answer：Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The sequence {a_n} is a non-constant arithmetic sequence, which means it has a common difference, right? And they told me that a_4 is 10. Also, a_3, a_6, and a_10 form a geometric sequence. I need to find the general term formula for {a_n} and then, for part (2), define b_n as 2^n times a_n and find the sum of the first n terms of {b_n}.Starting with part (1). Since it's an arithmetic sequence, the general term is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. They told me that a_4 = 10, so plugging into the formula: a_4 = a_1 + 3d = 10. That's one equation.Now, the other condition is that a_3, a_6, a_10 form a geometric sequence. In a geometric sequence, each term is multiplied by a common ratio. So, the ratio between a_6 and a_3 should be the same as the ratio between a_10 and a_6. That means (a_6 / a_3) = (a_10 / a_6). So, if I write that out, it's (a_6)^2 = a_3 * a_10.Let me express a_3, a_6, and a_10 using the arithmetic sequence formula. a_3 = a_1 + 2da_6 = a_1 + 5da_10 = a_1 + 9dSo, plugging these into the geometric sequence condition: (a_1 + 5d)^2 = (a_1 + 2d)(a_1 + 9d)Let me expand both sides:Left side: (a_1 + 5d)^2 = a_1^2 + 10a_1d + 25d^2Right side: (a_1 + 2d)(a_1 + 9d) = a_1^2 + 11a_1d + 18d^2So, setting left side equal to right side:a_1^2 + 10a_1d + 25d^2 = a_1^2 + 11a_1d + 18d^2Subtracting a_1^2 from both sides:10a_1d + 25d^2 = 11a_1d + 18d^2Now, subtract 10a_1d and 18d^2 from both sides:25d^2 - 18d^2 = 11a_1d - 10a_1dSimplify:7d^2 = a_1dAssuming d ≠ 0 (since it's a non-constant sequence), we can divide both sides by d:7d = a_1So, a_1 = 7d.Now, from the first equation, a_1 + 3d = 10. Since a_1 = 7d, substitute:7d + 3d = 1010d = 10So, d = 1.Then, a_1 = 7d = 7*1 = 7.Therefore, the general term is a_n = a_1 + (n-1)d = 7 + (n-1)*1 = n + 6.Wait, let me check that. If a_1 is 7, then a_2 is 8, a_3 is 9, a_4 is 10, which matches the given a_4 = 10. Then, a_3 = 9, a_6 = 12, a_10 = 16. Let's see if these form a geometric sequence.Check the ratios: 12 / 9 = 4/3, and 16 / 12 = 4/3. Yes, so the common ratio is 4/3. So that works. So, the general term is a_n = n + 6.Okay, part (1) seems done.Now, part (2): Define b_n = 2^n * a_n. So, b_n = 2^n * (n + 6). I need to find the sum of the first n terms, S_n.So, S_n = sum_{k=1}^n b_k = sum_{k=1}^n [2^k (k + 6)].Hmm, this looks like a sum involving both an exponential and a linear term. I remember that for sums like sum_{k=1}^n k * r^k, there is a formula. Maybe I can split this into two sums: sum_{k=1}^n 2^k * k + 6 * sum_{k=1}^n 2^k.Yes, let's do that.So, S_n = sum_{k=1}^n k * 2^k + 6 * sum_{k=1}^n 2^k.I know that sum_{k=1}^n 2^k is a geometric series. The sum is 2(2^n - 1)/(2 - 1) = 2^{n+1} - 2.Now, the other sum, sum_{k=1}^n k * 2^k. I think the formula for sum_{k=1}^n k r^k is r(1 - (n+1) r^n + n r^{n+1}) / (1 - r)^2. Let me verify that.Yes, for r ≠ 1, sum_{k=1}^n k r^k = r(1 - (n+1) r^n + n r^{n+1}) ) / (1 - r)^2.In this case, r = 2. So, plugging in:sum_{k=1}^n k * 2^k = 2(1 - (n+1)2^n + n 2^{n+1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1.So, numerator: 2(1 - (n+1)2^n + n 2^{n+1}) = 2 - 2(n+1)2^n + 2n 2^{n+1}Wait, let me compute step by step:First, expand numerator:2 * [1 - (n+1)2^n + n 2^{n+1}]= 2*1 - 2*(n+1)2^n + 2*n 2^{n+1}= 2 - (n+1)2^{n+1} + n 2^{n+2}Wait, that seems complicated. Maybe I made a mistake in expanding.Wait, let me re-express the formula:sum_{k=1}^n k r^k = r(1 - (n+1) r^n + n r^{n+1}) / (1 - r)^2So, for r=2, it's:2(1 - (n+1)2^n + n 2^{n+1}) / (1 - 2)^2Denominator: (1 - 2)^2 = 1, so denominator is 1.So, numerator: 2[1 - (n+1)2^n + n 2^{n+1}]= 2*1 - 2*(n+1)2^n + 2*n 2^{n+1}= 2 - (n+1)2^{n+1} + n 2^{n+2}Wait, that seems messy. Maybe I should factor 2^{n+1}:= 2 + 2^{n+1}[-(n+1) + 2n]= 2 + 2^{n+1}(n - 1)Wait, let's compute:2 - (n+1)2^{n+1} + n 2^{n+2}= 2 - (n+1)2^{n+1} + n*2*2^{n+1}= 2 - (n+1)2^{n+1} + 2n 2^{n+1}= 2 + [ - (n+1) + 2n ] 2^{n+1}= 2 + (n - 1)2^{n+1}So, sum_{k=1}^n k*2^k = 2 + (n - 1)2^{n+1}Wait, let me check for small n.Let n=1: sum is 1*2 = 2. Formula: 2 + (1 -1)2^{2} = 2 + 0 = 2. Correct.n=2: sum is 1*2 + 2*4 = 2 + 8 = 10. Formula: 2 + (2 -1)2^{3} = 2 + 8 = 10. Correct.n=3: sum is 2 + 8 + 24 = 34. Formula: 2 + (3 -1)2^{4} = 2 + 2*16 = 2 +32=34. Correct.Okay, so the formula seems correct.So, sum_{k=1}^n k*2^k = 2 + (n -1)2^{n+1}Therefore, going back to S_n:S_n = [2 + (n -1)2^{n+1}] + 6*(2^{n+1} - 2)Let me compute each part:First part: 2 + (n -1)2^{n+1}Second part: 6*(2^{n+1} - 2) = 6*2^{n+1} - 12So, adding them together:2 + (n -1)2^{n+1} + 6*2^{n+1} - 12Combine like terms:2 - 12 = -10(n -1)2^{n+1} + 6*2^{n+1} = (n -1 + 6)2^{n+1} = (n +5)2^{n+1}So, altogether:S_n = (n +5)2^{n+1} -10Let me check with n=1:b_1 = 2^1*(1 +6)=14. So, S_1=14. Formula: (1 +5)2^{2} -10=6*4 -10=24 -10=14. Correct.n=2:b_1=14, b_2=2^2*(2 +6)=4*8=32. So, S_2=14 +32=46.Formula: (2 +5)2^{3} -10=7*8 -10=56 -10=46. Correct.n=3:b_3=2^3*(3 +6)=8*9=72. S_3=14 +32 +72=118.Formula: (3 +5)2^{4} -10=8*16 -10=128 -10=118. Correct.Looks good. So, the sum S_n is (n +5)2^{n+1} -10.**Final Answer**(1) The general term is boxed{a_n = n + 6}.(2) The sum of the first ( n ) terms is boxed{S_n = (n + 5) cdot 2^{n+1} - 10}.

answer：Okay, so I need to find the remainder when ( x^2 + 5x - 3 ) divides ( 3x^4 - 8x^3 + 20x^2 - 7x + 13 ). Hmm, I remember that when dividing polynomials, the remainder has a degree less than the divisor. Since the divisor is a quadratic (degree 2), the remainder should be a linear polynomial (degree 1) or a constant. Wait, but in the previous answer, the remainder was a quadratic. That doesn't make sense because the degree of the remainder should be less than the divisor's degree. Maybe I made a mistake there.Let me try this again. I think I should use polynomial long division. So, I'll set it up with ( 3x^4 - 8x^3 + 20x^2 - 7x + 13 ) as the dividend and ( x^2 + 5x - 3 ) as the divisor.First, I divide the leading term of the dividend, which is ( 3x^4 ), by the leading term of the divisor, ( x^2 ). That gives me ( 3x^2 ). Then, I multiply the entire divisor by ( 3x^2 ):( 3x^2 times (x^2 + 5x - 3) = 3x^4 + 15x^3 - 9x^2 ).Next, I subtract this from the original dividend:( (3x^4 - 8x^3 + 20x^2 - 7x + 13) - (3x^4 + 15x^3 - 9x^2) = -23x^3 + 29x^2 - 7x + 13 ).Now, I repeat the process with the new polynomial ( -23x^3 + 29x^2 - 7x + 13 ). I divide ( -23x^3 ) by ( x^2 ) to get ( -23x ). Multiply the divisor by ( -23x ):( -23x times (x^2 + 5x - 3) = -23x^3 - 115x^2 + 69x ).Subtract this from the current polynomial:( (-23x^3 + 29x^2 - 7x + 13) - (-23x^3 - 115x^2 + 69x) = 144x^2 - 76x + 13 ).Now, I have ( 144x^2 - 76x + 13 ). I divide ( 144x^2 ) by ( x^2 ) to get ( 144 ). Multiply the divisor by ( 144 ):( 144 times (x^2 + 5x - 3) = 144x^2 + 720x - 432 ).Subtract this from the current polynomial:( (144x^2 - 76x + 13) - (144x^2 + 720x - 432) = -796x + 445 ).Now, the degree of the remainder ( -796x + 445 ) is 1, which is less than the degree of the divisor ( x^2 + 5x - 3 ) (which is 2). So, this should be the final remainder.Wait, but in the previous attempt, I ended up with a quadratic remainder, which was incorrect. I think I made a mistake in the multiplication or subtraction steps. Let me double-check my calculations.Starting again, after the first subtraction, I had ( -23x^3 + 29x^2 - 7x + 13 ). Dividing ( -23x^3 ) by ( x^2 ) gives ( -23x ). Multiplying the divisor by ( -23x ) gives ( -23x^3 - 115x^2 + 69x ). Subtracting this from the current polynomial:( (-23x^3 + 29x^2 - 7x + 13) - (-23x^3 - 115x^2 + 69x) ).Let's compute term by term:- ( -23x^3 - (-23x^3) = 0 ).- ( 29x^2 - (-115x^2) = 144x^2 ).- ( -7x - 69x = -76x ).- ( 13 - 0 = 13 ).So, the result is ( 144x^2 - 76x + 13 ). That seems correct.Next, dividing ( 144x^2 ) by ( x^2 ) gives 144. Multiplying the divisor by 144:( 144x^2 + 720x - 432 ).Subtracting this from ( 144x^2 - 76x + 13 ):( 144x^2 - 76x + 13 - (144x^2 + 720x - 432) ).Term by term:- ( 144x^2 - 144x^2 = 0 ).- ( -76x - 720x = -796x ).- ( 13 - (-432) = 13 + 432 = 445 ).So, the remainder is ( -796x + 445 ). This makes sense because the degree is less than the divisor's degree.Wait, but in the initial attempt, I had a different remainder. I think I must have made an error in the multiplication or subtraction steps before. Let me check again.Alternatively, maybe I can use the Remainder Theorem. Since the divisor is quadratic, I can express the dividend as ( (x^2 + 5x - 3)(Q(x)) + R(x) ), where ( R(x) ) is linear or constant. To find ( R(x) ), I can set ( x^2 + 5x - 3 = 0 ), so ( x^2 = -5x + 3 ). Then, I can express higher powers of x in terms of lower powers.Let's try that. Let me express ( x^4 ) in terms of lower degrees.From ( x^2 = -5x + 3 ), then ( x^3 = x times x^2 = x(-5x + 3) = -5x^2 + 3x ).But ( x^2 = -5x + 3 ), so ( x^3 = -5(-5x + 3) + 3x = 25x - 15 + 3x = 28x - 15 ).Similarly, ( x^4 = x times x^3 = x(28x - 15) = 28x^2 - 15x ).Again, substitute ( x^2 = -5x + 3 ):( x^4 = 28(-5x + 3) - 15x = -140x + 84 - 15x = -155x + 84 ).Now, substitute ( x^4 ) and ( x^3 ) into the dividend:( 3x^4 - 8x^3 + 20x^2 - 7x + 13 ).Replace ( x^4 ) with ( -155x + 84 ):( 3(-155x + 84) = -465x + 252 ).Replace ( x^3 ) with ( 28x - 15 ):( -8(28x - 15) = -224x + 120 ).Replace ( x^2 ) with ( -5x + 3 ):( 20(-5x + 3) = -100x + 60 ).Now, combine all these:( (-465x + 252) + (-224x + 120) + (-100x + 60) - 7x + 13 ).Combine like terms:- ( -465x - 224x - 100x - 7x = -796x ).- ( 252 + 120 + 60 + 13 = 445 ).So, the dividend simplifies to ( -796x + 445 ) when expressed in terms of the divisor. Therefore, the remainder is ( -796x + 445 ).This matches the result from the polynomial long division. So, the correct remainder is ( -796x + 445 ).I think in the initial attempt, I might have made an error in the multiplication or subtraction steps, leading to an incorrect quadratic remainder. By using the Remainder Theorem approach, I arrived at the correct linear remainder.

answer：Alright, so I've got this problem here about a piecewise function f(x). It's defined differently for x > 0 and x < 0. The function is:f(x) = { e^x + a x, when x > 0 { 1/e^x - a x, when x < 0And the question is asking about the range of the real number a such that f(x) has four zeros. The options are A: (-∞, -1/e), B: (-∞, -e), C: (e, +∞), D: (1/e, +∞).Hmm, okay. So, first, I need to understand what it means for f(x) to have four zeros. A zero of a function is a point where the function crosses the x-axis, meaning f(x) = 0. Since f(x) is defined differently for positive and negative x, I probably need to analyze each piece separately and then see how they contribute to the total number of zeros.Looking at the function, I notice that for x > 0, it's e^x + a x, and for x < 0, it's 1/e^x - a x. Wait a second, 1/e^x is the same as e^{-x}, right? So, for x < 0, the function is e^{-x} - a x.Is this function even or odd? Let me check. For even functions, f(-x) = f(x). Let's see:If x > 0, then f(x) = e^x + a x.If I plug in -x, which would be negative, so f(-x) = e^{-(-x)} - a (-x) = e^{x} + a x. Wait, that's the same as f(x). So, f(-x) = f(x). That means f(x) is an even function. So, it's symmetric about the y-axis.Therefore, if I can find the number of zeros for x > 0, the number of zeros for x < 0 will be the same. So, if f(x) has four zeros in total, that would mean two zeros for x > 0 and two zeros for x < 0. But since it's even, actually, each zero for x > 0 corresponds to a zero for x < 0. So, if there are two zeros for x > 0, there will be two zeros for x < 0, making a total of four zeros.Wait, but actually, hold on. If x = 0 is a zero, that would count once, but in this case, f(0) isn't defined because the function isn't defined at x = 0. So, all zeros must be either positive or negative, but not at zero.So, yeah, if there are two zeros for x > 0, then there will be two zeros for x < 0, giving four zeros in total. So, the key is to find the values of a such that f(x) = 0 has two solutions for x > 0.So, let's focus on x > 0. The equation is e^x + a x = 0. Let's solve for a:a x = -e^xSo, a = -e^x / xLet me define a function g(x) = -e^x / x for x > 0. Then, the equation a = g(x) will have solutions when a is in the range of g(x). So, to have two solutions, the horizontal line y = a must intersect the graph of g(x) at two points.Therefore, I need to analyze the function g(x) = -e^x / x for x > 0. Let's find its critical points to understand its behavior.First, compute the derivative of g(x):g(x) = -e^x / xSo, g'(x) = derivative of (-e^x / x). Let's use the quotient rule.If we have h(x) = u(x)/v(x), then h'(x) = (u'(x)v(x) - u(x)v'(x)) / [v(x)]^2.Here, u(x) = -e^x, so u'(x) = -e^x.v(x) = x, so v'(x) = 1.Therefore,g'(x) = [(-e^x)(x) - (-e^x)(1)] / x^2Simplify numerator:(-e^x x) + e^x = e^x (-x + 1) = e^x (1 - x)So,g'(x) = [e^x (1 - x)] / x^2So, g'(x) = e^x (1 - x) / x^2Now, let's analyze the sign of g'(x):e^x is always positive, x^2 is always positive for x ≠ 0. So, the sign of g'(x) depends on (1 - x).So, when is g'(x) positive? When (1 - x) > 0, which is when x < 1.When is g'(x) negative? When (1 - x) < 0, which is when x > 1.Therefore, g(x) is increasing on (0, 1) and decreasing on (1, ∞). So, it has a maximum at x = 1.Let's compute g(1):g(1) = -e^1 / 1 = -eSo, the maximum value of g(x) is -e at x = 1.Now, let's analyze the behavior of g(x) as x approaches 0+ and as x approaches ∞.As x approaches 0 from the right:e^x approaches 1, so g(x) = -e^x / x ≈ -1 / x, which tends to -∞.As x approaches ∞:e^x grows exponentially, and x grows linearly, so e^x / x tends to ∞, but with the negative sign, g(x) tends to -∞.So, the graph of g(x) starts at -∞ as x approaches 0, increases to a maximum of -e at x = 1, and then decreases back to -∞ as x approaches ∞.Therefore, the function g(x) has a single peak at x = 1 with value -e, and it goes to -∞ on both ends.So, if we draw a horizontal line y = a, the number of intersections with g(x) depends on the value of a.If a is greater than -e, say a = -1, which is greater than -e (since e ≈ 2.718, so -e ≈ -2.718), then y = a would intersect g(x) only once on the decreasing part, right? Because the maximum is at -e, so if a is above that, it can't intersect on the increasing part, only once on the decreasing.Wait, no. Wait, actually, since the function goes from -∞ to -e, then back to -∞. So, if a is less than -e, meaning more negative, then y = a would intersect the graph twice: once on the increasing part (left of x=1) and once on the decreasing part (right of x=1). If a is equal to -e, it only touches the peak, so one intersection. If a is greater than -e, meaning less negative, then y = a doesn't intersect the graph at all because the maximum of g(x) is -e, which is lower than a.Wait, hold on, let me think again.Wait, g(x) is always negative because it's -e^x / x, which is negative for all x > 0.So, a is equal to g(x) must be negative as well.So, if a is less than -e, meaning more negative, then y = a is below the maximum of g(x), which is -e. So, in that case, the horizontal line y = a would intersect the graph of g(x) twice: once on the increasing part (x < 1) and once on the decreasing part (x > 1). If a is equal to -e, it only touches at x = 1. If a is greater than -e, meaning less negative, then y = a does not intersect g(x) at all because the maximum of g(x) is -e, which is lower than a.Wait, that seems contradictory to my earlier thought. Let me clarify.Since g(x) is negative everywhere, and its maximum is -e. So, if a is less than -e, meaning a is more negative, then y = a is below the maximum of g(x). So, the graph of g(x) goes from -∞, rises to -e, then falls back to -∞. So, if a is less than -e, the line y = a would intersect the graph twice: once on the way up (x < 1) and once on the way down (x > 1). If a is equal to -e, it only intersects once at x = 1. If a is greater than -e (i.e., closer to zero), then y = a doesn't intersect the graph because the graph never reaches that high.Therefore, to have two solutions for x > 0, a must be less than -e. So, a ∈ (-∞, -e).But wait, the options are A: (-∞, -1/e), B: (-∞, -e), C: (e, +∞), D: (1/e, +∞). So, B is (-∞, -e), which matches our conclusion.But let me verify again because sometimes it's easy to get confused with the signs.So, g(x) = -e^x / x. So, for x > 0, g(x) is negative. The maximum value is at x = 1, which is -e. So, the graph is a curve that starts at -∞, peaks at -e when x = 1, and then goes back to -∞ as x increases.Therefore, for a horizontal line y = a to intersect this graph twice, a must be less than -e because that's the peak. If a is less than -e, the line will cut through the graph once on the left side of the peak and once on the right side.If a is equal to -e, it just touches the peak. If a is greater than -e, meaning closer to zero, the line doesn't intersect the graph at all.Therefore, to have two solutions for x > 0, a must be less than -e. So, a ∈ (-∞, -e), which is option B.But just to make sure, let's think about the negative side. Since f(x) is even, the number of zeros for x < 0 will mirror those for x > 0. So, if there are two zeros for x > 0, there will be two zeros for x < 0, making a total of four zeros.Therefore, the correct range for a is (-∞, -e), which is option B.**Final Answer**boxed{B}

answer：Okay, so Didi is trying to raise money for her local soup kitchen, and she got some cakes from her family and friends. There are 10 cakes, and each cake is cut into 8 slices. She sells each slice for 1. That seems straightforward. But then, there are two business owners who want to help out by donating money for each slice she sells. One donates 50 cents per slice, and the other donates 25 cents per slice. So, I need to figure out how much money Didi raised in total by selling all the slices and including these donations.First, I should calculate the total number of slices. If there are 10 cakes and each cake is cut into 8 slices, then the total number of slices is 10 times 8. Let me write that down: 10 cakes * 8 slices per cake = 80 slices. Okay, so there are 80 slices in total.Next, Didi sells each slice for 1. So, if she sells all 80 slices, she'll make 80 * 1 = 80 from the sales alone. That's the first part of the money she raised.Now, there are two business owners contributing as well. The first business owner donates 50 cents for each slice sold. So, for each of the 80 slices, that's an additional 50 cents. Let me calculate that: 80 slices * 0.50 = 40. So, the first business owner donates 40.The second business owner donates a quarter for each slice sold. A quarter is 25 cents, so that's 80 slices * 0.25 = 20. So, the second business owner donates 20.Now, to find the total amount of money Didi raised, I need to add up all these amounts: the money from the sales, the donation from the first business owner, and the donation from the second business owner. That would be 80 + 40 + 20.Let me add those up step by step to make sure I don't make a mistake. First, 80 plus 40 is 120. Then, 120 plus 20 is 140. So, altogether, Didi raised 140.Wait, let me double-check my calculations to be sure. Total slices: 10 cakes * 8 slices = 80 slices. Sales: 80 * 1 = 80. First donation: 80 * 0.50 = 40. Second donation: 80 * 0.25 = 20. Total: 80 + 40 + 20 = 140. Yep, that seems correct.I think I've covered all the parts of the problem. She sold all the slices, and both business owners donated based on the number of slices sold. Adding up all the contributions gives the total amount raised. So, Didi successfully raised 140 for her soup kitchen.

answer：Okay, so I have this problem here with two sets, M and N. Set M is {a, 0}, and set N is defined as all integers x such that 2x² - 5x < 0. The question is asking, if the intersection of M and N is not empty, then what does a equal? The options are A: 1, B: 2, C: 1 or 2.5, and D: 1 or 2.Alright, let's break this down step by step. First, I need to understand what set N is. It's defined by the inequality 2x² - 5x < 0, and x has to be an integer. So, I need to solve this inequality to find all integer values of x that satisfy it.Let me write down the inequality: 2x² - 5x < 0. Hmm, this is a quadratic inequality. To solve it, I can factor it or find its roots. Let's try factoring first. I can factor out an x, so it becomes x(2x - 5) < 0. Okay, that looks better.Now, the roots of the equation x(2x - 5) = 0 are x = 0 and x = 5/2, which is 2.5. These roots divide the number line into intervals. To determine where the inequality holds, I need to test each interval.The intervals are:1. x < 02. 0 < x < 2.53. x > 2.5Let me test each interval.First interval: x < 0. Let's pick x = -1. Plugging into the inequality: (-1)(2*(-1) - 5) = (-1)(-2 - 5) = (-1)(-7) = 7, which is greater than 0. So, the inequality doesn't hold here.Second interval: 0 < x < 2.5. Let's pick x = 1. Plugging in: 1*(2*1 - 5) = 1*(2 - 5) = 1*(-3) = -3, which is less than 0. So, the inequality holds here.Third interval: x > 2.5. Let's pick x = 3. Plugging in: 3*(2*3 - 5) = 3*(6 - 5) = 3*(1) = 3, which is greater than 0. So, the inequality doesn't hold here.Therefore, the solution to the inequality 2x² - 5x < 0 is 0 < x < 2.5. But since x has to be an integer, the possible integer values in this interval are x = 1 and x = 2. So, set N is {1, 2}.Now, set M is {a, 0}. We need to find a such that the intersection of M and N is not empty. That means there has to be at least one common element between M and N.Set N is {1, 2}, and set M is {a, 0}. The only common element could be either 1 or 2 because 0 is not in N. So, for M ∩ N to be non-empty, a must be either 1 or 2.Looking at the options, option D is "1 or 2," which matches our conclusion.Wait, let me double-check. If a were 2.5, that's not an integer, and since N only contains integers, 2.5 wouldn't be in N. So, option C is out. Options A and B only give one value each, but since a can be either 1 or 2, D is the correct choice.Yeah, I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be D: 1 or 2.

answer：First, determine the total number of girls by subtracting the number of boys from the total number of people at the track meet.Next, calculate the number of girls with long hair by finding three fifths of the total number of girls.Finally, subtract the number of girls with long hair from the total number of girls to find the number of girls with short hair.