answer：Okay, so I have this problem here: For how many integers ( x ) is ( |15x^2 - 32x - 28| ) a prime number? The options are from a to e, with e being "None of above." Hmm, let me try to figure this out step by step.First, I know that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So, the absolute value of the quadratic expression ( 15x^2 - 32x - 28 ) must be a prime number. That means ( |15x^2 - 32x - 28| ) should be equal to some prime number ( p ), where ( p ) is prime.Since we're dealing with integers ( x ), I can try plugging in different integer values for ( x ) and see if the result is a prime number. But before I start plugging in numbers, maybe I can factor the quadratic expression or find its roots to understand its behavior better.Let me write down the quadratic: ( 15x^2 - 32x - 28 ). I wonder if this can be factored. Let me try to factor it. To factor ( 15x^2 - 32x - 28 ), I need two numbers that multiply to ( 15 times (-28) = -420 ) and add up to -32.Hmm, let's see. The factors of 420 are numerous, but let me think of pairs that multiply to 420 and see if their difference is 32. Wait, actually, since the product is negative, one number is positive and the other is negative. So, I need two numbers whose product is -420 and sum is -32.Let me list the factor pairs of 420:1 and 4202 and 2103 and 1404 and 1055 and 846 and 707 and 6010 and 4212 and 3514 and 3015 and 2820 and 21Now, considering the negative counterparts as well. Let's see which pair subtracts to give 32. Wait, actually, since the sum is -32, one number is larger in magnitude and negative, and the other is positive.Let me try 14 and -30: 14 + (-30) = -16, which is not -32.How about 20 and -21: 20 + (-21) = -1, nope.Wait, maybe 28 and -15: 28 + (-15) = 13, nope.Wait, 35 and -12: 35 + (-12) = 23, nope.Wait, 42 and -10: 42 + (-10) = 32. Oh, but we need the sum to be -32. So, if I take -42 and 10: (-42) + 10 = -32. Yes, that works.So, the two numbers are -42 and 10. Therefore, I can rewrite the middle term using these numbers:( 15x^2 - 42x + 10x - 28 )Now, let's factor by grouping:Group the first two terms and the last two terms:( (15x^2 - 42x) + (10x - 28) )Factor out the greatest common factor (GCF) from each group:From the first group, GCF of 15x^2 and -42x is 3x:( 3x(5x - 14) )From the second group, GCF of 10x and -28 is 2:( 2(5x - 14) )So, now we have:( 3x(5x - 14) + 2(5x - 14) )Now, factor out the common binomial factor ( (5x - 14) ):( (5x - 14)(3x + 2) )So, the quadratic factors as ( (5x - 14)(3x + 2) ). Therefore, the expression inside the absolute value is ( |(5x - 14)(3x + 2)| ).Since we're taking the absolute value, it's the same as ( |5x - 14| times |3x + 2| ). For this product to be a prime number, one of the factors must be 1 or -1, and the other must be a prime or its negative counterpart. Because primes have only two positive divisors: 1 and themselves.So, let me set up the equations:Case 1: ( |5x - 14| = 1 ) and ( |3x + 2| ) is prime.Case 2: ( |3x + 2| = 1 ) and ( |5x - 14| ) is prime.Let me handle each case separately.**Case 1: ( |5x - 14| = 1 )**This equation can be rewritten as two possibilities:1a. ( 5x - 14 = 1 ) => ( 5x = 15 ) => ( x = 3 )1b. ( 5x - 14 = -1 ) => ( 5x = 13 ) => ( x = 13/5 )But since ( x ) must be an integer, ( x = 13/5 ) is not an integer, so only ( x = 3 ) is a solution here.Now, let's compute ( |3x + 2| ) when ( x = 3 ):( |3*3 + 2| = |9 + 2| = |11| = 11 ), which is a prime number. So, ( x = 3 ) is a valid solution.**Case 2: ( |3x + 2| = 1 )**Similarly, this gives two possibilities:2a. ( 3x + 2 = 1 ) => ( 3x = -1 ) => ( x = -1/3 )2b. ( 3x + 2 = -1 ) => ( 3x = -3 ) => ( x = -1 )Again, ( x = -1/3 ) is not an integer, so only ( x = -1 ) is a candidate.Now, compute ( |5x - 14| ) when ( x = -1 ):( |5*(-1) - 14| = |-5 -14| = |-19| = 19 ), which is a prime number. So, ( x = -1 ) is also a valid solution.So, from both cases, we have two integer solutions: ( x = 3 ) and ( x = -1 ). Therefore, it seems like the answer is 2, which is option c.But wait, hold on. I should check if there are any other possibilities where the product ( |5x -14| times |3x + 2| ) is prime without one of the factors being 1. Because primes have only two positive divisors, 1 and themselves, so unless one of the factors is 1 or -1, the product can't be prime. Because if both factors are greater than 1 in absolute value, their product would be composite.But let me double-check. Suppose both ( |5x -14| ) and ( |3x + 2| ) are primes, but neither is 1. Then their product would be a product of two primes, which is a semiprime, not a prime. So, the only way the product is prime is if one of them is 1 or -1.Therefore, I think I have considered all possible cases. So, only ( x = 3 ) and ( x = -1 ) satisfy the condition. Therefore, there are 2 integers ( x ) for which ( |15x^2 -32x -28| ) is prime.Wait, just to be thorough, maybe I should test some other integer values around these solutions to see if any others might work.Let me try ( x = 0 ):( |15*0 - 32*0 -28| = |-28| = 28 ), which is not prime.( x = 1 ):( |15 -32 -28| = |-45| = 45 ), not prime.( x = 2 ):( |15*4 -32*2 -28| = |60 -64 -28| = |-32| = 32, not prime.( x = 4 ):( |15*16 -32*4 -28| = |240 -128 -28| = |84| = 84, not prime.( x = -2 ):( |15*4 -32*(-2) -28| = |60 +64 -28| = |96| = 96, not prime.( x = -3 ):( |15*9 -32*(-3) -28| = |135 +96 -28| = |203| = 203, which is 7*29, not prime.So, none of these other integer values near the solutions give a prime number. So, I think my initial conclusion is correct: only ( x = 3 ) and ( x = -1 ) work.Therefore, the number of integers ( x ) is 2, which is option c.**Final Answer**boxed{C}

answer：Alright, so I have this geometry problem here. It says that in triangle ABC, side AB is divided into 4 equal segments, so AB1 = B1B2 = B2B3 = B3B. Similarly, side AC is divided into 5 equal segments, so AC1 = C1C2 = C2C3 = C3C4 = C4C. The question is asking by how many times the area of triangle ABC is greater than the sum of the areas of the triangles C1B1C2, C2B2C3, C3B3C4, and C4BC.Hmm, okay. Let me visualize this. Triangle ABC, with AB divided into four equal parts and AC divided into five equal parts. So, points B1, B2, B3 are on AB, each segment equal. Similarly, points C1, C2, C3, C4 are on AC, each segment equal.The triangles mentioned are C1B1C2, C2B2C3, C3B3C4, and C4BC. So, each of these triangles is formed by connecting points on AB and AC. For example, the first triangle is connecting C1 on AC, B1 on AB, and C2 on AC. Wait, no, actually, C1 is on AC, B1 is on AB, and C2 is on AC? Or is it C2 on AC and B1 on AB? Wait, no, C1B1C2 would be a triangle with vertices at C1, B1, and C2. So, C1 is on AC, B1 is on AB, and C2 is on AC. So, that triangle is formed by two points on AC and one point on AB.Similarly, C2B2C3 is another triangle with two points on AC and one on AB, and so on. The last triangle is C4BC, which is connecting C4 on AC, B on AB, and C on AC. So, that's a triangle with two points on AC and one on AB as well.So, all these triangles have two vertices on AC and one on AB, except maybe the last one, which connects to point B and C. Wait, point C is the vertex of the original triangle, so C4BC is a triangle with vertices at C4, B, and C.I need to find the sum of the areas of these four triangles and then compare it to the area of ABC. The question is asking by how many times the area of ABC is greater than this sum.So, first, maybe I should find the coordinates of these points to calculate the areas. Let me assign coordinates to triangle ABC to make it easier. Let me place point A at (0, 0), point B at (4, 0) since AB is divided into 4 equal segments, so each segment is 1 unit. Similarly, point C can be at (0, 5) since AC is divided into 5 equal segments, each of length 1 unit. Wait, but actually, the lengths might not be 1 unit each, but the divisions are equal. So, if AB is divided into 4 equal segments, each segment is AB/4, and AC is divided into 5 equal segments, each segment is AC/5.But for simplicity, maybe I can assign coordinates such that AB is along the x-axis from (0,0) to (4,0), and AC is along the y-axis from (0,0) to (0,5). Then, point B is at (4,0), point C is at (0,5). Then, points B1, B2, B3 are at (1,0), (2,0), (3,0). Similarly, points C1, C2, C3, C4 are at (0,1), (0,2), (0,3), (0,4).Wait, but then triangle ABC is a right triangle with legs of length 4 and 5, so its area is (4*5)/2 = 10.Now, let's find the coordinates of the triangles mentioned:1. Triangle C1B1C2: C1 is (0,1), B1 is (1,0), C2 is (0,2).2. Triangle C2B2C3: C2 is (0,2), B2 is (2,0), C3 is (0,3).3. Triangle C3B3C4: C3 is (0,3), B3 is (3,0), C4 is (0,4).4. Triangle C4BC: C4 is (0,4), B is (4,0), C is (0,5).So, each of these triangles has vertices at two points on the y-axis (AC) and one point on the x-axis (AB), except the last one, which connects to point B and C.Now, let's compute the area of each triangle.Starting with triangle C1B1C2: points (0,1), (1,0), (0,2).I can use the shoelace formula to compute the area.Formula is: For points (x1,y1), (x2,y2), (x3,y3),Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|So, plugging in:x1=0, y1=1; x2=1, y2=0; x3=0, y3=2.Area = |0*(0 - 2) + 1*(2 - 1) + 0*(1 - 0)| / 2= |0 + 1*(1) + 0| / 2= |1| / 2 = 0.5So, area is 0.5.Next, triangle C2B2C3: points (0,2), (2,0), (0,3).Using the same formula:x1=0, y1=2; x2=2, y2=0; x3=0, y3=3.Area = |0*(0 - 3) + 2*(3 - 2) + 0*(2 - 0)| / 2= |0 + 2*(1) + 0| / 2= |2| / 2 = 1.So, area is 1.Third triangle: C3B3C4: points (0,3), (3,0), (0,4).Applying the formula:x1=0, y1=3; x2=3, y2=0; x3=0, y3=4.Area = |0*(0 - 4) + 3*(4 - 3) + 0*(3 - 0)| / 2= |0 + 3*(1) + 0| / 2= |3| / 2 = 1.5.So, area is 1.5.Fourth triangle: C4BC: points (0,4), (4,0), (0,5).Using the formula:x1=0, y1=4; x2=4, y2=0; x3=0, y3=5.Area = |0*(0 - 5) + 4*(5 - 4) + 0*(4 - 0)| / 2= |0 + 4*(1) + 0| / 2= |4| / 2 = 2.So, area is 2.Now, summing up the areas:0.5 + 1 + 1.5 + 2 = 5.So, the total area of the four triangles is 5.The area of triangle ABC is 10, so 10 is greater than 5 by 2 times.Wait, but hold on. The problem says "by how many times is the area of triangle ABC greater than the sum of the areas...". So, 10 is 2 times greater than 5? Wait, actually, 10 is twice as big as 5, so it's 2 times greater. But sometimes, people interpret "times greater" differently, but in mathematical terms, it's usually multiplicative. So, 10 is 2 times 5, so it's 2 times greater.But let me double-check my calculations because sometimes when I do shoelace formula, I might have messed up.First triangle: C1B1C2: (0,1), (1,0), (0,2). Shoelace:List the points in order: (0,1), (1,0), (0,2), (0,1).Compute sum of x_i y_{i+1}: 0*0 + 1*2 + 0*1 = 0 + 2 + 0 = 2.Compute sum of y_i x_{i+1}: 1*1 + 0*0 + 2*0 = 1 + 0 + 0 = 1.Area = |2 - 1| / 2 = 0.5. Correct.Second triangle: C2B2C3: (0,2), (2,0), (0,3).Shoelace:Points: (0,2), (2,0), (0,3), (0,2).Sum x_i y_{i+1}: 0*0 + 2*3 + 0*2 = 0 + 6 + 0 = 6.Sum y_i x_{i+1}: 2*2 + 0*0 + 3*0 = 4 + 0 + 0 = 4.Area = |6 - 4| / 2 = 1. Correct.Third triangle: C3B3C4: (0,3), (3,0), (0,4).Shoelace:Points: (0,3), (3,0), (0,4), (0,3).Sum x_i y_{i+1}: 0*0 + 3*4 + 0*3 = 0 + 12 + 0 = 12.Sum y_i x_{i+1}: 3*3 + 0*0 + 4*0 = 9 + 0 + 0 = 9.Area = |12 - 9| / 2 = 1.5. Correct.Fourth triangle: C4BC: (0,4), (4,0), (0,5).Shoelace:Points: (0,4), (4,0), (0,5), (0,4).Sum x_i y_{i+1}: 0*0 + 4*5 + 0*4 = 0 + 20 + 0 = 20.Sum y_i x_{i+1}: 4*4 + 0*0 + 5*0 = 16 + 0 + 0 = 16.Area = |20 - 16| / 2 = 4 / 2 = 2. Correct.So, the areas are indeed 0.5, 1, 1.5, 2, which sum to 5. The area of ABC is 10, so 10 is twice as large as 5. So, the area of ABC is 2 times greater than the sum of those four triangles.But wait, hold on. Let me think again. Is there another way to approach this problem without coordinates? Maybe using ratios or similar triangles?Since AB is divided into 4 equal parts and AC into 5 equal parts, perhaps we can use the concept of similar triangles or area ratios.Each of these smaller triangles is formed by connecting points on AB and AC. So, for example, triangle C1B1C2 is formed by points that are 1/5 along AC and 1/4 along AB.In general, when you have two points dividing sides of a triangle proportionally, the area of the triangle formed can be found using the formula: area = (base ratio * height ratio) * area of original triangle.But in this case, each of these triangles is not similar to ABC, but they are smaller triangles within ABC.Wait, perhaps we can think in terms of determinants or vectors, but since I already solved it with coordinates, maybe that's sufficient.Alternatively, maybe using the concept of dividing the sides proportionally and then using the formula for the area of a triangle with vertices on the sides.Wait, another approach is to consider that each of these triangles is a trapezoid minus some areas, but I think the coordinate method was straightforward.But just to confirm, let me see if there's a pattern in the areas.Looking at the areas: 0.5, 1, 1.5, 2. Each subsequent triangle has an area that increases by 0.5. So, the areas are in an arithmetic sequence with a common difference of 0.5.So, sum of these areas is n/2 * (first term + last term). Here, n=4, first term=0.5, last term=2.Sum = 4/2 * (0.5 + 2) = 2 * 2.5 = 5. Which matches what I got earlier.So, the total area is 5, and ABC is 10, so the ratio is 2.Therefore, the area of ABC is 2 times greater than the sum of the areas of those four triangles.Wait, but just to make sure, is there a possibility that I missed something? For example, is the triangle C4BC correctly calculated?Point C4 is at (0,4), point B is at (4,0), and point C is at (0,5). So, the triangle is between (0,4), (4,0), and (0,5). So, it's a triangle with base along AC from (0,4) to (0,5), which is 1 unit, and then a line from (0,5) to (4,0) and back to (0,4). So, the area is 2, as calculated.Alternatively, maybe I can think of the entire figure as being divided into smaller regions, and the sum of these four triangles is 5, which is half of ABC's area. So, ABC is twice as big.Alternatively, maybe I can use vectors or parametric equations, but since I already have the coordinate method working, I think that's solid.So, in conclusion, the area of triangle ABC is 2 times greater than the sum of the areas of those four triangles.**Final Answer**The area of triangle (ABC) is greater than the sum of the areas of the given triangles by a factor of boxed{2}.

answer：Alright, so I've got two expressions to compute here. Let me take them one by one. I'll start with the first one.**Problem (1):** Compute the expression (frac{1}{ sqrt{2}-1}-left(frac{3}{5} right)^{0}+left(frac{9}{4} right)^{-0.5}+ sqrt[4]{( sqrt{2}-e)^{4}}).Hmm, okay. Let me break this down term by term.First term: (frac{1}{ sqrt{2}-1}). I remember that when you have a denominator with a radical, it's often helpful to rationalize it. So, I can multiply the numerator and denominator by the conjugate of the denominator, which is (sqrt{2} + 1). Let me do that:[frac{1}{ sqrt{2}-1} times frac{sqrt{2} + 1}{sqrt{2} + 1} = frac{sqrt{2} + 1}{(sqrt{2})^2 - (1)^2} = frac{sqrt{2} + 1}{2 - 1} = sqrt{2} + 1]Okay, so the first term simplifies to (sqrt{2} + 1).Second term: (-left(frac{3}{5} right)^{0}). Any number to the power of 0 is 1, so this term is just (-1).Third term: (left(frac{9}{4} right)^{-0.5}). Negative exponents mean reciprocals, so this is the same as (left(frac{4}{9}right)^{0.5}). The 0.5 exponent is the square root. So, let's compute that:[sqrt{frac{4}{9}} = frac{sqrt{4}}{sqrt{9}} = frac{2}{3}]So, the third term is (frac{2}{3}).Fourth term: (sqrt[4]{( sqrt{2}-e)^{4}}). The fourth root of something to the fourth power is the absolute value of that something because even roots return non-negative results. So, this simplifies to (| sqrt{2} - e |). Now, since (e) is approximately 2.718 and (sqrt{2}) is about 1.414, so (sqrt{2} - e) is negative. Therefore, the absolute value is (e - sqrt{2}).Putting it all together:First term: (sqrt{2} + 1)Second term: (-1)Third term: (frac{2}{3})Fourth term: (e - sqrt{2})So, let's add them up:[(sqrt{2} + 1) - 1 + frac{2}{3} + (e - sqrt{2})]Let me simplify step by step.First, (sqrt{2} + 1 - 1) simplifies to (sqrt{2}).Then, adding (frac{2}{3}) gives (sqrt{2} + frac{2}{3}).Then, adding (e - sqrt{2}) gives (sqrt{2} + frac{2}{3} + e - sqrt{2}).The (sqrt{2}) and (-sqrt{2}) cancel each other out, leaving us with (frac{2}{3} + e).So, the entire expression simplifies to (e + frac{2}{3}).Wait, let me double-check that. So, step by step:1. (frac{1}{sqrt{2} - 1} = sqrt{2} + 1). Correct.2. (-left(frac{3}{5}right)^0 = -1). Correct.3. (left(frac{9}{4}right)^{-0.5} = left(frac{4}{9}right)^{0.5} = frac{2}{3}). Correct.4. (sqrt[4]{(sqrt{2} - e)^4} = | sqrt{2} - e | = e - sqrt{2}). Correct.Adding them all:[(sqrt{2} + 1) - 1 + frac{2}{3} + (e - sqrt{2}) = sqrt{2} + 1 - 1 + frac{2}{3} + e - sqrt{2}]Simplify term by term:- (sqrt{2} - sqrt{2} = 0)- (1 - 1 = 0)- So, we are left with (frac{2}{3} + e)Yes, that seems right. So, the result is (e + frac{2}{3}).But wait, is that the case? Let me compute the numerical value to verify.Compute each term numerically:1. (frac{1}{sqrt{2} - 1}):(sqrt{2} approx 1.4142), so denominator is 1.4142 - 1 = 0.4142. So, 1 / 0.4142 ≈ 2.4142.2. (-left(frac{3}{5}right)^0 = -1).3. (left(frac{9}{4}right)^{-0.5} = left(frac{4}{9}right)^{0.5} ≈ 0.6667).4. (sqrt[4]{(sqrt{2} - e)^4}):Compute (sqrt{2} - e ≈ 1.4142 - 2.718 ≈ -1.3038). The fourth power is positive, and the fourth root is the absolute value, so 1.3038.So, adding them up:2.4142 - 1 + 0.6667 + 1.3038 ≈ 2.4142 - 1 = 1.4142; 1.4142 + 0.6667 ≈ 2.0809; 2.0809 + 1.3038 ≈ 3.3847.Now, compute (e + frac{2}{3}):(e ≈ 2.718), so (2.718 + 0.6667 ≈ 3.3847). That matches. So, yes, the expression simplifies to (e + frac{2}{3}).Alright, so problem (1) is done. Now, moving on to problem (2).**Problem (2):** Compute the expression (lg 500 + lg frac{8}{5} - frac{1}{2}lg 64 + 50 times (lg 2 + lg 5)^2).Hmm, logarithms. Let me recall that (lg) is base 10 logarithm. So, I can use logarithm properties to simplify this expression.Let me write down each term:1. (lg 500)2. (lg frac{8}{5})3. (-frac{1}{2}lg 64)4. (50 times (lg 2 + lg 5)^2)Let me handle each term step by step.First, let's combine the first three terms using logarithm properties.Recall that (lg a + lg b = lg(ab)) and (c lg a = lg(a^c)).So, let's compute the sum of the first three terms:[lg 500 + lg frac{8}{5} - frac{1}{2}lg 64]Let me rewrite the subtraction as addition of a negative:[lg 500 + lg frac{8}{5} + lg 64^{-1/2}]Because (-frac{1}{2}lg 64 = lg 64^{-1/2}).So, now we can combine all three terms:[lg 500 + lg frac{8}{5} + lg 64^{-1/2} = lg left(500 times frac{8}{5} times 64^{-1/2}right)]Compute the product inside the logarithm.First, compute each part:500 × (8/5) = 500 × 1.6 = 800.Then, 64^{-1/2} is the same as 1 / 64^{1/2} = 1 / 8.So, 800 × (1/8) = 100.Therefore, the first three terms simplify to (lg 100).Since (lg 100 = 2), because 10^2 = 100.Okay, so the first three terms add up to 2.Now, moving on to the fourth term: (50 times (lg 2 + lg 5)^2).Let me compute (lg 2 + lg 5) first.(lg 2 + lg 5 = lg(2 times 5) = lg 10 = 1).So, the fourth term becomes (50 times (1)^2 = 50 times 1 = 50).Therefore, the entire expression is:First three terms: 2Fourth term: 50So, total expression: 2 + 50 = 52.Wait, is that correct? Let me double-check.Compute each step:First three terms:[lg 500 + lg frac{8}{5} - frac{1}{2}lg 64]Convert to multiplication:500 × (8/5) = 800.64^{-1/2} = 1/8.So, 800 × 1/8 = 100.(lg 100 = 2). Correct.Fourth term:(lg 2 + lg 5 = lg 10 = 1), so squared is 1, multiplied by 50 is 50. Correct.Therefore, total is 2 + 50 = 52.Wait, but let me compute each term numerically to verify.Compute each term numerically:1. (lg 500): (lg 500 ≈ 2.69897)2. (lg frac{8}{5} = lg 1.6 ≈ 0.20412)3. (-frac{1}{2}lg 64): (lg 64 ≈ 1.80618), so half is ≈ 0.90309, negative is ≈ -0.903094. (50 times (lg 2 + lg 5)^2): (lg 2 ≈ 0.3010), (lg 5 ≈ 0.6990), sum is ≈ 1.0, squared is 1.0, multiplied by 50 is 50.Now, add all four terms:2.69897 + 0.20412 - 0.90309 + 50 ≈First, 2.69897 + 0.20412 ≈ 2.90309Then, 2.90309 - 0.90309 ≈ 2.0Then, 2.0 + 50 = 52.0Yes, that matches. So, the expression simplifies to 52.So, summarizing:Problem (1) simplifies to (e + frac{2}{3}), and Problem (2) simplifies to 52.**Final Answer**(1) (boxed{e + dfrac{2}{3}})(2) (boxed{52})

answer：First, I will calculate the product of -7 and 45, which equals -315.Next, I will calculate the product of -7 and 55, which equals -385.Finally, I will add the two results together: -315 plus -385 equals -700.

answer：Okay, so I have this problem about a cube, and I need to find the number of different 3-edge trips from point A to point B. The cube is shown in the Asymptote figure, but since I can't see the image, I'll assume it's a standard cube with vertices labeled A and B. The problem mentions that the shortest trip from A to B is along 3 edges, so I need to figure out how many such paths exist.First, I should visualize the cube. A cube has 8 vertices and 12 edges. Each vertex is connected to three other vertices. If A and B are opposite corners of the cube (i.e., they are diagonally opposite), then the shortest path between them would indeed be 3 edges long because you can't get from one corner to the opposite corner in fewer than 3 moves along the edges.So, let me think about how to count the number of such paths. Each path consists of moving along three edges, changing direction at each vertex. Since the cube is a regular structure, I can use symmetry to help me count.One approach is to model the cube as a graph, where each vertex is a node and each edge is a connection between two nodes. Then, the problem reduces to finding the number of shortest paths from A to B in this graph.In graph theory, the number of shortest paths between two nodes can be found using breadth-first search (BFS), but since the cube is a well-known structure, I can probably figure it out without getting too technical.Let me consider the cube's structure. Each vertex has three neighbors. Starting from A, I can go to any of the three adjacent vertices. From each of those, I can go to two new vertices (since one edge leads back to A, which we don't want to revisit in the shortest path). Then, from each of those two, there should be one edge leading to B.Wait, is that correct? Let me think step by step.1. From A, there are 3 possible edges to take. Let's call these edges E1, E2, E3.2. From each of these adjacent vertices, say after taking E1, we arrive at vertex V1. From V1, we can go to two other vertices, say V1a and V1b, because one edge leads back to A, which we shouldn't take if we're trying to get to B in the shortest path.3. From V1a, how many ways can we get to B? Similarly, from V1b, how many ways?Wait, maybe it's better to think in terms of coordinates. Let me assign coordinates to the cube's vertices. Let's say A is at (0,0,0) and B is at (1,1,1). Each edge corresponds to moving one unit along one of the axes.So, to get from (0,0,0) to (1,1,1) in three moves, each move must be along a different axis. So, the path is a permutation of moving along x, y, and z axes.But wait, in the cube, moving along an edge can be either in the positive or negative direction, but since we're going from (0,0,0) to (1,1,1), we only move in the positive direction along each axis.So, the number of shortest paths is the number of permutations of the sequence of moves along x, y, and z. Since each path is a sequence of three moves, each along a different axis, the number of such paths is 3! = 6.But wait, hold on. Is that the case? Because in the cube, each move is along an edge, but depending on the cube's structure, sometimes you might have to make turns that aren't just along the axes.Wait, no. If we model the cube as a graph where each vertex is connected to three others, then moving along an edge is equivalent to changing one coordinate from 0 to 1 or 1 to 0. So, starting at (0,0,0), each move flips one coordinate. To get to (1,1,1), we need to flip each coordinate exactly once. The number of ways to do this is the number of permutations of the three coordinates, which is 3! = 6.Therefore, there are 6 shortest paths from A to B.But wait, let me verify this because sometimes in cube problems, the number of paths can be different. For example, sometimes people count paths that might revisit edges or vertices, but in this case, since we're looking for the shortest path, which doesn't revisit any vertices, the count should be 6.Alternatively, I can think of it as a combinatorial problem. Each shortest path corresponds to a sequence of three moves, each along a different axis. The number of such sequences is 3! = 6.But to be thorough, let me try to enumerate them.Starting at A (0,0,0):1. Move along x-axis to (1,0,0), then y-axis to (1,1,0), then z-axis to (1,1,1).2. Move along x-axis to (1,0,0), then z-axis to (1,0,1), then y-axis to (1,1,1).3. Move along y-axis to (0,1,0), then x-axis to (1,1,0), then z-axis to (1,1,1).4. Move along y-axis to (0,1,0), then z-axis to (0,1,1), then x-axis to (1,1,1).5. Move along z-axis to (0,0,1), then x-axis to (1,0,1), then y-axis to (1,1,1).6. Move along z-axis to (0,0,1), then y-axis to (0,1,1), then x-axis to (1,1,1).So that's 6 different paths. Each corresponds to a permutation of the order in which we traverse the x, y, and z axes.Therefore, the number of different 3-edge trips from A to B is 6.Wait, but hold on a second. Is this considering all possible paths, or are there more?Wait, in the cube, each vertex is connected to three others, but when moving from A, you have three choices. From each of those, you have two choices (since you can't go back to A). Then, from each of those, you have only one choice to go to B.So, that would be 3 * 2 * 1 = 6, which matches the earlier result.Therefore, I think 6 is the correct answer.But just to make sure, let me think about another way. The cube is a bipartite graph with two sets of vertices: those with even number of 1s in their coordinates and those with odd. A is (0,0,0), which is even, and B is (1,1,1), which is odd. The shortest path from A to B must have length 3, which is odd, so it makes sense.In a bipartite graph, the number of shortest paths can be calculated using the number of ways to traverse from one partition to the other. But in this case, since it's a cube, the number is 6.Alternatively, I can think about the number of paths as the number of ways to arrange the three moves in different orders, which is 3 factorial, so 6.Therefore, I think the answer is 6.**Final Answer**The number of different 3-edge trips from A to B is boxed{6}.

answer：To determine the motorcyclist's average speed, I'll start by defining the total distance traveled as ( D ) kilometers. Since the motorcyclist travels the first half of the distance at 30 km/h and the second half at 60 km/h, each half of the journey is ( frac{D}{2} ) kilometers.Next, I'll calculate the time taken for each half of the trip. For the first half, the time is ( frac{frac{D}{2}}{30} = frac{D}{60} ) hours. For the second half, the time is ( frac{frac{D}{2}}{60} = frac{D}{120} ) hours.Adding these two times together gives the total time for the entire journey: ( frac{D}{60} + frac{D}{120} = frac{D}{40} ) hours.Finally, the average speed is calculated by dividing the total distance by the total time: ( frac{D}{frac{D}{40}} = 40 ) km/h.