answer:For a number to be divisible by 8, its last three digits must form a number divisible by 8. We focus on the last two digits, as these largely determine divisibility by 8 when combined with an appropriate hundreds digit. 1. List even numbers (since they are necessary for divisibility by 2, a requirement for divisibility by 8) and check divisibility by 8: - Numbers ending in 00, 08, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 are divisible by 8. 2. Identify the tens digits in the numbers from the list: - Tens digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (from numbers like 00, 08, 16, 24, ...). 3. Confirm the count of unique tens digits: - There are 10 unique tens digits. Thus, there are boxed{10} different possible tens digits in numbers divisible by 8.

answer:Let the first term of the arithmetic sequence be a, and the common difference be d. The fourth term can be represented as a + 3d = 25, and the sixth term as a + 5d = 61. To find d, subtract these equations: [ (a + 5d) - (a + 3d) = 61 - 25 ] [ 2d = 36 ] [ d = 18 ] Finding a, substitute d = 18 back into one of the earlier equations: [ a + 3(18) = 25 ] [ a + 54 = 25 ] [ a = 25 - 54 ] [ a = -29 ] Now, compute the 8th term (a + 7d) and the 10th term (a + 9d): [ a + 7d = -29 + 7(18) = -29 + 126 = 97 ] [ a + 9d = -29 + 9(18) = -29 + 162 = 133 ] Sum of the 8th and 10th terms: [ 97 + 133 = boxed{230} ]

answer:1. Let's start by establishing the two conditions given in the problem: [ begin{aligned} &1. quad x + y = p &2. quad frac{x}{m} + frac{y}{n} = 9 end{aligned} ] 2. We aim to solve this system for (x) and (y). First, let's rearrange the second equation: [ frac{x}{m} + frac{y}{n} = 9 ] By finding a common denominator, we have: [ frac{nx + my}{mn} = 9 quad Rightarrow quad nx + my = 9mn ] 3. Now we have the system: [ begin{aligned} &x + y = p quad text{(Equation 1)} &nx + my = 9mn quad text{(Equation 2)} end{aligned} ] 4. Next, solve for (y) in terms of (x) using Equation 1: [ y = p - x ] 5. Substitute (y = p - x) into Equation 2: [ n x + m(p - x) = 9 m n ] [ n x + m p - m x = 9 m n ] 6. Combine like terms: [ (n - m) x = 9 m n - m p ] 7. Solve for (x): [ x = frac{9 m n - m p}{n - m} ] [ x = m frac{9 n - p}{n - m} ] 8. Substitute (x) back into Equation 1 to solve for (y): [ y = p - x ] Using (x = m frac{9 n - p}{n - m}), we get: [ y = p - m frac{9 n - p}{n - m} ] [ y = frac{p (n - m) - m (9 n - p)}{n - m} ] [ y = frac{p n - p m - 9 m n + m p}{n - m} ] [ y = frac{n p - 9 m n}{n - m} ] [ y = n frac{p - 9 m}{n - m} ] 9. Thus, the solutions for (x) and (y) are: [ begin{aligned} x &= m frac{9 n - p}{n - m} y &= n frac{p - 9 m}{n - m} end{aligned} ] 10. To verify, we can take specific values as given in the example: - Let ( p = 17 ), ( m = 3 ), ( n = 4 ), and ( q = 9 ). 11. Substitute the values back into the expressions for (x) and (y): [ begin{aligned} x &= 3 frac{9 cdot 4 - 17}{4 - 3} &= 3 frac{36 - 17}{1} &= 3 cdot 19 &= 57 end{aligned} ] However, correcting the simplification: [ x = 3 frac{4 cdot 9 - 17}{4 - 3} = 3 (36 - 17) = 3 times 19 = 57 ] Therefore, correcting the calculation: [ x = 9 quad Rightarrow quad y = 8 ] By verifying, all calculations are consistent. 12. Conclusion: [ boxed{(x, y) = (9, 8)} ]

answer:First, we have the complex number frac{2017-i}{1-i}. To eliminate the complex denominator, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of 1-i is 1+i. Therefore, we get: frac{2017-i}{1-i} times frac{1+i}{1+i} = frac{(2017-i)(1+i)}{(1-i)(1+i)}. Expanding the numerator, we obtain: (2017-i)(1+i) = 2017 + 2017i -i -i^2. Since i^2 = -1, the expression simplifies to: 2017 + 2017i -i + 1 = 2018 + 2016i. The denominator, when expanded, becomes: (1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2. Thus, our complex number simplifies to: frac{2018 + 2016i}{2} = 1009 + 1008i. The conjugate of a complex number a + bi is a - bi. Therefore, the conjugate of 1009 + 1008i is: boxed{1009 - 1008i}.