answer：Okay, so I've got this problem here where Al, Betty, and Clare split 1000 among themselves. Each of them starts with a different amount. After one year, the total amount they have together is 1500. Betty and Clare both doubled their money, while Al lost 100. I need to figure out how much Al originally had. Hmm, let me try to break this down step by step.First, let's denote the original amounts each person had. Let me assign variables to make it easier. Let's say:- Let A be the amount Al originally had.- Let B be the amount Betty originally had.- Let C be the amount Clare originally had.From the problem, we know that the total amount they split is 1000. So, we can write the equation:A + B + C = 1000That's straightforward. Now, after one year, the total amount they have is 1500. The problem states that Betty and Clare both doubled their money, while Al lost 100. So, let's express their amounts after one year.- Al's amount after one year: A - 100 (since he lost 100)- Betty's amount after one year: 2B (since she doubled her money)- Clare's amount after one year: 2C (since she also doubled her money)So, the total amount after one year is:(A - 100) + 2B + 2C = 1500Now, I have two equations:1. A + B + C = 10002. (A - 100) + 2B + 2C = 1500I need to solve these equations to find A, which is Al's original portion.Let me simplify the second equation first. Let's expand it:A - 100 + 2B + 2C = 1500Now, let's rearrange it:A + 2B + 2C = 1500 + 100A + 2B + 2C = 1600So, now we have:1. A + B + C = 10002. A + 2B + 2C = 1600Hmm, maybe I can subtract the first equation from the second to eliminate A and find a relationship between B and C.Subtracting equation 1 from equation 2:(A + 2B + 2C) - (A + B + C) = 1600 - 1000Simplify the left side:A - A + 2B - B + 2C - C = 600Which simplifies to:B + C = 600So, B + C = 600But from equation 1, we know that A + B + C = 1000. If B + C = 600, then A must be:A = 1000 - (B + C) = 1000 - 600 = 400Wait, so does that mean Al originally had 400? Let me check if that makes sense.If A is 400, then B + C is 600. After one year, Al has 400 - 100 = 300. Betty and Clare each doubled their money, so Betty has 2B and Clare has 2C. The total after one year is 300 + 2B + 2C.But 2B + 2C is 2*(B + C) = 2*600 = 1200. So, 300 + 1200 = 1500, which matches the given total. So, that seems to check out.But wait, the problem says that each of them began with a different amount. So, A, B, and C are all different. If A is 400, then B + C is 600, but we don't know the exact values of B and C. But since they are different from each other and from A, that's okay as long as B ≠ C and neither is 400.But let me think, is there a way to find the exact values of B and C? The problem doesn't ask for them, just for A, so maybe we don't need to. But just to be thorough, let's see if we can find B and C.We have B + C = 600, but without more information, we can't find exact values for B and C. However, the problem doesn't specify any constraints on how much more or less Betty and Clare had compared to each other, just that they are different from each other and from Al. So, as long as B and C are different and not equal to 400, it's fine.Wait, but let me double-check my equations because sometimes when subtracting equations, I might have made a mistake.Original equations:1. A + B + C = 10002. A + 2B + 2C = 1600Subtracting equation 1 from equation 2:(A + 2B + 2C) - (A + B + C) = 1600 - 1000Which is:0A + B + C = 600So, yes, that's correct. So, B + C = 600, which leads to A = 400.Therefore, Al's original portion was 400.But just to make sure, let me plug the numbers back into the original problem.Al starts with 400, Betty and Clare together start with 600. After one year, Al has 300, Betty has 2B, and Clare has 2C. The total is 300 + 2B + 2C.Since B + C = 600, 2B + 2C = 1200. So, 300 + 1200 = 1500, which matches the total given.Also, each started with a different amount. So, A is 400, B and C are different from each other and from 400. For example, if B is 200, then C is 400, but wait, C can't be 400 because A is already 400. So, actually, B and C must each be different from 400 and from each other.So, if B + C = 600, and neither B nor C is 400, then possible values could be, for example, B = 250 and C = 350. Then, all three amounts are different: 400, 250, 350.Alternatively, B = 300 and C = 300, but that would mean B and C are the same, which is not allowed because each begins with a different amount. So, B and C must be different, but both not equal to 400.Therefore, as long as B and C are different and sum to 600, and neither is 400, the conditions are satisfied.So, I think my solution is correct. Al's original portion was 400.**Final Answer**Al's original portion was boxed{400} dollars.

answer：Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. I remember that in a geometric sequence, each term is multiplied by a common ratio to get the next term. So, if the first term is ( a_1 ), then the second term is ( a_1 times r ), the third term is ( a_1 times r^2 ), and so on. The problem says that ( a_1 + a_2 = 20 ). Let me write that down:( a_1 + a_2 = 20 )Since it's a geometric sequence, ( a_2 = a_1 times r ). So, substituting that in, we get:( a_1 + a_1 r = 20 )I can factor out ( a_1 ):( a_1 (1 + r) = 20 ) ...(1)Okay, that's the first equation. Now, the problem also gives me ( a_3 + a_4 = 40 ). Let me write that down too:( a_3 + a_4 = 40 )Again, using the geometric sequence formula, ( a_3 = a_1 r^2 ) and ( a_4 = a_1 r^3 ). So, substituting these in:( a_1 r^2 + a_1 r^3 = 40 )Factor out ( a_1 r^2 ):( a_1 r^2 (1 + r) = 40 ) ...(2)Hmm, now I have two equations: equation (1) is ( a_1 (1 + r) = 20 ) and equation (2) is ( a_1 r^2 (1 + r) = 40 ). Maybe I can divide equation (2) by equation (1) to eliminate ( a_1 (1 + r) ). Let me try that.Dividing equation (2) by equation (1):( frac{a_1 r^2 (1 + r)}{a_1 (1 + r)} = frac{40}{20} )Simplify the left side: ( a_1 ) and ( (1 + r) ) cancel out, leaving ( r^2 ). On the right side, 40 divided by 20 is 2. So:( r^2 = 2 )Therefore, ( r = sqrt{2} ) or ( r = -sqrt{2} ). Hmm, so the common ratio could be positive or negative square root of 2. I need to consider both possibilities.Let me first consider ( r = sqrt{2} ). Then, from equation (1):( a_1 (1 + sqrt{2}) = 20 )So, solving for ( a_1 ):( a_1 = frac{20}{1 + sqrt{2}} )I can rationalize the denominator by multiplying numerator and denominator by ( 1 - sqrt{2} ):( a_1 = frac{20 (1 - sqrt{2})}{(1 + sqrt{2})(1 - sqrt{2})} )The denominator becomes ( 1 - (sqrt{2})^2 = 1 - 2 = -1 ). So,( a_1 = frac{20 (1 - sqrt{2})}{-1} = -20 (1 - sqrt{2}) = -20 + 20 sqrt{2} )Wait, that's a negative first term. Is that possible? Well, geometric sequences can have negative terms, so I guess that's okay. Let me just note that down.Now, let's check the other possibility where ( r = -sqrt{2} ). Then, equation (1) becomes:( a_1 (1 - sqrt{2}) = 20 )Solving for ( a_1 ):( a_1 = frac{20}{1 - sqrt{2}} )Again, rationalizing the denominator by multiplying numerator and denominator by ( 1 + sqrt{2} ):( a_1 = frac{20 (1 + sqrt{2})}{(1 - sqrt{2})(1 + sqrt{2})} )Denominator is ( 1 - 2 = -1 ), so:( a_1 = frac{20 (1 + sqrt{2})}{-1} = -20 (1 + sqrt{2}) = -20 - 20 sqrt{2} )Hmm, so in both cases, ( a_1 ) is negative. That's fine, but let me see if that affects the rest of the problem.The question is asking for ( a_5 + a_6 ). Let me express these terms in terms of ( a_1 ) and ( r ).( a_5 = a_1 r^4 )( a_6 = a_1 r^5 )So, ( a_5 + a_6 = a_1 r^4 + a_1 r^5 = a_1 r^4 (1 + r) )Hmm, interesting. So, similar to equation (2), which was ( a_1 r^2 (1 + r) = 40 ). So, if I can find ( a_1 r^4 (1 + r) ), that will give me ( a_5 + a_6 ).Alternatively, maybe I can find a pattern or ratio between the sums.Looking back, equation (1): ( a_1 (1 + r) = 20 )Equation (2): ( a_1 r^2 (1 + r) = 40 )So, equation (2) is equation (1) multiplied by ( r^2 ). So, 20 multiplied by ( r^2 ) is 40, which is consistent with ( r^2 = 2 ).So, if I continue this pattern, the next sum ( a_5 + a_6 ) would be equation (2) multiplied by ( r^2 ). Let me check:Equation (2): ( a_1 r^2 (1 + r) = 40 )Multiply both sides by ( r^2 ):( a_1 r^4 (1 + r) = 40 r^2 )But ( r^2 = 2 ), so:( a_1 r^4 (1 + r) = 40 times 2 = 80 )Therefore, ( a_5 + a_6 = 80 )Wait, so regardless of the value of ( r ), whether positive or negative square root of 2, ( r^2 ) is 2, so the multiplication by ( r^2 ) would still give 80. So, the answer is 80.But just to make sure, let me compute it using the value of ( a_1 ) I found earlier.First, let's take ( r = sqrt{2} ) and ( a_1 = -20 + 20 sqrt{2} ).Compute ( a_5 + a_6 ):( a_5 = a_1 r^4 = (-20 + 20 sqrt{2}) (sqrt{2})^4 )( (sqrt{2})^4 = (2)^{2} = 4 )So, ( a_5 = (-20 + 20 sqrt{2}) times 4 = (-80 + 80 sqrt{2}) )Similarly, ( a_6 = a_1 r^5 = (-20 + 20 sqrt{2}) (sqrt{2})^5 )( (sqrt{2})^5 = (2)^{2} times sqrt{2} = 4 sqrt{2} )So, ( a_6 = (-20 + 20 sqrt{2}) times 4 sqrt{2} )Let me compute that:First, distribute:( (-20)(4 sqrt{2}) + (20 sqrt{2})(4 sqrt{2}) )Compute each term:( -80 sqrt{2} + 80 times 2 ) because ( sqrt{2} times sqrt{2} = 2 )So, that's ( -80 sqrt{2} + 160 )Therefore, ( a_6 = 160 - 80 sqrt{2} )Now, adding ( a_5 + a_6 ):( (-80 + 80 sqrt{2}) + (160 - 80 sqrt{2}) )Combine like terms:-80 + 160 = 8080√2 - 80√2 = 0So, total is 80. Perfect, that's consistent.Now, let me check with the other value of ( r = -sqrt{2} ) and ( a_1 = -20 - 20 sqrt{2} ).Compute ( a_5 + a_6 ):( a_5 = a_1 r^4 = (-20 - 20 sqrt{2}) (-sqrt{2})^4 )( (-sqrt{2})^4 = (sqrt{2})^4 = 4 )So, ( a_5 = (-20 - 20 sqrt{2}) times 4 = (-80 - 80 sqrt{2}) )Similarly, ( a_6 = a_1 r^5 = (-20 - 20 sqrt{2}) (-sqrt{2})^5 )( (-sqrt{2})^5 = - (sqrt{2})^5 = -4 sqrt{2} )So, ( a_6 = (-20 - 20 sqrt{2}) times (-4 sqrt{2}) )Compute that:Multiply each term:( (-20)(-4 sqrt{2}) + (-20 sqrt{2})(-4 sqrt{2}) )Which is:( 80 sqrt{2} + 80 times 2 ) because ( sqrt{2} times sqrt{2} = 2 )So, that's ( 80 sqrt{2} + 160 )Therefore, ( a_6 = 160 + 80 sqrt{2} )Now, adding ( a_5 + a_6 ):( (-80 - 80 sqrt{2}) + (160 + 80 sqrt{2}) )Combine like terms:-80 + 160 = 80-80√2 + 80√2 = 0So, total is 80. Again, consistent.Therefore, regardless of whether ( r ) is positive or negative square root of 2, the sum ( a_5 + a_6 ) is 80.So, I think that's the answer. It makes sense because each pair of terms is doubling each time. From ( a_1 + a_2 = 20 ), then ( a_3 + a_4 = 40 ), which is double, and then ( a_5 + a_6 ) should be double of 40, which is 80. So, that's another way to see it.Yeah, I think 80 is the right answer.**Final Answer**The value of ( a_5 + a_6 ) is boxed{80}.

answer：Okay, so I need to figure out if this big sum is rational or irrational. The sum is:frac{1}{2 sqrt{1}+1 sqrt{2}}+frac{1}{3 sqrt{2}+2 sqrt{3}}+frac{1}{4 sqrt{3}+3 sqrt{4}}+cdots+frac{1}{100 sqrt{99}+99 sqrt{100}}Hmm, let's see. Each term in the sum looks similar. The general term seems to be:frac{1}{(n+1)sqrt{n} + nsqrt{n+1}}where n starts at 1 and goes up to 99. So, the first term is when n=1: 1/(2√1 + 1√2), which matches the given sum. The last term is when n=99: 1/(100√99 + 99√100), which also matches. So, that's the general term.Now, I need to figure out if the sum of all these terms is rational or irrational. I remember that sometimes when you have denominators with square roots, you can rationalize them. Maybe that will help here.Let me try rationalizing the denominator for a general term. So, take:frac{1}{(n+1)sqrt{n} + nsqrt{n+1}}To rationalize, I can multiply the numerator and denominator by the conjugate of the denominator. The conjugate would be:(n+1)sqrt{n} - nsqrt{n+1}So, multiplying numerator and denominator by that:frac{1 times [(n+1)sqrt{n} - nsqrt{n+1}]}{[(n+1)sqrt{n} + nsqrt{n+1}][(n+1)sqrt{n} - nsqrt{n+1}]}Let me compute the denominator first. It's a difference of squares, so:[(n+1)sqrt{n}]^2 - [nsqrt{n+1}]^2Calculating each part:First term: [(n+1)√n]^2 = (n+1)^2 * n = n(n+1)^2Second term: [n√(n+1)]^2 = n^2(n+1)So, denominator becomes:n(n+1)^2 - n^2(n+1) = n(n+1)[(n+1) - n] = n(n+1)(1) = n(n+1)So, denominator simplifies to n(n+1). Nice.Now, the numerator is:(n+1)√n - n√(n+1)So, putting it all together, the term becomes:frac{(n+1)sqrt{n} - nsqrt{n+1}}{n(n+1)}We can split this fraction into two parts:frac{(n+1)sqrt{n}}{n(n+1)} - frac{nsqrt{n+1}}{n(n+1)}Simplify each term:First term: (n+1) cancels out, so we have √n / n = 1/√nSecond term: n cancels out, so we have √(n+1)/(n+1) = 1/√(n+1)So, each term simplifies to:frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}}Wow, that's a telescoping series! That means when we add up all these terms from n=1 to n=99, most of the terms will cancel out.Let me write out the first few terms to see the pattern.For n=1:1/√1 - 1/√2For n=2:1/√2 - 1/√3For n=3:1/√3 - 1/√4...For n=99:1/√99 - 1/√100So, when we add all these up, the -1/√2 from the first term cancels with the +1/√2 from the second term. Similarly, the -1/√3 from the second term cancels with the +1/√3 from the third term, and so on. This pattern continues all the way to the end.So, after all the cancellations, what's left is the first term of the first expression and the last term of the last expression. That is:1/√1 - 1/√100Simplify that:1/1 - 1/10 = 1 - 1/10 = 9/10So, the entire sum simplifies to 9/10.Wait, 9/10 is a rational number because it can be expressed as a fraction of two integers. So, the sum is rational.But let me double-check my steps to make sure I didn't make a mistake.1. I identified the general term correctly: 1/[(n+1)√n + n√(n+1)].2. I rationalized the denominator by multiplying numerator and denominator by the conjugate. That step seems correct.3. Calculated the denominator as n(n+1), which seems right.4. Split the fraction into two terms: 1/√n - 1/√(n+1). That also looks correct.5. Recognized it as a telescoping series, which makes sense because each negative term cancels the next positive term.6. Calculated the sum as 1 - 1/10 = 9/10. That seems correct.So, all steps check out. Therefore, the sum is 9/10, which is rational.**Final Answer**The number is rational, and the final answer is boxed{dfrac{9}{10}}.

answer：Okay, so I have this problem here about a right triangle ABC, where angle C is 90 degrees. They tell me that BC is 6 units long, and the sine of angle A is 3/4. I need to find the length of AB. Hmm, let's see. I remember that in a right triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. So, maybe I can use that to figure out the lengths of the sides.First, let me visualize the triangle. Since angle C is 90 degrees, sides AB, BC, and AC form a right triangle with AB as the hypotenuse. So, angle A is one of the acute angles, and angle B is the other. They've given me BC, which is one of the legs, and the sine of angle A. I think I can use the sine function to find another side, maybe AC or AB.Wait, sine of angle A is opposite over hypotenuse. In this case, the side opposite angle A would be BC, right? Because in triangle ABC, angle A is at vertex A, so the side opposite to it is BC. So, if sin A = 3/4, then BC / AB = 3/4. Because BC is opposite angle A and AB is the hypotenuse.So, plugging in the values I have: BC is 6, so 6 / AB = 3/4. Hmm, so I can solve for AB. Let me write that down:6 / AB = 3 / 4To solve for AB, I can cross-multiply:3 * AB = 6 * 43AB = 24Then, divide both sides by 3:AB = 24 / 3AB = 8Wait, so AB is 8 units long? That seems straightforward. Let me just verify if that makes sense. If AB is 8, then AC can be found using the Pythagorean theorem, right? So, AC squared plus BC squared equals AB squared.So, AC² + 6² = 8²AC² + 36 = 64Subtract 36 from both sides:AC² = 64 - 36AC² = 28So, AC is the square root of 28, which simplifies to 2 times the square root of 7. That seems okay. Let me check if the sine of angle A is indeed 3/4.Angle A is at vertex A, so the opposite side is BC = 6, and the hypotenuse is AB = 8. So, sin A = 6 / 8, which simplifies to 3/4. Yes, that matches the given information. So, that checks out.Alternatively, I could have used the cosine of angle A to find AC, but since they gave me the sine, it was more straightforward to use that. Let me just recap what I did:1. Recognized that in a right triangle, sine of an angle is opposite over hypotenuse.2. Identified that BC is opposite angle A, so sin A = BC / AB.3. Plugged in the known values: 3/4 = 6 / AB.4. Solved for AB by cross-multiplying and dividing.5. Verified the result using the Pythagorean theorem to ensure consistency.I think that covers all the steps. I don't see any mistakes in my reasoning, so I feel confident that AB is indeed 8 units long.**Final Answer**The length of ( AB ) is boxed{8}.

answer：Alright, so I have this problem here: given the equation ( a^5 - a^4b - a^4 + a - b - 1 = 0 ) and another equation ( 2a - 3b = 1 ). I need to find the value of ( a^3 + b^3 ). Hmm, okay, let me try to figure out how to approach this.First, I notice that there are two equations and two variables, so it should be possible to solve for ( a ) and ( b ) and then compute ( a^3 + b^3 ). The second equation is linear, which might be easier to handle, so maybe I can express one variable in terms of the other and substitute into the first equation.Looking at the second equation: ( 2a - 3b = 1 ). Let me solve for ( a ) in terms of ( b ). So, adding ( 3b ) to both sides gives ( 2a = 3b + 1 ), and then dividing both sides by 2, we get ( a = frac{3b + 1}{2} ). Okay, so now I can substitute this expression for ( a ) into the first equation.Substituting ( a = frac{3b + 1}{2} ) into the first equation: ( left( frac{3b + 1}{2} right)^5 - left( frac{3b + 1}{2} right)^4 b - left( frac{3b + 1}{2} right)^4 + left( frac{3b + 1}{2} right) - b - 1 = 0 ). Wow, that looks complicated. Maybe expanding all of that isn't the best approach. It might get too messy with all the exponents.Alternatively, perhaps I can factor the first equation to simplify it before substituting. Let me look at the first equation again: ( a^5 - a^4b - a^4 + a - b - 1 = 0 ). Hmm, I notice that the first three terms have ( a^4 ) in them. Let me factor ( a^4 ) out of those terms: ( a^4(a - b - 1) + a - b - 1 = 0 ). Oh, wait, that's interesting. So, factoring ( a^4 ) from the first three terms gives ( a^4(a - b - 1) ), and then the remaining terms are ( a - b - 1 ). So, actually, both parts have a common factor of ( (a - b - 1) ).So, factoring ( (a - b - 1) ) out, we get: ( (a - b - 1)(a^4 + 1) = 0 ). That's a much simpler expression! So, the first equation simplifies to ( (a - b - 1)(a^4 + 1) = 0 ). Therefore, either ( a - b - 1 = 0 ) or ( a^4 + 1 = 0 ).But ( a^4 + 1 = 0 ) would imply that ( a^4 = -1 ). However, since ( a ) is a real number (I assume, because the problem doesn't specify otherwise), ( a^4 ) is always non-negative, so ( a^4 + 1 ) is always positive. Therefore, ( a^4 + 1 = 0 ) has no real solutions. So, the only possibility is ( a - b - 1 = 0 ), which simplifies to ( a = b + 1 ).Alright, so now I have two equations: ( a = b + 1 ) and ( 2a - 3b = 1 ). Let me substitute ( a = b + 1 ) into the second equation.Substituting into ( 2a - 3b = 1 ): ( 2(b + 1) - 3b = 1 ). Let's expand that: ( 2b + 2 - 3b = 1 ). Combining like terms: ( -b + 2 = 1 ). Subtracting 2 from both sides: ( -b = -1 ), so ( b = 1 ).Now that I have ( b = 1 ), I can substitute back into ( a = b + 1 ) to find ( a ): ( a = 1 + 1 = 2 ). So, ( a = 2 ) and ( b = 1 ).Now, the question asks for ( a^3 + b^3 ). Let's compute that: ( 2^3 + 1^3 = 8 + 1 = 9 ). So, the value of ( a^3 + b^3 ) is 9.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the first equation, I factored it correctly to ( (a - b - 1)(a^4 + 1) = 0 ), which makes sense because ( a^4 + 1 ) can't be zero for real ( a ). So, ( a = b + 1 ). Then substituting into the second equation, I correctly solved for ( b ) and found ( b = 1 ), leading to ( a = 2 ). Then, computing ( a^3 + b^3 ) gives 9. That seems correct.Alternatively, just to be thorough, I can plug ( a = 2 ) and ( b = 1 ) back into the original equations to ensure they satisfy both.First equation: ( 2^5 - 2^4(1) - 2^4 + 2 - 1 - 1 ). Calculating each term:- ( 2^5 = 32 )- ( 2^4(1) = 16 )- ( 2^4 = 16 )- ( 2 = 2 )- ( -1 = -1 )- ( -1 = -1 )So, putting it all together: ( 32 - 16 - 16 + 2 - 1 - 1 = 32 - 16 = 16; 16 - 16 = 0; 0 + 2 = 2; 2 - 1 = 1; 1 - 1 = 0 ). So, the first equation is satisfied.Second equation: ( 2(2) - 3(1) = 4 - 3 = 1 ), which is correct. So, both equations are satisfied with ( a = 2 ) and ( b = 1 ). Therefore, my solution seems solid.Thus, the value of ( a^3 + b^3 ) is indeed 9.**Final Answer**The value of ( a^3 + b^3 ) is boxed{9}.

answer：Okay, so I have this problem about two wheels rotating, and each rotation results in two pointers pointing to two different numbers. I need to find the probability that the sum of these two numbers is even. The options are given from (A) to (E), with (C) being 1/2, which is a common probability, but I need to figure it out step by step.First, I should visualize what's happening. There are two wheels, each with numbers, and when they rotate, each wheel points to a number. The sum of these two numbers needs to be even. So, probability is about favorable outcomes over total possible outcomes. I need to figure out how many possible outcomes there are and how many of those result in an even sum.But wait, the problem doesn't specify how many numbers are on each wheel or what the numbers are. Hmm, that's a bit confusing. Maybe it's a standard problem where each wheel has numbers from 1 to some number, but since it's not specified, perhaps it's a general case or maybe the wheels have the same numbers? Or maybe each wheel has numbers from 1 to 7? Wait, looking at the answer choices, the denominators are 6, 7, 2, 3, 7. So maybe 7 is involved. Maybe each wheel has numbers 1 through 7? Or maybe one wheel has 6 numbers and the other has 7? Hmm, not sure.Wait, let me think again. The problem says each rotation results in two pointers pointing to two different numbers. So, does that mean each wheel has numbers, and each wheel is equally likely to point to any number? So, if each wheel has, say, n numbers, then the total number of possible outcomes is n squared, right? Because for each number on the first wheel, there are n possibilities on the second wheel.But without knowing n, how can I compute the probability? Maybe the wheels have numbers such that half are even and half are odd? Or maybe it's a different distribution. Wait, perhaps the wheels have numbers from 1 to 7? Let me check the answer choices: (E) is 5/7, which is a bit more than half. Hmm.Alternatively, maybe each wheel has numbers 1 through 6? Then the total outcomes would be 36. But 36 isn't in the denominators of the answer choices. The denominators are 6, 7, 2, 3, 7. So maybe one wheel has 6 numbers and the other has 7? Or maybe each wheel has numbers such that the number of even and odd numbers is different.Wait, perhaps the wheels are like roulette wheels with numbers, but again, without specific info, it's hard. Maybe the problem is more about the parity of the numbers rather than their specific values. So, maybe it's about whether each number is even or odd, regardless of their actual values.So, if I can figure out the number of even and odd numbers on each wheel, I can compute the probability. Let me assume that each wheel has an equal number of even and odd numbers. If that's the case, then the probability that a number is even is 1/2, and the same for odd.But wait, if both wheels have equal numbers of even and odd, then the probability that their sum is even would be the probability that both are even or both are odd. So, that would be (1/2 * 1/2) + (1/2 * 1/2) = 1/4 + 1/4 = 1/2. So, probability 1/2, which is option (C). But wait, is that the case?But hold on, the problem says "each rotation results in two pointers pointing to two different numbers." So, does that mean that the two numbers must be different? Or does it just mean that each wheel points to a number, which could be the same or different? Hmm, the wording is a bit unclear.Wait, if it's two different numbers, that might change things. So, if the two numbers have to be different, then the total number of possible outcomes is not n squared, but n squared minus n, because we exclude the cases where both pointers point to the same number. Hmm, but without knowing n, it's tricky.Alternatively, maybe the wheels have numbers such that each wheel has numbers from 1 to 7, with some even and some odd. Let me think: numbers 1 to 7 have four odd numbers (1,3,5,7) and three even numbers (2,4,6). So, if each wheel has numbers 1 to 7, then each wheel has 4 odd and 3 even numbers.So, if each wheel has 7 numbers, the total number of possible outcomes is 7*7=49. But since the pointers point to two different numbers, does that mean we have to exclude the cases where both pointers point to the same number? So, 49 total, minus 7 same-number cases, so 42 possible outcomes.But wait, the problem says "each rotation results in two pointers pointing to two different numbers." So, maybe each outcome is two different numbers, so the total number of possible outcomes is 7*6=42, since for the first wheel, 7 choices, and for the second wheel, 6 choices (excluding the number already chosen by the first wheel). So, total 42.But then, how many of these 42 result in an even sum? So, the sum is even if both numbers are even or both are odd. So, let's compute the number of ways to pick two different numbers where both are even or both are odd.First, number of even numbers on each wheel: 3 (2,4,6). Number of odd numbers: 4 (1,3,5,7).So, number of ways to pick two different even numbers: 3 choices for the first wheel, then 2 choices for the second wheel (since it has to be different). So, 3*2=6.Similarly, number of ways to pick two different odd numbers: 4 choices for the first wheel, then 3 choices for the second wheel. So, 4*3=12.So, total favorable outcomes: 6 + 12 = 18.Total possible outcomes: 42.So, probability is 18/42, which simplifies to 3/7. So, that's option (B). Hmm, but wait, is that correct?Wait, hold on. When we compute the number of ways to pick two different even numbers, is it 3*2=6? Or is it 3 choices for the first wheel and 2 for the second, but since the wheels are distinguishable, meaning that (even on first wheel, even on second wheel) is different from (even on second wheel, even on first wheel). But in the total outcomes, we have 7*6=42, which already accounts for order.Wait, actually, in the total possible outcomes, order matters because the two wheels are separate. So, when I compute the favorable outcomes, I also need to consider order.So, for two different even numbers: first wheel can be even in 3 ways, second wheel can be even in 2 ways (since it has to be different). So, 3*2=6.Similarly, for two different odd numbers: first wheel can be odd in 4 ways, second wheel can be odd in 3 ways. So, 4*3=12.Total favorable: 6+12=18.Total possible: 42.So, 18/42=3/7, which is option (B). So, that seems to be the answer.But wait, let me think again. Is the total number of possible outcomes 42? Because if the two wheels are independent, meaning that each can point to any number regardless of the other, then the total number of outcomes is 7*7=49. But the problem says "two different numbers," so does that mean that the two numbers must be different? If so, then total outcomes are 49 - 7=42.But if the problem doesn't specify that the numbers must be different, then the total outcomes would be 49, and the favorable outcomes would be the number of pairs where both are even or both are odd, regardless of being the same number.Wait, but the problem says "each rotation results in two pointers pointing to two different numbers." So, that implies that the two numbers are different. So, the total number of possible outcomes is 42.Therefore, the probability is 18/42=3/7, which is option (B).But wait, another way to think about it: the probability that the sum is even is equal to the probability that both numbers are even or both are odd.Since the two numbers are different, the probability that both are even is (number of ways to choose two different evens) divided by total number of different pairs.Similarly, probability both are odd is (number of ways to choose two different odds) divided by total.So, number of ways to choose two different evens: 3*2=6.Number of ways to choose two different odds: 4*3=12.Total favorable: 18.Total possible: 42.So, 18/42=3/7.Alternatively, if the numbers didn't have to be different, then total outcomes would be 49, and number of favorable would be (3*3)+(4*4)=9+16=25. So, 25/49≈0.51, which is roughly 1/2, but not exactly. But 25/49 is approximately 0.51, which is close to 1/2, but not exactly.But in this case, since the numbers must be different, it's 18/42=3/7≈0.428, which is less than 1/2.Wait, but let me think again. Is the probability that the sum is even equal to the probability that both are even or both are odd. So, if the two numbers are independent, then the probability is P(both even) + P(both odd). If the numbers are dependent (i.e., they have to be different), then it's a bit different.Wait, maybe another approach: the probability that the sum is even is equal to the probability that both numbers are even or both are odd. So, if the two numbers are independent, then the probability is (number of evens / total)^2 + (number of odds / total)^2.But in this case, the numbers are dependent because they have to be different. So, the probability that the first number is even is 3/7, then the probability that the second number is also even is 2/6=1/3. So, P(both even)= (3/7)*(2/6)= (3/7)*(1/3)=1/7.Similarly, P(both odd)= (4/7)*(3/6)= (4/7)*(1/2)=2/7.So, total P=1/7 + 2/7=3/7.Yes, that's another way to get 3/7.So, that confirms it. So, the probability is 3/7, which is option (B).But just to make sure, let me think about another scenario. Suppose each wheel has numbers 1 to 6. So, 3 even and 3 odd. Then, if the numbers must be different, total outcomes would be 6*5=30.Number of ways to pick two different evens: 3*2=6.Number of ways to pick two different odds: 3*2=6.Total favorable:12.Probability:12/30=2/5, which is 0.4.But in our case, since one wheel has more odd numbers, the probability is higher, 3/7≈0.428.So, yeah, 3/7 is correct.Alternatively, if the wheels had different numbers, say one wheel has 1-4 and the other has 1-5, but the problem doesn't specify, so I think the assumption is that both wheels have numbers 1-7, each with 4 odd and 3 even numbers.Therefore, the probability is 3/7, which is option (B).**Final Answer**boxed{B}