🧠:<think>Alright, so I have this problem about a football league that uses a double round-robin format. That means each team plays every other team twice, right? Once at home and once away. Okay, so there are ( m ) teams participating, and after the tournament, they played a total of ( 9n^2 + 6n + 32 ) matches, where ( n ) is an integer. My task is to find the value of ( m ).First, I need to understand how many matches are played in a double round-robin tournament. Since each pair of teams plays twice, the total number of matches should be ( m(m - 1) ), because each team plays ( m - 1 ) other teams twice. Wait, no, actually, in a single round-robin, each pair plays once, so it's ( frac{m(m - 1)}{2} ) matches. For double round-robin, it should be double that, right? So that would be ( m(m - 1) ) matches in total.So, the total number of matches is ( m(m - 1) ), and according to the problem, this equals ( 9n^2 + 6n + 32 ). So, I can set up the equation:[m(m - 1) = 9n^2 + 6n + 32]Hmm, this is a quadratic in terms of ( m ), but it's also involving ( n ). Maybe I can rearrange this equation to solve for ( m ) in terms of ( n ), or vice versa.Let me write it as:[m^2 - m - (9n^2 + 6n + 32) = 0]This is a quadratic equation in ( m ). The standard form is ( am^2 + bm + c = 0 ), so here ( a = 1 ), ( b = -1 ), and ( c = -(9n^2 + 6n + 32) ).Using the quadratic formula:[m = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Plugging in the values:[m = frac{1 pm sqrt{1 + 4(9n^2 + 6n + 32)}}{2}]Simplify inside the square root:[m = frac{1 pm sqrt{1 + 36n^2 + 24n + 128}}{2}][m = frac{1 pm sqrt{36n^2 + 24n + 129}}{2}]Hmm, this square root needs to be an integer because ( m ) is the number of teams, which has to be an integer. So, ( 36n^2 + 24n + 129 ) must be a perfect square.Let me denote ( k = sqrt{36n^2 + 24n + 129} ), so ( k ) must be an integer.So, ( k^2 = 36n^2 + 24n + 129 )Let me see if I can rewrite this quadratic in ( n ) to complete the square or factor it somehow.Factor out 36 from the first two terms:[k^2 = 36left(n^2 + frac{24}{36}nright) + 129]Simplify:[k^2 = 36left(n^2 + frac{2}{3}nright) + 129]To complete the square inside the parentheses:The coefficient of ( n ) is ( frac{2}{3} ), so half of that is ( frac{1}{3} ), and squaring it gives ( frac{1}{9} ).So,[k^2 = 36left(n^2 + frac{2}{3}n + frac{1}{9}right) + 129 - 36 times frac{1}{9}]Simplify:[k^2 = 36left(n + frac{1}{3}right)^2 + 129 - 4][k^2 = 36left(n + frac{1}{3}right)^2 + 125]Hmm, this doesn't seem to help much because of the fraction. Maybe another approach.Alternatively, perhaps factor the quadratic in ( n ). Let me see:( 36n^2 + 24n + 129 - k^2 = 0 )Wait, maybe instead, I can think of ( k^2 - 36n^2 - 24n = 129 )Hmm, perhaps write ( k^2 - (6n)^2 - 24n = 129 )Wait, ( (6n)^2 = 36n^2 ), so[k^2 - (6n)^2 - 24n = 129]Not sure if that helps.Alternatively, maybe express ( k^2 = 36n^2 + 24n + 129 ) as ( k^2 = (6n + a)^2 + b ) and see if I can find integers ( a ) and ( b ).Let's try:( (6n + a)^2 = 36n^2 + 12a n + a^2 )Comparing with ( 36n^2 + 24n + 129 ):So, ( 12a = 24 ) implies ( a = 2 )Then, ( a^2 = 4 ), so:[(6n + 2)^2 = 36n^2 + 24n + 4]Thus,[k^2 = (6n + 2)^2 + 125]So,[k^2 - (6n + 2)^2 = 125]This is a difference of squares:[(k - (6n + 2))(k + (6n + 2)) = 125]Now, since ( k ) and ( n ) are integers, ( k - (6n + 2) ) and ( k + (6n + 2) ) must be integer factors of 125.The factors of 125 are:1 and 1255 and 25Negative factors as well: -1 and -125, -5 and -25.So, let's set up equations:Case 1:[k - (6n + 2) = 1k + (6n + 2) = 125]Adding both equations:[2k = 126 implies k = 63Subtracting the first from the second:[12n + 4 = 124 implies 12n = 120 implies n = 10So, n = 10Then, k = 63Now, recall that ( m = frac{1 pm k}{2} )Since ( m ) must be positive, we take the positive root:[m = frac{1 + 63}{2} = frac{64}{2} = 32So, m = 32Case 2:[k - (6n + 2) = 5k + (6n + 2) = 25]Adding both equations:[2k = 30 implies k = 15Subtracting:[12n + 4 = 20 implies 12n = 16 implies n = frac{16}{12} = frac{4}{3}But n must be an integer, so this case is invalid.Case 3:[k - (6n + 2) = -1k + (6n + 2) = -125]Adding:[2k = -126 implies k = -63Since k is a square root, it must be non-negative, so this case is invalid.Case 4:[k - (6n + 2) = -5k + (6n + 2) = -25]Adding:[2k = -30 implies k = -15Again, k must be non-negative, so invalid.So, the only valid solution is from Case 1: n = 10, k = 63, m = 32.Wait, but in the problem statement, it's mentioned that ( n ) is an integer, which is satisfied here with n=10.But, just to double-check, let's compute the total number of matches with m=32.Total matches in double round-robin: ( 32 times 31 = 992 )Now, ( 9n^2 + 6n + 32 ) with n=10:( 9(100) + 6(10) + 32 = 900 + 60 + 32 = 992 )Yes, that matches.But wait, earlier in my thought process, I thought maybe m could be 8. Let me check if that's possible.Suppose m=8, then total matches would be 8*7=56Let's see if 56 can be expressed as ( 9n^2 + 6n + 32 )So,[9n^2 + 6n + 32 = 56implies 9n^2 + 6n - 24 = 0Divide by 3:3n^2 + 2n - 8 = 0Using quadratic formula:n = [-2 ± sqrt(4 + 96)] / 6 = [-2 ± sqrt(100)] /6 = [-2 ±10]/6So,n=(8)/6=4/3 or n=(-12)/6=-2n must be integer, so n=-2Check if n=-2 gives m=8From earlier, with n=-2:Going back, the equation for k:k^2 = 36n^2 +24n +129Plug n=-2:k^2 = 36(4) +24(-2) +129 = 144 -48 +129 = 225k=15Then, m=(1 +15)/2=16/2=8Yes, so m=8 is also a solution when n=-2So, both m=8 and m=32 are solutions, corresponding to n=-2 and n=10 respectively.Therefore, the possible values of m are 8 and 32.Wait, but in the initial quadratic formula, when I had ( m = frac{1 pm k}{2} ), both solutions are possible as long as they result in positive integers.So, both m=8 and m=32 satisfy the equation with integer n.Therefore, the answer is m=8 or m=32.But the problem states "determine the value of m", implying perhaps multiple possible values.Hence, the values are 8 and 32.</think>

🧠:<think>Alright, so I have this expression to simplify:[frac{2 cos ^{2} 2 alpha - 1}{2 operatorname{tg}left(frac{pi}{4}-2 alpharight) sin ^{2}left(frac{3}{4} pi-2 alpharight)} - operatorname{tg} 2 alpha + cos 2 alpha - sin 2 alpha.]Hmm, okay. Let me try to break this down step by step. First, I should recall some trigonometric identities to see if I can simplify any parts of this expression.Starting with the numerator of the fraction: (2 cos^2 2alpha - 1). I remember there's a double-angle identity for cosine that might help here. The identity is:[cos 2theta = 2 cos^2 theta - 1]So if I let (theta = 2alpha), then:[2 cos^2 2alpha - 1 = cos 4alpha]Okay, that simplifies the numerator to (cos 4alpha). Good, that's progress.Now, looking at the denominator: (2 operatorname{tg}left(frac{pi}{4} - 2alpharight) sin^2left(frac{3pi}{4} - 2alpharight)). This looks a bit complicated, but maybe I can break it down.First, let's look at (operatorname{tg}left(frac{pi}{4} - 2alpharight)). The tangent of a difference can be expressed using the identity:[tan(A - B) = frac{tan A - tan B}{1 + tan A tan B}]So, plugging in (A = frac{pi}{4}) and (B = 2alpha):[tanleft(frac{pi}{4} - 2alpharight) = frac{tan frac{pi}{4} - tan 2alpha}{1 + tan frac{pi}{4} tan 2alpha}]Since (tan frac{pi}{4} = 1), this simplifies to:[frac{1 - tan 2alpha}{1 + tan 2alpha}]Alright, so (operatorname{tg}left(frac{pi}{4} - 2alpharight)) simplifies to (frac{1 - tan 2alpha}{1 + tan 2alpha}).Next, let's tackle (sin^2left(frac{3pi}{4} - 2alpharight)). I know that (sinleft(frac{3pi}{4} - 2alpharight)) can be rewritten using co-function identities or angle subtraction formulas.Recall that:[sinleft(frac{3pi}{4} - 2alpharight) = sinleft(pi - frac{pi}{4} - 2alpharight) = sinleft(frac{pi}{4} + 2alpharight)]Wait, no, that doesn't seem right. Let's double-check.Actually, (frac{3pi}{4} - 2alpha) is just an angle, and I can use the identity for (sin(A - B)):[sin(A - B) = sin A cos B - cos A sin B]So, applying this to (sinleft(frac{3pi}{4} - 2alpharight)):[sinleft(frac{3pi}{4}right)cos 2alpha - cosleft(frac{3pi}{4}right)sin 2alpha]I know that:[sinleft(frac{3pi}{4}right) = frac{sqrt{2}}{2}, quad cosleft(frac{3pi}{4}right) = -frac{sqrt{2}}{2}]So substituting these values in:[frac{sqrt{2}}{2} cos 2alpha - left(-frac{sqrt{2}}{2}right) sin 2alpha = frac{sqrt{2}}{2} cos 2alpha + frac{sqrt{2}}{2} sin 2alpha]Factor out (frac{sqrt{2}}{2}):[frac{sqrt{2}}{2} (cos 2alpha + sin 2alpha)]Therefore, (sinleft(frac{3pi}{4} - 2alpharight) = frac{sqrt{2}}{2} (cos 2alpha + sin 2alpha)).Now, squaring this:[sin^2left(frac{3pi}{4} - 2alpharight) = left(frac{sqrt{2}}{2}right)^2 (cos 2alpha + sin 2alpha)^2 = frac{1}{2} (cos 2alpha + sin 2alpha)^2]Expanding the square:[frac{1}{2} (cos^2 2alpha + 2 cos 2alpha sin 2alpha + sin^2 2alpha)]We know that (cos^2 x + sin^2 x = 1), so this becomes:[frac{1}{2} (1 + 2 cos 2alpha sin 2alpha)]And (2 cos 2alpha sin 2alpha = sin 4alpha), so:[frac{1}{2} (1 + sin 4alpha)]So, (sin^2left(frac{3pi}{4} - 2alpharight) = frac{1}{2} (1 + sin 4alpha)).Putting this back into the denominator:[2 cdot frac{1 - tan 2alpha}{1 + tan 2alpha} cdot frac{1}{2} (1 + sin 4alpha)]The 2 and 1/2 cancel out, so we're left with:[frac{1 - tan 2alpha}{1 + tan 2alpha} (1 + sin 4alpha)]So now, the entire fraction simplifies to:[frac{cos 4alpha}{frac{1 - tan 2alpha}{1 + tan 2alpha} (1 + sin 4alpha)}]This looks a bit messy. Maybe I can rewrite it as:[cos 4alpha cdot frac{1 + tan 2alpha}{1 - tan 2alpha} cdot frac{1}{1 + sin 4alpha}]Hmm. Let's see if I can simplify (frac{1 + tan 2alpha}{1 - tan 2alpha}). Recall that (tan 2alpha = frac{sin 2alpha}{cos 2alpha}), so:[frac{1 + frac{sin 2alpha}{cos 2alpha}}{1 - frac{sin 2alpha}{cos 2alpha}} = frac{cos 2alpha + sin 2alpha}{cos 2alpha - sin 2alpha}]So, substituting back in:[cos 4alpha cdot frac{cos 2alpha + sin 2alpha}{cos 2alpha - sin 2alpha} cdot frac{1}{1 + sin 4alpha}]Now, let's see if we can simplify (cos 4alpha) and (1 + sin 4alpha). I remember that:[cos 4alpha = 1 - 2 sin^2 2alpha quad text{or} quad 2 cos^2 2alpha - 1]But not sure if that helps. Alternatively, (1 + sin 4alpha) can be written using the identity:[1 + sin x = 2 sinleft(frac{pi}{4} + frac{x}{2}right) cosleft(frac{pi}{4} - frac{x}{2}right)]Wait, maybe that's too complicated. Alternatively, I recall that:[1 + sin x = (sin frac{x}{2} + cos frac{x}{2})^2]Let me verify that:[(sin frac{x}{2} + cos frac{x}{2})^2 = sin^2 frac{x}{2} + 2 sin frac{x}{2} cos frac{x}{2} + cos^2 frac{x}{2} = 1 + sin x]Yes, that's correct. So, applying this to (1 + sin 4alpha):[1 + sin 4alpha = left(sin 2alpha + cos 2alpharight)^2]Therefore, the expression becomes:[cos 4alpha cdot frac{cos 2alpha + sin 2alpha}{cos 2alpha - sin 2alpha} cdot frac{1}{left(sin 2alpha + cos 2alpharight)^2}]Simplifying, we can cancel out (cos 2alpha + sin 2alpha) in the numerator and denominator:[cos 4alpha cdot frac{1}{cos 2alpha - sin 2alpha} cdot frac{1}{sin 2alpha + cos 2alpha}]Notice that (cos 2alpha - sin 2alpha) and (cos 2alpha + sin 2alpha) are conjugates. Their product is:[(cos 2alpha - sin 2alpha)(cos 2alpha + sin 2alpha) = cos^2 2alpha - sin^2 2alpha = cos 4alpha]So, substituting back:[cos 4alpha cdot frac{1}{cos 4alpha} = 1]Wait, that simplifies to 1? That's nice. So the entire fraction simplifies to 1.So, going back to the original expression:[frac{2 cos ^{2} 2 alpha - 1}{2 operatorname{tg}left(frac{pi}{4}-2 alpharight) sin ^{2}left(frac{3}{4} pi-2 alpharight)} - operatorname{tg} 2 alpha + cos 2 alpha - sin 2 alpha = 1 - tan 2alpha + cos 2alpha - sin 2alpha]Now, let's simplify (1 - tan 2alpha + cos 2alpha - sin 2alpha).First, recall that (tan 2alpha = frac{sin 2alpha}{cos 2alpha}). So:[1 - frac{sin 2alpha}{cos 2alpha} + cos 2alpha - sin 2alpha]Let me combine the terms over a common denominator. Let's express everything in terms of (cos 2alpha):[frac{cos 2alpha}{cos 2alpha} - frac{sin 2alpha}{cos 2alpha} + cos 2alpha - sin 2alpha]Combine the first two terms:[frac{cos 2alpha - sin 2alpha}{cos 2alpha} + cos 2alpha - sin 2alpha]Now, let's write the entire expression as:[frac{cos 2alpha - sin 2alpha}{cos 2alpha} + cos 2alpha - sin 2alpha]To combine these, I'll express (cos 2alpha - sin 2alpha) as (frac{(cos 2alpha - sin 2alpha)cos 2alpha}{cos 2alpha}):[frac{cos 2alpha - sin 2alpha}{cos 2alpha} + frac{(cos 2alpha - sin 2alpha)cos 2alpha}{cos 2alpha}]Now, combine the numerators:[frac{cos 2alpha - sin 2alpha + (cos 2alpha - sin 2alpha)cos 2alpha}{cos 2alpha}]Factor out (cos 2alpha - sin 2alpha):[frac{(cos 2alpha - sin 2alpha)(1 + cos 2alpha)}{cos 2alpha}]Hmm, this looks a bit more manageable. Let me see if I can simplify this further.First, multiply out the numerator:[(cos 2alpha - sin 2alpha)(1 + cos 2alpha) = cos 2alpha (1 + cos 2alpha) - sin 2alpha (1 + cos 2alpha)]Expand each term:[= cos 2alpha + cos^2 2alpha - sin 2alpha - sin 2alpha cos 2alpha]Now, let's see if we can combine like terms or use identities. Notice that (cos^2 2alpha = frac{1 + cos 4alpha}{2}), but that might complicate things further. Alternatively, (sin 2alpha cos 2alpha = frac{1}{2} sin 4alpha), so:[cos 2alpha + frac{1 + cos 4alpha}{2} - sin 2alpha - frac{1}{2} sin 4alpha]But this seems to be getting more complex. Maybe there's another approach.Wait, perhaps instead of expanding, I can factor differently. Let's look back at the expression:[frac{(cos 2alpha - sin 2alpha)(1 + cos 2alpha)}{cos 2alpha}]Let me factor out (cos 2alpha) from the first term in the numerator:[frac{cos 2alpha(1 + cos 2alpha) - sin 2alpha(1 + cos 2alpha)}{cos 2alpha}]This can be rewritten as:[frac{cos 2alpha(1 + cos 2alpha)}{cos 2alpha} - frac{sin 2alpha(1 + cos 2alpha)}{cos 2alpha}]Simplify the first term:[1 + cos 2alpha - frac{sin 2alpha(1 + cos 2alpha)}{cos 2alpha}]So, the expression becomes:[1 + cos 2alpha - frac{sin 2alpha(1 + cos 2alpha)}{cos 2alpha}]Let me write this as:[1 + cos 2alpha - left(frac{sin 2alpha}{cos 2alpha} (1 + cos 2alpha)right)]Which is:[1 + cos 2alpha - tan 2alpha (1 + cos 2alpha)]Hmm, this seems cyclic since we had terms like (1 - tan 2alpha) earlier. Maybe another approach is needed.Alternatively, perhaps I can consider the entire expression:[frac{(cos 2alpha - sin 2alpha)(1 + cos 2alpha)}{cos 2alpha}]Let me factor out (cos 2alpha - sin 2alpha):Wait, actually, another idea: maybe express (cos 2alpha - sin 2alpha) as (sqrt{2} cosleft(2alpha + frac{pi}{4}right)). Let me verify:Using the identity:[A cos x + B sin x = C cos(x - phi)]Where (C = sqrt{A^2 + B^2}) and (tan phi = frac{B}{A}).In this case, (cos 2alpha - sin 2alpha) can be written as:[sqrt{2} cosleft(2alpha + frac{pi}{4}right)]Because:[cos 2alpha - sin 2alpha = sqrt{2} left(frac{1}{sqrt{2}} cos 2alpha - frac{1}{sqrt{2}} sin 2alpharight) = sqrt{2} cosleft(2alpha + frac{pi}{4}right)]Similarly, (1 + cos 2alpha) can be written using the identity:[1 + cos 2alpha = 2 cos^2 alpha]So, substituting back into our expression:[frac{sqrt{2} cosleft(2alpha + frac{pi}{4}right) cdot 2 cos^2 alpha}{cos 2alpha}]Simplify the constants:[frac{2sqrt{2} cosleft(2alpha + frac{pi}{4}right) cos^2 alpha}{cos 2alpha}]Now, let's see if we can simplify (cosleft(2alpha + frac{pi}{4}right)). Using the angle addition formula:[cosleft(2alpha + frac{pi}{4}right) = cos 2alpha cos frac{pi}{4} - sin 2alpha sin frac{pi}{4} = frac{sqrt{2}}{2} (cos 2alpha - sin 2alpha)]So, substituting back:[frac{2sqrt{2} cdot frac{sqrt{2}}{2} (cos 2alpha - sin 2alpha) cos^2 alpha}{cos 2alpha}]Simplify the constants:[2sqrt{2} cdot frac{sqrt{2}}{2} = 2]So, now we have:[frac{2 (cos 2alpha - sin 2alpha) cos^2 alpha}{cos 2alpha}]Factor out (cos 2alpha - sin 2alpha):Wait, actually, let's split the fraction:[2 cdot frac{cos 2alpha - sin 2alpha}{cos 2alpha} cdot cos^2 alpha]Which is:[2 left(1 - frac{sin 2alpha}{cos 2alpha}right) cos^2 alpha = 2 (1 - tan 2alpha) cos^2 alpha]But this seems to bring us back to earlier steps. Maybe there's a smarter way.Wait, going back to the expression:[frac{(cos 2alpha - sin 2alpha)(1 + cos 2alpha)}{cos 2alpha}]I can write this as:[(cos 2alpha - sin 2alpha) cdot frac{1 + cos 2alpha}{cos 2alpha}]Which is:[(cos 2alpha - sin 2alpha) left( frac{1}{cos 2alpha} + 1 right )]Simplify (frac{1}{cos 2alpha} = sec 2alpha):[(cos 2alpha - sin 2alpha)(sec 2alpha + 1)]Expanding this:[(cos 2alpha - sin 2alpha)sec 2alpha + (cos 2alpha - sin 2alpha)]Which is:[1 - tan 2alpha + cos 2alpha - sin 2alpha]Wait, that's exactly the expression we had earlier. So, we're going in circles here. Maybe it's time to accept that the simplified expression is:[frac{2 sqrt{2} sin left(frac{pi}{4} - 2alpha right) cos^2 alpha}{cos 2alpha}]But how did we get here? Let me recall earlier steps. We had:[frac{(cos 2alpha - sin 2alpha)(1 + cos 2alpha)}{cos 2alpha}]Expressed as:[frac{sqrt{2} cosleft(2alpha + frac{pi}{4}right) cdot 2 cos^2 alpha}{cos 2alpha}]Which simplifies to:[frac{2 sqrt{2} cosleft(2alpha + frac{pi}{4}right) cos^2 alpha}{cos 2alpha}]But (cosleft(2alpha + frac{pi}{4}right)) can be written as (sinleft(frac{pi}{4} - 2alpharight)), since (cos(x + frac{pi}{2}) = -sin x). Wait, let's verify:[cosleft(2alpha + frac{pi}{4}right) = sinleft(frac{pi}{2} - (2alpha + frac{pi}{4})right) = sinleft(frac{pi}{4} - 2alpharight)]Yes, that's correct because (cos x = sinleft(frac{pi}{2} - xright)). So:[cosleft(2alpha + frac{pi}{4}right) = sinleft(frac{pi}{4} - 2alpharight)]Substituting back:[frac{2 sqrt{2} sinleft(frac{pi}{4} - 2alpharight) cos^2 alpha}{cos 2alpha}]And that seems to be as simplified as it gets. Therefore, the entire original expression simplifies to:[frac{2 sqrt{2} sin left(frac{pi}{4} - 2alpha right) cos^2 alpha}{cos 2alpha}]I think this is the final simplified form. It might be possible to express this differently, but I can't see a more straightforward simplification at this point.Final Answerboxed{frac{2 sqrt{2} sin left(frac{pi}{4} - 2alpha right) cos^2 alpha}{cos 2alpha}}</think>

🧠:<think>Okay, so Linda traveled for five consecutive days. Each day, she traveled for two hours, and her speed was such that she could cover one mile in an integer number of minutes. The number of minutes required to travel one mile doubled each day, starting from the first day. Also, the distance she traveled each day was an integer number of miles. I need to find the total number of miles she traveled over these five days.Alright, let's break this down. First, she travels for two hours each day. Since speed is distance over time, and she covers one mile in a certain number of minutes, her speed is 1 mile per x minutes on the first day. On the second day, it's 1 mile per 2x minutes, then 4x, 8x, and 16x minutes for the subsequent days. Since speed is inversely proportional to the time taken to cover a mile, her speed halves each day, right? So, if she takes more minutes per mile, her speed decreases. Now, she travels for two hours each day, which is 120 minutes. So, on the first day, she can cover 120 / x miles because she spends x minutes per mile. On the second day, she can cover 120 / (2x) miles, since it takes her twice as long per mile. Similarly, on the third day, it's 120 / (4x), fourth day 120 / (8x), and fifth day 120 / (16x).Each of these distances needs to be an integer. So, 120 / x, 120 / (2x), 120 / (4x), 120 / (8x), and 120 / (16x) must all be integers. That means x, 2x, 4x, 8x, and 16x must all be divisors of 120.Let me list the divisors of 120 to see possible values for x. The divisors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.Now, x has to be such that x, 2x, 4x, 8x, and 16x are all in this list. Let's test possible x values.If x = 1: Then 2x = 2, 4x = 4, 8x = 8, 16x = 16. All these are divisors of 120.If x = 2: 2x = 4, 4x = 8, 8x = 16, 16x = 32. But 32 is not a divisor of 120, so x=2 is invalid.x=3: 2x=6, 4x=12, 8x=24, 16x=48. 48 is not a divisor, so invalid.x=4: 2x=8, 4x=16, 8x=32, 16x=64. 32 and 64 are not divisors, invalid.x=5: 2x=10, 4x=20, 8x=40, 16x=80. 80 is not a divisor, invalid.x=6: 2x=12, 4x=24, 8x=48, 16x=96. None of these beyond 24 are divisors, invalid.x=8: 2x=16, 4x=32, which isn't a divisor, so invalid.Similarly, higher x values don't work because 16x quickly exceeds the divisors of 120. So, the only valid x is 1.Wait, but let me double-check. Maybe I missed something. For x=1, the number of minutes per mile on each day would be 1, 2, 4, 8, 16 minutes per mile. That means her speed is 60 mph on the first day, 30 mph on the second, 15 mph on the third, 7.5 mph on the fourth, and 3.75 mph on the fifth day.But wait, the problem says the number of minutes required to travel one mile is an integer, which is satisfied here. However, the distance each day must also be an integer. Let's calculate the distances:First day: 120 minutes / 1 minute per mile = 120 miles.Second day: 120 / 2 = 60 miles.Third day: 120 / 4 = 30 miles.Fourth day: 120 / 8 = 15 miles.Fifth day: 120 / 16 = 7.5 miles.Oh, wait, the fifth day's distance is 7.5 miles, which isn't an integer. That's a problem because the problem states that the distance each day must be an integer. So, x=1 doesn't work because on the fifth day, the distance is not an integer.Hmm, so maybe I need to reconsider. Perhaps x isn't 1. Maybe x is a different value such that 16x divides 120. Let's see. 16x must divide 120, so x must be a divisor of 120 / 16, which is 7.5. But x must be an integer, so 7.5 isn't an integer. That suggests there's no integer x that satisfies all these conditions. But that can't be right because the problem states that such an x exists.Wait, maybe I made a mistake in assuming x must be an integer. Let me check the problem again. It says the number of minutes required to travel one mile is an integer each day, and the distance is also an integer. So, x must be an integer, yes, because the minutes per mile are integers. So, x must be an integer, and 16x must divide 120. But 16x must be a divisor of 120, meaning 16x divides 120, so x must be 120 divided by a multiple of 16. But 120 isn't a multiple of 16. 120 divided by 16 is 7.5, which isn't an integer. So, this seems impossible.Wait, but maybe I'm misunderstanding the problem. It says the number of minutes required to travel one mile doubled each day, starting from an initial time on the first day. So, the first day is x minutes per mile, then 2x, 4x, 8x, 16x. All these must result in integer distances when dividing 120 minutes by each of these.So, 120 / x, 120 / (2x), 120 / (4x), 120 / (8x), 120 / (16x) must all be integers. That means x must be a divisor of 120, 2x must be a divisor of 120, and so on. So, x must be such that x, 2x, 4x, 8x, 16x all divide 120.The divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.So, x must be a divisor of 120, and 2x, 4x, 8x, 16x must also be divisors of 120.Let's test x=1: 2x=2, 4x=4, 8x=8, 16x=16. All these are divisors except 16 is a divisor? Wait, 16 isn't a divisor of 120. 120 divided by 16 is 7.5, which isn't an integer. So, x=1 is invalid.Next, x=2: 2x=4, 4x=8, 8x=16, 16x=32. 32 isn't a divisor of 120, so invalid.x=3: 2x=6, 4x=12, 8x=24, 16x=48. 48 isn't a divisor, invalid.x=4: 2x=8, 4x=16, which is invalid as before.x=5: 2x=10, 4x=20, 8x=40, 16x=80. 80 isn't a divisor, invalid.x=6: 2x=12, 4x=24, 8x=48, invalid.x=8: 2x=16, invalid.x=10: 2x=20, 4x=40, 8x=80, invalid.x=12: 2x=24, 4x=48, invalid.x=15: 2x=30, 4x=60, 8x=120, 16x=240. 240 isn't a divisor of 120, invalid.x=20: 2x=40, 4x=80, invalid.x=24: 2x=48, invalid.x=30: 2x=60, 4x=120, 8x=240, invalid.x=40: 2x=80, invalid.x=60: 2x=120, 4x=240, invalid.x=120: 2x=240, invalid.Hmm, so none of these x values satisfy the condition that x, 2x, 4x, 8x, 16x are all divisors of 120. That's a problem because the problem states that such an x exists.Wait, maybe I'm missing something. Perhaps x doesn't have to be an integer? But the problem says the number of minutes required to travel one mile is an integer each day. So, x must be an integer.Alternatively, maybe the distances don't have to be whole numbers? But the problem says the distance she traveled each day was also an integer number of miles. So, both x and the distances must be integers.Wait, maybe I'm overcomplicating this. Let's think differently. If 120 / (16x) must be an integer, then 16x must divide 120. So, 16x | 120. Therefore, x must be a divisor of 120 / 16, which is 7.5. But x must be an integer, so the only way is if x is a divisor of 7.5, but 7.5 isn't an integer. So, there's no integer x that satisfies this condition.But the problem says such an x exists, so I must have made a mistake.Wait, maybe the problem doesn't require that 16x divides 120, but that 120 divided by 16x is an integer. So, 120 / (16x) must be an integer, meaning 16x must be a divisor of 120.So, 16x must divide 120. Therefore, 16x must be one of the divisors of 120, which are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.So, 16x must be one of these. Therefore, x must be 16x divided by 16, so x must be 1/16, 1/8, 3/16, 1/4, 5/16, 3/8, 1/2, 5/8, 3/4, 15/16, 5/4, 3/2, 15/8, 5/2, 15/4, 15/2.But x must be an integer, so the only possible x is when 16x is 16, which is a divisor of 120? Wait, 16 isn't a divisor of 120, as we saw earlier.Wait, 120 divided by 16 is 7.5, which isn't an integer. So, there's no integer x such that 16x divides 120. Therefore, this seems impossible.But the problem states that such an x exists, so I must have made a mistake in my approach.Let me try another angle. Maybe the speed is such that the time per mile doubles each day, but the total distance is still an integer.Wait, if we consider that the time per mile doubles each day, starting from x minutes on day 1, then day 2 is 2x, day 3 is 4x, day 4 is 8x, day 5 is 16x.So, the distance each day is 120 / x, 120 / (2x), 120 / (4x), 120 / (8x), 120 / (16x).All these must be integers. So, x must be a divisor of 120, 2x must be a divisor of 120, and so on.Wait, but if x is a divisor of 120, then 2x, 4x, etc., must also be divisors. So, x must be such that x, 2x, 4x, 8x, 16x are all divisors of 120.But as we saw earlier, the only x that satisfies this is x=1, but that leads to a non-integer distance on day 5.Wait, maybe x isn't 1. Let's try x=15.If x=15, then 2x=30, 4x=60, 8x=120, 16x=240.But 240 isn't a divisor of 120, so that's invalid.x=10: 2x=20, 4x=40, 8x=80, 16x=160. 80 and 160 aren't divisors.x=5: 2x=10, 4x=20, 8x=40, 16x=80. 40 and 80 aren't divisors.x=3: 2x=6, 4x=12, 8x=24, 16x=48. 24 isn't a divisor.x=4: 2x=8, 4x=16, which isn't a divisor.x=2: 2x=4, 4x=8, 8x=16, 16x=32. 16 and 32 aren't divisors.x=1: 2x=2, 4x=4, 8x=8, 16x=16. 16 isn't a divisor.Wait, so none of these x values work. That's confusing because the problem says such an x exists.Maybe I'm misunderstanding the problem. Perhaps the number of minutes per mile doesn't have to be an integer on each day, but the total distance must be an integer. But no, the problem says the number of minutes required to travel one mile is an integer each day.Alternatively, maybe the speed is such that the distance is an integer, but the minutes per mile can be a fraction. But no, the problem states that the minutes per mile are integers.Wait, perhaps I'm overcomplicating it. Let's think about it differently. Since the time per mile doubles each day, the distances she travels each day form a geometric sequence with a common ratio of 1/2.So, the distances are D, D/2, D/4, D/8, D/16.But she travels for five days, so the total distance is D + D/2 + D/4 + D/8 + D/16.This sum is D*(1 + 1/2 + 1/4 + 1/8 + 1/16) = D*(31/16).But we also know that each day's distance is 120 / (x*2^(n-1)), where n is the day number.Wait, maybe the key is to find x such that 120 / x, 120 / (2x), etc., are all integers.So, x must be a divisor of 120, and 2x, 4x, 8x, 16x must also be divisors of 120.But as we saw, the only x that works is x=1, but that leads to a non-integer on day 5.Wait, maybe x isn't 1, but a fraction. But x must be an integer, so that's not possible.Alternatively, maybe the problem allows for x to be a fraction, but the minutes per mile must be an integer. So, x must be a rational number such that 16x is an integer.Wait, but x must be an integer because the minutes per mile are integers.I'm stuck here. Maybe I need to look at the problem again.Wait, the problem says that the number of minutes required to travel one mile doubled each day, starting from an initial time on the first day. The distance she traveled each day was also an integer number of miles.So, if we let x be the minutes per mile on day 1, then x, 2x, 4x, 8x, 16x must all be integers, and 120 / x, 120 / (2x), 120 / (4x), 120 / (8x), 120 / (16x) must all be integers.So, x must be a divisor of 120, and 2x, 4x, 8x, 16x must also be divisors of 120.But as we saw, there's no such x that satisfies this condition because 16x would have to divide 120, and 120 isn't a multiple of 16.Wait, unless x is a fraction. Let me consider x as a fraction.If x is a fraction, say x = 15/16, then 16x = 15, which is a divisor of 120. Then 8x = 15/2, which isn't an integer, so that doesn't work.Alternatively, x = 5/4: 16x = 20, which is a divisor. 8x=10, 4x=5, 2x=2.5, which isn't an integer. So, x=5/4 is invalid.x=3/2: 16x=24, which is a divisor. 8x=12, 4x=6, 2x=3, x=1.5. But x must be an integer, so invalid.x=15/8: 16x=30, which is a divisor. 8x=15, 4x=7.5, which isn't an integer. Invalid.x=5/2: 16x=40, which is a divisor. 8x=20, 4x=10, 2x=5, x=2.5. Not an integer.x=15/4: 16x=60, which is a divisor. 8x=30, 4x=15, 2x=7.5, invalid.x=15/2: 16x=120, which is a divisor. 8x=60, 4x=30, 2x=15, x=7.5. Not an integer.So, none of these fractions work because x must be an integer.Wait, maybe I'm approaching this wrong. Let's consider that the distance each day is an integer, but the minutes per mile can be a fraction, as long as it's an integer. Wait, no, the minutes per mile must be an integer.Wait, maybe the initial x is such that x divides 120, and 16x divides 120. So, x must be a common divisor of 120 and 120/16.But 120/16=7.5, which isn't an integer, so x must divide 7.5, but x must be an integer, so x must be 1 or 3/2, but 3/2 isn't an integer.This seems impossible. Maybe the problem has a typo, or I'm misinterpreting it.Wait, let's try to see what the answer choices are. They are 232.5, 232, 230, 235.If x=1, the total distance is 120 + 60 + 30 + 15 + 7.5 = 232.5, which is option A.But the problem states that the distance each day must be an integer, so 7.5 is invalid. Therefore, x=1 is invalid.But maybe the problem allows for non-integer distances on the fifth day, but the problem says the distance each day is an integer. So, that can't be.Wait, maybe the initial x is not 1, but a different value that allows all distances to be integers.Wait, if x=15, then:First day: 120 /15=8 miles.Second day: 120 /30=4 miles.Third day: 120 /60=2 miles.Fourth day: 120 /120=1 mile.Fifth day: 120 /240=0.5 miles, which is not an integer.So, invalid.If x=10:First day:12 miles.Second day:6 miles.Third day:3 miles (since 120/40=3).Fourth day:1.5 miles (invalid).x=5:First day:24 miles.Second day:12 miles.Third day:6 miles.Fourth day:3 miles.Fifth day:1.5 miles (invalid).x=3:First day:40 miles.Second day:20 miles.Third day:10 miles.Fourth day:5 miles.Fifth day:2.5 miles (invalid).x=4:First day:30 miles.Second day:15 miles.Third day:7.5 miles (invalid).x=2:First day:60 miles.Second day:30 miles.Third day:15 miles.Fourth day:7.5 miles (invalid).x=6:First day:20 miles.Second day:10 miles.Third day:5 miles.Fourth day:2.5 miles (invalid).x=8:First day:15 miles.Second day:7.5 miles (invalid).x=12:First day:10 miles.Second day:5 miles.Third day:2.5 miles (invalid).x=15:As above, fifth day is 0.5.x=20:First day:6 miles.Second day:3 miles.Third day:1.5 miles (invalid).x=24:First day:5 miles.Second day:2.5 miles (invalid).x=30:First day:4 miles.Second day:2 miles.Third day:1 mile.Fourth day:0.5 miles (invalid).x=40:First day:3 miles.Second day:1.5 miles (invalid).x=60:First day:2 miles.Second day:1 mile.Third day:0.5 miles (invalid).x=120:First day:1 mile.Second day:0.5 miles (invalid).So, none of these x values result in all five days having integer distances. The only x that gets close is x=1, but the fifth day is 7.5, which is not an integer. However, the answer choice includes 232.5, which is the sum when x=1. Maybe the problem allows for the fifth day to have a non-integer distance, but the problem states that the distance each day is an integer. Therefore, this is confusing.Alternatively, maybe I made a mistake in interpreting the problem. Perhaps the number of minutes per mile doesn't have to be an integer on each day, but the total time is two hours, and the distance is an integer. But the problem specifically says the number of minutes required to travel one mile is an integer each day.Wait, maybe the key is that the speed allows her to cover one mile in an integer number of minutes, but the total distance doesn't have to be an integer. But no, the problem says the distance she traveled each day was also an integer number of miles.I'm really stuck here. Maybe the answer is 232.5, even though the fifth day's distance is not an integer, but perhaps the problem allows for that. Or maybe I'm missing something in the problem statement.Wait, let me read the problem again carefully."Linda traveled for five consecutive days. On each day, she traveled for two hours at a speed that allowed her to cover one mile in an integer number of minutes. The number of minutes required to travel one mile doubled each day, starting from an initial time on the first day. The distance she traveled each day was also an integer number of miles. Calculate the total number of miles Linda traveled over these five days."So, the key points are:- 2 hours (120 minutes) each day.- Speed: 1 mile per x minutes, where x is an integer.- Each subsequent day, x doubles: x, 2x, 4x, 8x, 16x.- Distance each day: 120 / x, 120 / (2x), etc., must be integers.So, x must be such that x, 2x, 4x, 8x, 16x all divide 120.But as we saw, no such x exists. Therefore, there's a mistake in the problem or in my understanding.Wait, maybe the problem allows for x to be a fraction, but the minutes per mile must be an integer. So, x must be such that x is an integer, but 2x, 4x, etc., can be fractions as long as the distance is an integer.Wait, no, because the minutes per mile must be integers each day. So, x, 2x, 4x, 8x, 16x must all be integers. Therefore, x must be an integer, and 2x, 4x, etc., are also integers.But as we saw, no such x exists because 16x must divide 120, and 120 isn't a multiple of 16.Wait, unless x is a fraction that when multiplied by 16 gives a divisor of 120. But x must be an integer, so that's not possible.I'm really confused. Maybe the answer is 232.5, even though it's not an integer, but the problem says the distance each day must be an integer. Alternatively, maybe the problem has a typo, and the number of days is four instead of five, but that's not the case.Wait, let's try to see what the answer is among the options. If x=1, the total is 232.5, which is option A. If x=15, the total is 8+4+2+1+0.5=15.5, which isn't an option. If x=5, the total is 24+12+6+3+1.5=46.5, not an option.Wait, maybe the problem allows for the fifth day's distance to be 7.5, even though it's not an integer, but the problem says it must be an integer. So, perhaps the answer is 232.5, accepting that the fifth day's distance is 7.5, but that contradicts the problem statement.Alternatively, maybe the problem is designed such that x is 1, and the fifth day's distance is 7.5, but the answer is rounded or something. But the options include 232.5 and 232, so maybe 232.5 is the intended answer.Alternatively, maybe I made a mistake in assuming x must be an integer. Let's think again.If x is a fraction such that x*2^(n-1) is an integer for n=1 to 5. So, x must be a multiple of 1/16, because on day 5, 16x must be an integer. So, x= k/16, where k is an integer.Then, 16x= k, which must divide 120, so k must be a divisor of 120.So, k can be 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.Therefore, x=1/16, 2/16=1/8, 3/16, 4/16=1/4, 5/16, 6/16=3/8, 8/16=1/2, 10/16=5/8, 12/16=3/4, 15/16, 20/16=5/4, 24/16=3/2, 30/16=15/8, 40/16=5/2, 60/16=15/4, 120/16=15/2.Now, for each x=k/16, we need to check if 120/(x*2^(n-1)) is an integer for n=1 to 5.Let's take x=1/16:n=1: 120/(1/16)=1920 miles (integer).n=2: 120/(2/16)=120/(1/8)=960 miles (integer).n=3: 120/(4/16)=120/(1/4)=480 miles (integer).n=4: 120/(8/16)=120/(1/2)=240 miles (integer).n=5: 120/(16/16)=120/1=120 miles (integer).So, x=1/16 works, and all distances are integers. Therefore, the total distance is 1920 + 960 + 480 + 240 + 120 = 3720 miles. But that's way higher than the answer choices.Wait, but 3720 isn't among the options. Maybe I'm misunderstanding something.Wait, no, because if x=1/16, then the minutes per mile on day 1 is 1/16 minutes per mile, which is 3.75 seconds per mile, which is incredibly fast, and the distance would be 1920 miles in two hours, which is 960 mph, which is unrealistic.But the problem doesn't specify realistic speeds, just that the minutes per mile are integers. Wait, no, x must be such that x is an integer number of minutes per mile. So, x=1/16 isn't an integer. So, that approach is invalid.I think I'm going in circles here. Maybe the answer is 232.5, even though it contradicts the problem's condition, because that's the only sum that matches the options when x=1.Alternatively, maybe the problem allows for the fifth day's distance to be a non-integer, but the problem says it must be an integer. Therefore, there's a mistake in the problem or in my understanding.Given the answer choices, and the fact that x=1 gives 232.5, which is option A, I'll go with that, assuming that perhaps the problem allows for the fifth day's distance to be a non-integer, even though it contradicts the problem statement.</think>

🧠:<think>Okay, so I have this problem about coloring the vertices of a cube. The cube has vertices at all combinations of (x, y, z) where each of x, y, and z is either 0 or 1. So, that's 8 vertices in total. I need to color each vertex either black or white. The condition is that for any vertex, if all of its neighbors are the same color, then it must also be that color. Two vertices are neighbors if they're connected by an edge of the cube.First, I need to understand the cube structure. Each vertex is connected to three others because in a cube, each vertex has three edges coming out of it. So, each vertex has three neighbors.The problem is about counting the number of valid colorings under this condition. That is, for any vertex, if all its neighbors are black, then it has to be black. Similarly, if all its neighbors are white, then it has to be white. But if the neighbors are a mix, then the vertex can be either color.I think this condition is similar to some kind of constraint satisfaction. Maybe it's related to graph colorings with specific rules. I remember something about graph colorings where certain conditions force the color of a vertex, like in the case of a bipartite graph where vertices alternate colors.But in this case, the condition is different. It's not about conflicting colors but about forcing a color if all neighbors are the same. So, maybe this can be thought of as a kind of closure or a dependency where colors propagate through the graph.I also recall that cubes can be split into two sets of vertices in a bipartition, like a checkerboard pattern. For example, you can color all vertices with even parity (even number of 1s in their coordinates) one color and odd parity the other. In a cube, this results in two sets of four vertices each, and each vertex is connected only to vertices of the opposite color.Wait, that might be useful here. Let me think. If I split the cube into two sets, say Set A and Set B, each containing four vertices, such that every edge connects a vertex in Set A to Set B. Then, maybe I can analyze the colorings based on these sets.So, Set A and Set B are independent sets, meaning no two vertices within each set are adjacent. Each vertex in Set A is connected only to vertices in Set B and vice versa.Given that, maybe I can consider the coloring of Set A and Set B separately, but with dependencies because the color of a vertex in Set A affects the color of its neighbors in Set B and vice versa.Let me try to formalize this. Suppose I have a vertex v in Set A. If all neighbors of v (which are in Set B) are the same color, then v must be that color. Similarly, for a vertex u in Set B, if all its neighbors (in Set A) are the same color, then u must be that color.So, the coloring of Set A affects the possible colorings of Set B and vice versa. It seems like a mutual dependency.Maybe I can approach this by considering possible colorings for Set A and then determining the valid colorings for Set B given the constraints.How many possible colorings are there for Set A? Since each vertex can be black or white, there are 2^4 = 16 possible colorings for Set A. Similarly, 16 for Set B. But with the constraints, the actual number will be less.Wait, but since the colorings are interdependent, maybe I can model this as something like a constraint satisfaction problem where the colorings of Set A and Set B have to satisfy each other's constraints.Alternatively, maybe I can consider the cube as a bipartite graph with partitions Set A and Set B, and the condition is that for each vertex, if all its neighbors are the same color, then it must be that color. So, in terms of Set A, for each vertex in Set A, if all its neighbors in Set B are black, then it must be black. If all its neighbors in Set B are white, then it must be white. Similarly, for each vertex in Set B, if all its neighbors in Set A are a single color, then it must be that color.This seems like a system of equations or constraints. Maybe I can model this as a system where the color of each vertex is a variable, and the constraints are based on the colors of their neighbors.But this might get complicated because each vertex in Set A has three neighbors in Set B, and each vertex in Set B has three neighbors in Set A.Alternatively, maybe I can consider that if a vertex in Set A has all neighbors in Set B of the same color, then it must match that color. Similarly for Set B. So, if I fix the colors of Set A, then the colors of Set B are constrained, and vice versa.Let me try this approach.Case 1: Suppose all vertices in Set A are black. Then, for each vertex in Set B, all their neighbors in Set A are black. So, each vertex in Set B must be black as well. So, the entire cube is black. Similarly, if all Set A are white, then all Set B must be white as well.So, these are two colorings: all black and all white.Case 2: Suppose Set A has three black and one white. Let's see how this affects Set B.Each vertex in Set B is connected to three vertices in Set A. If a vertex in Set B is connected to three black vertices in Set A, it must be black. However, if a vertex in Set B is connected to two black and one white in Set A, it can be either color because the neighbors are not all the same.Wait, but actually, the condition only applies if all neighbors are the same color. So, if a vertex has neighbors of mixed colors, it can be either color.But in the case where Set A has three black and one white, each vertex in Set B is connected to three vertices in Set A. Depending on which vertex in Set A is white, the connections to Set B will vary.Let me think of Set A as four vertices, one of which is white, and the rest black. Each vertex in Set B is connected to three in Set A. So, depending on which vertex in Set A is white, each vertex in Set B will be connected to either three black, or two black and one white.Wait, no. Each vertex in Set B is connected to three in Set A. If one vertex in Set A is white, then each vertex in Set B is connected to either all three black or two black and one white.Wait, actually, each vertex in Set B is connected to three in Set A, but if one vertex in Set A is white, then how does that distribute across Set B?It depends on how Set A and Set B are connected. In a cube, each vertex in Set A is connected to three in Set B, but each vertex in Set B is connected to three in Set A. So, if one vertex in Set A is white, each vertex in Set B is connected to three in Set A, so some of them might be connected to the white one.Wait, actually, since each vertex in Set B is connected to three in Set A, and only one vertex in Set A is white, then each vertex in Set B is connected to either 0 or 1 white vertex in Set A.Wait, no. Because each vertex in Set B is connected to three in Set A, and there is only one white in Set A. So, only three vertices in Set B are connected to the white vertex in Set A, right? Because each vertex in Set A is connected to three in Set B, so the white vertex is connected to three in Set B, each of which is connected to that white vertex and two others.So, in Set B, three vertices are connected to the white vertex in Set A, and one vertex in Set B is connected only to black vertices in Set A.Wait, no, actually, each vertex in Set B is connected to three in Set A, but since only one in Set A is white, each vertex in Set B can be connected to at most one white vertex in Set A.But since the white vertex in Set A is connected to three in Set B, those three in Set B are each connected to one white and two black in Set A. The remaining vertex in Set B is connected to three black in Set A.Wait, let me clarify.If Set A has one white vertex, say vertex W, and three black vertices, B1, B2, B3.Each vertex in Set B is connected to three in Set A. Since vertex W is connected to three in Set B, let's call them S1, S2, S3. So, S1, S2, S3 are each connected to W and two other black vertices in Set A. The remaining vertex in Set B, let's call it S4, is connected to three black vertices in Set A.Therefore, S4 is connected only to black vertices in Set A, so S4 must be black, because all its neighbors are black.For S1, S2, S3, each is connected to one white (W) and two black vertices in Set A. Since their neighbors are not all the same color, they can be either black or white.So, S1, S2, S3 can each be black or white, independently. So, each has two choices.Therefore, the number of colorings in this case is the number of ways to choose the white vertex in Set A (which is 4 choices), multiplied by the number of colorings for Set B.But for each choice of the white vertex in Set A, Set B has one vertex (S4) that must be black, and three vertices (S1, S2, S3) that can be black or white.So, the number of colorings for Set B is 2^3 = 8. But wait, is there any restriction on S1, S2, S3? Let me see.No, because their neighbors are a mix of black and white. The condition only applies if all neighbors are the same color. Since S1, S2, S3 each have two black and one white neighbor, they don't have all neighbors the same, so they can be any color.Therefore, for each choice of the white vertex in Set A (4 choices), we have 8 colorings for Set B.But wait, hold on. If Set B has S4 black, and S1, S2, S3 each can be black or white, that's 8 possibilities. However, is there a constraint from the other direction?Meaning, if we color S1, S2, S3 in Set B, does that affect the colors in Set A? Well, in this case, Set A was already assigned with one white and three black. So, as long as the assignments in Set B don't create a situation where a vertex in Set A has all neighbors of the same color, but it's already fixed.Wait, but in Set A, we have one white and three black. So, the white vertex in Set A is connected to three in Set B, which could be any color. But since the white vertex in Set A is already white, it doesn't impose any constraints on its neighbors in Set B, except that if all its neighbors in Set B are the same color, then it must be that color. But since the vertex in Set A is already white, if all its neighbors in Set B are white, then it must be white, which it is. Similarly, if all its neighbors in Set B are black, then it must be black, but it's white, so that's a contradiction.Wait, that's a problem. If the white vertex in Set A has all its neighbors in Set B black, then according to the condition, the white vertex must be black, which it's not. So, that would be an invalid coloring.Therefore, in this case, when we choose the colors for S1, S2, S3 (the neighbors of the white vertex in Set A), we must ensure that not all of them are black. Because if they were all black, the white vertex in Set A would have to be black, which it isn't.So, in addition to S4 being black, we have to ensure that S1, S2, S3 are not all black. Therefore, the number of valid colorings for Set B is 2^3 - 1 = 7.So, for each choice of the white vertex in Set A (4 choices), we have 7 valid colorings for Set B.Therefore, the total number of colorings in this case is 4 * 7 = 28.Case 3: Set A has two black and two white vertices. Hmm, this seems more complicated.First, how many ways are there to color Set A with two black and two white? It's C(4,2) = 6 ways.Now, for each such coloring, we need to determine the valid colorings of Set B.Each vertex in Set B is connected to three vertices in Set A. Depending on how the two white vertices in Set A are arranged, the connections in Set B will vary.Wait, but in a cube, the two white vertices in Set A could be adjacent or not. Wait, no, in Set A, which is an independent set, so no two vertices in Set A are adjacent. Therefore, in Set A, all vertices are non-adjacent.So, in Set A, which is an independent set, the two white vertices are not connected by an edge. So, they are at least two edges apart.Therefore, each vertex in Set B is connected to three in Set A. Given that two of the Set A vertices are white, how does this affect the neighbors in Set B?Each vertex in Set B is connected to three in Set A. Since two are white and two are black, each vertex in Set B could be connected to 0, 1, 2, or all 3 white vertices, but given that Set A has only two white vertices, the maximum a vertex in Set B can be connected to is two white vertices.Wait, actually, each vertex in Set B is connected to three in Set A, and Set A has two white and two black. So, the number of white vertices connected to a vertex in Set B can be 0, 1, or 2.But let's think about how the two white vertices in Set A are connected to Set B.Each white vertex in Set A is connected to three vertices in Set B. Since there are two white vertices in Set A, together they are connected to 3 + 3 = 6 vertices in Set B. However, Set B only has four vertices, so there must be overlap.Therefore, each vertex in Set B is connected to either one or two white vertices in Set A.Wait, because the two white vertices in Set A are connected to three each, but there are only four in Set B. So, the overlap is such that some vertices in Set B are connected to both white vertices in Set A.Let me calculate how many vertices in Set B are connected to both white vertices in Set A.Each white vertex in Set A is connected to three in Set B. So, total connections from white vertices to Set B are 6. Since Set B has four vertices, the average number of connections per vertex is 6/4 = 1.5. So, some vertices in Set B are connected to one white vertex, and some are connected to two.Specifically, let’s denote the two white vertices in Set A as W1 and W2. Each is connected to three in Set B.The overlap between W1 and W2 in their connections to Set B can be calculated.The total number of unique vertices in Set B connected to W1 or W2 is 3 + 3 - overlap.Since Set B has four vertices, 3 + 3 - overlap <= 4.So, 6 - overlap <= 4 => overlap >= 2.Therefore, the overlap is at least two. So, at least two vertices in Set B are connected to both W1 and W2.Therefore, in Set B, two vertices are connected to both white vertices in Set A, and the other two are connected to only one white vertex each.So, in Set B, two vertices have two white neighbors in Set A, and two vertices have one white neighbor in Set A.Therefore, for each vertex in Set B:- If a vertex has two white neighbors in Set A, then its remaining neighbor in Set A is black. So, it's connected to two white and one black.- If a vertex has one white neighbor in Set A, then its remaining two neighbors are black. So, it's connected to one white and two black.Therefore, for the two vertices in Set B with two white neighbors, since their neighbors are not all the same color, they can be either black or white.Similarly, for the two vertices in Set B with one white neighbor, their neighbors are not all the same color, so they can be either black or white.Wait, but hold on. The condition is only triggered if all neighbors are the same color. So, for vertices in Set B with two white and one black neighbor, since not all neighbors are the same, they can be either color.Similarly, for vertices with one white and two black neighbors, they can be either color.Therefore, all four vertices in Set B can be colored independently, except for one thing: we have to ensure that the coloring doesn't cause a conflict in Set A.Wait, because the coloring in Set B affects the constraints for Set A.In Set A, we have two white and two black vertices. Each vertex in Set A is connected to three in Set B.For the two white vertices in Set A, each is connected to three in Set B. If all three neighbors in Set B are the same color, then the white vertex must be that color. But since the white vertex is already white, if all its neighbors in Set B are white, then it's fine. If all its neighbors in Set B are black, then the white vertex must be black, which it isn't, so that would be invalid.Similarly, for the two black vertices in Set A, each is connected to three in Set B. If all neighbors in Set B are the same color, then the black vertex must be that color. If all neighbors are black, it's fine; if all are white, it must be white, which it isn't, so that's invalid.Therefore, for each white vertex in Set A, we must ensure that not all its neighbors in Set B are black.Similarly, for each black vertex in Set A, we must ensure that not all its neighbors in Set B are white.So, going back to Set B: Each white vertex in Set A is connected to three in Set B. We need to ensure that for each white vertex in Set A, not all three neighbors in Set B are black.Similarly, each black vertex in Set A is connected to three in Set B, and we need to ensure that not all three neighbors in Set B are white.Given the structure of Set B, where two vertices have two white neighbors in Set A and two have one white neighbor in Set A.Wait, actually, no. Earlier, we had that in Set B, two vertices are connected to both white vertices in Set A, so they have two white neighbors, and two vertices are connected to only one white vertex, so they have one white neighbor.But each vertex in Set B is connected to three in Set A. So, the two vertices in Set B connected to two white vertices in Set A have two white and one black neighbor in Set A.The other two vertices in Set B connected to one white vertex in Set A have one white and two black neighbors in Set A.Therefore, for each vertex in Set B:- The two with two white neighbors can be colored either black or white.- The two with one white neighbor can be colored either black or white.But we have constraints:1. For each white vertex in Set A, its three neighbors in Set B cannot all be black.2. For each black vertex in Set A, its three neighbors in Set B cannot all be white.So, let's denote the two white vertices in Set A as W1 and W2. Each is connected to three in Set B.Given that in Set B, two vertices are connected to both W1 and W2 (let's call them S1 and S2), and two vertices are connected to only one white vertex each (S3 connected to W1, S4 connected to W2).So, S1 and S2 are connected to W1 and W2, and each has a third connection to a black vertex in Set A.Similarly, S3 is connected to W1 and two black vertices in Set A.S4 is connected to W2 and two black vertices in Set A.Now, for the constraints:- For W1, its neighbors in Set B are S1, S2, and S3. We must ensure that not all of S1, S2, S3 are black.- Similarly, for W2, its neighbors in Set B are S1, S2, and S4. We must ensure that not all of S1, S2, S4 are black.- For the black vertices in Set A, let's call them B1 and B2. Each is connected to three in Set B.Wait, actually, each black vertex in Set A is connected to three in Set B, but since Set A has two black vertices, each connected to three in Set B, we have to make sure that for each black vertex, not all its neighbors in Set B are white.But given the structure, each black vertex in Set A is connected to S1, S2, and one other. Let me think.Wait, actually, each black vertex in Set A is connected to three in Set B. Since Set A has two black vertices, B1 and B2, each connected to three in Set B.But in Set B, we have four vertices: S1, S2, S3, S4.Each of B1 and B2 must be connected to three in Set B, but given that Set B's connections are:- S1 connected to W1, W2, and B1.- S2 connected to W1, W2, and B2.- S3 connected to W1, B1, B2.- S4 connected to W2, B1, B2.Wait, this might not be accurate. Let me think about the cube's structure.In the cube, each vertex in Set A is connected to three in Set B. So, if we have two white vertices in Set A, W1 and W2, each connected to three in Set B. Similarly, the two black vertices in Set A, B1 and B2, are each connected to three in Set B.But given that each vertex in Set B is connected to three in Set A, which includes either one or two white vertices.Wait, perhaps it's better to think in terms of the cube's adjacency.Let me try to imagine the cube. Let's label the vertices in Set A as follows:- A1: (0,0,0)- A2: (0,1,1)- A3: (1,0,1)- A4: (1,1,0)And Set B as:- B1: (0,0,1)- B2: (0,1,0)- B3: (1,0,0)- B4: (1,1,1)This is a standard bipartition of the cube.Now, each vertex in Set A is connected to three in Set B:- A1: connected to B1, B2, B3- A2: connected to B1, B2, B4- A3: connected to B1, B3, B4- A4: connected to B2, B3, B4Similarly, each vertex in Set B is connected to three in Set A:- B1: connected to A1, A2, A3- B2: connected to A1, A2, A4- B3: connected to A1, A3, A4- B4: connected to A2, A3, A4Okay, so if we have two white vertices in Set A, say A1 and A2 are white, and A3 and A4 are black.Then, for each vertex in Set B:- B1 is connected to A1 (white), A2 (white), A3 (black). So, B1 has two white and one black neighbor.- B2 is connected to A1 (white), A2 (white), A4 (black). So, B2 has two white and one black neighbor.- B3 is connected to A1 (white), A3 (black), A4 (black). So, B3 has one white and two black neighbors.- B4 is connected to A2 (white), A3 (black), A4 (black). So, B4 has one white and two black neighbors.Therefore, in this case, B1 and B2 each have two white neighbors, and B3 and B4 each have one white neighbor.Now, considering the constraints:1. For each white vertex in Set A (A1 and A2), their neighbors in Set B (B1, B2, B3 for A1; B1, B2, B4 for A2) must not all be black.So, for A1: B1, B2, B3 cannot all be black.For A2: B1, B2, B4 cannot all be black.2. For each black vertex in Set A (A3 and A4), their neighbors in Set B (for A3: B1, B3, B4; for A4: B2, B3, B4) must not all be white.So, for A3: B1, B3, B4 cannot all be white.For A4: B2, B3, B4 cannot all be white.Therefore, when coloring Set B, we have the following constraints:- B1 and B2 cannot both be black if B3 is black (from A1's constraint).Wait, actually, more precisely:From A1's constraint: B1, B2, B3 cannot all be black.Similarly, from A2's constraint: B1, B2, B4 cannot all be black.From A3's constraint: B1, B3, B4 cannot all be white.From A4's constraint: B2, B3, B4 cannot all be white.Therefore, these are four constraints on the coloring of Set B.So, how many colorings of Set B satisfy all these constraints?Each vertex in Set B can be black or white, so 16 total possibilities.But we need to subtract the colorings that violate any of the four constraints.Let me enumerate the invalid colorings.First, the colorings where B1, B2, B3 are all black: This violates A1's constraint.Second, the colorings where B1, B2, B4 are all black: This violates A2's constraint.Third, the colorings where B1, B3, B4 are all white: This violates A3's constraint.Fourth, the colorings where B2, B3, B4 are all white: This violates A4's constraint.Additionally, we need to check if any colorings violate multiple constraints, to avoid double-counting.So, let's calculate the number of invalid colorings:1. B1, B2, B3 all black: 1 coloring (B1=B, B2=B, B3=B; B4 can be anything? Wait, no, because the constraint is only about B1, B2, B3 being black. So, actually, for this constraint, B4 can be black or white. So, actually, this is 2 colorings: B1=B, B2=B, B3=B, B4=B or W.Wait, no, hold on. The constraint is that B1, B2, B3 cannot all be black. So, if B1, B2, B3 are all black, regardless of B4, it's invalid. So, how many colorings have B1=B, B2=B, B3=B? For each such coloring, B4 can be B or W, so two colorings.Similarly, the constraint for B1, B2, B4 all black: two colorings (B4 is fixed as B, and B1, B2, B4 are B; B3 can be B or W, but wait, no, B1, B2, B4 must be all black, so B3 can be anything. Wait, actually, no, the constraint is about B1, B2, B4 being all black, so regardless of B3, it's invalid.Wait, no, if B1, B2, B4 are all black, then regardless of B3, it's invalid. So, for B1, B2, B4 being all black, B3 can be B or W, so two colorings.Similarly, the constraint for B1, B3, B4 all white: two colorings (B1, B3, B4 are W, B2 can be B or W).And the constraint for B2, B3, B4 all white: two colorings (B2, B3, B4 are W, B1 can be B or W).So, in total, we have 2 + 2 + 2 + 2 = 8 invalid colorings.But wait, we need to check if any of these invalid colorings overlap, i.e., a coloring that violates more than one constraint.For example, is there a coloring where both B1, B2, B3 are black and B1, B2, B4 are black? That would require B1, B2, B3, B4 all being black, which is one coloring.Similarly, is there a coloring where both B1, B3, B4 are white and B2, B3, B4 are white? That would require B1, B2, B3, B4 all being white, which is another coloring.Also, is there a coloring that violates both a black and a white constraint? For example, B1, B2, B3 all black and B1, B3, B4 all white. Is that possible? Let's see: If B1, B2, B3 are black, and B1, B3, B4 are white, then B1 must be both black and white, which is impossible. So, no overlap between black and white constraints.Similarly, other overlaps: If a coloring violates two black constraints, like B1, B2, B3 all black and B1, B2, B4 all black, that requires all B1, B2, B3, B4 black, as above. Similarly, violating two white constraints requires all white.Therefore, the total number of invalid colorings is 8, but we have two colorings that were counted twice: all black and all white. So, we need to subtract those overlaps.So, total invalid colorings = 8 - 2 = 6.Wait, let me think again.Each of the four constraints individually gives 2 invalid colorings, totaling 8. However, the all-black coloring violates two constraints (B1,B2,B3 and B1,B2,B4 all black), and the all-white coloring violates two constraints (B1,B3,B4 and B2,B3,B4 all white). So, these two colorings are each counted twice in the 8.Therefore, the actual number of unique invalid colorings is 8 - 2 = 6.Therefore, the number of valid colorings for Set B is 16 - 6 = 10.So, for each coloring of Set A with two white and two black vertices, there are 10 valid colorings for Set B.Since there are C(4,2) = 6 ways to choose the two white vertices in Set A, the total number of colorings in this case is 6 * 10 = 60.Case 4: Set A has one black and three white vertices. This is symmetric to Case 2, where Set A had one white and three black. So, by symmetry, the number of colorings should be the same.In Case 2, we had 4 * 7 = 28 colorings. Therefore, here, we should also have 4 * 7 = 28 colorings.Case 5: Set A has all four vertices white. Then, similar to Case 1, all vertices in Set B must be white as well. So, only one coloring: all white.Similarly, if Set A is all white, Set B must be all white.Wait, but in Case 1, we considered Set A being all black, leading to all Set B being black. Similarly, Set A all white leads to all Set B being white.So, Case 1 and Case 5 are symmetric, each giving one coloring.Therefore, summing up all the cases:- Case 1: 1 coloring- Case 2: 28 colorings- Case 3: 60 colorings- Case 4: 28 colorings- Case 5: 1 coloringTotal colorings: 1 + 28 + 60 + 28 + 1 = 118.Therefore, the total number of valid colorings is 118.</think>