answer：Okay, so I have this problem here: I need to find the difference between the sum of the first 1500 even counting numbers and the sum of the first 1500 odd counting numbers. The options are A) 1000, B) 1500, C) 2000, D) 2500, and E) 3000. Hmm, let's try to figure this out step by step.First, I remember that counting numbers are just the numbers we use to count things, like 1, 2, 3, and so on. So, the first 1500 odd counting numbers would be 1, 3, 5, ..., up to the 1500th odd number. Similarly, the first 1500 even counting numbers would be 2, 4, 6, ..., up to the 1500th even number.I think the key here is to find the sum of each sequence and then subtract the smaller sum from the larger one to find the difference. Let me recall how to find the sum of an arithmetic series because both odd and even numbers form arithmetic sequences.An arithmetic sequence has a common difference between consecutive terms. For odd numbers, the difference is 2, and for even numbers, the difference is also 2. The formula for the sum of the first n terms of an arithmetic sequence is:[ S_n = frac{n}{2} times (a_1 + a_n) ]where ( S_n ) is the sum, ( n ) is the number of terms, ( a_1 ) is the first term, and ( a_n ) is the nth term.So, I need to find the sum of the first 1500 odd numbers and the sum of the first 1500 even numbers using this formula.Let's start with the odd numbers. The first term ( a_1 ) is 1, and the common difference ( d ) is 2. To find the 1500th term, ( a_{1500} ), I can use the formula for the nth term of an arithmetic sequence:[ a_n = a_1 + (n - 1) times d ]Plugging in the values:[ a_{1500} = 1 + (1500 - 1) times 2 ][ a_{1500} = 1 + 1499 times 2 ][ a_{1500} = 1 + 2998 ][ a_{1500} = 2999 ]So, the 1500th odd number is 2999. Now, let's find the sum of the first 1500 odd numbers:[ S_{text{odd}} = frac{1500}{2} times (1 + 2999) ][ S_{text{odd}} = 750 times 3000 ][ S_{text{odd}} = 2,250,000 ]Okay, that's the sum of the odd numbers. Now, let's do the same for the even numbers. The first term ( a_1 ) is 2, and the common difference ( d ) is also 2. Let's find the 1500th term:[ a_{1500} = 2 + (1500 - 1) times 2 ][ a_{1500} = 2 + 1499 times 2 ][ a_{1500} = 2 + 2998 ][ a_{1500} = 3000 ]So, the 1500th even number is 3000. Now, let's find the sum of the first 1500 even numbers:[ S_{text{even}} = frac{1500}{2} times (2 + 3000) ][ S_{text{even}} = 750 times 3002 ][ S_{text{even}} = 750 times 3000 + 750 times 2 ][ S_{text{even}} = 2,250,000 + 1,500 ][ S_{text{even}} = 2,251,500 ]Alright, so the sum of the even numbers is 2,251,500, and the sum of the odd numbers is 2,250,000. To find the difference between these two sums:[ Delta S = S_{text{even}} - S_{text{odd}} ][ Delta S = 2,251,500 - 2,250,000 ][ Delta S = 1,500 ]So, the difference is 1,500. Looking back at the options, that's option B.Wait, let me double-check my calculations to make sure I didn't make a mistake.For the odd numbers:- First term: 1- 1500th term: 2999- Sum: 750 * (1 + 2999) = 750 * 3000 = 2,250,000For the even numbers:- First term: 2- 1500th term: 3000- Sum: 750 * (2 + 3000) = 750 * 3002 = 2,251,500Difference: 2,251,500 - 2,250,000 = 1,500Yes, that seems correct. So, the difference is indeed 1,500, which is option B.

answer：Okay, so I have this problem here where four positive integers, when divided by 15, leave remainders of 11, 12, 13, and 14 respectively. I need to find the remainder when the sum of these four integers is divided by 15. Hmm, let me try to figure this out step by step.First, I remember that when a number is divided by another number, say 15, the remainder is what's left after you subtract the largest multiple of 15 that fits into it. So, if a number leaves a remainder of 11 when divided by 15, that means it can be expressed as 15k + 11, where k is some integer. Similarly, the other numbers can be written in the same way with their respective remainders.So, let me denote the four integers as a, b, c, and d. Based on the problem, I can write:a = 15k₁ + 11 b = 15k₂ + 12 c = 15k₃ + 13 d = 15k₄ + 14 Here, k₁, k₂, k₃, and k₄ are integers representing how many times 15 fits into each number before the remainder.Now, the problem asks for the remainder when the sum of these four integers is divided by 15. So, I need to find (a + b + c + d) mod 15.Let me write out the sum:a + b + c + d = (15k₁ + 11) + (15k₂ + 12) + (15k₃ + 13) + (15k₄ + 14)If I simplify this, I can group the terms with 15 together and the remainders together:= 15k₁ + 15k₂ + 15k₃ + 15k₄ + 11 + 12 + 13 + 14 = 15(k₁ + k₂ + k₃ + k₄) + (11 + 12 + 13 + 14)So, the sum is equal to 15 times some integer plus the sum of the remainders. Now, when I take this sum modulo 15, the term 15(k₁ + k₂ + k₃ + k₄) will be divisible by 15, so it will leave no remainder. Therefore, the remainder will solely depend on the sum of the remainders.Let me calculate the sum of the remainders:11 + 12 + 13 + 14 = ?Let's add them step by step:11 + 12 = 23 23 + 13 = 36 36 + 14 = 50So, the sum of the remainders is 50. Now, I need to find 50 mod 15, which is the remainder when 50 is divided by 15.To find this, I can perform the division:15 × 3 = 45 50 - 45 = 5So, 50 divided by 15 is 3 with a remainder of 5. Therefore, 50 mod 15 is 5.Putting it all together, the remainder when the sum of the four integers is divided by 15 is 5.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Adding the remainders: 11 + 12 is 23, plus 13 is 36, plus 14 is indeed 50. Dividing 50 by 15: 15 × 3 is 45, subtracting that from 50 gives 5. Yep, that seems correct.I think I've got it right. The key was recognizing that the multiples of 15 in each number don't affect the remainder when the sum is divided by 15, so I only needed to add up the remainders and then find that sum modulo 15.

answer：First, the rope is folded into 3 equal parts, creating 3 segments.Next, the rope is folded in half again, doubling the number of segments to 6.Finally, cutting the rope in the middle will divide each of the 6 segments into 2 parts, resulting in a total of 12 segments.

answer：First, I need to convert the repeating decimals (0.overline{54}) and (0.overline{18}) into fractions. For (0.overline{54}), I recognize that the repeating part has two digits, so I can write it as (frac{54}{99}).Similarly, (0.overline{18}) can be expressed as (frac{18}{99}).Next, I set up the division of these two fractions:[frac{frac{54}{99}}{frac{18}{99}}]When dividing fractions, I multiply by the reciprocal of the denominator:[frac{54}{99} times frac{99}{18}]The (99) in the numerator and denominator cancel out, simplifying the expression to:[frac{54}{18}]Finally, I simplify (frac{54}{18}) by dividing both the numerator and the denominator by their greatest common divisor, which is 18:[frac{54 div 18}{18 div 18} = frac{3}{1} = 3]

answer：Alright, so I have this problem about a triangle with sides ( n + 8 ), ( 2n + 5 ), and ( 5n - 4 ). I need to find the number of possible integer values for ( n ) such that all sides are unequal and ( n ) is a positive integer. Hmm, okay, let's break this down step by step.First, I remember that for any triangle, the sum of any two sides must be greater than the third side. This is called the triangle inequality theorem. So, I need to apply this theorem to the given side lengths. Also, since all sides must be unequal, I have to make sure that none of the expressions ( n + 8 ), ( 2n + 5 ), and ( 5n - 4 ) are equal to each other.Let me start by checking the distinctness of the sides. I'll set each pair of sides equal to each other and solve for ( n ) to see if there are any values that make them equal.1. ( n + 8 = 2n + 5 ) Subtract ( n ) from both sides: ( 8 = n + 5 ) Subtract 5 from both sides: ( n = 3 ) So, if ( n = 3 ), the first and second sides are equal. But the problem states all sides are unequal, so ( n ) cannot be 3.2. ( n + 8 = 5n - 4 ) Subtract ( n ) from both sides: ( 8 = 4n - 4 ) Add 4 to both sides: ( 12 = 4n ) Divide both sides by 4: ( n = 3 ) Again, ( n = 3 ) makes the first and third sides equal, which is not allowed.3. ( 2n + 5 = 5n - 4 ) Subtract ( 2n ) from both sides: ( 5 = 3n - 4 ) Add 4 to both sides: ( 9 = 3n ) Divide both sides by 3: ( n = 3 ) Once more, ( n = 3 ) causes two sides to be equal. So, definitely, ( n ) cannot be 3.Alright, so ( n ) cannot be 3. Now, moving on to the triangle inequalities. I need to ensure that the sum of any two sides is greater than the third side. Let's write down all three inequalities.1. ( (n + 8) + (2n + 5) > (5n - 4) )2. ( (n + 8) + (5n - 4) > (2n + 5) )3. ( (2n + 5) + (5n - 4) > (n + 8) )Let's solve each inequality one by one.Starting with the first inequality:( (n + 8) + (2n + 5) > (5n - 4) )Combine like terms:( 3n + 13 > 5n - 4 )Subtract ( 3n ) from both sides:( 13 > 2n - 4 )Add 4 to both sides:( 17 > 2n )Divide both sides by 2:( 8.5 > n )So, ( n < 8.5 ). Since ( n ) is a positive integer, ( n ) can be at most 8.Now, the second inequality:( (n + 8) + (5n - 4) > (2n + 5) )Combine like terms:( 6n + 4 > 2n + 5 )Subtract ( 2n ) from both sides:( 4n + 4 > 5 )Subtract 4 from both sides:( 4n > 1 )Divide both sides by 4:( n > 0.25 )Since ( n ) is a positive integer, this means ( n geq 1 ).Third inequality:( (2n + 5) + (5n - 4) > (n + 8) )Combine like terms:( 7n + 1 > n + 8 )Subtract ( n ) from both sides:( 6n + 1 > 8 )Subtract 1 from both sides:( 6n > 7 )Divide both sides by 6:( n > frac{7}{6} )Which is approximately ( n > 1.1667 ). Since ( n ) is an integer, ( n geq 2 ).So, combining the results from the inequalities:- From the first inequality: ( n < 8.5 ) → ( n leq 8 )- From the second inequality: ( n > 0.25 ) → ( n geq 1 )- From the third inequality: ( n > frac{7}{6} ) → ( n geq 2 )So, the most restrictive condition is ( 2 leq n leq 8 ). But we also know that ( n ) cannot be 3 because that would make two sides equal. Therefore, the possible integer values for ( n ) are 2, 4, 5, 6, 7, and 8.Let me count them: 2, 4, 5, 6, 7, 8. That's six values.Wait, just to make sure I didn't miss anything, let me verify each value of ( n ) from 2 to 8, excluding 3, to ensure that all sides are positive and satisfy the triangle inequality.For ( n = 2 ):Sides: ( 2 + 8 = 10 ), ( 2*2 + 5 = 9 ), ( 5*2 - 4 = 6 )Check triangle inequality:10 + 9 > 6 → 19 > 6 ✔️10 + 6 > 9 → 16 > 9 ✔️9 + 6 > 10 → 15 > 10 ✔️All good.For ( n = 4 ):Sides: ( 4 + 8 = 12 ), ( 2*4 + 5 = 13 ), ( 5*4 - 4 = 16 )Check triangle inequality:12 + 13 > 16 → 25 > 16 ✔️12 + 16 > 13 → 28 > 13 ✔️13 + 16 > 12 → 29 > 12 ✔️All good.For ( n = 5 ):Sides: ( 5 + 8 = 13 ), ( 2*5 + 5 = 15 ), ( 5*5 - 4 = 21 )Check triangle inequality:13 + 15 > 21 → 28 > 21 ✔️13 + 21 > 15 → 34 > 15 ✔️15 + 21 > 13 → 36 > 13 ✔️All good.For ( n = 6 ):Sides: ( 6 + 8 = 14 ), ( 2*6 + 5 = 17 ), ( 5*6 - 4 = 26 )Check triangle inequality:14 + 17 > 26 → 31 > 26 ✔️14 + 26 > 17 → 40 > 17 ✔️17 + 26 > 14 → 43 > 14 ✔️All good.For ( n = 7 ):Sides: ( 7 + 8 = 15 ), ( 2*7 + 5 = 19 ), ( 5*7 - 4 = 31 )Check triangle inequality:15 + 19 > 31 → 34 > 31 ✔️15 + 31 > 19 → 46 > 19 ✔️19 + 31 > 15 → 50 > 15 ✔️All good.For ( n = 8 ):Sides: ( 8 + 8 = 16 ), ( 2*8 + 5 = 21 ), ( 5*8 - 4 = 36 )Check triangle inequality:16 + 21 > 36 → 37 > 36 ✔️16 + 36 > 21 → 52 > 21 ✔️21 + 36 > 16 → 57 > 16 ✔️All good.So, all these values satisfy the triangle inequality and ensure all sides are unequal. Therefore, the number of possible values for ( n ) is 6.**Final Answer**The number of possible values for ( n ) is boxed{6}.

answer：Okay, so I have this problem about finding the equations of ellipses that satisfy certain conditions. There are two parts: part (1) and part (2). Let me tackle them one by one.Starting with part (1): The ellipse is centered at the origin, with the coordinate axes as its axes of symmetry. It passes through two points, P₁(√6, 1) and P₂(-√3, -√2). Hmm, okay. So, since the ellipse is centered at the origin and symmetric about the coordinate axes, its general equation should be something like (x²/a²) + (y²/b²) = 1, right? Where a and b are the semi-major and semi-minor axes. But I'm not sure which one is which because the problem doesn't specify if it's a horizontal or vertical ellipse. Maybe I don't need to worry about that right now.The problem says it passes through P₁ and P₂, so I can plug these points into the equation to get two equations with two unknowns, a² and b². Let me write that down.For point P₁(√6, 1):( (√6)² ) / a² + (1²) / b² = 1Which simplifies to 6/a² + 1/b² = 1.For point P₂(-√3, -√2):( (-√3)² ) / a² + ( (-√2)² ) / b² = 1Which simplifies to 3/a² + 2/b² = 1.So now I have a system of two equations:1. 6/a² + 1/b² = 12. 3/a² + 2/b² = 1I need to solve for a² and b². Let me denote 1/a² as m and 1/b² as n for simplicity. So the equations become:1. 6m + n = 12. 3m + 2n = 1Now I can solve this system. Let's use substitution or elimination. Maybe elimination is easier here. If I multiply the first equation by 2, I get:12m + 2n = 2Now subtract the second equation from this:(12m + 2n) - (3m + 2n) = 2 - 1Which simplifies to 9m = 1, so m = 1/9.Now plug m back into the first equation:6*(1/9) + n = 1Which is 2/3 + n = 1So n = 1 - 2/3 = 1/3.Therefore, m = 1/9 and n = 1/3. Remembering that m = 1/a² and n = 1/b², so:1/a² = 1/9 => a² = 91/b² = 1/3 => b² = 3So the equation of the ellipse is (x²)/9 + (y²)/3 = 1. That seems straightforward. Let me just check if both points satisfy this equation.For P₁(√6, 1):( (√6)² )/9 + (1²)/3 = 6/9 + 1/3 = 2/3 + 1/3 = 1. Good.For P₂(-√3, -√2):( (-√3)² )/9 + ( (-√2)² )/3 = 3/9 + 2/3 = 1/3 + 2/3 = 1. Perfect.So part (1) is done. The equation is (x²)/9 + (y²)/3 = 1.Moving on to part (2): The ellipse has the coordinate axes as its axes of symmetry, and its major axis is three times the length of its minor axis. It passes through point P(3, 0). Hmm, okay.First, the major axis is three times the minor axis. So if the major axis is along the x-axis, then 2a = 3*(2b), so a = 3b. Alternatively, if the major axis is along the y-axis, then 2a = 3*(2b), so a = 3b as well. Wait, but in that case, if the major axis is along the y-axis, then actually a would be the semi-major axis, so 2a = 3*(2b), so a = 3b. So regardless of the orientation, a = 3b.But wait, actually, if the major axis is three times the minor axis, then the length of the major axis is 3 times the length of the minor axis. So if the major axis is along the x-axis, then 2a = 3*(2b), so a = 3b. Similarly, if the major axis is along the y-axis, then 2a = 3*(2b), so a = 3b. So in both cases, a = 3b.But I need to figure out whether the major axis is along the x-axis or y-axis. Since the ellipse passes through point P(3, 0), which is on the x-axis, that might give a clue. If the major axis is along the x-axis, then the ellipse will extend to x = a, which is 3b. But since it passes through (3, 0), that would mean that a must be at least 3. Similarly, if the major axis is along the y-axis, then the ellipse would extend to y = a, but since it passes through (3, 0), which is on the x-axis, that would mean that b must be at least 3.Wait, let me think. If the major axis is along the x-axis, then the standard equation is (x²/a²) + (y²/b²) = 1, with a > b. Since it passes through (3, 0), plugging in, we get (9)/a² + 0 = 1, so a² = 9, so a = 3. Since a = 3b, then 3 = 3b => b = 1. So the equation would be (x²)/9 + y² = 1.Alternatively, if the major axis is along the y-axis, then the standard equation is (x²/b²) + (y²/a²) = 1, with a > b. Again, passing through (3, 0), we plug in: (9)/b² + 0 = 1, so b² = 9, so b = 3. Since a = 3b, then a = 9. So the equation would be (x²)/9 + (y²)/81 = 1.Wait, but in this case, if the major axis is along the y-axis, then a = 9, which is the semi-major axis, and b = 3, which is the semi-minor axis. So both cases are possible, depending on whether the major axis is along x or y.But does the problem specify whether the major axis is along x or y? Let me check. It just says the major axis is three times the length of the minor axis. So both possibilities exist. Therefore, there are two ellipses satisfying the conditions: one with major axis along x, and another with major axis along y.So for part (2), there are two equations:1. (x²)/9 + y² = 1 (major axis along x)2. (x²)/9 + (y²)/81 = 1 (major axis along y)Let me verify both equations with point P(3, 0).For the first equation: (3²)/9 + 0 = 9/9 + 0 = 1. Correct.For the second equation: (3²)/9 + 0 = 9/9 + 0 = 1. Correct.So both equations satisfy the condition of passing through (3, 0). Therefore, both are valid.Wait, but in the first case, a = 3, b = 1, so the major axis is along x, and in the second case, a = 9, b = 3, so major axis is along y. Both satisfy the condition that major axis is three times the minor axis.So, in conclusion, for part (2), there are two possible ellipses.But let me make sure I didn't make a mistake in interpreting the major axis. The major axis is three times the minor axis. So if major axis is along x, then 2a = 3*(2b) => a = 3b. Similarly, if major axis is along y, then 2a = 3*(2b) => a = 3b. So in both cases, a = 3b.But in the first case, when major axis is along x, a = 3, b = 1, so 2a = 6, 2b = 2, and 6 = 3*2, which is correct.In the second case, major axis is along y, a = 9, b = 3, so 2a = 18, 2b = 6, and 18 = 3*6, which is also correct.Therefore, both equations are valid.So, summarizing:Part (1): (x²)/9 + (y²)/3 = 1Part (2): Two possibilities:1. (x²)/9 + y² = 12. (x²)/9 + (y²)/81 = 1I think that's it. Let me just write them in the standard form with denominators.For part (1), it's already in standard form.For part (2), the first equation is (x²)/9 + y² = 1, and the second is (x²)/9 + (y²)/81 = 1.I think that's all.