answer:1. Given the sequence ( a_1, a_2, dots, a_{1999} ) where ( a_n - a_{n-1} - a_{n-2} ) is divisible by ( 100 ) for ( 3 leq n leq 1999 ), we start by noting that this implies: [ a_n equiv a_{n-1} + a_{n-2} pmod{100} ] We are given ( a_1 = 19 ) and ( a_2 = 99 ). 2. To find the remainder of ( a_1^2 + a_2^2 + dots + a_{1999}^2 ) modulo ( 8 ), we first reduce the sequence modulo ( 4 ): [ a_1 equiv 19 equiv 3 pmod{4} ] [ a_2 equiv 99 equiv 3 pmod{4} ] Using the recurrence relation ( a_n equiv a_{n-1} + a_{n-2} pmod{4} ), we compute the next few terms: [ a_3 equiv a_2 + a_1 equiv 3 + 3 equiv 6 equiv 2 pmod{4} ] [ a_4 equiv a_3 + a_2 equiv 2 + 3 equiv 5 equiv 1 pmod{4} ] [ a_5 equiv a_4 + a_3 equiv 1 + 2 equiv 3 pmod{4} ] [ a_6 equiv a_5 + a_4 equiv 3 + 1 equiv 4 equiv 0 pmod{4} ] [ a_7 equiv a_6 + a_5 equiv 0 + 3 equiv 3 pmod{4} ] We observe that the sequence is periodic with a period of 6: ( 3, 3, 2, 1, 3, 0 ). 3. Next, we calculate the sum of the squares of one period modulo ( 8 ): [ 3^2 + 3^2 + 2^2 + 1^2 + 3^2 + 0^2 = 9 + 9 + 4 + 1 + 9 + 0 = 32 equiv 0 pmod{8} ] 4. Since the sequence is periodic with a period of 6, we need to determine how many complete periods fit into 1999 terms and the remainder: [ 1999 div 6 = 333 text{ complete periods with a remainder of } 1 ] Therefore, the sequence up to ( a_{1999} ) consists of 333 complete periods plus one additional term. 5. The sum of the squares of the complete periods is: [ 333 times 0 equiv 0 pmod{8} ] The additional term is ( a_1^2 equiv 3^2 equiv 9 equiv 1 pmod{8} ). 6. Adding the contributions from the complete periods and the additional term: [ 0 + 1 equiv 1 pmod{8} ] The final answer is ( boxed{1} ).

answer:Firstly, let's understand these complex numbers by their positions on the unit circle. Each number is of the form e^{itheta} where theta is the angle from the positive real axis. By symmetry, we observe: 1. e^{8pi i/40} and e^{33pi i/40} are symmetric about the real axis. 2. e^{13pi i/40} and e^{28pi i/40} are also symmetric. 3. e^{18pi i/40} and e^{23pi i/40} are symmetric. Let's calculate the average of each symmetric pair's arguments: 1. frac{8pi}{40} + frac{33pi}{40} = frac{41pi}{40} rightarrow text{average} = frac{41pi}{80} = frac{41pi}{80} 2. frac{13pi}{40} + frac{28pi}{40} = frac{41pi}{40} rightarrow text{average} = frac{41pi}{80} 3. frac{18pi}{40} + frac{23pi}{40} = frac{41pi}{40} rightarrow text{average} = frac{41pi}{80} Each pair contributes to the total vector sum having the same average argument, frac{41pi}{80}. Since all pairs sum along the same line in the complex plane, the argument of the total sum is simply theta = frac{41pi}{80}. Therefore, the argument theta of the sum is boxed{frac{41pi}{80}}.

answer:**Analysis**: This problem tests our understanding of the equations of ellipses and circles and our ability to analyze and solve problems. We find the maximum distance between a point on the ellipse and the center of the circle, then add the radius of the circle to obtain the maximum distance between points P and Q. **Solution**: Let's denote a point on the ellipse as (x, y). Given that the center of the circle ({x}^{2}+{{left( y-6 right)}^{2}}=2) is (0, 6) and its radius is sqrt{2}, the distance between point (x, y) on the ellipse and the center of the circle is: sqrt{{x}^{2}+(y−6{)}^{2}} = sqrt{10(1−{y}^{2})+(y−6{)}^{2}} = sqrt{−9(y+ dfrac{2}{3}{)}^{2}+50}leqslant 5 sqrt{2} Thus, the maximum distance between points P and Q is 5 sqrt{2} + sqrt{2} = boxed{6 sqrt{2}}.

answer:Let's denote the cost of a couple ticket as ( x ). We know that 16 couple tickets were sold, so the total revenue from couple tickets is ( 16x ). Since a single ticket costs 20, and 128 people attended, we can calculate the number of single tickets sold by subtracting the number of people who used couple tickets from the total attendance. Each couple ticket accounts for 2 people, so the number of people who used couple tickets is ( 16 times 2 = 32 ). Therefore, the number of single tickets sold is ( 128 - 32 = 96 ). The total revenue from single tickets is ( 96 times 20 ). The total revenue from all tickets is the sum of the revenue from single tickets and couple tickets, which is ( 96 times 20 + 16x ). We know the total revenue is 2280, so we can set up the equation: ( 96 times 20 + 16x = 2280 ) Now we can solve for ( x ): ( 1920 + 16x = 2280 ) Subtract 1920 from both sides: ( 16x = 2280 - 1920 ) ( 16x = 360 ) Divide both sides by 16: ( x = frac{360}{16} ) ( x = 22.5 ) Therefore, the cost of a couple ticket is boxed{22.50} .