answer:1. Assume the contrary, i.e., it is not possible to move any of the n-1 pieces to an adjacent cell such that none of the pieces are on neighboring cells. 2. In a n times n table, there are n columns. Since there are n-1 pieces, at least one column must be empty. 3. Notice that an empty column cannot be an edge column. To illustrate, if the rightmost column is empty, then from the farthest non-empty column to the left, we can move a piece to the right. 4. Similarly, there cannot be two consecutive empty columns. This is because, from the leftmost of these empty columns, the piece in the neighboring column can be moved to the right into the empty column. 5. Therefore, to the left of every empty column, there is a piece. If this piece cannot be moved to the right, there must be another piece in the same row that prevents this movement. Thus, each row containing a "left" piece must have another piece in the same row. 6. Let the number of empty columns be k. Since there must be at least one piece to the left of every empty column, the corresponding number of "left" pieces is at least k. 7. Consequently, the n-1-k pieces occupy at most n-1-k rows. Thus, there are at least k+1 empty rows, which is more than the k empty columns. 8. However, by similar reasoning, if we consider the empty rows, there must be more empty columns than empty rows, which contradicts the previous conclusion. 9. Therefore, the assumption that it is not possible to move any piece to an adjacent cell without adjacency is false. Conclusion. blacksquare

answer:First, we convert all the fractions to have a common denominator. A suitable common denominator for 40, 30, 60, and 120 is 120. We perform the conversion as follows: begin{align*} text{Dodgeball: } &frac{13}{40} = frac{13 times 3}{40 times 3} = frac{39}{120}, text{Karaoke: } &frac{9}{30} = frac{9 times 4}{30 times 4} = frac{36}{120}, text{Magic Show: } &frac{17}{60} = frac{17 times 2}{60 times 2} = frac{34}{120}, text{Quiz Bowl: } &frac{23}{120}. end{align*} Comparing these values: - Dodgeball: frac{39}{120} - Karaoke: frac{36}{120} - Magic Show: frac{34}{120} - Quiz Bowl: frac{23}{120} The order from most popular to least popular is boxed{text{Dodgeball, Karaoke, Magic Show, Quiz Bowl}}.

answer:1. **Expression simplification and factorization**: First, express the given equation in terms of its prime factorization: [ 2^{10} cdot 7^3 ] We explore if this can be written as a^b by looking for possible values of a such that a^b = 2^{10} cdot 7^3. 2. **Combining bases**: Consider making a a product of both bases present in the factorization: [ a = 2^2 cdot 7 = 4 cdot 7 = 28 ] Hence, a^b = 28^b. 3. **Determining the power b**: Now find b such that 28^b = 2^{10} cdot 7^3: [ 28^b = (2^2 cdot 7)^b = 2^{2b} cdot 7^b ] Setting 2^{2b} = 2^{10} and 7^b = 7^3, we find 2b = 10 and b = 3. 4. **Calculate a + b**: [ a + b = 28 + 3 = 31 ] 5. **Conclusion**: The smallest possible value of a + b is 31. The final answer is boxed{31} (Choice B).

answer:(1) From ρ= frac {2cosθ}{sin^{2}theta }, we obtain ρ^2sin^2θ=2ρcosθ, so the rectangular coordinate equation of curve C is y^2=2x. The parametric equations of line l are begin{cases} x= frac {1}{2}+ frac {1}{2}t y= frac { sqrt {3}}{2}tend{cases}. Eliminating the parameter, we get the ordinary equation of line l as 2sqrt{3}x - 2y - sqrt{3} = 0. (2) Substituting the parametric equations of line l into the rectangular coordinate equation of curve C, we have 3t^2 - 4t - 4 = 0. Let t_1 and t_2 be the parameters corresponding to points A and B, respectively. Then, we have t_1 + t_2 = frac{4}{3} and t_1t_2 = -frac{4}{3}. Thus, |PA| + |PB| = |t_1 - t_2| = sqrt{(t_1 + t_2)^2 - 4t_1t_2} = boxed{frac{8}{3}}.