answer:We have S_n=n^2-5n (n∈N^{}). Thus, a_1 = S_1 = -4. For n geq 2, we have S_{n-1} = (n-1)^2 - 5(n-1) = n^2 - 7n + 6. Since a_n = S_n - S_{n-1}, we obtain a_n = 2n - 6. When n = 1, we get a_1 = -4. Notice that a_n - a_{n-1} = 2 is constant, so {a_n} is an arithmetic sequence with first term a_1 = -4 and common difference d = 2. Given p - q = 4, let q = 1. Then p = 5, and so a_5 - a_1 = 8. Therefore, the answer is boxed{8}.

answer:Let's denote the number of photos Claire has taken as C. According to the information given, Lisa has taken 3 times as many photos as Claire, so Lisa has taken 3C photos. Robert has taken 10 more photos than Claire, so Robert has taken C + 10 photos. Since Lisa and Robert have taken the same number of photos, we can set their quantities equal to each other: 3C = C + 10 Now, we can solve for C: 3C - C = 10 2C = 10 C = 10 / 2 C = 5 Therefore, Claire has taken boxed{5} photos.

answer:1. **Board Setup and Initial Conditions**: - We start by coloring the cells of the 100 times 100 board in a checkerboard pattern. - The lower-left cell (first row, first column) is black. Due to the checkerboard coloring, the leftmost upper cell (first row, hundredth column) is white. 2. **Movement Analysis**: - The first move is horizontal, and it is known that from a black cell, moves alternate between horizontal and vertical. - Specifically, from a black cell, the move must be horizontal; from a white cell, it must be vertical, and so forth. 3. **Color Transition**: - The transition between cell colors accordingly: - From black: horizontal move to adjacent column, and the next cell is white. - From white: vertical move to adjacent row, and the next cell is black. 4. **Path Observation**: - The piece starts from the bottom left black cell and reaches the top left white cell. This requires an even number of moves (alternating between horizontal and vertical). - Next, it should traverse to the top right cell (also white), continuing alternating horizontal and vertical moves. 5. **Contradiction Assumption**: - Assume that no cell is visited more than once horizontally and vertically. - This implies that no side of any cell in any direction is crossed twice by the piece. 6. **Wall Construction**: - Since the piece's path avoids any repetition of moves, it effectively builds a "wall" along its path. - Each move divides the board into distinct sections demarcated by the wall. 7. **Separation of Target Cells**: - When the piece reaches the upper left cell (white), the constructed wall ensures that it separates the lower side and upper side of the board. - Thus, without retracing steps, the piece transitions into a new, previously unvisited portion of the board by the top move. 8. **Inconsistency in Reaching Final Destination**: - When considering the above segments: - The wall created during these moves should naturally segregate the cells into sections. - Specifically, the lower-left to upper-left cells are within one region, requiring another altogether for the piece’s continuation to the upper-right. - Such segmentation implies non-overlapping which precludes the possibility of directly linking the top-left to the top-right without revisiting some paths. 9. **Final Conclusion**: - The contradiction found (i.e., the requirement of revisiting or passing through previously traversed pathways to achieve the transition) establishes: - It verifies that at least two moves join two identical cells in traversing horizontally back and forth. - Consequently, demonstrating that it is impossible to reach the final required destination without duplicity of movement. [ boxed{text{Thus, the piece must visit at least two cells more than once during its path as demonstrated.}} ]

answer:To construct a triangle ( triangle ABC ) around the acute-angled triangle ( triangle A_0B_0C_0 ) such that ( triangle ABC ) is similar to another given triangle ( triangle A'B'C' ), we need to ensure that the vertices ( A, B, ), and ( C ) of ( triangle ABC ) correspond respectively to ( A', B', ), and ( C' ) in ( triangle A'B'C' ). Also, the point ( C_0 ) must be an interior point of side ( AB ), ( A_0 ) must be an interior point of side ( BC ), and ( B_0 ) must be an interior point of side ( CA ). We then aim to find the triangle ( triangle ABC ) with the largest possible area among all such triangles. Step-by-Step Construction: 1. **Construct Angle Arcs:** - For the given triangle ( triangle A_0B_0C_0 ), construct an arc over side ( A_0B_0 ) such that it subtends an angle (gamma) at all points on the arc. This arc is denoted by ( i_c ). - Similarly, construct an arc over side ( B_0C_0 ) that subtends an angle (alpha). This arc is denoted by ( i_a ). 2. **Draw Line through ( B_0 ):** - Through ( B_0 ), draw a line that intersects both arcs ( i_c ) and ( i_a ) at points ( C ) and ( A ) respectively. This ensures that the vertices ( A ) and ( C ) form angles (alpha) and (gamma) with their corresponding sides. 3. **Find Intersection Point ( B ):** - Identify the intersection of line segments ( CA_0 ) and ( AC_0 ) to locate point ( B ). 4. **Ensure Similarity and Internal Points:** - Ensure that ( triangle ABC ) is similar to ( triangle A'B'C' ) by verifying the angles ( angle BAC = alpha, angle ABC = beta, angle BCA = gamma ). 5. **Special Case with Largest Area:** - To maximize the area of ( triangle ABC ), note that among the similar triangles, the one with the longest side ( CA ) will have the largest area. - Consider the complete circle ( k_a ) passing through ( A_0 ) and the complete circle ( k_c ) passing through ( C_0 ). We maximize the area by selecting points ( C ) and ( A ) such that ( M ), where ( k_a ) and ( k_c ) intersect, is as far as possible on the chosen arcs ( i_a ) and ( i_c ). Conclusion: By following the above steps, we can construct the triangle ( triangle ABC ), which not only meets the given conditions but also is the triangle with the largest area. [ boxed{} ]