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question:Consider an inifinte sequence x_1, x_2,dots of positive integers such that, for every integer ngeq 1 : If x_n is even, x_{n+1}=dfrac{x_n}{2} ; If x_n is odd, x_{n+1}=dfrac{x_n1}{2}+2^{k1} , where 2^{k1}leq x_n<2^k . Determine the smaller possible value of x_1 for which 2020 is in the sequence.
answer:To determine the smallest possible value of ( x_1 ) for which ( 2020 ) is in the sequence, we need to analyze the sequence generation rules and backtrack from ( 2020 ). 1. **Understanding the sequence rules:** - If ( x_n ) is even, then ( x_{n+1} = frac{x_n}{2} ). - If ( x_n ) is odd, then ( x_{n+1} = frac{x_n - 1}{2} + 2^{k-1} ), where ( 2^{k-1} leq x_n < 2^k ). 2. **Backtracking from ( 2020 ):** - Since ( 2020 ) is even, the previous term ( x_{n-1} ) must be ( 4040 ) (which is impossible since ( 4040 ) is not less than ( 2020 )). - Therefore, ( 2020 ) must be generated from an odd number using the second rule. 3. **Finding the previous term for ( 2020 ):** - We need ( x_{n-1} ) such that ( 2020 = frac{x_{n-1} - 1}{2} + 2^{k-1} ). - Rearranging, we get ( x_{n-1} = 2 times (2020 - 2^{k-1}) + 1 ). 4. **Determining ( k ):** - Since ( 2^{k-1} leq 2020 < 2^k ), we find ( k ) such that ( 2^{k-1} leq 2020 < 2^k ). - The smallest ( k ) satisfying this is ( k = 11 ) because ( 2^{10} = 1024 leq 2020 < 2048 = 2^{11} ). 5. **Calculating ( x_{n-1} ):** - Using ( k = 11 ), we have ( 2^{10} = 1024 ). - Therefore, ( x_{n-1} = 2 times (2020 - 1024) + 1 = 2 times 996 + 1 = 1993 ). 6. **Repeating the process:** - Continue backtracking using the same rules until we find the smallest ( x_1 ). 7. **Sequence generation:** - ( 2020 leftarrow 1993 leftarrow 1939 leftarrow 1831 leftarrow 1615 leftarrow 1183 ). 8. **Checking the smallest ( x_1 ):** - We need to ensure that ( 1183 ) is the smallest possible starting value that can generate ( 2020 ) in the sequence. - Testing smaller values and verifying the sequence rules confirm that ( 1183 ) is indeed the smallest value. The final answer is ( boxed{1183} ).
question:Given an arithmetic sequence {a_n} with a common difference of d, if frac{1}{2} < d < frac{1}{3} and (a_1 1)^2 = (a_{11} 1)^2, find the maximum value of n that maximizes the sum of the first n terms, S_n.
answer:Since (a_{11} - 1)^2 = (a_1 - 1)^2, we have: (a_{11} - 1 + a_1 - 1)(a_{11} - 1 - a_1 + 1) = 0, (12a_1 + 10d - 2)10d = 0, 12a_1 + 10d - 2 = 0, a_1 = frac{1 - 5d}{6}, a_n = frac{1 - 5d}{6} + (n - 1)d = frac{1}{6} + left(n - frac{11}{6}right)d. Since a_n geq 0, we have: frac{1}{6} + left(n - frac{11}{6}right)d geq 0, 6n - 11 leq -frac{1}{d}. Given that -frac{1}{2} < d < -frac{1}{3}, we have: 2 < frac{1}{d} < 3, 6n - 11 leq -3, n leq frac{8}{3}. Thus, the number of terms n that maximizes the sum of the first n terms S_n is 2. Therefore, the answer is: boxed{2}.
question:Semicircles of diameter 4 inches are aligned in a linear pattern, with a second row staggered under the first such that the flat edges of the semicircles in the second row touch the midpoints of the arcs in the first row. What is the area, in square inches, of the shaded region in an 18inch length of this pattern? Express your answer in terms of pi.
answer:1. **Calculate Number of Semicircles in Each Row**: Each semicircle has a diameter of 4 inches, so each has a radius of 2 inches. In an 18-inch length, the number of semicircles in each row is frac{18}{4} = 4.5. Since semicircles can't be half in a continuous pattern, we consider 4 full semicircles in each row. 2. **Calculate Total Number of Semicircles**: There are two rows, each contributing 4 full semicircles. Thus, there are 4 times 2 = 8 semicircles. 3. **Calculate Equivalent Full Circles**: Since one full circle is composed of two semicircles, the number of full circles formed by 8 semicircles is frac{8}{2} = 4 full circles. 4. **Calculate the Area**: The area of one full circle with radius 2 inches is pi times 2^2 = 4pi square inches. Therefore, the total area for 4 full circles is 4 times 4pi = 16pi square inches. boxed{16pi} square inches.
question:An artist constructs a sculpture using 20 cubes, each with an edge of 1 meter. She arranges them on the ground to form a structure where the bottom layer consists of 12 cubes in a 4x3 rectangle, the middle layer consists of 6 cubes forming a Tshape, and the top layer consists of 2 cubes placed side by side at the center of the Tshape. She then paints all the exposed surfaces. Calculate the total area she paints. A) 54 square meters B) 58 square meters C) 62 square meters D) 66 square meters E) 70 square meter
answer:1. **Bottom Layer (Layer 1)**: - **12 cubes**: All top faces exposed, side faces exposed only for cubes on the perimeter (8 cubes). - **Exposed Faces Calculation**: - Top faces: 12 times 1 = 12 square meters. - Side faces: 8 times 3 = 24 (each perimeter cube has 3 exposed sides). 2. **Middle Layer (Layer 2)**: - **6 cubes** forming a T-shape: Top faces exposed, side and bottom faces vary. - **Exposed Faces Calculation**: - Top faces: 6 times 1 = 6 square meters. - Side faces: 10 (edges of T-shape). 3. **Top Layer (Layer 3)**: - **2 cubes** side by side: All faces exposed except the bottom and one adjoining side. - **Exposed Faces Calculation**: - Top faces: 2 times 1 = 2 square meters. - Side faces: 2 times 4 = 8 (each cube has 4 exposed sides). 4. **Total Exposed Surface Area**: - **Total**: 12 + 24 + 6 + 10 + 2 + 8 = 62 square meters. Conclusion: The total area that the artist paints is 62 square meters. The final answer is boxed{C}