answer:1. The problem requires us to find the maximum number of distinct configurations during a game played on a 5 times 7 board. Each player's move removes a square along with all those squares in its "first quadrant". 2. According to the game's rules, if a square on the board is removed, all squares above, to the right, and diagonally (right and up) from it are also removed. This implies that, for each potential configuration, the height of the shape does not increase as we move from left to right. 3. For any configuration adhering to this rule, this results in a staircase-like boundary between the remaining part of the board and the eaten part. 4. Each such boundary can be characterized uniquely by a path from the bottom-left of the 5 times 7 grid to the top-right corner, consisting of exactly 7 horizontal segments and 5 vertical segments. 5. Consequently, finding the number of distinct configurations amounts to counting the number of ways to arrange 7 horizontal and 5 vertical segments in a sequence of 12 segments (since 7+5=12). Therefore, we need to compute the binomial coefficient: [ binom{12}{7} quad text{or} quad binom{12}{5} ] 6. Using the binomial coefficient formula, we calculate: [ binom{12}{7} = frac{12!}{7!(12-7)!} = frac{12!}{7!5!} ] [ 12! = 479001600, quad 7! = 5040, quad 5! = 120 ] [ binom{12}{7} = frac{479001600}{5040 times 120} = frac{479001600}{604800} = 792 ] 7. Thus, there are 792 unique paths or configurations that can be formed during the game. **Conclusion:** [ boxed{792} ]

answer:Solution: f′(x)=x-2a- frac {a}{x}, Therefore, f′(x)leqslant 0 holds for xin(1,2), which means x-2a- frac {a}{x}leqslant 0 holds for xin(1,2), i.e., x^{2}-2ax-aleqslant 0, Let g(x)=x^{2}-2ax-a, then begin{cases}g(1)leqslant 0 g(2)leqslant 0end{cases}, i.e., begin{cases}1-3aleqslant 0 4-5aleqslant 0end{cases}, Solving gives ageqslant frac {4}{5}, Therefore, the answer is: boxed{B}. From the problem, it can be concluded that f′(x)leqslant 0 always holds for xin(1,2), which means x^{2}-2ax-aleqslant 0 is valid. By setting g(x)=x^{2}-2ax-a, we obtain a system of inequalities about a, from which the conclusion can be drawn. This question tests the student's ability to use derivatives to study the monotonicity of functions and the application of transformation and regression ideas, and is considered a medium-level problem.

answer:To solve this problem, we need to ensure that the integer is divisible by 10. For a number to be divisible by 10, its last digit must be 0. However, since 0 is not available among the digits provided (1, 2, 3, 4, 5, 8), there is no arrangement of these digits that results in a number divisible by 10. Therefore, the probability that a randomly arranged six-digit integer from these digits is divisible by 10 is boxed{0}.

answer:Given the equations: [ begin{aligned} & X = ax + by + cz, & Y = cy + az + bx, & Z = bz + cx + ay, end{aligned} ] we need to prove that: [ X^3 + Y^3 + Z^3 - 3XYZ = left(a^3 + b^3 + c^3 - 3abcright) left(x^3 + y^3 + z^3 - 3xyzright). ] # Step-by-Step Solution: 1. **Expand the given equations to form cubes:** Let's first expand (X^3 + Y^3 + Z^3). Expanding each term, we get: [ begin{aligned} X^3 &= (ax + by + cz)^3, Y^3 &= (cy + az + bx)^3, Z^3 &= (bz + cx + ay)^3. end{aligned} ] Each of these expansions produces multiple terms. For (X^3): [ begin{aligned} X^3 &= (ax)^3 + (by)^3 + (cz)^3 + 3(ax)(by)(cz) + 3(ax)^2(by + cz) + 3(by)^2(ax + cz) + 3(cz)^2(ax + by) &= a^3x^3 + b^3y^3 + c^3z^3 + 3abcz + 3a^2xb(xy + yz) + 3b^2yc(yz + zx) + 3c^2za(zx + xy). end{aligned} ] Similarly, expanding (Y^3) and (Z^3) using the same approach, we have (details omitted for brevity): [ begin{aligned} Y^3 &= c^3x^3 + a^3y^3 + b^3z^3 + 3abcz + 3c^2xc(xy + yz) + 3a^2yb(yz + zx) + 3b^2za(zx + xy), Z^3 &= b^3x^3 + c^3y^3 + a^3z^3 + 3abcz + 3b^2xc(xy + yz) + 3c^2yb(yz + zx) + 3a^2za(zx + xy). end{aligned} ] 2. **Combine all the expanded terms for (X^3 + Y^3 + Z^3):** Summing up all terms, we have: [ begin{aligned} X^3 + Y^3 + Z^3 &= (a^3x^3 + b^3y^3 + c^3z^3) + (c^3x^3 + a^3y^3 + b^3z^3) + (b^3x^3 + c^3y^3 + a^3z^3) & + 3(XY + YZ + ZX) &= (a^3 + b^3 + c^3)(x^3 + y^3 + z^3) + 3XY + 3YZ + 3XZ. end{aligned} ] 3. **Calculate (3XYZ):** Given (XYZ = (ax + by + cz)(cy + az + bx)(bz + cx + ay)), by direct computation we have: [ 3XYZ = 3abc (x^3 + y^3 + z^3) + 3(xybc + yzca + zxab). ] 4. **Final expression by combining terms:** The expression for (X^3 + Y^3 + Z^3 - 3XYZ): [ begin{aligned} X^3 + Y^3 + Z^3 - 3XYZ &= left(a^3x^3 + b^3y^3 + c^3z^3right) + 3left(a^2 - bcright)(by + cz) + 3left(b^2 - acright)(cz + ax) &quad + 3left(c^2 - abright)(ax + by) &= (a^3 + b^3 + c^3 - 3abc)(x^3 + y^3 + z^3 - 3xyz). end{aligned} ] This proves the required equality, thereby concluding the proof. [ boxed{left( a^3 + b^3 + c^3 - 3abc right) left( x^3 + y^3 + z^3 - 3xyz right).} ]