answer:We are given that a_2=3 and S_4=16. 1. First, we use the formula for the n-th term of an arithmetic sequence, which is a_n = a_1 + (n-1)d. Since a_2 = a_1 + (2-1)d = 3, we can rewrite this as a_1 + d = 3. 2. Next, we use the formula for the sum of the first n terms of an arithmetic sequence, which is S_n = frac{n}{2}(2a_1 + (n-1)d). Since S_4 = frac{4}{2}(2a_1 + (4-1)d) = 16, we can rewrite this as 4a_1 + 6d = 16. 3. Now, we solve the system of linear equations: [ begin{cases} a_1 + d = 3 4a_1 + 6d = 16 end{cases} ] Solving this system, we find a_1 = 1 and d = 2. 4. Therefore, the common difference of the arithmetic sequence is d = boxed{2}.

answer:1. **Identify the circles and points:** Let ( omega_1 ) be the circle with radius 15, and ( omega_2 ) be the circle with radius 13. Let the centers of ( omega_1 ) and ( omega_2 ) be ( O_1 ) and ( O_2 ) respectively. The circles intersect at points ( P ) and ( Q ), and the segment ( PQ ) is given to be 24 units long. 2. **Determine the relationship between ( O_1O_2 ) and ( PQ ):** Since ( O_1 ) and ( O_2 ) are the centers of two intersecting circles, ( O_1O_2 ) is perpendicular to ( PQ ) at their midpoint ( X ). Thus, ( PX = XQ = frac{24}{2} = 12 ). 3. **Apply the Pythagorean Theorem to ( Delta O_1PX ) and ( Delta O_2PX ):** - In ( Delta O_1PX ): [ O_1X = sqrt{O_1P^2 - PX^2} = sqrt{15^2 - 12^2} = sqrt{225 - 144} = sqrt{81} = 9 ] - In ( Delta O_2PX ): [ O_2X = sqrt{O_2P^2 - PX^2} = sqrt{13^2 - 12^2} = sqrt{169 - 144} = sqrt{25} = 5 ] 4. **Calculate the distance ( O_1O_2 ):** Since point ( X ) is on both circles and ( O_1O_2 = O_1X + O_2X ): [ O_1O_2 = O_1X + O_2X = 9 + 5 = 14 ] 5. **Set up to find the tangent lengths and position of point ( A ):** - Let ( S ) and ( R ) be the points of tangency from ( A ) to ( omega_1 ) and ( omega_2 ), respectively. - Assume ( SAR ) forms a right angle at ( A ), implying ( SA text{ and } RA ) form a square (each side length ( s )). Thus, the distances from ( O_1 ) and ( O_2 ) to ( Y ), where ( S ) and ( R ) meet: [ O_1Y = s - 15 quad text{and} quad O_2Y = s - 13 ] 6. **Use the Pythagorean Theorem on the distance ( O_1O_2Y ):** Since ( O_1O_2Y ) is a right triangle with sides ( 14, s-15, text{and } s-13 ): [ (s-15)^2 + (s-13)^2 = 14^2 ] 7. **Solve for ( s ):** Expand and simplify the equation: [ (s-15)^2 + (s-13)^2 = 196 ] [ (s^2 - 30s + 225) + (s^2 - 26s + 169) = 196 ] [ 2s^2 - 56s + 394 = 196 ] [ 2s^2 - 56s + 198 = 0 ] [ s^2 - 28s + 99 = 0 ] 8. **Use the quadratic formula to solve for ( s ):** [ s = frac{28 pm sqrt{28^2 - 4 cdot 99}}{2} ] [ s = frac{28 pm sqrt{784 - 396}}{2} ] [ s = frac{28 pm sqrt{388}}{2} ] [ s = frac{28 pm sqrt{4 cdot 97}}{2} ] [ s = 14 pm sqrt{97} ] Since ( 14 - sqrt{97} < 15 ) and it must be greater than the radius of ( omega_1 ): [ s = 14 + sqrt{97} ] 9. **Conclusion:** The length of ( AR ) is: [ boxed{14 + sqrt{97}} ]

answer:1. Let the number of teams be ( T ). We need to determine the averages used in the problem. For the average involving the size ( 9 ) subsets, note that each team is a part of ( binom{n-5}{4} ) size ( 9 ) subsets, since we need to pick ( 4 ) other participants to be a part of the subset. Thus, this average is [ frac{T cdot binom{n-5}{4}}{binom{n}{9}}. ] 2. Similarly, the average for the size ( 8 ) subsets is [ frac{T cdot binom{n-5}{3}}{binom{n}{8}}. ] 3. According to the problem, these two averages are reciprocals of each other: [ frac{T cdot binom{n-5}{4}}{binom{n}{9}} = frac{binom{n}{8}}{T cdot binom{n-5}{3}}. ] 4. Isolating ( T ) gives: [ T^2 = frac{binom{n}{8} cdot binom{n}{9}}{binom{n-5}{4} cdot binom{n-5}{3}}. ] 5. Expanding the right-hand side using the explicit formulas for the binomial coefficients and simplifying slightly gives: [ T^2 = frac{n^2 (n-1)^2 (n-2)^2 (n-3)^2 (n-4)^2 cdot 4}{8^2 cdot 7^2 cdot 6^2 cdot 5^2 cdot 4^2 cdot 9}. ] 6. Simplifying further: [ T^2 = left( frac{2 cdot n (n-1) (n-2) (n-3) (n-4)}{8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3} right)^2. ] 7. Since ( T ) must be an integer, we know that: [ 8 cdot 7 cdot 6 cdot 5 cdot 2 cdot 3 mid n(n-1)(n-2)(n-3)(n-4). ] 8. We now determine all possible residues for ( n ) modulo ( 32, 9, 5, 7 ) for which this is an integer. - (pmod{7}): One of the 5 terms in the product must be divisible by ( 7 ), so ( n pmod{7} in {0, 1, 2, 3, 4} ), which is ( 5 ) possibilities. - (pmod{5}): One of the 5 terms in the product must be divisible by ( 5 ), but this always occurs, so we do not need to worry about this case. - (pmod{9}): ( v_3 ) of the product must be at least ( 2 ), which gives the ( 7 ) possibilities for ( n pmod{9} ) of ( 0, 1, 2, 3, 4, 6, 7 ). - (pmod{32}): ( v_2 ) of the product must be at least ( 5 ), but since there are always at least two even terms in the product, we actually only need to deal with (pmod{16}). There are ( 8 ) possibilities: ( n pmod{16} in {0, 1, 2, 3, 4, 8, 10, 12} ). 9. By the Chinese Remainder Theorem, we have ( 5 cdot 7 cdot 8 = 280 ) possible residues for ( n ) modulo ( 7 cdot 9 cdot 16 = 1008 ), giving ( 560 ) possible values for ( n ) in the range ( [1, 2016] ). 10. We then account for the slight difference in the given range: - ( 2017 ) is ( 1 ) mod 1008, so it is a valid residue which we must add. - Of the numbers from ( 1 ) to ( 8 ), only ( 1, 2, 3, 4, 8 ) work mod 16, and ( 8 ) does not work mod 9, so we must subtract only ( 1, 2, 3, 4 ), all of which clearly work for all three moduli. 11. Thus, we have: [ 560 + 1 - 4 = 557. ] The final answer is (boxed{557}).

answer:To solve the problem where b=sqrt{3-a}+sqrt{a-3}+2, and we need to find a^{b}, we follow these steps: 1. First, we analyze the conditions for the square roots to be real numbers. For sqrt{3-a} to be real, we need 3-a geqslant 0. Similarly, for sqrt{a-3} to be real, we need a-3 geqslant 0. Combining these, we get: [ 3-a geqslant 0 quad text{and} quad a-3 geqslant 0 ] 2. From 3-a geqslant 0, we deduce that a leqslant 3. And from a-3 geqslant 0, we deduce that a geqslant 3. Combining these two inequalities, we find that the only value that satisfies both conditions is a=3: [ a leqslant 3 quad text{and} quad a geqslant 3 quad Rightarrow quad a=3 ] 3. Substituting a=3 into the expression for b, we get: [ b = sqrt{3-3} + sqrt{3-3} + 2 = 0 + 0 + 2 = 2 ] 4. Finally, we calculate a^{b} with a=3 and b=2: [ a^{b} = 3^{2} = 9 ] Therefore, the final answer is boxed{9}.