answer:1. **Choosing appropriate primes**: We start by selecting primes ( p ) such that ( p equiv 1 pmod{4} ) and ( p equiv 2 pmod{n} ). This ensures that ( gcd(n, p-1) = 1 ). This choice is crucial because it implies that ( n ) and ( p-1 ) are coprime, which will be useful in the subsequent steps. 2. **Behavior of ( k^{2n} ) modulo ( p )**: Since ( gcd(n, p-1) = 1 ), the map ( k mapsto k^{2n} ) is a permutation of the set ( {1, 2, ldots, frac{p-1}{2}} ). This means that as ( k ) ranges from ( 1 ) to ( frac{p-1}{2} ), ( k^{2n} ) will take on each quadratic residue modulo ( p ) exactly once. 3. **Quadratic residues and their properties**: For ( p equiv 1 pmod{4} ), (-1) is a quadratic residue modulo ( p ). The quadratic residues modulo ( p ) are symmetric around ( frac{p}{2} ). Specifically, if ( r ) is a quadratic residue, then ( p-r ) is also a quadratic residue. 4. **Sum of quadratic residues**: The sum of all quadratic residues modulo ( p ) is given by: [ sum_{k=1}^{frac{p-1}{2}} k^2 equiv frac{p(p-1)(2p-1)}{6} pmod{p} ] Simplifying this modulo ( p ), we get: [ sum_{k=1}^{frac{p-1}{2}} k^2 equiv 0 pmod{p} ] This implies that the average of the quadratic residues is: [ frac{1}{frac{p-1}{2}} sum_{k=1}^{frac{p-1}{2}} k^2 equiv frac{p}{2} pmod{p} ] 5. **Fractional part calculation**: The fractional part ( left{frac{k^{2n}}{p}right} ) is given by: [ left{frac{k^{2n}}{p}right} = frac{k^{2n}}{p} - leftlfloor frac{k^{2n}}{p} rightrfloor ] Since ( k^{2n} ) hits every nonzero quadratic residue once, the sum of the fractional parts over all ( k ) from ( 1 ) to ( frac{p-1}{2} ) is: [ sum_{k=1}^{frac{p-1}{2}} left{frac{k^{2n}}{p}right} = sum_{k=1}^{frac{p-1}{2}} left( frac{k^{2n}}{p} - leftlfloor frac{k^{2n}}{p} rightrfloor right) ] Since the sum of ( frac{k^{2n}}{p} ) over all ( k ) is ( frac{1}{p} sum_{k=1}^{frac{p-1}{2}} k^{2n} ) and the sum of the floor terms is an integer, the average value of the fractional parts is: [ a_p = frac{1}{p-1} sum_{k=1}^{frac{p-1}{2}} left{frac{k^{2n}}{p}right} = frac{1}{p-1} cdot frac{p-1}{2} = frac{1}{2} ] 6. **Conclusion**: Therefore, for the chosen primes ( p ), we have: [ a_p = frac{1}{4} ] This shows that there is a real number ( c = frac{1}{4} ) such that ( a_p = c ) for infinitely many primes ( p ). (blacksquare)

answer:From the equation of circle C: (x-1)^{2}+(y-2)^{2}=1, we obtain the center of the circle C(1,2). Given point P(-1,0), The coordinates of the center of the required circle are left( frac{1-1}{2}, frac{2+0}{2} right), which is (0,1). The radius of the circle r= sqrt {(1-0)^{2}+(0-1)^{2}}= sqrt {2}. Hence, the equation of the circle passing through points A, B, and C is: x^{2}+(y-1)^{2}=2. Thus, the answer is boxed{A}. According to the properties of tangents, PA is perpendicular to CA, and PB is perpendicular to CB. Therefore, the circle passing through points A, B, and C is the circumcircle of the quadrilateral PACB. Moreover, segment AC is the diameter of the circumcircle. So, we can find the center of the circumcircle using the midpoint formula and calculate the radius using the distance formula. Finally, we can write the standard equation of the circle using the obtained center and radius. This problem tests the student's understanding of the properties of tangents and circles, the ability to use the midpoint formula and distance formula to simplify and find values, and the skill to write the standard equation of a circle given its center and radius. It is a comprehensive problem.

answer:Let's denote the breadth of the rectangular prism as ( b ), the height as ( h ), and the length as ( l ). According to the problem, we have the following relationships: 1. ( l = 3b ) (length is thrice the breadth) 2. ( l = 2h ) (length is twice the height) The volume ( V ) of a rectangular prism is given by the formula: [ V = l times b times h ] Given that the volume is ( 12168 ) cubic meters, we can write: [ 12168 = l times b times h ] Using the relationships from above, we can express ( l ) and ( h ) in terms of ( b ): [ l = 3b ] [ h = frac{l}{2} = frac{3b}{2} ] Now, we can substitute ( l ) and ( h ) in terms of ( b ) into the volume equation: [ 12168 = (3b) times b times left(frac{3b}{2}right) ] [ 12168 = 3b^2 times frac{3b}{2} ] [ 12168 = frac{9b^3}{2} ] To find ( b ), we multiply both sides by ( 2 ) and then divide by ( 9 ): [ 2 times 12168 = 9b^3 ] [ 24336 = 9b^3 ] [ b^3 = frac{24336}{9} ] [ b^3 = 2704 ] Now, we take the cube root of both sides to find ( b ): [ b = sqrt[3]{2704} ] [ b = 14 ] Now that we have ( b ), we can find ( l ) and ( h ): [ l = 3b = 3 times 14 = 42 ] [ h = frac{l}{2} = frac{42}{2} = 21 ] Therefore, the dimensions of the rectangular prism are: - Length (( l )) = 42 meters - Breadth (( b )) = 14 meters - Height (( h )) = boxed{21} meters

answer:To determine the value of x, we can use the property that the scalar (dot) product of two perpendicular vectors is zero. Since BC is the hypotenuse of the right triangle, overrightarrow{AB} is perpendicular to overrightarrow{AC}. First, we find the vector overrightarrow{AB} as follows: overrightarrow{AB} = B - A = (2 - (-1), -4 - 2, 1 - 3) = (3, -6, -2). Next, we find the vector overrightarrow{AC}: overrightarrow{AC} = C - A = (x - (-1), -1 - 2, -3 - 3) = (x + 1, -3, -6). The two vectors are perpendicular, so their scalar product equals zero: overrightarrow{AB} cdot overrightarrow{AC} = 3(x+1) + (-6)(-3) + (-2)(-6) = 0. Simplifying the equation, we get: 3x + 3 + 18 + 12 = 0. 3x + 33 = 0. 3x = -33. x = boxed{-11}. Therefore, the value of x for which C would form a right angle at B is x=-11.