{1}{2}$, so this triple forms a triangle. Also, the triple formed by Alice's segments and Bob's smaller segment forms a triangle. Two more suitable triples are obtained when we take the segments $1,1-x$ and $\frac{1}{2}$.