answer:1. Define the twin primes as p and p+2. We need these twin primes' sum to be a power of a prime number q. This gives us the equation: [ p + (p+2) = q^k quad text{for some integer } k geq 1. ] 2. Simplify the equation: [ p + p + 2 = q^k implies 2p + 2 = q^k. ] 3. Factor out the common terms: [ 2(p + 1) = q^k. ] 4. Since q is a prime number and must divide the left-hand side (which is even), q has to be 2 (the only even prime): [ 2(p + 1) = 2^k. ] 5. Divide both sides by 2: [ p + 1 = 2^{k-1}. ] 6. Therefore: [ p = 2^{k-1} - 1, ] and [ p + 2 = 2^{k-1} + 1. ] 7. For both p and p+2 to be prime, at least one of them must be divisible by 3, since three consecutive integers cannot all be non-divisible by 3. Hence, one of 2^{k-1} - 1, 2^{k-1}, or 2^{k-1} + 1 must be divisible by 3. Since 2^{k-1} itself is a power of 2, it cannot be divisible by 3: [ 3 mid (2^{k-1} - 1) quad text{or} quad 3 mid (2^{k-1} + 1) quad text{must hold.} ] 8. Checking the values: - If 2^{k-1} = 3, then k-1 = 1 which gives k = 2. Substituting: [ p = 2^{2-1} - 1 = 2 - 1 = 1, ] which is not a prime. - If k-1 > 1, then let's find the first valid k value to satisfy both p and p+2 being primes: Let: [ p = 3 quad text{prime} quad text{and} quad p + 2 = 5 quad text{prime} ] Both 3 and 5 are prime: [ p = 3 = 2^{3-1} - 1 = 3, quad text{and} p + 2 = 5 = 2^{3-1} + 1 = 5. ] 9. Conclusion: There is only one pair of twin primes (3, 5) whose sum (3 + 5 = 8) is a power of 2 (2^3). [ boxed{(3,5)} ]

answer:Calculate the circumference of each circle: - Smaller circle ( C_1 = 2 times 3pi = 6pi ) inches. - Larger circle ( C_2 = 2 times 7pi = 14pi ) inches. Determine the time each bug takes to crawl its respective circumference: - Time for bug on ( C_1 ) (smaller circle) = ( frac{6pi}{4pi} = frac{3}{2} ) minutes. - Time for bug on ( C_2 ) (larger circle) = ( frac{14pi}{6pi} = frac{7}{3} ) minutes. The bugs will meet again at point P when the time ( t ) in minutes satisfies that ( t div frac{3}{2} ) and ( t div frac{7}{3} ) are both integers. We must find the least common multiple (LCM) of ( frac{3}{2} ) and ( frac{7}{3} ): - Convert ( frac{3}{2} ) and ( frac{7}{3} ) to have a common denominator: ( frac{9}{6} ) and ( frac{14}{6} ). - LCM of ( 9 ) and ( 14 ) is 126. - LCM of denominators (6) is 6. Thus, LCM of ( frac{9}{6} ) and ( frac{14}{6} ) in minutes is: [ frac{126}{6} = 21 ] Therefore, the two bugs will meet again at point P after ( boxed{21} ) minutes.

answer:**Analysis:** Given that the equation of curve C in Cartesian coordinates is x^2 + (y - 2)^2 = 4, with the center of the circle at C(0, 2) and the radius BC = 2. The Cartesian coordinates of the point (4, frac{pi}{6}) are A(2, 2). As shown in the diagram, in the right triangle ABC, The length of the tangent line AB^2 = AC^2 - BC^2 = (2)^2 - 2^2 = 8. Therefore, the answer is: 2, choose boxed{C}.

answer:To solve this problem, we need to determine how many 2-digit numbers can be written as the sum of exactly six different powers of 2, including (2^0). 1. **Identify the powers of 2 within the appropriate range:** Since (2^7 = 128) is greater than the largest 2-digit number (100), we consider the first six powers of 2: [ 2^0 = 1, ; 2^1 = 2, ; 2^2 = 4, ; 2^3 = 8, ; 2^4 = 16, ; 2^5 = 32, ; 2^6 = 64 ] 2. **Sum of all seven chosen powers:** By summing all seven different powers of 2, we have: [ 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127 ] 3. **Eliminating one power to stay within the 2-digit range:** To form a 2-digit number, we need to remove one of these powers to ensure the remaining total is less than 100. Consider the results when each power of 2 is excluded: - **Removing (2^5 = 32):** [ 1 + 2 + 4 + 8 + 16 + 64 = 95 ] This is a 2-digit number, so 95 is one valid option. - **Removing (2^6 = 64):** [ 1 + 2 + 4 + 8 + 16 + 32 = 63 ] This is another 2-digit number, so 63 is another valid option. 4. **Consider all other removals (Optional but to show full elimination process):** - **Removing (2^0 = 1):** [ 2 + 4 + 8 + 16 + 32 + 64 = 126 ] This exceeds the 2-digit range, thus invalid. - **Removing (2^1 = 2):** [ 1 + 4 + 8 + 16 + 32 + 64 = 125 ] Again, this exceeds the range. - **Removing (2^2 = 4):** [ 1 + 2 + 8 + 16 + 32 + 64 = 123 ] This exceeds the range. - **Removing (2^3 = 8):** [ 1 + 2 + 4 + 16 + 32 + 64 = 119 ] This exceeds the range. - **Removing (2^4 = 16):** [ 1 + 2 + 4 + 8 + 32 + 64 = 111 ] This exceeds the range. 5. **Conclusion:** Only two sums, 63 and 95, are valid 2-digit numbers that can be formed from the sum of exactly six different powers of 2, including (2^0). (boxed{C} )