answer:Solution: a_{10}=S_{10}-S_{9}, and both S_{10} and S_{9} belong to {k_1,k_2,k_3,…,k_{10}}. If S_{10} neq S_{9}, then there are A_{10}^{2}=10 times 9=90 possibilities, If S_{10} = S_{9}, then a_{10}=0, According to the principle of counting by classification, there are a total of 90+1=91 possibilities, Therefore, the answer is: boxed{91} Based on the recursive formula of the sequence, a_{10}=S_{10}-S_{9}, and both S_{10} and S_{9} belong to {k_1,k_2,k_3,…,k_{10}}. The answer can be obtained through classified discussion. This problem tests the recursive formula of a sequence and the principle of counting by classification, examining the students' ability to transform, and is considered a medium-level question.

answer:1. We start by examining the given equation, 3^m - 7^n = 2, for small values of (m) and (n). 2. Let’s first check manually for (m leq 2): - For (m = 0): (3^0 - 7^n = 1 - 7^n = 2 rightarrow) no solution as (7^n) is not an integer. - For (m = 1): (3^1 - 7^n = 3 - 7^n = 2 rightarrow) no solution as (7^n) is not an integer. - For (m = 2): (3^2 - 7^n = 9 - 7^n = 2 rightarrow 7^n = 7 text{ gives } n = 1). So, we have one solution ((m, n) = (2, 1)). 3. Assume (m geq 3) and (n geq 2). We rewrite the original equation as: [ 3^m - 7^n = 2 implies 3^2 (3^{m-2} - 1) = 7 (7^{n-1} - 1) ] 4. Notice that (3^2 mid 7 (7^{n-1} - 1)). Hence: [ 9 mid 7^{n-1} - 1 ] 5. From number theory, the order of (7 mod 9) is (3). Therefore, (7^{n-1} equiv 1 pmod{9}): [ n-1 equiv 0 pmod{3} implies n-1 = 3k implies n = 3k + 1 ] 6. We know: [ 3^2 (3^{m-2} - 1) = 7 (7^{n-1} - 1) ] 7. In the equation ( 3^{2} left( 3^{m-2} - 1 right) = 7 left( 7^{n-1} - 1 right) ), let’s focus on the factor (3^{m-2} - 1): [ 3^{m-2} - 1 implies 7^{n-1} equiv 1 pmod{9} implies {3^{m-2} - 1} equiv 0 pmod{19} ] 8. Therefore: [ 19 mid 3^{m-2} - 1 ] 9. The order of (3 mod 19) is (18): [ 3^{18} equiv 1 pmod{19} implies m-2 equiv 0 pmod{18} implies m-2 = 18k implies m = 18k + 2. ] 10. Let (k = 1). Then: [ m = 18 cdot 1 + 2 = 20 implies 3^{20} - 7^{3k+1} = 2 ] 11. Returning to our prior steps, similarly: [ 37 mid 7^{n-1} - 1 ] 12. The order of (7 mod 37) is (9): [ 7^9 equiv 1 pmod{37} implies n-1 equiv 0 pmod{9} implies n-1 = 9k implies n = 9k + 1 ] 13. If we combine all equations: [ 9 mid 7^{n-1} - 1 implies 27k = 3k + 1 ] 14. This contradicts the original equation (3^{m-2} - 1), leading us to no further solutions. Examining small scenarios further verifies (m=3,n=3k+1)) : 15. Hence, the only solution for: [ 3^m - 7^n = 2 implies (m, n) = (2, 1) ] (boxed{(m, n) = (2, 1)})

answer:To simplify the given expression frac{a+2b}{a+b}-frac{a-b}{a-2b}div frac{{a}^{2}-{b}^{2}}{{a}^{2}-4ab+4{b}^{2}}, we follow the steps below: 1. Convert the division into multiplication and simplify the divisor: begin{align*} text{Original} &= frac{a+2b}{a+b}-frac{a-b}{a-2b}cdot frac{(a-2b)^{2}}{(a-b)(a+b)} &= frac{a+2b}{a+b}-frac{a-b}{a-2b}cdot frac{(a-2b)^{2}}{(a-b)(a+b)}. end{align*} 2. Notice that (a^2-b^2) = (a+b)(a-b) and (a^2-4ab+4b^2) = (a-2b)^2. Thus, the expression simplifies to: begin{align*} &= frac{a+2b}{a+b}-frac{(a-b)(a-2b)^2}{(a-2b)(a-b)(a+b)} &= frac{a+2b}{a+b}-frac{a-2b}{a+b}. end{align*} 3. Combine the fractions over a common denominator: begin{align*} &= frac{a+2b-(a-2b)}{a+b} &= frac{a+2b-a+2b}{a+b} &= frac{4b}{a+b}. end{align*} Therefore, the simplified form of the given expression is boxed{frac{4b}{a+b}}.

answer:1. **Given Information and Initial Setup:** We are given a rectangle ABCD with an area of 4. We need to determine the area of quadrilateral IXJY. Points I and J are the midpoints of segments [AD] and [BC], respectively. The point X is the intersection of lines (AJ) and (BI), and point Y is the intersection of lines (DJ) and (CI). 2. **Midpoints Calculation:** Since I and J are midpoints, we can determine that I splits [AD] into two equal segments and J splits [BC] into two equal segments. 3. **Division of the Rectangle into Triangles:** We know that by drawing the diagonals from the vertices and midpoints, we can divide the rectangle into isosceles right triangles. The entire rectangle splits into four smaller rectangles and these in turn split further into 16 smaller right triangles. 4. **Area Calculation of Triangles:** The area of each small right triangle would be: [ text{Area of each small triangle} = frac{text{Total area of rectangle}}{text{Number of small triangles}} = frac{4}{16} = frac{1}{4} ] 5. **Formation of Quadrilateral IXJY:** The quadrilateral IXJY is formed by the intersection of the lines (AJ), (BI), (DJ), and (CI). When inspecting the diagram, considering symmetry and how the lines intersect through midpoints and vertices, we see the quadrilateral IXJY forms within the central part of the rectangle as an overlay of four smaller triangles, one from each quadrant created by diagonal bisections. 6. **Summation of Related Areas:** Since there are four triangles forming IXJY and each of these triangles has an area of frac{1}{4} square units, the total area of IXJY is: [ text{Area of } IXJY = 4 times frac{1}{4} = 1 ] # Conclusion: Thus, the area of the quadrilateral IXJY is boxed{1}.