💡First, calculate the total number of pages Robert can read in 8 hours:[ 100 text{ pages/hour} times 8 text{ hours} = 800 text{ pages} ]Next, determine how many 400-page books he can read:[ frac{800 text{ pages}}{400 text{ pages/book}} = 2 text{ books} ]Therefore, Robert can read 2 full 400-page books in 8 hours.

💡Okay, so I have this problem about a cubic function, f(x) = 2x³ + ax² + bx + m. The derivative of this function, f'(x), has a graph that's symmetric about the line x = -1/2, and also f'(1) = 0. I need to find the values of a and b first, and then determine the range of m such that f(x) has exactly three zeros.Starting with part (1): finding a and b.First, I know that the derivative of f(x) will be a quadratic function because the original function is cubic. So, f'(x) should be 6x² + 2ax + b. That's straightforward—just take the derivative term by term.Now, the graph of f'(x) is symmetric about x = -1/2. For a quadratic function, the axis of symmetry is given by x = -B/(2A), where the quadratic is in the form Ax² + Bx + C. In this case, A is 6, and B is 2a. So, the axis of symmetry should be x = -(2a)/(2*6) = -a/6.According to the problem, this axis of symmetry is x = -1/2. So, setting -a/6 equal to -1/2:-a/6 = -1/2Multiplying both sides by 6:-a = -3So, a = 3.Alright, so we've found a = 3. Now, let's substitute that back into f'(x):f'(x) = 6x² + 2*3*x + b = 6x² + 6x + b.Now, we also know that f'(1) = 0. So, let's plug x = 1 into f'(x):f'(1) = 6*(1)² + 6*(1) + b = 6 + 6 + b = 12 + b.But f'(1) is given as 0, so:12 + b = 0Therefore, b = -12.So, that gives us a = 3 and b = -12.Wait, let me double-check that. If a is 3, then f'(x) is 6x² + 6x + b. Plugging in x = 1, we get 6 + 6 + b = 12 + b. Setting that equal to 0, so yes, b = -12. That seems correct.So, part (1) is done. a is 3, b is -12.Moving on to part (2): determining the range of m such that f(x) has exactly three zeros.First, let's write out f(x) with the known values of a and b:f(x) = 2x³ + 3x² - 12x + m.We need to find the range of m for which this cubic function has exactly three real zeros. For a cubic function, having exactly three real zeros means that it must cross the x-axis three times. This typically happens when the function has a local maximum above the x-axis and a local minimum below the x-axis, or vice versa.To find this, we can analyze the critical points of f(x). We've already found f'(x) in part (1):f'(x) = 6x² + 6x - 12.We can factor this derivative to find the critical points:f'(x) = 6(x² + x - 2) = 6(x + 2)(x - 1).So, the critical points are at x = -2 and x = 1.These are the points where the function changes from increasing to decreasing or vice versa. Let's determine the intervals where f(x) is increasing or decreasing.Since f'(x) is a quadratic opening upwards (because the coefficient of x² is positive), the derivative will be positive outside the interval (-2, 1) and negative inside that interval.So, f(x) is increasing on (-∞, -2), decreasing on (-2, 1), and increasing again on (1, ∞).Therefore, x = -2 is a local maximum, and x = 1 is a local minimum.To have three real zeros, the function must cross the x-axis three times. This requires that the local maximum is above the x-axis and the local minimum is below the x-axis. So, f(-2) > 0 and f(1) < 0.Let's compute f(-2) and f(1):First, f(-2):f(-2) = 2*(-2)³ + 3*(-2)² - 12*(-2) + mCalculating each term:2*(-8) = -163*(4) = 12-12*(-2) = 24So, adding these up: -16 + 12 + 24 + m = (-16 + 12) + 24 + m = (-4) + 24 + m = 20 + m.So, f(-2) = 20 + m.Similarly, f(1):f(1) = 2*(1)³ + 3*(1)² - 12*(1) + mCalculating each term:2*1 = 23*1 = 3-12*1 = -12So, adding these up: 2 + 3 - 12 + m = (2 + 3) - 12 + m = 5 - 12 + m = -7 + m.So, f(1) = m - 7.Now, for the function to have three real zeros, we need f(-2) > 0 and f(1) < 0.So, setting up the inequalities:20 + m > 0 => m > -20andm - 7 < 0 => m < 7Therefore, combining these two inequalities, we get:-20 < m < 7.So, the range of m is (-20, 7).Wait, let me make sure I didn't make a mistake in computing f(-2) and f(1). Let me recalculate:For f(-2):2*(-2)^3 = 2*(-8) = -163*(-2)^2 = 3*4 = 12-12*(-2) = 24So, total is -16 + 12 + 24 = (-16 + 12) = -4; -4 + 24 = 20. So, 20 + m. Correct.For f(1):2*(1)^3 = 23*(1)^2 = 3-12*(1) = -12So, 2 + 3 - 12 = (2 + 3) = 5; 5 - 12 = -7. So, -7 + m. Correct.Thus, the conditions are 20 + m > 0 and m - 7 < 0, leading to m > -20 and m < 7. So, m must be between -20 and 7.Therefore, the range of m is (-20, 7).I think that's solid. Let me just visualize the graph for a second. The function is a cubic, so it goes from negative infinity to positive infinity. It has a local maximum at x = -2, which is 20 + m, and a local minimum at x = 1, which is m - 7. For it to cross the x-axis three times, the maximum must be above the x-axis and the minimum must be below. So, yes, m has to be greater than -20 to make the maximum positive and less than 7 to make the minimum negative.Therefore, the range is (-20, 7).**Final Answer**(1) The values of the real numbers are boxed{a = 3} and boxed{b = -12}.(2) The range of the real number m is boxed{(-20, 7)}.

💡Alright, so I have this problem here: I need to show that (a^{2} - 2b) can always be expressed as the sum of two perfect squares, and (3ab - a^{3}) as the sum of two perfect cubes, given that (a^{2} - 4b) is a perfect square. Hmm, okay, let me try to unpack this step by step.First off, the condition given is that (a^{2} - 4b) is a perfect square. That means there exists some integer (k) such that (a^{2} - 4b = k^{2}). So, I can write that as:[a^{2} - 4b = k^{2}]From this equation, I can solve for (b):[4b = a^{2} - k^{2}][b = frac{a^{2} - k^{2}}{4}]Okay, so (b) is expressed in terms of (a) and (k). Now, I need to express (a^{2} - 2b) as the sum of two perfect squares. Let me substitute (b) from the above equation into (a^{2} - 2b):[a^{2} - 2b = a^{2} - 2left(frac{a^{2} - k^{2}}{4}right)][= a^{2} - frac{2a^{2} - 2k^{2}}{4}][= a^{2} - frac{a^{2}}{2} + frac{k^{2}}{2}][= frac{a^{2}}{2} + frac{k^{2}}{2}]Hmm, so (a^{2} - 2b) simplifies to (frac{a^{2}}{2} + frac{k^{2}}{2}). Now, I need this to be the sum of two perfect squares. But right now, it's half of (a^{2}) plus half of (k^{2}). For this to be an integer (since perfect squares are integers), both (frac{a^{2}}{2}) and (frac{k^{2}}{2}) must be integers. That implies that (a) and (k) must both be even numbers because if (a) is even, (a^{2}) is divisible by 4, and similarly for (k).Let me denote (a = 2m) and (k = 2n), where (m) and (n) are integers. Substituting back into the expression:[frac{a^{2}}{2} + frac{k^{2}}{2} = frac{(2m)^{2}}{2} + frac{(2n)^{2}}{2} = frac{4m^{2}}{2} + frac{4n^{2}}{2} = 2m^{2} + 2n^{2}]Wait, that's not quite the sum of two perfect squares; it's twice the sum of two perfect squares. Hmm, did I make a mistake somewhere?Let me go back. Maybe I should approach this differently. Since (a^{2} - 4b = k^{2}), perhaps I can think of (a^{2} - 2b) in terms of (k). Let's see:From (a^{2} - 4b = k^{2}), we have:[a^{2} - 2b = (a^{2} - 4b) + 2b = k^{2} + 2b]But that doesn't seem helpful. Maybe I should express (b) in terms of (a) and (k) again:[b = frac{a^{2} - k^{2}}{4}]So,[a^{2} - 2b = a^{2} - 2left(frac{a^{2} - k^{2}}{4}right) = a^{2} - frac{a^{2} - k^{2}}{2} = frac{2a^{2} - a^{2} + k^{2}}{2} = frac{a^{2} + k^{2}}{2}]Okay, so (a^{2} - 2b = frac{a^{2} + k^{2}}{2}). For this to be an integer, (a^{2} + k^{2}) must be even, which means both (a) and (k) must be of the same parity (both even or both odd). But since (a^{2} - 4b = k^{2}), and (4b) is always even, (a^{2}) and (k^{2}) must have the same parity. So, if (a) is even, (k) is even; if (a) is odd, (k) is odd.Now, to express (frac{a^{2} + k^{2}}{2}) as the sum of two perfect squares. Let me think about how to decompose this. Maybe I can write it as:[frac{a^{2} + k^{2}}{2} = left(frac{a + k}{2}right)^{2} + left(frac{a - k}{2}right)^{2}]Let me check:[left(frac{a + k}{2}right)^{2} + left(frac{a - k}{2}right)^{2} = frac{(a + k)^{2} + (a - k)^{2}}{4} = frac{a^{2} + 2ak + k^{2} + a^{2} - 2ak + k^{2}}{4} = frac{2a^{2} + 2k^{2}}{4} = frac{a^{2} + k^{2}}{2}]Yes! That works. So, (a^{2} - 2b) can indeed be expressed as the sum of two perfect squares:[a^{2} - 2b = left(frac{a + k}{2}right)^{2} + left(frac{a - k}{2}right)^{2}]Great, that takes care of the first part. Now, onto the second part: showing that (3ab - a^{3}) can be expressed as the sum of two perfect cubes.Again, starting from the condition (a^{2} - 4b = k^{2}), we have (b = frac{a^{2} - k^{2}}{4}). Let's substitute this into (3ab - a^{3}):[3ab - a^{3} = 3aleft(frac{a^{2} - k^{2}}{4}right) - a^{3} = frac{3a^{3} - 3ak^{2}}{4} - a^{3}][= frac{3a^{3} - 3ak^{2} - 4a^{3}}{4} = frac{-a^{3} - 3ak^{2}}{4}]Hmm, that seems messy. Maybe I should factor out (-a):[= -aleft(frac{a^{2} + 3k^{2}}{4}right)]Not sure if that helps. Let me think differently. Maybe I can express (3ab - a^{3}) in terms of (k). From (b = frac{a^{2} - k^{2}}{4}), we have:[3ab - a^{3} = 3aleft(frac{a^{2} - k^{2}}{4}right) - a^{3} = frac{3a^{3} - 3ak^{2}}{4} - a^{3}][= frac{3a^{3} - 3ak^{2} - 4a^{3}}{4} = frac{-a^{3} - 3ak^{2}}{4}]Same result. Maybe I need to find a different approach. Let's recall that (a^{2} - 4b = k^{2}), so (a^{2} = k^{2} + 4b). Let's substitute this into (3ab - a^{3}):[3ab - a^{3} = 3ab - a(a^{2}) = 3ab - a(k^{2} + 4b) = 3ab - ak^{2} - 4ab = -ab - ak^{2}]Hmm, that's simpler:[3ab - a^{3} = -ab - ak^{2} = -a(b + k^{2})]But I'm not sure if that helps. Let me think about expressing this as the sum of two cubes. Maybe I can find integers (m) and (n) such that:[3ab - a^{3} = m^{3} + n^{3}]But I'm not sure how to find such (m) and (n). Maybe I can relate this to the earlier expression involving (k). From (a^{2} - 4b = k^{2}), we have (b = frac{a^{2} - k^{2}}{4}). Let's substitute this back into (3ab - a^{3}):[3ab - a^{3} = 3aleft(frac{a^{2} - k^{2}}{4}right) - a^{3} = frac{3a^{3} - 3ak^{2}}{4} - a^{3}][= frac{3a^{3} - 3ak^{2} - 4a^{3}}{4} = frac{-a^{3} - 3ak^{2}}{4}]Again, same result. Maybe I need to factor this differently. Let's factor out (-a):[= -aleft(frac{a^{2} + 3k^{2}}{4}right)]Hmm, not helpful. Maybe I need to think about specific values. Let's pick some small integers for (a) and (k) that satisfy (a^{2} - 4b = k^{2}), and see if (3ab - a^{3}) can be expressed as the sum of two cubes.For example, let's take (a = 2). Then, (a^{2} = 4). Let's choose (k = 0), so (4 - 4b = 0), which gives (b = 1). Then, (3ab - a^{3} = 3*2*1 - 8 = 6 - 8 = -2). Is (-2) the sum of two perfect cubes? Well, (-2 = (-1)^3 + (-1)^3), so yes, it is.Another example: (a = 3). Then, (a^{2} = 9). Let's choose (k = 1), so (9 - 4b = 1), which gives (b = 2). Then, (3ab - a^{3} = 3*3*2 - 27 = 18 - 27 = -9). Is (-9) the sum of two perfect cubes? Yes, (-9 = (-2)^3 + (-1)^3 = -8 -1 = -9).Another example: (a = 4). Then, (a^{2} = 16). Let's choose (k = 2), so (16 - 4b = 4), which gives (b = 3). Then, (3ab - a^{3} = 3*4*3 - 64 = 36 - 64 = -28). Is (-28) the sum of two perfect cubes? Let's see: (-28 = (-3)^3 + 1^3 = -27 + 1 = -26), which is not (-28). Hmm, not quite. Maybe another combination: (-28 = (-4)^3 + 6^3 = -64 + 216 = 152), nope. Wait, maybe I need to think differently. Perhaps (-28 = (-2)^3 + (-2)^3 + (-2)^3 = -8 -8 -8 = -24), still not. Hmm, maybe this approach isn't working.Wait, maybe I made a mistake in choosing (k). For (a = 4), (a^{2} - 4b = k^{2}), so (16 - 4b = k^{2}). Let's choose (k = 0), then (b = 4). Then, (3ab - a^{3} = 3*4*4 - 64 = 48 - 64 = -16). Is (-16) the sum of two perfect cubes? Yes, (-16 = (-2)^3 + (-2)^3 = -8 -8 = -16).Okay, so in this case, it works. Maybe my previous choice of (k = 2) didn't give a valid sum, but choosing (k = 0) does. So, perhaps the choice of (k) affects whether (3ab - a^{3}) can be expressed as the sum of two perfect cubes.Wait, but the problem states "if (a^{2} - 4b) is a perfect square," so (k) can be any integer such that (a^{2} - 4b = k^{2}). So, for a given (a), there might be multiple (k) values, and for each, (b) is determined. So, perhaps for some (k), (3ab - a^{3}) can be expressed as the sum of two cubes, and for others, it might not. But the problem says "can always be expressed," so I need a general proof, not just examples.Let me go back to the expression:[3ab - a^{3} = -aleft(frac{a^{2} + 3k^{2}}{4}right)]Hmm, not helpful. Maybe I need to think about expressing this as the sum of two cubes. Let's recall that the sum of two cubes can be factored as:[m^{3} + n^{3} = (m + n)(m^{2} - mn + n^{2})]But I'm not sure if that helps here. Alternatively, maybe I can express (3ab - a^{3}) in terms of (k). From (a^{2} - 4b = k^{2}), we have (b = frac{a^{2} - k^{2}}{4}). Substituting back:[3ab - a^{3} = 3aleft(frac{a^{2} - k^{2}}{4}right) - a^{3} = frac{3a^{3} - 3ak^{2}}{4} - a^{3}][= frac{3a^{3} - 3ak^{2} - 4a^{3}}{4} = frac{-a^{3} - 3ak^{2}}{4}]Again, same result. Maybe I need to factor this differently. Let's factor out (-a):[= -aleft(frac{a^{2} + 3k^{2}}{4}right)]Hmm, still not helpful. Maybe I need to think about specific forms of (a) and (k). For instance, if (a = 2m) and (k = 2n), as before, then:[3ab - a^{3} = 3*(2m)*b - (2m)^{3} = 6mb - 8m^{3}]But (b = frac{a^{2} - k^{2}}{4} = frac{4m^{2} - 4n^{2}}{4} = m^{2} - n^{2}). So:[6mb - 8m^{3} = 6m(m^{2} - n^{2}) - 8m^{3} = 6m^{3} - 6mn^{2} - 8m^{3} = -2m^{3} - 6mn^{2}]Hmm, not helpful. Maybe I need to think about expressing this as the sum of two cubes. Let's consider that (3ab - a^{3}) might be equal to ((m - n)^3 + (m + n)^3) or something similar. Let's try expanding ((m - n)^3 + (m + n)^3):[(m - n)^3 + (m + n)^3 = m^{3} - 3m^{2}n + 3mn^{2} - n^{3} + m^{3} + 3m^{2}n + 3mn^{2} + n^{3} = 2m^{3} + 6mn^{2}]Hmm, that's (2m^{3} + 6mn^{2}). Comparing this to our expression (-2m^{3} - 6mn^{2}), it's the negative of that. So, if I take (- (m - n)^3 - (m + n)^3), I get:[- (m - n)^3 - (m + n)^3 = -2m^{3} - 6mn^{2}]Which matches our expression for (3ab - a^{3}) when (a = 2m) and (k = 2n). So, we can write:[3ab - a^{3} = - (m - n)^3 - (m + n)^3]But the problem asks for the sum of two perfect cubes, not the negative sum. Hmm, but perfect cubes can be negative as well, right? So, if I write:[3ab - a^{3} = (- (m - n))^3 + (- (m + n))^3]Which simplifies to:[3ab - a^{3} = (-m + n)^3 + (-m - n)^3]Yes, that works. So, (3ab - a^{3}) can be expressed as the sum of two perfect cubes, specifically ((-m + n)^3) and ((-m - n)^3).Putting it all together, we've shown that:1. (a^{2} - 2b) can be expressed as the sum of two perfect squares: (left(frac{a + k}{2}right)^{2} + left(frac{a - k}{2}right)^{2}).2. (3ab - a^{3}) can be expressed as the sum of two perfect cubes: ((-m + n)^3 + (-m - n)^3), where (a = 2m) and (k = 2n).Therefore, both expressions can indeed be represented as the sum of two perfect squares and the sum of two perfect cubes, respectively, given that (a^{2} - 4b) is a perfect square.

💡Okay, so I have this problem where a positive integer m greater than 1 can split its cube into a sum of consecutive odd numbers. For example, 2³ is 3 + 5, 3³ is 7 + 9 + 11, and 4³ is 13 + 15 + 17 + 19, and so on. The question is, if m³ is split according to this rule, and one of the odd numbers is 2017, what is the position of 2017 among these consecutive odd numbers? The options are A: 16, B: 17, C: 18, D: 19.Alright, so first, I need to understand how exactly the cube of m is being split into consecutive odd numbers. Let me look at the examples given.For m=2, 2³=8, which is split into 3 + 5. So, two consecutive odd numbers. For m=3, 3³=27, which is 7 + 9 + 11, three consecutive odd numbers. For m=4, 4³=64, which is 13 + 15 + 17 + 19, four consecutive odd numbers. So, it seems that for each m, m³ is expressed as the sum of m consecutive odd numbers.So, in general, m³ can be written as the sum of m consecutive odd numbers. That makes sense because the sum of the first n odd numbers is n², but here we're starting from a different odd number each time.Wait, so if the sum of the first n odd numbers is n², then how is m³ being expressed as the sum of m consecutive odd numbers? Maybe it's starting from a different point in the sequence of odd numbers.Let me think about the examples. For m=2, starting at 3, which is the second odd number. For m=3, starting at 7, which is the fourth odd number. For m=4, starting at 13, which is the seventh odd number. Hmm, is there a pattern here?Let me list the starting odd numbers for each m:- m=2: starts at 3 (which is the 2nd odd number)- m=3: starts at 7 (which is the 4th odd number)- m=4: starts at 13 (which is the 7th odd number)Wait, let me see the positions:- For m=2, starting at position 2- For m=3, starting at position 4- For m=4, starting at position 7Hmm, so the starting position seems to be increasing by 2, then 3, then maybe 4? Let me check:From m=2 to m=3: 4 - 2 = 2From m=3 to m=4: 7 - 4 = 3If this pattern continues, from m=4 to m=5, the starting position would increase by 4, so starting at 7 + 4 = 11, which would be the 11th odd number, which is 21. Let me check if 5³=125 can be expressed as the sum of five consecutive odd numbers starting at 21.21 + 23 + 25 + 27 + 29 = 21+23=44, 44+25=69, 69+27=96, 96+29=125. Yes, that works. So the starting position for m=5 is 11, which is the 11th odd number.So, the starting position for m is the (1 + 2 + 3 + ... + (m-1)) + 1? Wait, let me think.Wait, for m=2, starting at position 2, which is 1 + 1For m=3, starting at position 4, which is 1 + 1 + 2For m=4, starting at position 7, which is 1 + 1 + 2 + 3For m=5, starting at position 11, which is 1 + 1 + 2 + 3 + 4Wait, that seems like the starting position for m is 1 + the sum from 1 to (m-1). So, starting position S(m) = 1 + sum_{k=1}^{m-1} k.Sum from 1 to n is n(n+1)/2, so sum from 1 to (m-1) is (m-1)m/2. Therefore, S(m) = 1 + (m-1)m/2.Let me verify this with the examples:For m=2: 1 + (1)(2)/2 = 1 + 1 = 2. Correct.For m=3: 1 + (2)(3)/2 = 1 + 3 = 4. Correct.For m=4: 1 + (3)(4)/2 = 1 + 6 = 7. Correct.For m=5: 1 + (4)(5)/2 = 1 + 10 = 11. Correct.Okay, so the starting position for m is S(m) = 1 + (m-1)m/2.Therefore, the starting odd number for m is the S(m)th odd number. The nth odd number is given by 2n - 1. So, the starting odd number is 2*S(m) - 1.So, starting odd number O(m) = 2*(1 + (m-1)m/2) - 1 = 2 + (m-1)m - 1 = (m-1)m + 1.Wait, let me compute that again:O(m) = 2*S(m) - 1 = 2*(1 + (m-1)m/2) - 1 = 2 + (m-1)m - 1 = (m-1)m + 1.Yes, that's correct.So, the starting odd number is (m-1)m + 1.Let me check with m=2: (2-1)*2 +1 = 1*2 +1=3. Correct.m=3: (3-1)*3 +1=2*3 +1=7. Correct.m=4: (4-1)*4 +1=3*4 +1=13. Correct.m=5: (5-1)*5 +1=4*5 +1=21. Correct.Great, so the starting odd number is (m-1)m +1.Now, the sequence of consecutive odd numbers for m³ is m terms long, starting from (m-1)m +1.So, the sequence is: (m-1)m +1, (m-1)m +3, ..., up to m terms.We need to find for which m, 2017 is one of these terms, and then find its position in that sequence.So, first, let's find m such that 2017 is in the sequence starting at (m-1)m +1, with m terms, each increasing by 2.So, 2017 must satisfy:(m-1)m +1 + 2*(k-1) = 2017, where k is the position in the sequence (1 ≤ k ≤ m).So, rearranged:2*(k-1) = 2017 - (m-1)m -1 = 2016 - (m² - m) = 2016 - m² + m.So, 2*(k-1) = -m² + m + 2016.Therefore,k = (-m² + m + 2016)/2 + 1.But k must be an integer between 1 and m.So, (-m² + m + 2016) must be even, and k must be integer.So, let's write:k = (-m² + m + 2016)/2 + 1 = (-m² + m + 2016 + 2)/2 = (-m² + m + 2018)/2.So, k = ( -m² + m + 2018 ) / 2.Since k must be an integer, (-m² + m + 2018) must be even.Let me see, m² and m have the same parity because m² ≡ m mod 2.So, m² - m is even, so -m² + m = -(m² - m) is also even. Therefore, -m² + m + 2018 is even + even = even. So, k is integer.So, now, we have k = ( -m² + m + 2018 ) / 2.We need to find m such that 1 ≤ k ≤ m.So,1 ≤ ( -m² + m + 2018 ) / 2 ≤ m.Multiply all parts by 2:2 ≤ -m² + m + 2018 ≤ 2m.So,Left inequality: -m² + m + 2018 ≥ 2 ⇒ -m² + m + 2016 ≥ 0 ⇒ m² - m - 2016 ≤ 0.Right inequality: -m² + m + 2018 ≤ 2m ⇒ -m² - m + 2018 ≤ 0 ⇒ m² + m - 2018 ≥ 0.So, we have two inequalities:1. m² - m - 2016 ≤ 02. m² + m - 2018 ≥ 0We need to find integer m >1 satisfying both.Let me solve the quadratic inequalities.First inequality: m² - m - 2016 ≤ 0.Find roots:m = [1 ± sqrt(1 + 4*2016)] / 2 = [1 ± sqrt(1 + 8064)] / 2 = [1 ± sqrt(8065)] / 2.Compute sqrt(8065):89² = 7921, 90²=8100. So sqrt(8065) is between 89 and 90.Compute 89²=7921, 89*89=7921, 89*90=8010, 89*91=8099, which is over 8065.Wait, 89²=7921, 89*90=8010, 89*91=8099, which is 8099, which is higher than 8065.Wait, 89*90=8010, 89*90 + 55=8065. So sqrt(8065)=89 + 55/(2*89) approximately, but exact value is not necessary.So, sqrt(8065) ≈ 89.8.Thus, m ≈ [1 + 89.8]/2 ≈ 45.4.Similarly, the other root is negative, so we can ignore it.Thus, the inequality m² - m - 2016 ≤ 0 holds for m between the two roots, so m ≤ approximately 45.4.Since m must be integer, m ≤45.Second inequality: m² + m - 2018 ≥ 0.Find roots:m = [ -1 ± sqrt(1 + 4*2018) ] / 2 = [ -1 ± sqrt(1 + 8072) ] / 2 = [ -1 ± sqrt(8073) ] / 2.Compute sqrt(8073):89²=7921, 90²=8100. So sqrt(8073) is between 89 and 90.Compute 89²=7921, 89*90=8010, 89*91=8099, which is 8099, which is higher than 8073.So, 89²=7921, 89*90=8010, 89*90 + 63=8073.So, sqrt(8073)=89 + 63/(2*89)≈89 + 0.35≈89.35.Thus, m ≈ [ -1 + 89.35 ] / 2 ≈ 88.35 /2≈44.175.The other root is negative, so we can ignore it.Thus, the inequality m² + m - 2018 ≥ 0 holds for m ≥ approximately 44.175.Since m must be integer, m ≥45.So, combining both inequalities:From first inequality: m ≤45From second inequality: m ≥45Thus, m=45.So, m=45.Therefore, the cube being split is 45³.Now, let's compute the starting odd number for m=45.Starting odd number O(m) = (m-1)m +1 = (44)(45) +1 = 1980 +1=1981.So, the sequence starts at 1981, and has 45 terms, each increasing by 2.So, the sequence is 1981, 1983, 1985, ..., up to 45 terms.We need to find the position of 2017 in this sequence.So, let's compute how many steps from 1981 to 2017.Each step increases by 2, so the difference is 2017 -1981=36.Number of steps is 36 /2=18.But since we start counting from 0, the position is 18 +1=19.Wait, let me think.Wait, the first term is 1981, which is position 1.Then, 1983 is position 2, 1985 is position 3, and so on.So, the nth term is 1981 + 2*(n-1).So, set 1981 + 2*(n-1)=2017.Solve for n:2*(n-1)=2017 -1981=36n-1=18n=19.So, 2017 is the 19th term in the sequence.Therefore, the position is 19, which is option D.But wait, let me double-check.Compute 1981 + 2*(19-1)=1981 + 36=2017. Yes, correct.So, the answer is D:19.**Final Answer**boxed{D}

💡Okay, so I have this problem about arithmetic and geometric sequences. Let me try to understand it step by step. First, the problem says that if {a_n} is an arithmetic sequence, then when we define b_n as the average of the first n terms of {a_n}, which is (a_1 + a_2 + ... + a_n)/n, then {b_n} is also an arithmetic sequence. That makes sense because the average of terms in an arithmetic sequence should also form an arithmetic sequence. Now, the question is asking about a similar property for geometric sequences. If {c_n} is a geometric sequence with all positive terms, then we need to find an expression for d_n such that {d_n} is also a geometric sequence. The options given are:A: d_n = (c_1 + c_2 + ... + c_n)/nB: d_n = (c_1 * c_2 * ... * c_n)/nC: d_n = nth root of (c_1 * c_2 * ... * c_n)D: d_n = n * (c_1^n * c_2^n * ... * c_n^n)/nHmm, so I need to figure out which of these expressions for d_n will make {d_n} a geometric sequence if {c_n} is already a geometric sequence.Let me recall what an arithmetic sequence is. In an arithmetic sequence, each term increases by a constant difference. So, if {a_n} is arithmetic, then a_{n+1} - a_n = d, where d is the common difference. Similarly, in a geometric sequence, each term is multiplied by a constant ratio. So, if {c_n} is geometric, then c_{n+1}/c_n = r, where r is the common ratio.Now, for the arithmetic sequence, taking the average of the first n terms gives another arithmetic sequence. So, what operation on the terms of a geometric sequence would preserve the geometric nature?In the arithmetic case, we used the arithmetic mean, which is a sum divided by n. For the geometric sequence, maybe we need to use the geometric mean instead of the arithmetic mean. The geometric mean of n numbers is the nth root of their product.Let me test this idea. If {c_n} is a geometric sequence, then c_n = c_1 * r^{n-1}. So, the product of the first n terms would be c_1 * c_2 * ... * c_n = c_1^n * r^{0+1+2+...+(n-1)}. The exponent of r is the sum of the first (n-1) integers, which is n(n-1)/2. So, the product is c_1^n * r^{n(n-1)/2}.Taking the nth root of this product would give us c_1 * r^{(n-1)/2}. So, d_n = c_1 * r^{(n-1)/2}. Now, let's see if {d_n} is a geometric sequence. The ratio between consecutive terms would be d_{n+1}/d_n = [c_1 * r^{(n)/2}]/[c_1 * r^{(n-1)/2}] = r^{1/2}. So, the ratio is constant, which means {d_n} is indeed a geometric sequence with common ratio sqrt(r).That seems to work. So, taking the geometric mean, which is the nth root of the product of the first n terms, gives us another geometric sequence.Let me check the other options to make sure.Option A: d_n = (c_1 + c_2 + ... + c_n)/n. This is the arithmetic mean of the geometric sequence. I don't think this would result in a geometric sequence. Let's test it with a simple geometric sequence. Suppose c_n = 2^n. Then, the sum of the first n terms is 2 + 4 + 8 + ... + 2^n = 2(2^n - 1). So, d_n = (2(2^n - 1))/n. Is this a geometric sequence? Let's compute d_1 = 2, d_2 = (2 + 4)/2 = 3, d_3 = (2 + 4 + 8)/3 = 14/3 ≈ 4.666. The ratios between terms are 3/2 = 1.5, and (14/3)/3 ≈ 1.555. These are not constant, so {d_n} is not a geometric sequence. So, option A is incorrect.Option B: d_n = (c_1 * c_2 * ... * c_n)/n. This is the product divided by n. Let's test with c_n = 2^n again. The product is 2 * 4 * 8 * ... * 2^n = 2^{1+2+3+...+n} = 2^{n(n+1)/2}. So, d_n = 2^{n(n+1)/2}/n. Is this a geometric sequence? Let's compute d_1 = 2/1 = 2, d_2 = (2*4)/2 = 4, d_3 = (2*4*8)/3 = 64/3 ≈ 21.333. The ratios are 4/2 = 2, and (64/3)/4 ≈ 5.333. Not constant, so {d_n} is not geometric. So, option B is incorrect.Option C: d_n = sqrt[n]{c_1 * c_2 * ... * c_n}. As I calculated earlier, this gives us a geometric sequence with common ratio sqrt(r). So, this works.Option D: d_n = n * (c_1^n * c_2^n * ... * c_n^n)/n. Simplifying, this is (c_1^n * c_2^n * ... * c_n^n). Let's see what this is. For c_n = 2^n, this would be (2^1 * 2^2 * ... * 2^n)^n = (2^{1+2+...+n})^n = (2^{n(n+1)/2})^n = 2^{n^2(n+1)/2}. So, d_n = 2^{n^2(n+1)/2}. Is this a geometric sequence? Let's compute d_1 = 2^{1*2/2} = 2^1 = 2, d_2 = 2^{4*3/2} = 2^6 = 64, d_3 = 2^{9*4/2} = 2^{18} = 262144. The ratios are 64/2 = 32, and 262144/64 = 4096. These are not constant, so {d_n} is not a geometric sequence. So, option D is incorrect.Therefore, the correct answer is option C.

💡Alright, so I have this problem where I need to find the remainder when |s(1012)| is divided by 100. Here, s(x) is the remainder when q(x) is divided by another polynomial, x³ + x² + x + 1. The polynomial q(x) is given as x¹⁰¹² + x¹⁰¹¹ + … + x + 1. First, I need to understand what q(x) is. It looks like a geometric series. The sum of a geometric series from x⁰ to xⁿ is (xⁿ⁺¹ - 1)/(x - 1). So, in this case, q(x) is the sum from x⁰ to x¹⁰¹², which would be (x¹⁰¹³ - 1)/(x - 1). That might be useful later.Next, I need to find the remainder when q(x) is divided by x³ + x² + x + 1. Polynomial division can be tricky, especially with such a high degree polynomial like q(x). Maybe there's a smarter way to do this without performing the entire division.I recall that when dividing by a polynomial, the remainder will have a degree less than the divisor. Since the divisor is degree 3, the remainder s(x) should be a quadratic polynomial, so it will look like ax² + bx + c.Another approach is to use the Remainder Factor Theorem. If I can find the roots of the divisor polynomial, I can evaluate q(x) at those roots and set up equations to solve for the coefficients of the remainder.Let me factor the divisor polynomial x³ + x² + x + 1. Maybe it factors into simpler polynomials. Trying to factor it:x³ + x² + x + 1 = x²(x + 1) + 1(x + 1) = (x² + 1)(x + 1). So, it factors into (x + 1)(x² + 1). That means the roots are x = -1, x = i, and x = -i, where i is the imaginary unit.Since the divisor factors into (x + 1) and (x² + 1), I can use the Chinese Remainder Theorem for polynomials. This theorem states that if I know the remainder of q(x) modulo each of the factors, I can find the remainder modulo the product.So, I need to find q(x) mod (x + 1) and q(x) mod (x² + 1). Then, I can combine these results to find q(x) mod (x³ + x² + x + 1).Starting with q(x) mod (x + 1):To find the remainder when q(x) is divided by (x + 1), I can substitute x = -1 into q(x). q(-1) = (-1)¹⁰¹² + (-1)¹⁰¹¹ + … + (-1) + 1.Let me compute this. Since the exponents go from 0 to 1012, which is 1013 terms. Each pair of terms (-1)^{2k} + (-1)^{2k+1} = 1 - 1 = 0. But since 1013 is odd, there will be an extra term at the end.Wait, actually, 1013 terms starting from x⁰ to x¹⁰¹². So, when x = -1, each term alternates between 1 and -1. Let's see:Number of terms: 1013. Since 1013 is odd, the number of pairs is 506 pairs and one extra term. Each pair sums to 0, and the extra term is 1 (since the last term is x⁰ = 1). So, q(-1) = 1.Wait, hold on, that doesn't seem right. Let me recount. The exponents go from 0 to 1012, which is 1013 terms. So, if I pair them as (x⁰ + x¹) + (x² + x³) + … + (x¹⁰¹²). Each pair is (1 + (-1)) = 0, except the last term if the number of terms is odd. But 1013 is odd, so the last term is x¹⁰¹², which is (-1)^1012 = 1. So, the total sum is 1.Wait, but in the original q(x), the constant term is 1, so when x = -1, q(-1) = 1 + (-1) + 1 + (-1) + … + 1. Since there are 1013 terms, starting and ending with 1, so the number of 1s is 507 and the number of -1s is 506. So, 507 - 506 = 1. So, q(-1) = 1.Wait, but in my initial thought, I thought it was 2. Hmm, maybe I made a mistake earlier. Let me check.Wait, if I write q(x) as (x¹⁰¹³ - 1)/(x - 1), then q(-1) = [(-1)¹⁰¹³ - 1]/(-1 - 1) = [(-1) - 1]/(-2) = (-2)/(-2) = 1. So, that's consistent. So, q(-1) = 1.But in the initial problem, the user had written that q(-1) = 2. Maybe that was a mistake? Hmm.Wait, let me recast. If q(x) is the sum from x⁰ to x¹⁰¹², then q(-1) is the sum from (-1)^0 to (-1)^1012. Since 1012 is even, the last term is 1. So, the number of terms is 1013, which is odd. So, the number of 1s is (1013 + 1)/2 = 507, and the number of -1s is 506. So, 507 - 506 = 1. So, q(-1) = 1.So, the remainder when q(x) is divided by (x + 1) is 1. So, s(x) ≡ 1 mod (x + 1).Now, moving on to q(x) mod (x² + 1). Since x² + 1 factors into (x - i)(x + i), the roots are i and -i. So, to find q(x) mod (x² + 1), I can evaluate q(i) and q(-i) and set up equations.But since x² ≡ -1 mod (x² + 1), I can reduce higher powers of x modulo x² + 1. Let's see:x² ≡ -1x³ ≡ x * x² ≡ x*(-1) = -xx⁴ ≡ x² * x² ≡ (-1)*(-1) = 1x⁵ ≡ x * x⁴ ≡ x*1 = xx⁶ ≡ x² * x⁴ ≡ (-1)*1 = -1x⁷ ≡ x * x⁶ ≡ x*(-1) = -xx⁸ ≡ x² * x⁶ ≡ (-1)*(-1) = 1So, the cycle repeats every 4 exponents: 1, x, -1, -x, 1, x, -1, -x, etc.Given that, I can find x^k mod (x² + 1) by looking at k mod 4.So, for q(x) = sum_{k=0}^{1012} x^k, we can group the terms in cycles of 4.Each cycle of 4 terms: x^{4m} + x^{4m+1} + x^{4m+2} + x^{4m+3} ≡ 1 + x + (-1) + (-x) = 0 mod (x² + 1).So, every 4 terms sum to 0 modulo (x² + 1). Now, how many complete cycles are there in 1013 terms (from 0 to 1012)? 1013 divided by 4 is 253 cycles with a remainder of 1 term.So, 253 cycles * 4 terms = 1012 terms. Then, the last term is x¹⁰¹².So, the sum q(x) ≡ 0 + x¹⁰¹² mod (x² + 1).Now, x¹⁰¹² mod (x² + 1). Let's compute 1012 divided by 4.1012 / 4 = 253, exactly, so x¹⁰¹² = (x⁴)^253 ≡ 1^253 = 1 mod (x² + 1).Therefore, q(x) ≡ 1 mod (x² + 1).So, putting it all together, we have:q(x) ≡ 1 mod (x + 1)q(x) ≡ 1 mod (x² + 1)We need to find a quadratic polynomial s(x) = ax² + bx + c such that:s(x) ≡ 1 mod (x + 1)s(x) ≡ 1 mod (x² + 1)We can use the Chinese Remainder Theorem for polynomials. Since (x + 1) and (x² + 1) are coprime, the solution will be unique modulo (x³ + x² + x + 1).Let me write s(x) as:s(x) = (x² + 1) * A(x) + 1But since s(x) must also satisfy s(-1) = 1, let's plug x = -1 into s(x):s(-1) = ((-1)² + 1) * A(-1) + 1 = (1 + 1) * A(-1) + 1 = 2 * A(-1) + 1 = 1So, 2 * A(-1) + 1 = 1 => 2 * A(-1) = 0 => A(-1) = 0This means that A(x) must have (x + 1) as a factor. So, A(x) = (x + 1) * B(x)Therefore, s(x) = (x² + 1)(x + 1) * B(x) + 1But since s(x) is a quadratic polynomial, the degree of s(x) is 2. The term (x² + 1)(x + 1) is degree 3, so B(x) must be a constant polynomial, because otherwise s(x) would have degree higher than 2.Let me denote B(x) = k, a constant. Then,s(x) = k(x² + 1)(x + 1) + 1But s(x) must be quadratic, so let's expand this:s(x) = k(x³ + x² + x + 1) + 1But s(x) is quadratic, so the coefficient of x³ must be zero. Therefore, k must be zero.Wait, that can't be right because then s(x) = 1, which is a constant, but we need a quadratic polynomial. Hmm, maybe I made a mistake in the approach.Alternatively, perhaps I should express s(x) as:s(x) = (x + 1) * C(x) + 1But s(x) must also satisfy s(x) ≡ 1 mod (x² + 1). So, substituting x = i and x = -i into s(x):s(i) = (i + 1) * C(i) + 1 = 1s(-i) = (-i + 1) * C(-i) + 1 = 1So, we have:(i + 1) * C(i) + 1 = 1 => (i + 1) * C(i) = 0Similarly, (-i + 1) * C(-i) = 0Since (i + 1) ≠ 0 and (-i + 1) ≠ 0, this implies that C(i) = 0 and C(-i) = 0.Therefore, C(x) must have both (x - i) and (x + i) as factors, which means C(x) is divisible by (x² + 1). So, C(x) = (x² + 1) * D(x)Therefore, s(x) = (x + 1)(x² + 1) * D(x) + 1But again, s(x) is quadratic, so (x + 1)(x² + 1) is cubic, so D(x) must be a constant. Let D(x) = k.Thus, s(x) = k(x³ + x² + x + 1) + 1But s(x) is quadratic, so the x³ term must cancel out. Therefore, k must be zero. But then s(x) = 1, which is a constant, not quadratic. This seems contradictory.Wait, perhaps my initial assumption is wrong. Maybe s(x) is not necessarily of degree 2, but just less than 3, so it can be quadratic, linear, or constant. In this case, since both remainders modulo (x + 1) and (x² + 1) are 1, perhaps s(x) is just 1.But let me check. If s(x) = 1, then s(x) ≡ 1 mod (x + 1) and s(x) ≡ 1 mod (x² + 1), which satisfies both conditions. So, s(x) = 1.But wait, in the initial problem, the user had written that s(x) ≡ 2 mod (x³ + x² + x + 1). But according to my calculations, it's 1. Maybe I made a mistake.Wait, let me go back. When I computed q(-1), I got 1, but the user had 2. Let me double-check.q(x) = x¹⁰¹² + x¹⁰¹¹ + … + x + 1At x = -1, q(-1) = (-1)^1012 + (-1)^1011 + … + (-1) + 1Since 1012 is even, (-1)^1012 = 1Similarly, (-1)^1011 = -1So, the terms alternate between 1 and -1, starting with 1 and ending with 1.Number of terms: 1013, which is odd. So, the number of 1s is (1013 + 1)/2 = 507, and the number of -1s is 506.So, total sum: 507*1 + 506*(-1) = 507 - 506 = 1So, q(-1) = 1, not 2. So, the user had a mistake there.Similarly, when evaluating q(i):q(i) = i¹⁰¹² + i¹⁰¹¹ + … + i + 1Since i⁴ = 1, the exponents cycle every 4.1012 divided by 4 is 253, so i¹⁰¹² = (i⁴)^253 = 1^253 = 1Similarly, i¹⁰¹¹ = i^(4*252 + 3) = i³ = -ii¹⁰¹⁰ = i² = -1i¹⁰¹³ = i¹ = iWait, but q(x) goes up to x¹⁰¹², so the exponents are from 0 to 1012.So, let's compute q(i):Sum_{k=0}^{1012} i^kThis is a geometric series with ratio i, starting from k=0 to k=1012.Sum = (i¹⁰¹³ - 1)/(i - 1)Compute i¹⁰¹³: 1013 divided by 4 is 253 with remainder 1, so i¹⁰¹³ = i^1 = iThus, Sum = (i - 1)/(i - 1) = 1Similarly, q(-i) = Sum_{k=0}^{1012} (-i)^kAgain, geometric series with ratio (-i), so Sum = [(-i)^1013 - 1]/[(-i) - 1]Compute (-i)^1013: 1013 divided by 4 is 253 with remainder 1, so (-i)^1013 = (-i)^1 = -iThus, Sum = (-i - 1)/(-i - 1) = 1So, q(i) = 1 and q(-i) = 1, which means q(x) ≡ 1 mod (x² + 1)Therefore, both remainders are 1, so s(x) = 1.Wait, but in the initial problem, the user had s(x) ≡ 2. So, I think the user made a mistake in their calculation.Therefore, the correct remainder s(x) is 1.Thus, |s(1012)| = |1| = 1Then, the remainder when 1 is divided by 100 is 1.But wait, the user's initial problem had s(x) ≡ 2, leading to |s(1012)| = 2, and 2 mod 100 is 2. But according to my calculations, it's 1.I think the confusion arises from whether q(x) is x¹⁰¹² + … + 1 or x¹⁰¹³ - 1 over x - 1. But in any case, the evaluation at x = -1 and x = i gives 1, not 2.Wait, perhaps I made a mistake in the number of terms. Let me recount.q(x) is from x⁰ to x¹⁰¹², which is 1013 terms. So, when x = -1, the sum is 1 + (-1) + 1 + (-1) + … + 1. Since 1013 is odd, the number of 1s is 507 and the number of -1s is 506. So, 507 - 506 = 1. So, q(-1) = 1.Similarly, for q(i), the sum is 1 + i + (-1) + (-i) + 1 + i + (-1) + (-i) + … + 1. Since 1013 terms, which is 253 full cycles of 4 terms (each summing to 0) plus one extra term, which is 1. So, q(i) = 1.Therefore, s(x) = 1.Thus, |s(1012)| = |1| = 1, and 1 mod 100 is 1.But the user's initial problem had s(x) ≡ 2, leading to 2 mod 100. So, I think the correct answer is 1, not 2.Wait, but let me double-check the initial problem statement."Let q(x) = x¹⁰¹² + x¹⁰¹¹ + … + x + 1, and let s(x) be the polynomial remainder when q(x) is divided by x³ + x² + x + 1. Determine the remainder when |s(1012)| is divided by 100."So, according to my calculations, s(x) = 1, so |s(1012)| = 1, and 1 mod 100 is 1.But the user's initial problem had s(x) ≡ 2, leading to 2 mod 100. So, perhaps the user made a mistake in their calculation.Alternatively, maybe I made a mistake in interpreting the problem.Wait, another approach: Since q(x) = (x¹⁰¹³ - 1)/(x - 1), and we need to find q(x) mod (x³ + x² + x + 1). Let me denote the divisor as D(x) = x³ + x² + x + 1.We can write q(x) = D(x) * Q(x) + s(x), where s(x) is the remainder, degree less than 3.We can find s(x) by evaluating q(x) at the roots of D(x). Since D(x) factors as (x + 1)(x² + 1), the roots are x = -1, x = i, x = -i.So, s(x) must satisfy:s(-1) = q(-1) = 1s(i) = q(i) = 1s(-i) = q(-i) = 1Since s(x) is a quadratic polynomial, we can write s(x) = ax² + bx + c.We have:s(-1) = a(-1)² + b(-1) + c = a - b + c = 1s(i) = a(i)² + b(i) + c = a(-1) + bi + c = -a + bi + c = 1s(-i) = a(-i)² + b(-i) + c = a(-1) - bi + c = -a - bi + c = 1So, we have the following system of equations:1. a - b + c = 12. -a + bi + c = 13. -a - bi + c = 1Let me write equations 2 and 3:From equation 2: -a + c + bi = 1From equation 3: -a + c - bi = 1If I add equations 2 and 3: (-a + c + bi) + (-a + c - bi) = 2*(-a + c) = 2So, -2a + 2c = 2 => -a + c = 1 => c = a + 1From equation 1: a - b + c = 1Substitute c = a + 1: a - b + (a + 1) = 1 => 2a - b + 1 = 1 => 2a - b = 0 => b = 2aNow, from equation 2: -a + c + bi = 1Substitute c = a + 1 and b = 2a:-a + (a + 1) + (2a)i = 1Simplify:(-a + a) + 1 + 2a i = 1 => 0 + 1 + 2a i = 1So, 1 + 2a i = 1 => 2a i = 0Since i ≠ 0, this implies that a = 0Then, from b = 2a, b = 0From c = a + 1, c = 1Therefore, s(x) = 0x² + 0x + 1 = 1So, s(x) = 1Thus, |s(1012)| = |1| = 1Then, 1 mod 100 is 1.Therefore, the remainder when |s(1012)| is divided by 100 is 1.But wait, the user's initial problem had s(x) ≡ 2, leading to 2 mod 100. So, I think the correct answer is 1.Alternatively, maybe I made a mistake in the number of terms. Let me check again.q(x) = x¹⁰¹² + x¹⁰¹¹ + … + x + 1Number of terms: from x⁰ to x¹⁰¹², which is 1013 terms.At x = -1, the sum is 1 + (-1) + 1 + (-1) + … + 1. Since 1013 is odd, the last term is 1, so the number of 1s is 507, and the number of -1s is 506. So, 507 - 506 = 1. So, q(-1) = 1.Similarly, for q(i), the sum is 1 + i + (-1) + (-i) + 1 + i + (-1) + (-i) + … + 1. Since 1013 terms, which is 253 full cycles of 4 terms (each summing to 0) plus one extra term, which is 1. So, q(i) = 1.Therefore, s(x) = 1.Thus, |s(1012)| = 1, and 1 mod 100 is 1.I think the user's initial problem had a mistake in their calculation, leading them to think s(x) ≡ 2. But according to my calculations, it's 1.Therefore, the correct answer is 1.