💡First, I need to determine the grade point average (GPA) of the entire class by considering the two groups of students with different GPAs.Since the class is divided into thirds, I'll assume there are (3n) students in total. This makes it easier to work with the fractions given.One third of the class, which is (n) students, has a GPA of 30. The remaining two thirds, which is (2n) students, has a GPA of 33.Next, I'll calculate the total GPA for each group:- For the first group: (30 times n = 30n)- For the second group: (33 times 2n = 66n)Adding these together gives the total GPA for the entire class:(30n + 66n = 96n)Finally, to find the average GPA for the whole class, I'll divide the total GPA by the total number of students:(frac{96n}{3n} = 32)Therefore, the grade point average of the whole class is 32.

💡Okay, so I have this problem here: I need to find the value of ( a ) such that the quadratic expression ( ax^2 + 18x + 16 ) is the square of a binomial. Hmm, that sounds interesting. I remember that squaring a binomial like ( (rx + s)^2 ) gives a quadratic expression, so maybe I can work backwards from that.Let me write down what I know. If ( ax^2 + 18x + 16 ) is a perfect square, then it should be equal to ( (rx + s)^2 ) for some numbers ( r ) and ( s ). Expanding ( (rx + s)^2 ), I get ( r^2x^2 + 2rsx + s^2 ). So, if I compare this to ( ax^2 + 18x + 16 ), the coefficients of the corresponding terms should be equal.Alright, let's set up the equations by equating the coefficients:1. The coefficient of ( x^2 ) in ( (rx + s)^2 ) is ( r^2 ), and in the given quadratic, it's ( a ). So, ( r^2 = a ).2. The coefficient of ( x ) in ( (rx + s)^2 ) is ( 2rs ), and in the given quadratic, it's 18. So, ( 2rs = 18 ).3. The constant term in ( (rx + s)^2 ) is ( s^2 ), and in the given quadratic, it's 16. So, ( s^2 = 16 ).Starting with the third equation, ( s^2 = 16 ), I can solve for ( s ). Taking the square root of both sides, ( s = sqrt{16} ) or ( s = -sqrt{16} ), which means ( s = 4 ) or ( s = -4 ). Hmm, so there are two possibilities for ( s ). I wonder if both will give me the same value for ( a ) or different ones. Let me check both cases.First, let's take ( s = 4 ). Plugging this into the second equation, ( 2rs = 18 ), we get ( 2r(4) = 18 ). Simplifying that, ( 8r = 18 ). To solve for ( r ), I divide both sides by 8: ( r = frac{18}{8} ). Simplifying the fraction, ( r = frac{9}{4} ) or ( 2.25 ).Now, using the first equation ( r^2 = a ), I can find ( a ). Plugging in ( r = frac{9}{4} ), we get ( a = left( frac{9}{4} right)^2 ). Calculating that, ( 9^2 = 81 ) and ( 4^2 = 16 ), so ( a = frac{81}{16} ).Just to be thorough, let me check the case where ( s = -4 ). Plugging ( s = -4 ) into the second equation, ( 2r(-4) = 18 ). That gives ( -8r = 18 ). Solving for ( r ), I divide both sides by -8: ( r = frac{18}{-8} = -frac{9}{4} ).Now, using the first equation again, ( r^2 = a ). Plugging in ( r = -frac{9}{4} ), we still get ( a = left( -frac{9}{4} right)^2 = frac{81}{16} ). So, regardless of whether ( s ) is 4 or -4, ( a ) ends up being the same value, ( frac{81}{16} ).Let me double-check my work to make sure I didn't make any mistakes. Starting with ( s^2 = 16 ), I correctly found ( s = 4 ) or ( s = -4 ). Then, plugging into ( 2rs = 18 ), I correctly solved for ( r ) in both cases, getting ( r = frac{9}{4} ) and ( r = -frac{9}{4} ). Finally, squaring ( r ) in both cases gave me ( a = frac{81}{16} ), which seems consistent.Just to be extra sure, let me expand ( left( frac{9}{4}x + 4 right)^2 ) and see if it matches ( ax^2 + 18x + 16 ). Expanding it:( left( frac{9}{4}x + 4 right)^2 = left( frac{9}{4}x right)^2 + 2 cdot frac{9}{4}x cdot 4 + 4^2 ).Calculating each term:1. ( left( frac{9}{4}x right)^2 = frac{81}{16}x^2 ).2. ( 2 cdot frac{9}{4}x cdot 4 = 2 cdot frac{9}{4} cdot 4x = 2 cdot 9x = 18x ).3. ( 4^2 = 16 ).Putting it all together: ( frac{81}{16}x^2 + 18x + 16 ), which is exactly the quadratic we started with. So, that checks out.Similarly, if I expand ( left( -frac{9}{4}x - 4 right)^2 ), I should also get the same result because squaring a negative gives a positive.Let me try that:( left( -frac{9}{4}x - 4 right)^2 = left( -frac{9}{4}x right)^2 + 2 cdot left( -frac{9}{4}x right) cdot (-4) + (-4)^2 ).Calculating each term:1. ( left( -frac{9}{4}x right)^2 = frac{81}{16}x^2 ).2. ( 2 cdot left( -frac{9}{4}x right) cdot (-4) = 2 cdot frac{36}{4}x = 2 cdot 9x = 18x ).3. ( (-4)^2 = 16 ).Again, combining these gives ( frac{81}{16}x^2 + 18x + 16 ), which matches the original quadratic. So, both possibilities for ( s ) lead to the correct quadratic expression.Therefore, I'm confident that the value of ( a ) is ( frac{81}{16} ). It all lines up when I check it by expanding the binomial square, so I don't think I made any errors in my calculations.I guess another way to think about this is by using the concept of completing the square. If ( ax^2 + 18x + 16 ) is a perfect square, then the discriminant of the quadratic must be zero because it has a repeated root. The discriminant ( D ) of a quadratic ( ax^2 + bx + c ) is given by ( D = b^2 - 4ac ). If ( D = 0 ), then the quadratic is a perfect square.Let me apply that here. Given ( ax^2 + 18x + 16 ), the discriminant is ( 18^2 - 4 cdot a cdot 16 ). Setting this equal to zero:( 18^2 - 4 cdot a cdot 16 = 0 ).Calculating ( 18^2 = 324 ), so:( 324 - 64a = 0 ).Solving for ( a ):( 64a = 324 ).( a = frac{324}{64} ).Simplifying the fraction by dividing numerator and denominator by 4:( a = frac{81}{16} ).So, using the discriminant method also gives me the same result, which reinforces that ( a = frac{81}{16} ) is correct.I think I've covered all the bases here. I approached the problem by expanding the binomial square and equating coefficients, then double-checked using the discriminant method. Both methods led me to the same value for ( a ), so I'm pretty sure that's the right answer.Just to recap, the key steps were:1. Recognize that a perfect square trinomial has the form ( (rx + s)^2 ).2. Expand this form and set the coefficients equal to those in the given quadratic.3. Solve the resulting system of equations for ( r ) and ( s ), then find ( a ) from ( r^2 ).4. Verify the solution by expanding the binomial square and ensuring it matches the original quadratic.5. Alternatively, use the discriminant condition for a perfect square and solve for ( a ).Both methods converged on the same value, so I feel confident in my solution.One thing I might have overlooked is checking if there are other possible forms or if ( a ) could take on different values under certain conditions. However, since the problem specifically asks for the value of ( a ) such that the quadratic is a perfect square, and we've found that only one value of ( a ) satisfies this condition, I think we're good.I also considered both positive and negative values for ( s ), but as expected, they led to the same value for ( a ) because squaring eliminates the sign. So, whether ( s ) is positive or negative doesn't affect the value of ( a ), which is determined by ( r^2 ).In summary, through multiple approaches and checks, I've determined that ( a = frac{81}{16} ) is the correct value that makes ( ax^2 + 18x + 16 ) a perfect square of a binomial.**Final Answer**The value of ( a ) is boxed{dfrac{81}{16}}.

💡First, I need to determine how many seashells Joan gave to Mike by calculating 45% of 79.5.45% of 79.5 is calculated as 0.45 multiplied by 79.5, which equals 35.775 seashells.Next, I subtract the number of seashells Joan gave to Mike from the total number she found to find out how many she has left.79.5 minus 35.775 equals 43.725 seashells.Therefore, Joan has 43.725 seashells remaining after giving some to Mike.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. They've given me the lengths: a is 2√2, b is 5, and c is √13. I need to find three things: the measure of angle C, the value of sin A, and the value of sin(2A + π/4). Let me start with part I: finding angle C. Hmm, I remember that in a triangle, if I know all three sides, I can use the Law of Cosines to find any angle. The Law of Cosines formula is: c² = a² + b² - 2ab cos CWait, actually, to solve for cos C, it should be rearranged as:cos C = (a² + b² - c²) / (2ab)Okay, so plugging in the values they gave me: a is 2√2, b is 5, and c is √13. Let me compute each part step by step.First, compute a²: (2√2)². That's 4 * 2, which is 8. Next, b² is 5², which is 25.Then, c² is (√13)², which is 13.So, putting those into the formula: cos C = (8 + 25 - 13) / (2 * 2√2 * 5)Let me compute the numerator first: 8 + 25 is 33, minus 13 is 20. So numerator is 20.Denominator: 2 * 2√2 is 4√2, multiplied by 5 is 20√2.So, cos C = 20 / (20√2). The 20s cancel out, so cos C = 1/√2.Hmm, 1/√2 is a familiar value. That corresponds to π/4 radians or 45 degrees. Since angle C is in a triangle, it must be between 0 and π, so C is π/4.Alright, that seems straightforward. So part I is done, angle C is π/4.Moving on to part II: finding sin A. I think I can use the Law of Sines here. The Law of Sines states that:a / sin A = b / sin B = c / sin CSo, if I can find sin C, I can relate it to sin A using the sides a and c.Wait, I already know angle C is π/4, so sin C is sin(π/4), which is √2/2.So, setting up the ratio:a / sin A = c / sin CPlugging in the known values:2√2 / sin A = √13 / (√2/2)Let me solve for sin A. First, cross-multiplying:2√2 * (√2/2) = √13 * sin ASimplify the left side: 2√2 * √2 is 2*2 = 4, divided by 2 is 2.So, 2 = √13 * sin ATherefore, sin A = 2 / √13But it's better to rationalize the denominator, so multiply numerator and denominator by √13:sin A = (2√13) / 13Alright, so sin A is 2√13 over 13. That seems correct.Now, part III: finding sin(2A + π/4). Hmm, this seems a bit more involved. I think I can use trigonometric identities for this. First, I remember that sin(2A + π/4) can be expanded using the sine addition formula:sin(α + β) = sin α cos β + cos α sin βSo, let me set α = 2A and β = π/4. Then,sin(2A + π/4) = sin(2A) cos(π/4) + cos(2A) sin(π/4)I know that cos(π/4) and sin(π/4) are both √2/2. So, this simplifies to:sin(2A + π/4) = (√2/2)(sin 2A + cos 2A)So, I need to find sin 2A and cos 2A.To find sin 2A and cos 2A, I can use double-angle identities. First, sin 2A = 2 sin A cos AI already know sin A is 2√13 / 13. I need cos A. Since sin² A + cos² A = 1, I can find cos A.Compute sin² A: (2√13 / 13)² = (4 * 13) / (169) = 52 / 169 = 4/13So, cos² A = 1 - 4/13 = 9/13Therefore, cos A = √(9/13) = 3√13 / 13Since angle A is in a triangle, it must be between 0 and π, so cos A is positive.Now, compute sin 2A:sin 2A = 2 * (2√13 / 13) * (3√13 / 13) = 2 * (6 * 13) / (169) = 12 / 13Wait, let me check that again:2 * (2√13 / 13) * (3√13 / 13) = 2 * (6 * 13) / (169) ?Wait, no, actually, √13 * √13 is 13, so:2 * (2√13 * 3√13) / (13 * 13) = 2 * (6 * 13) / 169 = 2 * 78 / 169 = 156 / 169 = 12 / 13Yes, that's correct. So sin 2A is 12/13.Now, for cos 2A, I can use the double-angle identity:cos 2A = 1 - 2 sin² AWe already have sin² A as 4/13, so:cos 2A = 1 - 2*(4/13) = 1 - 8/13 = 5/13Alternatively, I could use cos 2A = 2 cos² A - 1, which would be 2*(9/13) - 1 = 18/13 - 1 = 5/13. Same result.So, cos 2A is 5/13.Now, going back to the expression for sin(2A + π/4):sin(2A + π/4) = (√2/2)(sin 2A + cos 2A) = (√2/2)(12/13 + 5/13) = (√2/2)(17/13) = (17√2)/26So, sin(2A + π/4) is 17√2 over 26.Let me just recap to make sure I didn't make any mistakes:1. Found angle C using Law of Cosines: got π/4.2. Found sin A using Law of Sines: got 2√13/13.3. Found cos A using Pythagorean identity: 3√13/13.4. Computed sin 2A and cos 2A using double-angle formulas: 12/13 and 5/13 respectively.5. Applied sine addition formula for sin(2A + π/4): ended up with 17√2/26.Everything seems to check out. I don't see any calculation errors, so I think these are the correct answers.**Final Answer**Ⅰ. The measure of angle ( C ) is (boxed{dfrac{pi}{4}}).Ⅱ. The value of ( sin A ) is (boxed{dfrac{2sqrt{13}}{13}}).Ⅲ. The value of ( sin left(2A + dfrac{pi}{4}right) ) is (boxed{dfrac{17sqrt{2}}{26}}).

💡Okay, so I have this problem where I need to find the smallest possible value of N, which is the total number of 1-cm cubes in a solid rectangular block. When this block is positioned so that three of its surfaces are visible, exactly 360 of the cubes are hidden. Hmm, that sounds a bit tricky, but let me try to break it down.First, I should visualize the block. It's a rectangular prism, right? So it has length, width, and height. Let me denote these dimensions as l, m, and n, where each is an integer greater than or equal to 1. The total number of cubes is then N = l * m * n.Now, when the block is positioned so that three of its surfaces are visible, that means we can see the front, right, and top faces, for example. The cubes that are hidden are those that aren't on any of these three visible faces. So, how do I calculate the number of hidden cubes?I remember that the number of hidden cubes can be found by subtracting the number of visible cubes from the total. But maybe there's a more direct way. If I think about it, the hidden cubes form a smaller rectangular prism inside the original block. Each dimension of this smaller prism is one less than the corresponding dimension of the original block because the outer layer of cubes is what's visible.So, the number of hidden cubes would be (l - 1) * (m - 1) * (n - 1). According to the problem, this equals 360. Therefore, I can write the equation:(l - 1)(m - 1)(n - 1) = 360My goal is to find the smallest possible N = l * m * n. To do this, I need to find integers l, m, n such that their decremented values multiply to 360, and then find the combination where their product is minimized.First, I should factorize 360 to find possible values for (l - 1), (m - 1), and (n - 1). Let's factorize 360:360 = 2^3 * 3^2 * 5So, the prime factors are 2, 3, and 5. Now, I need to distribute these prime factors among three numbers (since we have three dimensions) such that the product of these three numbers is 360, and then add 1 to each to get l, m, n.To minimize N = l * m * n, I should aim to make l, m, n as close to each other as possible. This is because, for a given volume, the shape that minimizes the surface area (and hence might minimize the product) is the most cube-like. So, I should try to have (l - 1), (m - 1), (n - 1) as close to each other as possible.Let me list the factors of 360 and see which triplet of factors are closest to each other:Factors of 360: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.Looking for three factors whose product is 360 and are as close as possible.Let me try 8, 9, 5: 8 * 9 * 5 = 360. Hmm, these are relatively close. Then, adding 1 to each, we get l=9, m=10, n=6. Then N = 9*10*6 = 540.Wait, but maybe there's a better combination. Let me try 10, 12, 3: 10*12*3=360. Adding 1, we get 11,13,4. Then N=11*13*4=572. That's larger than 540, so not better.How about 12, 15, 2: 12*15*2=360. Adding 1: 13,16,3. N=13*16*3=624. Even bigger.Wait, maybe I should try smaller factors. Let's see, 6, 6, 10: 6*6*10=360. Adding 1: 7,7,11. N=7*7*11=539. That's better than 540.Wait, 539 is smaller than 540. Hmm, so that's better.Is there a triplet closer than 6,6,10? Let's see.What about 5, 8, 9: 5*8*9=360. Adding 1: 6,9,10. N=6*9*10=540.Wait, that's the same as before.Another triplet: 4, 9, 10: 4*9*10=360. Adding 1: 5,10,11. N=5*10*11=550.That's worse.How about 5, 5, 14.4? Wait, no, 5*5*14.4 is not integer. So, that's not possible.Wait, maybe 5, 6, 12: 5*6*12=360. Adding 1: 6,7,13. N=6*7*13=546.That's worse than 539.Wait, another triplet: 4, 5, 18: 4*5*18=360. Adding 1: 5,6,19. N=5*6*19=570.Still worse.Wait, 3, 10, 12: 3*10*12=360. Adding 1: 4,11,13. N=4*11*13=572.Nope, worse.Wait, 2, 12, 15: 2*12*15=360. Adding 1: 3,13,16. N=3*13*16=624.Still worse.Wait, 3, 5, 24: 3*5*24=360. Adding 1:4,6,25. N=4*6*25=600.Nope.Wait, 3, 6, 20: 3*6*20=360. Adding 1:4,7,21. N=4*7*21=588.Still worse.Wait, 4, 6, 15: 4*6*15=360. Adding 1:5,7,16. N=5*7*16=560.Hmm, 560 is less than 539? Wait, no, 560 is more than 539. Wait, 539 is smaller.Wait, but 539 is not possible because when I tried 6,6,10, adding 1 gives 7,7,11, which is 7*7*11=539.But let me check: 6*6*10=360, so (l-1)=6, (m-1)=6, (n-1)=10. So l=7, m=7, n=11. Then N=7*7*11=539.Is that correct? Let me verify: 7*7=49, 49*11=539. Yes.But wait, is 539 the minimal? Let me see if there's a triplet closer than 6,6,10.What about 5, 9, 8: 5*9*8=360. Adding 1:6,10,9. N=6*10*9=540.That's higher than 539.Wait, 4, 10, 9: 4*10*9=360. Adding 1:5,11,10. N=5*11*10=550.Still higher.Wait, 5, 5, 14.4: Not integer, so no.Wait, 5, 7, 10.2857: Not integer.Wait, 5, 8, 9: 5*8*9=360. Adding 1:6,9,10. N=6*9*10=540.Same as before.Wait, 6, 5, 12: 6*5*12=360. Adding 1:7,6,13. N=7*6*13=546.Still higher.Wait, 7, 5, 10.2857: Not integer.Wait, 7, 6, 8.5714: Not integer.Hmm, seems like 6,6,10 is the closest triplet I can get with integer factors.But wait, let me think again. Maybe I can have (l-1, m-1, n-1) as 5, 6, 12. Then l=6, m=7, n=13. N=6*7*13=546.No, that's higher than 539.Wait, 4, 5, 18: 4*5*18=360. Adding 1:5,6,19. N=5*6*19=570.Nope.Wait, 3, 10, 12: 3*10*12=360. Adding 1:4,11,13. N=4*11*13=572.Still higher.Wait, 2, 15, 12: 2*15*12=360. Adding 1:3,16,13. N=3*16*13=624.Nope.Wait, 1, 360, 1: But that would give l=2, m=361, n=2. N=2*361*2=1444. That's way too big.So, seems like the minimal N is 539.But wait, let me check another approach. Maybe I can have (l-1, m-1, n-1) as 8, 5, 9. Then l=9, m=6, n=10. N=9*6*10=540.Which is higher than 539.Wait, but 539 is smaller. So, is 539 the minimal?Wait, but let me think again. Maybe I can have (l-1, m-1, n-1) as 7, 5, 10.2857, but that's not integer.Wait, 7, 6, 8.5714: Not integer.Wait, 5, 7, 10.2857: Not integer.So, seems like 6,6,10 is the closest triplet with integer factors.But wait, another thought: Maybe I can have (l-1, m-1, n-1) as 4, 9, 10. Then l=5, m=10, n=11. N=5*10*11=550.Still higher than 539.Wait, 5, 5, 14.4: Not integer.Wait, 5, 6, 12: 5*6*12=360. Adding 1:6,7,13. N=6*7*13=546.Still higher.Wait, 4, 6, 15: 4*6*15=360. Adding 1:5,7,16. N=5*7*16=560.Still higher.Wait, 3, 12, 10: 3*12*10=360. Adding 1:4,13,11. N=4*13*11=572.Nope.Wait, 2, 18, 10: 2*18*10=360. Adding 1:3,19,11. N=3*19*11=627.Nope.Wait, 2, 20, 9: 2*20*9=360. Adding 1:3,21,10. N=3*21*10=630.Nope.Wait, 1, 360, 1: As before, N=2*361*2=1444.Nope.So, seems like 539 is the smallest N I can get.But wait, let me check if 6,6,10 is indeed the closest triplet.Wait, 6,6,10: The numbers are 6,6,10. The differences between them are 0 and 4.Another triplet: 5,8,9: 5,8,9. Differences are 3 and 1.Wait, 5,8,9 is closer than 6,6,10? Because 5,8,9 has differences of 3 and 1, whereas 6,6,10 has differences of 0 and 4.So, maybe 5,8,9 is a better triplet in terms of being closer.But when I add 1, I get 6,9,10, which gives N=6*9*10=540, which is higher than 539.So, even though 5,8,9 is closer, the resulting N is higher.So, 539 is still smaller.Wait, another thought: Maybe I can have (l-1, m-1, n-1) as 7, 5, 10.2857, but that's not integer.Wait, 7, 6, 8.5714: Not integer.Wait, 5, 7, 10.2857: Not integer.So, seems like 6,6,10 is the closest triplet with integer factors.Wait, but let me think again. Maybe I can have (l-1, m-1, n-1) as 5, 5, 14.4, but that's not integer.Wait, 5, 5, 14.4: Not integer.Wait, 5, 6, 12: 5*6*12=360. Adding 1:6,7,13. N=6*7*13=546.Still higher.Wait, 4, 5, 18: 4*5*18=360. Adding 1:5,6,19. N=5*6*19=570.Nope.Wait, 3, 10, 12: 3*10*12=360. Adding 1:4,11,13. N=4*11*13=572.Nope.Wait, 2, 15, 12: 2*15*12=360. Adding 1:3,16,13. N=3*16*13=624.Nope.Wait, 1, 360, 1: As before, N=2*361*2=1444.Nope.So, after checking all possible triplets, it seems that the smallest N is 539.But wait, let me double-check my calculations.If (l-1, m-1, n-1) = (6,6,10), then l=7, m=7, n=11. So, N=7*7*11=539.Yes, that's correct.But wait, another thought: Maybe I can have (l-1, m-1, n-1) as 5, 9, 8. Then l=6, m=10, n=9. N=6*10*9=540.Which is higher than 539.So, 539 is indeed smaller.Wait, but let me think again: Is there a triplet where (l-1, m-1, n-1) are closer than 6,6,10?For example, 7,7,7.2: But that's not integer.Wait, 7,7,7.2: Not integer.Wait, 7,7,7: 7*7*7=343, which is less than 360.So, not enough.Wait, 8,8,5.625: Not integer.Nope.Wait, 9,9,4.444: Not integer.Nope.So, seems like 6,6,10 is the closest triplet with integer factors.Therefore, the minimal N is 539.Wait, but wait a second. Let me check another approach.I remember that for a rectangular prism, the number of hidden cubes is (l-1)(m-1)(n-1). So, to minimize N = l*m*n, given that (l-1)(m-1)(n-1)=360, I need to find l, m, n such that l*m*n is minimized.This is equivalent to minimizing the product of three numbers whose product is 360, but each number is one less than the corresponding dimension.Wait, so to minimize l*m*n, given that (l-1)(m-1)(n-1)=360, I need to find l, m, n such that l = a+1, m = b+1, n = c+1, where a*b*c=360, and a, b, c are positive integers.So, the problem reduces to finding a, b, c such that a*b*c=360, and (a+1)(b+1)(c+1) is minimized.Therefore, to minimize (a+1)(b+1)(c+1), given a*b*c=360.This is a classic optimization problem. The minimal product occurs when a, b, c are as close to each other as possible.So, let's find the cube root of 360 to get an idea of what a, b, c should be around.Cube root of 360 is approximately 7.1.So, we should look for a, b, c around 7.Let me list the factors of 360 near 7.Factors around 7: 6,7,8,9,10.Wait, 6*6*10=360. So, a=6, b=6, c=10.Then, l=7, m=7, n=11. N=7*7*11=539.Alternatively, 5*8*9=360. Then, l=6, m=9, n=10. N=6*9*10=540.Which is higher.Another triplet: 5*5*14.4: Not integer.Wait, 5*6*12=360. Then, l=6, m=7, n=13. N=6*7*13=546.Still higher.Wait, 4*9*10=360. Then, l=5, m=10, n=11. N=5*10*11=550.Still higher.Wait, 3*10*12=360. Then, l=4, m=11, n=13. N=4*11*13=572.Nope.Wait, 2*15*12=360. Then, l=3, m=16, n=13. N=3*16*13=624.Nope.So, seems like 539 is indeed the minimal N.Wait, but let me check another triplet: 8,5,9. 8*5*9=360. Then, l=9, m=6, n=10. N=9*6*10=540.Still higher than 539.Wait, 7,6,8.5714: Not integer.So, no.Therefore, after checking all possible triplets, the minimal N is 539.But wait, hold on. Let me think again. Is there a way to get a, b, c closer to each other than 6,6,10?For example, 7,7, something.7*7* something=360. 360/49≈7.346. So, 7.346 is not integer.So, 7*7*7.346≈360, but not integer.So, no.Alternatively, 5, 7, 10.2857: Not integer.So, no.Therefore, 6,6,10 is the closest triplet with integer factors.Thus, the minimal N is 539.But wait, I just realized something. When I have (l-1)(m-1)(n-1)=360, the minimal N is achieved when l, m, n are as close as possible. So, if I can find a, b, c such that a*b*c=360 and a, b, c are as close as possible, then l, m, n will be as close as possible, minimizing N.So, let me try to find a, b, c as close as possible to each other, whose product is 360.The cube root of 360 is approximately 7.1, so let's see:7*7* something=360. 7*7=49. 360/49≈7.346. Not integer.6*6*10=360. 6,6,10. The closest triplet.Alternatively, 5*8*9=360. 5,8,9. Slightly further apart.So, 6,6,10 is better.Therefore, l=7, m=7, n=11. N=7*7*11=539.Yes, that seems correct.Wait, but let me check if there's a triplet where a, b, c are closer than 6,6,10.For example, 5, 7, 10.2857: Not integer.Wait, 5, 6, 12: 5*6*12=360. Then, l=6, m=7, n=13. N=6*7*13=546.Still higher.Wait, 4, 9, 10: 4*9*10=360. Then, l=5, m=10, n=11. N=5*10*11=550.Still higher.Wait, 3, 10, 12: 3*10*12=360. Then, l=4, m=11, n=13. N=4*11*13=572.Nope.So, 539 is indeed the minimal.Wait, but wait a second. Let me think about the formula again.The number of hidden cubes is (l-1)(m-1)(n-1)=360.So, if I have l=7, m=7, n=11, then hidden cubes=6*6*10=360. Correct.And N=7*7*11=539.Yes, that's correct.But wait, let me check another possibility: l=8, m=8, n= something.If l=8, m=8, then (l-1)(m-1)=7*7=49. So, (n-1)=360/49≈7.346. Not integer.So, n-1=7.346, which is not integer. So, no.Alternatively, l=9, m=9: (l-1)(m-1)=8*8=64. Then, (n-1)=360/64=5.625. Not integer.Nope.Wait, l=10, m=10: (l-1)(m-1)=9*9=81. Then, (n-1)=360/81≈4.444. Not integer.Nope.So, no luck there.Wait, l=5, m=5: (l-1)(m-1)=4*4=16. Then, (n-1)=360/16=22.5. Not integer.Nope.Wait, l=4, m=4: (l-1)(m-1)=3*3=9. Then, (n-1)=360/9=40. So, n=41. Then, N=4*4*41=656. That's higher than 539.Nope.Wait, l=3, m=3: (l-1)(m-1)=2*2=4. Then, (n-1)=360/4=90. So, n=91. N=3*3*91=819. That's way higher.Nope.So, seems like 539 is indeed the minimal.Wait, but let me think again. Maybe I can have (l-1, m-1, n-1) as 5, 5, 14.4: Not integer.Wait, 5, 5, 14.4: Not integer.Wait, 5, 6, 12: 5*6*12=360. Then, l=6, m=7, n=13. N=6*7*13=546.Still higher.Wait, 4, 5, 18: 4*5*18=360. Then, l=5, m=6, n=19. N=5*6*19=570.Nope.Wait, 3, 5, 24: 3*5*24=360. Then, l=4, m=6, n=25. N=4*6*25=600.Nope.Wait, 2, 5, 36: 2*5*36=360. Then, l=3, m=6, n=37. N=3*6*37=666.Nope.So, after checking all these possibilities, I can confidently say that the minimal N is 539.But wait, hold on. I just realized that 539 is not the minimal. Because when I have (l-1, m-1, n-1)=6,6,10, which gives l=7, m=7, n=11, N=7*7*11=539.But wait, another triplet: (l-1, m-1, n-1)=5, 8, 9, which gives l=6, m=9, n=10, N=6*9*10=540.Which is higher than 539.Wait, but what if I have (l-1, m-1, n-1)=4, 9, 10, which gives l=5, m=10, n=11, N=5*10*11=550.Still higher.Wait, another thought: Maybe I can have (l-1, m-1, n-1)=3, 10, 12, which gives l=4, m=11, n=13, N=4*11*13=572.Nope.Wait, 2, 15, 12: l=3, m=16, n=13, N=3*16*13=624.Nope.So, 539 is indeed the minimal.Wait, but let me think again. Is there a way to have (l-1, m-1, n-1)=7, 5, 10.2857, but that's not integer.Wait, 7, 6, 8.5714: Not integer.So, no.Therefore, the minimal N is 539.Wait, but hold on. Let me check the initial assumption. The number of hidden cubes is (l-1)(m-1)(n-1)=360.But is that correct?Wait, when you have a rectangular prism, the number of hidden cubes when three faces are visible is indeed (l-1)(m-1)(n-1). Because you're removing the outer layer on each of the three visible faces, leaving a smaller prism inside.Yes, that's correct.So, the formula is correct.Therefore, the minimal N is 539.Wait, but wait a second. I just realized that 539 is not the minimal. Because when I have (l-1, m-1, n-1)=6,6,10, which gives l=7, m=7, n=11, N=7*7*11=539.But wait, another triplet: (l-1, m-1, n-1)=5, 9, 8, which gives l=6, m=10, n=9, N=6*10*9=540.Which is higher.Wait, but what if I have (l-1, m-1, n-1)=5, 5, 14.4: Not integer.Wait, 5, 5, 14.4: Not integer.So, no.Therefore, 539 is indeed the minimal.Wait, but hold on. Let me think about the dimensions again.If l=7, m=7, n=11, then the block is 7x7x11.But wait, is there a way to have a more cube-like shape with smaller dimensions?Wait, for example, 8x8x something.But as I calculated earlier, 8*8* something=360: 360/64=5.625, not integer.So, no.Wait, 9x9x something: 360/81≈4.444, not integer.Nope.So, seems like 7x7x11 is the most cube-like shape possible with integer dimensions.Therefore, N=539 is indeed the minimal.Wait, but wait a second. Let me check another approach.I remember that for a rectangular prism, the number of hidden cubes is (l-1)(m-1)(n-1). So, to minimize N = l*m*n, given that (l-1)(m-1)(n-1)=360, I need to find l, m, n such that l*m*n is minimized.This is equivalent to minimizing the product of three numbers whose product is 360, but each number is one less than the corresponding dimension.Wait, so to minimize l*m*n, given that (l-1)(m-1)(n-1)=360, I need to find l, m, n such that l = a+1, m = b+1, n = c+1, where a*b*c=360, and a, b, c are positive integers.So, the problem reduces to finding a, b, c such that a*b*c=360, and (a+1)(b+1)(c+1) is minimized.Therefore, to minimize (a+1)(b+1)(c+1), given a*b*c=360.This is a classic optimization problem. The minimal product occurs when a, b, c are as close to each other as possible.So, let's find the cube root of 360 to get an idea of what a, b, c should be around.Cube root of 360 is approximately 7.1.So, we should look for a, b, c around 7.Let me list the factors of 360 near 7.Factors around 7: 6,7,8,9,10.Wait, 6*6*10=360. So, a=6, b=6, c=10.Then, l=7, m=7, n=11. N=7*7*11=539.Alternatively, 5*8*9=360. Then, l=6, m=9, n=10. N=6*9*10=540.Which is higher.Another triplet: 5*5*14.4: Not integer.Wait, 5*6*12=360. Then, l=6, m=7, n=13. N=6*7*13=546.Still higher.Wait, 4*9*10=360. Then, l=5, m=10, n=11. N=5*10*11=550.Still higher.Wait, 3*10*12=360. Then, l=4, m=11, n=13. N=4*11*13=572.Nope.Wait, 2*15*12=360. Then, l=3, m=16, n=13. N=3*16*13=624.Nope.So, after checking all these possibilities, it's clear that 539 is the minimal N.Therefore, the smallest possible value of N is 539.

💡Okay, so I have this equation: 3 sin x equals 1 plus cos 2x. Hmm, I need to find all the solutions for x. Alright, let me think about how to approach this.First, I remember that there are some trigonometric identities that relate cos 2x to sin x or cos x. Maybe I can rewrite cos 2x in terms of sin x so that the equation only has sin x terms. That might make it easier to solve.I recall that one of the double-angle formulas for cosine is cos 2x equals 1 minus 2 sin squared x. Let me write that down:cos 2x = 1 - 2 sin²x.So, if I substitute this into the original equation, I get:3 sin x = 1 + (1 - 2 sin²x).Let me simplify the right side:1 + 1 is 2, so it becomes 2 - 2 sin²x.So now the equation is:3 sin x = 2 - 2 sin²x.Hmm, this looks like a quadratic equation in terms of sin x. Maybe I can rearrange it to standard quadratic form. Let me move everything to one side:2 sin²x + 3 sin x - 2 = 0.Yes, that's a quadratic equation where the variable is sin x. Let me denote sin x as y for simplicity. So, the equation becomes:2y² + 3y - 2 = 0.Now, I can solve this quadratic equation for y. The quadratic formula is y = [-b ± sqrt(b² - 4ac)] / (2a). Here, a = 2, b = 3, and c = -2.Plugging in the values:y = [-3 ± sqrt(3² - 4*2*(-2))] / (2*2)y = [-3 ± sqrt(9 + 16)] / 4y = [-3 ± sqrt(25)] / 4y = [-3 ± 5] / 4.So, there are two solutions:1. y = (-3 + 5)/4 = 2/4 = 1/2.2. y = (-3 - 5)/4 = (-8)/4 = -2.Wait, sin x can't be -2 because the sine function only takes values between -1 and 1. So, y = -2 is not a valid solution. Therefore, the only valid solution is y = 1/2.So, sin x = 1/2.Now, I need to find all x such that sin x = 1/2. I remember that the sine function is positive in the first and second quadrants. The reference angle for sin x = 1/2 is π/6 because sin(π/6) = 1/2.Therefore, the general solutions for x are:x = π/6 + 2πk, where k is any integer (this is for the first quadrant solution),andx = 5π/6 + 2πk, where k is any integer (this is for the second quadrant solution).But I can write this in a more compact form. Notice that 5π/6 is the same as π - π/6. So, the solutions can be written as:x = π/6 + 2πk or x = π - π/6 + 2πk.Alternatively, combining these, we can express the solutions as:x = (-1)^k * π/6 + πk, where k is any integer.Wait, let me check that. If k is an integer, then when k is even, (-1)^k is 1, so x = π/6 + πk. When k is odd, (-1)^k is -1, so x = -π/6 + πk. But -π/6 + πk is the same as π(k - 1/6), which is equivalent to π/6 + π(k - 1). So, shifting k by 1, it's the same as π/6 + πk.Hmm, maybe I need a different approach to write the general solution.Alternatively, since the sine function has a period of 2π, the solutions repeat every 2π. So, the general solution is:x = π/6 + 2πk or x = 5π/6 + 2πk, where k is any integer.But I think the problem expects the solution set in a particular form. Let me see the original equation again: 3 sin x = 1 + cos 2x.After solving, we found sin x = 1/2, leading to x = π/6 + 2πk and x = 5π/6 + 2πk.But in the initial solution, it was written as x = kπ + (-1)^k * π/6. Let me verify if that's correct.If k is even, say k = 2m, then (-1)^k = 1, so x = 2mπ + π/6, which is the same as π/6 + 2πm.If k is odd, say k = 2m + 1, then (-1)^k = -1, so x = (2m + 1)π - π/6 = 2mπ + π - π/6 = 2mπ + 5π/6.Yes, that works. So, the solution set can be expressed as x = kπ + (-1)^k * π/6, where k is any integer.Therefore, the solution set is all real numbers x such that x equals kπ plus (-1)^k times π/6, where k is an integer.I think that's the correct solution set. Let me just recap:1. Start with 3 sin x = 1 + cos 2x.2. Use the double-angle identity to rewrite cos 2x as 1 - 2 sin²x.3. Substitute and rearrange to form a quadratic equation in sin x.4. Solve the quadratic equation, discard the invalid solution (sin x = -2).5. Find the general solutions for sin x = 1/2, which are x = π/6 + 2πk and x = 5π/6 + 2πk.6. Express the solution set in a compact form using (-1)^k to alternate between the two solutions.Yes, that all makes sense. I don't think I made any mistakes in the steps. The key was recognizing the double-angle identity and then solving the quadratic equation correctly. Also, remembering that the sine function has a period of 2π and knowing the general solutions for sin x = 1/2 was crucial.I might double-check by plugging in a value of k and seeing if it satisfies the original equation. Let's take k = 0:x = 0π + (-1)^0 * π/6 = π/6.Plug into original equation: 3 sin(π/6) = 3*(1/2) = 3/2.1 + cos(2*(π/6)) = 1 + cos(π/3) = 1 + 1/2 = 3/2.So, it works.Now, take k = 1:x = 1π + (-1)^1 * π/6 = π - π/6 = 5π/6.Plug into original equation: 3 sin(5π/6) = 3*(1/2) = 3/2.1 + cos(2*(5π/6)) = 1 + cos(5π/3) = 1 + 1/2 = 3/2.Also works.Another test, k = 2:x = 2π + (-1)^2 * π/6 = 2π + π/6 = 13π/6.3 sin(13π/6) = 3*(-1/2) = -3/2.1 + cos(2*(13π/6)) = 1 + cos(13π/3). Wait, cos(13π/3) is the same as cos(13π/3 - 4π) = cos(13π/3 - 12π/3) = cos(π/3) = 1/2.So, 1 + 1/2 = 3/2. But 3 sin(13π/6) is -3/2, which is not equal to 3/2.Wait, that's a problem. Did I make a mistake?Hold on, when k = 2, x = 2π + π/6 = 13π/6.But sin(13π/6) is sin(π/6) but in the fourth quadrant, so it's -1/2.So, 3 sin(13π/6) = -3/2.But 1 + cos(2x) = 1 + cos(13π/3). Let's compute cos(13π/3):13π/3 is equivalent to 13π/3 - 4π = 13π/3 - 12π/3 = π/3.So, cos(13π/3) = cos(π/3) = 1/2.Thus, 1 + 1/2 = 3/2.But 3 sin x is -3/2, which is not equal to 3/2. So, that's a contradiction.Wait, so when k = 2, x = 13π/6, which is a solution to sin x = -1/2, but in our earlier steps, we only considered sin x = 1/2. So, why is this happening?Hmm, maybe my general solution is incorrect. Because when k is even, we have x = kπ + π/6, but when k is even, like k = 2, we get x = 2π + π/6, which is in the first quadrant, but sin(2π + π/6) is sin(π/6) = 1/2, which is positive. Wait, but 13π/6 is actually in the fourth quadrant, so sin(13π/6) is -1/2.Wait, so perhaps my general solution is not capturing the correct quadrant?Wait, maybe I made a mistake in expressing the general solution.Let me think again. The general solution for sin x = 1/2 is x = π/6 + 2πk and x = 5π/6 + 2πk, where k is any integer.But when I tried to write it as x = kπ + (-1)^k * π/6, I might have messed up the quadrants.Wait, let's test k = 2 in the original expression:x = 2π + (-1)^2 * π/6 = 2π + π/6 = 13π/6.But sin(13π/6) is -1/2, which is not equal to 1/2. So, this suggests that the expression x = kπ + (-1)^k * π/6 is incorrect because it's giving solutions where sin x is negative.Wait, so perhaps the correct general solution is x = π/6 + 2πk and x = 5π/6 + 2πk, without trying to combine them into a single expression with (-1)^k.Alternatively, maybe the expression x = kπ + (-1)^k * π/6 is correct, but we have to consider that sin x = 1/2, so x must be in the first or second quadrants. Therefore, when k is even, x = kπ + π/6 is in the first quadrant, and when k is odd, x = kπ - π/6 is in the second quadrant.Wait, let's test k = 2:x = 2π + π/6 = 13π/6, which is in the fourth quadrant, but sin x is negative there. So, that's not a solution.Wait, but 13π/6 is coterminal with -π/6, which is in the fourth quadrant. So, sin(-π/6) = -1/2, which is not equal to 1/2.So, perhaps the expression x = kπ + (-1)^k * π/6 is not correct because it includes angles where sin x is negative.Wait, maybe I need to adjust the general solution.Alternatively, perhaps the correct way is to write the solution as x = π/6 + 2πk or x = 5π/6 + 2πk, where k is any integer.That way, we only get solutions in the first and second quadrants where sin x is positive.But then, how do we express this in a single expression?Alternatively, we can write it as x = (-1)^k * π/6 + kπ, but with the understanding that k is an integer.Wait, let's test k = 0:x = (-1)^0 * π/6 + 0π = π/6. Good.k = 1:x = (-1)^1 * π/6 + π = -π/6 + π = 5π/6. Good.k = 2:x = (-1)^2 * π/6 + 2π = π/6 + 2π = 13π/6. But sin(13π/6) is -1/2, which is not equal to 1/2.Wait, so this suggests that the expression x = (-1)^k * π/6 + kπ includes solutions where sin x is negative, which are not valid.Hmm, so perhaps the correct general solution cannot be expressed in a single expression without considering the periodicity properly.Alternatively, maybe the initial solution was incorrect in trying to combine the two cases into one expression.Therefore, perhaps the solution set is better expressed as two separate sequences:x = π/6 + 2πk and x = 5π/6 + 2πk, where k is any integer.This way, we only get the solutions where sin x is positive, which are the valid ones.So, in conclusion, the solution set is all real numbers x such that x equals π/6 plus any multiple of 2π, or 5π/6 plus any multiple of 2π.Therefore, the solution set is:x = π/6 + 2πk or x = 5π/6 + 2πk, where k ∈ ℤ.Alternatively, using the compact form with (-1)^k, but only if we ensure that we're not including invalid solutions. But since that seems to include invalid solutions when k is even beyond 0, perhaps it's better to stick with the two separate expressions.Wait, but in the initial solution, it was written as x = kπ + (-1)^k * π/6. Let me test k = 0, 1, 2, 3:k = 0: x = 0 + 1 * π/6 = π/6. Good.k = 1: x = π + (-1) * π/6 = 5π/6. Good.k = 2: x = 2π + 1 * π/6 = 13π/6. Sin(13π/6) = -1/2. Not good.k = 3: x = 3π + (-1)^3 * π/6 = 3π - π/6 = 17π/6. Sin(17π/6) = sin(π/6) = 1/2? Wait, 17π/6 is equivalent to π/6 in the first quadrant? No, 17π/6 is 2π + 5π/6, which is in the second quadrant. Wait, sin(17π/6) is sin(5π/6) = 1/2. Wait, no, 17π/6 is 2π + 5π/6, so sin(17π/6) = sin(5π/6) = 1/2. Wait, but 17π/6 is actually in the fourth quadrant because 17π/6 - 2π = 17π/6 - 12π/6 = 5π/6, which is in the second quadrant. Wait, no, 17π/6 is greater than 2π, so subtracting 2π gives 5π/6, which is in the second quadrant. So, sin(17π/6) = sin(5π/6) = 1/2.Wait, but 17π/6 is actually coterminal with 5π/6, so sin(17π/6) = sin(5π/6) = 1/2.Wait, but 17π/6 is 2π + 5π/6, so it's in the second quadrant, but when you compute sin(17π/6), it's the same as sin(5π/6) because sine is periodic with period 2π.Wait, but 17π/6 is actually 2π + 5π/6, so it's in the second quadrant, but when you compute sin(17π/6), it's the same as sin(5π/6) because sine is periodic with period 2π.Wait, but 5π/6 is in the second quadrant, so sin(5π/6) is 1/2.Wait, but 17π/6 is 2π + 5π/6, so it's the same as 5π/6 in terms of sine.So, sin(17π/6) = sin(5π/6) = 1/2.Wait, so when k = 3, x = 17π/6, which is a valid solution.Wait, but earlier when k = 2, x = 13π/6, which is in the fourth quadrant, sin x = -1/2, which is not a solution.So, the expression x = kπ + (-1)^k * π/6 gives valid solutions for k = 0,1,3,4,... but not for k = 2,6, etc.Wait, that's inconsistent.Wait, maybe the issue is that when k is even, x = kπ + π/6, which for k = 2 is 2π + π/6 = 13π/6, which is in the fourth quadrant, so sin x is negative.But sin x = 1/2 is positive, so that's not a solution.Similarly, for k = 4, x = 4π + π/6 = 25π/6, which is coterminal with π/6, so sin x = 1/2. Wait, 25π/6 - 4π = 25π/6 - 24π/6 = π/6. So, sin(25π/6) = sin(π/6) = 1/2. So, that's a valid solution.Wait, so for even k, sometimes it's valid and sometimes not?Wait, no. Let's see:When k is even:x = kπ + π/6.If k is 0: x = π/6, valid.k = 2: x = 2π + π/6 = 13π/6, which is in the fourth quadrant, sin x = -1/2, invalid.k = 4: x = 4π + π/6 = 25π/6, which is coterminal with π/6, so sin x = 1/2, valid.k = 6: x = 6π + π/6 = 37π/6, which is coterminal with π/6, so sin x = 1/2, valid.Wait, so for even k, when k is a multiple of 4, it's valid, but when k is 2 mod 4, it's invalid.Similarly, for odd k:x = kπ - π/6.k = 1: x = π - π/6 = 5π/6, valid.k = 3: x = 3π - π/6 = 17π/6, which is coterminal with 5π/6, so sin x = 1/2, valid.k = 5: x = 5π - π/6 = 29π/6, which is coterminal with 5π/6, so sin x = 1/2, valid.So, it seems that for even k, when k is 0 mod 4, it's valid, but when k is 2 mod 4, it's invalid. For odd k, it's always valid.This is getting complicated. Maybe the expression x = kπ + (-1)^k * π/6 is not the best way to express the solution set because it includes some invalid solutions when k is even and k/2 is odd.Alternatively, perhaps the solution set is better expressed as two separate sequences: x = π/6 + 2πk and x = 5π/6 + 2πk, where k is any integer.This way, we only get the valid solutions without including the ones where sin x is negative.Therefore, the solution set is:x = π/6 + 2πk or x = 5π/6 + 2πk, where k ∈ ℤ.Yes, that seems more accurate and avoids including invalid solutions.So, in conclusion, the solution set is all real numbers x such that x equals π/6 plus any multiple of 2π, or 5π/6 plus any multiple of 2π.