💡Okay, so I have this equation to solve: (sin^2 x + sin^2 2x + sin^2 3x = 2). Hmm, that looks a bit complicated, but maybe I can break it down step by step. First, I remember that there are some trigonometric identities that can help simplify expressions involving squares of sine functions. One of them is the identity for (sin^2 t), which is (sin^2 t = frac{1 - cos 2t}{2}). Maybe I can apply this identity to each term in the equation.Let me try that. If I rewrite each sine squared term using the identity, I get:[sin^2 x = frac{1 - cos 2x}{2},][sin^2 2x = frac{1 - cos 4x}{2},][sin^2 3x = frac{1 - cos 6x}{2}.]So, substituting these back into the original equation, it becomes:[frac{1 - cos 2x}{2} + frac{1 - cos 4x}{2} + frac{1 - cos 6x}{2} = 2.]Now, let's simplify this. If I add up all the fractions, since they all have the same denominator, I can combine the numerators:[frac{1 - cos 2x + 1 - cos 4x + 1 - cos 6x}{2} = 2.]Adding the constants in the numerator:[frac{3 - cos 2x - cos 4x - cos 6x}{2} = 2.]To make this easier, I'll multiply both sides of the equation by 2 to eliminate the denominator:[3 - cos 2x - cos 4x - cos 6x = 4.]Hmm, now I need to solve for (x). Let's rearrange the equation to isolate the cosine terms:[cos 2x + cos 4x + cos 6x = -1.]This looks tricky. Maybe I can use another trigonometric identity to combine these cosine terms. I recall that the sum of cosines can sometimes be expressed as a product. Specifically, there's an identity for the sum of cosines with angles in arithmetic progression. Let me see if that applies here.The angles here are 2x, 4x, and 6x, which are in an arithmetic sequence with a common difference of 2x. The identity for the sum of three cosines in such a sequence is:[cos A + cos (A + D) + cos (A + 2D) = 3 cos left(A + Dright) cos D.]In this case, (A = 2x) and (D = 2x), so plugging these in:[cos 2x + cos 4x + cos 6x = 3 cos 4x cos 2x.]So, substituting back into our equation:[3 cos 4x cos 2x = -1.]Hmm, that simplifies things a bit. Let me write that as:[3 cos 4x cos 2x = -1.]I can divide both sides by 3 to make it simpler:[cos 4x cos 2x = -frac{1}{3}.]This still seems a bit complicated. Maybe I can express (cos 4x) in terms of (cos 2x). I remember that (cos 4x = 2 cos^2 2x - 1). Let me substitute that in:[(2 cos^2 2x - 1) cos 2x = -frac{1}{3}.]Expanding this:[2 cos^3 2x - cos 2x = -frac{1}{3}.]Let me bring all terms to one side:[2 cos^3 2x - cos 2x + frac{1}{3} = 0.]This is a cubic equation in terms of (cos 2x). Let me set (t = cos 2x) to make it easier:[2t^3 - t + frac{1}{3} = 0.]Hmm, solving a cubic equation might be a bit involved. Maybe I can multiply through by 3 to eliminate the fraction:[6t^3 - 3t + 1 = 0.]Now, I need to solve (6t^3 - 3t + 1 = 0). Let me see if I can factor this or find rational roots. Using the Rational Root Theorem, possible rational roots are (pm1, pmfrac{1}{2}, pmfrac{1}{3}, pmfrac{1}{6}).Let me test (t = -1):[6(-1)^3 - 3(-1) + 1 = -6 + 3 + 1 = -2 neq 0.]How about (t = frac{1}{2}):[6left(frac{1}{2}right)^3 - 3left(frac{1}{2}right) + 1 = 6left(frac{1}{8}right) - frac{3}{2} + 1 = frac{3}{4} - frac{3}{2} + 1 = frac{3}{4} - frac{6}{4} + frac{4}{4} = frac{1}{4} neq 0.]How about (t = -frac{1}{2}):[6left(-frac{1}{2}right)^3 - 3left(-frac{1}{2}right) + 1 = 6left(-frac{1}{8}right) + frac{3}{2} + 1 = -frac{3}{4} + frac{3}{2} + 1 = -frac{3}{4} + frac{6}{4} + frac{4}{4} = frac{7}{4} neq 0.]How about (t = frac{1}{3}):[6left(frac{1}{3}right)^3 - 3left(frac{1}{3}right) + 1 = 6left(frac{1}{27}right) - 1 + 1 = frac{2}{9} - 1 + 1 = frac{2}{9} neq 0.]Hmm, none of these seem to work. Maybe this cubic doesn't have rational roots. Perhaps I need to use the cubic formula or numerical methods. Alternatively, maybe I made a mistake earlier in simplifying.Wait, let me go back a few steps. After substituting (cos 4x = 2 cos^2 2x - 1), I got:[2 cos^3 2x - cos 2x + frac{1}{3} = 0.]But maybe instead of substituting (cos 4x), I can use another identity or approach. Let me think.Alternatively, maybe I can use the identity for the sum of cosines:[cos A + cos B = 2 cos left( frac{A+B}{2} right) cos left( frac{A-B}{2} right).]So, perhaps I can pair (cos 2x) and (cos 6x) together:[cos 2x + cos 6x = 2 cos 4x cos 2x.]So, substituting back into the equation:[2 cos 4x cos 2x + cos 4x = -1.]Factor out (cos 4x):[cos 4x (2 cos 2x + 1) = -1.]Ah, that's a bit simpler. So, we have:[cos 4x (2 cos 2x + 1) = -1.]Now, this equation is a product of two terms equal to -1. Let me denote (t = cos 2x). Then, (cos 4x = 2t^2 - 1). Substituting, we get:[(2t^2 - 1)(2t + 1) = -1.]Let me expand the left side:[(2t^2 - 1)(2t + 1) = 4t^3 + 2t^2 - 2t - 1.]So, the equation becomes:[4t^3 + 2t^2 - 2t - 1 = -1.]Adding 1 to both sides:[4t^3 + 2t^2 - 2t = 0.]Factor out a 2t:[2t(2t^2 + t - 1) = 0.]So, either (2t = 0) or (2t^2 + t - 1 = 0).Case 1: (2t = 0) implies (t = 0). Since (t = cos 2x), this gives:[cos 2x = 0.]The solutions to this are:[2x = frac{pi}{2} + pi n, quad n in mathbb{Z},]so,[x = frac{pi}{4} + frac{pi n}{2}.]Case 2: (2t^2 + t - 1 = 0). Let's solve this quadratic equation for (t):Using the quadratic formula:[t = frac{-b pm sqrt{b^2 - 4ac}}{2a} = frac{-1 pm sqrt{1 + 8}}{4} = frac{-1 pm 3}{4}.]So, the solutions are:[t = frac{-1 + 3}{4} = frac{2}{4} = frac{1}{2},]and[t = frac{-1 - 3}{4} = frac{-4}{4} = -1.]So, (t = frac{1}{2}) or (t = -1).Substituting back (t = cos 2x):For (t = frac{1}{2}):[cos 2x = frac{1}{2}.]The solutions are:[2x = pm frac{pi}{3} + 2pi n, quad n in mathbb{Z},]so,[x = pm frac{pi}{6} + pi n.]For (t = -1):[cos 2x = -1.]The solutions are:[2x = pi + 2pi n, quad n in mathbb{Z},]so,[x = frac{pi}{2} + pi n.]Now, let's compile all the solutions we've found:1. From Case 1: (x = frac{pi}{4} + frac{pi n}{2}).2. From Case 2, (t = frac{1}{2}): (x = pm frac{pi}{6} + pi n).3. From Case 2, (t = -1): (x = frac{pi}{2} + pi n).I should check if these solutions satisfy the original equation. Let me test a couple of them.First, let's test (x = frac{pi}{4}):Compute each term:[sin^2 left( frac{pi}{4} right) = left( frac{sqrt{2}}{2} right)^2 = frac{1}{2},][sin^2 left( 2 cdot frac{pi}{4} right) = sin^2 left( frac{pi}{2} right) = 1,][sin^2 left( 3 cdot frac{pi}{4} right) = sin^2 left( frac{3pi}{4} right) = left( frac{sqrt{2}}{2} right)^2 = frac{1}{2}.]Adding them up:[frac{1}{2} + 1 + frac{1}{2} = 2.]Good, it works.Next, let's test (x = frac{pi}{2}):Compute each term:[sin^2 left( frac{pi}{2} right) = 1,][sin^2 left( 2 cdot frac{pi}{2} right) = sin^2 (pi) = 0,][sin^2 left( 3 cdot frac{pi}{2} right) = sin^2 left( frac{3pi}{2} right) = 1.]Adding them up:[1 + 0 + 1 = 2.]Also works.Lastly, let's test (x = frac{pi}{6}):Compute each term:[sin^2 left( frac{pi}{6} right) = left( frac{1}{2} right)^2 = frac{1}{4},][sin^2 left( 2 cdot frac{pi}{6} right) = sin^2 left( frac{pi}{3} right) = left( frac{sqrt{3}}{2} right)^2 = frac{3}{4},][sin^2 left( 3 cdot frac{pi}{6} right) = sin^2 left( frac{pi}{2} right) = 1.]Adding them up:[frac{1}{4} + frac{3}{4} + 1 = 2.]Perfect, that works too.So, all the solutions we found satisfy the original equation. Therefore, the complete set of solutions is:1. (x = frac{pi}{4} + frac{pi n}{2}),2. (x = pm frac{pi}{6} + pi n),3. (x = frac{pi}{2} + pi n),where (n) is any integer.I think that's all the solutions. I don't see any mistakes in my steps, and I verified a few solutions, so I'm confident this is correct.

💡First, I need to determine the areas of the smaller square, the larger square, and the right triangle in terms of ( y ).The area of the smaller square is ( (3y)^2 = 9y^2 ) square inches.The area of the larger square is ( (6y)^2 = 36y^2 ) square inches.The area of the right triangle is ( frac{1}{2} times 3y times 6y = 9y^2 ) square inches.Next, I'll add these areas together to find the total area:[9y^2 + 36y^2 + 9y^2 = 54y^2]According to the problem, the total area is 1300 square inches:[54y^2 = 1300]Solving for ( y^2 ):[y^2 = frac{1300}{54} approx 24.0741]Taking the square root to find ( y ):[y = sqrt{24.0741} approx 4.907]Since ( y ) must be positive, the final value is approximately 4.907 inches.

💡Okay, let me try to figure out this problem step by step. So, we have a sequence ( a_n ) where the first term is ( a_1 = 2 ), and each subsequent term is defined by the recurrence relation ( a_{n+1} = 3a_n + 2 ) for ( n in mathbb{N}^* ). Then, we define another sequence ( b_n = log_{3}(a_n + 1) ). We need to find the sum of the first 2019 terms of the sequence ( left{ frac{1}{b_{2n-1}b_{2n+1}} right} ), denoted as ( S_{2019} ).Alright, let's start by understanding the sequence ( a_n ). The recurrence relation is ( a_{n+1} = 3a_n + 2 ). This looks like a linear recurrence relation. I remember that for such recursions, especially non-homogeneous ones, we can sometimes find a closed-form solution by finding the homogeneous solution and a particular solution.But before diving into that, maybe I can compute the first few terms to see if I can spot a pattern. Let's do that.Given ( a_1 = 2 ).Then,( a_2 = 3a_1 + 2 = 3*2 + 2 = 6 + 2 = 8 ).( a_3 = 3a_2 + 2 = 3*8 + 2 = 24 + 2 = 26 ).( a_4 = 3a_3 + 2 = 3*26 + 2 = 78 + 2 = 80 ).Hmm, 2, 8, 26, 80... It seems like each term is roughly tripling, but with an added 2 each time. Maybe it's a geometric sequence with some modification.Wait, the problem mentions ( b_n = log_{3}(a_n + 1) ). Maybe if I look at ( a_n + 1 ), it could simplify things.Let me compute ( a_n + 1 ) for the first few terms:( a_1 + 1 = 2 + 1 = 3 ).( a_2 + 1 = 8 + 1 = 9 ).( a_3 + 1 = 26 + 1 = 27 ).( a_4 + 1 = 80 + 1 = 81 ).Oh! These are powers of 3: 3, 9, 27, 81... So, ( a_n + 1 = 3^n ).Let me check if this holds for the recurrence relation.If ( a_{n+1} + 1 = 3(a_n + 1) ), then:( a_{n+1} + 1 = 3a_n + 3 ).But the given recurrence is ( a_{n+1} = 3a_n + 2 ), so:( a_{n+1} + 1 = 3a_n + 3 = 3(a_n + 1) ).Yes, that works! So, ( a_n + 1 ) is a geometric sequence with the first term ( a_1 + 1 = 3 ) and common ratio 3. Therefore, ( a_n + 1 = 3^n ).Therefore, ( a_n = 3^n - 1 ).Now, let's find ( b_n ).Given ( b_n = log_{3}(a_n + 1) ).Since ( a_n + 1 = 3^n ), then ( b_n = log_{3}(3^n) = n ).So, ( b_n = n ). That simplifies things a lot!Now, we need to consider the sequence ( frac{1}{b_{2n-1}b_{2n+1}} ).Given that ( b_n = n ), then ( b_{2n-1} = 2n - 1 ) and ( b_{2n+1} = 2n + 1 ).Therefore, ( frac{1}{b_{2n-1}b_{2n+1}} = frac{1}{(2n - 1)(2n + 1)} ).Hmm, this looks like a telescoping series. Maybe I can use partial fractions to decompose this term.Let me recall that partial fractions can be used for expressions like ( frac{1}{(2n - 1)(2n + 1)} ).Let me set:( frac{1}{(2n - 1)(2n + 1)} = frac{A}{2n - 1} + frac{B}{2n + 1} ).Multiplying both sides by ( (2n - 1)(2n + 1) ), we get:( 1 = A(2n + 1) + B(2n - 1) ).Expanding the right side:( 1 = A*2n + A*1 + B*2n - B*1 ).Combine like terms:( 1 = (2A + 2B)n + (A - B) ).This must hold for all n, so the coefficients of n and the constant term must be equal on both sides.Therefore, we have the system of equations:1. Coefficient of n: ( 2A + 2B = 0 ).2. Constant term: ( A - B = 1 ).Let me solve this system.From equation 1: ( 2A + 2B = 0 ) => ( A + B = 0 ) => ( A = -B ).Substitute into equation 2: ( (-B) - B = 1 ) => ( -2B = 1 ) => ( B = -frac{1}{2} ).Then, ( A = -B = frac{1}{2} ).So, the partial fraction decomposition is:( frac{1}{(2n - 1)(2n + 1)} = frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ).Great! So, each term in our sequence can be written as:( frac{1}{(2n - 1)(2n + 1)} = frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ).Therefore, the sum ( S_{2019} ) is the sum from ( n = 1 ) to ( n = 2019 ) of ( frac{1}{(2n - 1)(2n + 1)} ), which is equal to:( S_{2019} = sum_{n=1}^{2019} frac{1}{(2n - 1)(2n + 1)} = frac{1}{2} sum_{n=1}^{2019} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ).Now, let's write out the terms of this sum to see if it telescopes.Expanding the sum:( S_{2019} = frac{1}{2} left[ left( frac{1}{1} - frac{1}{3} right) + left( frac{1}{3} - frac{1}{5} right) + left( frac{1}{5} - frac{1}{7} right) + dots + left( frac{1}{2*2019 - 1} - frac{1}{2*2019 + 1} right) right] ).Simplify the terms:Each negative fraction cancels with the positive fraction in the next term. So, most terms cancel out, and we're left with the first term of the first expression and the last term of the last expression.Therefore, the sum simplifies to:( S_{2019} = frac{1}{2} left( 1 - frac{1}{2*2019 + 1} right) ).Compute ( 2*2019 + 1 ):( 2*2019 = 4038 ), so ( 4038 + 1 = 4039 ).Therefore,( S_{2019} = frac{1}{2} left( 1 - frac{1}{4039} right) ).Compute ( 1 - frac{1}{4039} ):( 1 = frac{4039}{4039} ), so ( frac{4039}{4039} - frac{1}{4039} = frac{4038}{4039} ).Thus,( S_{2019} = frac{1}{2} * frac{4038}{4039} = frac{4038}{2*4039} = frac{2019}{4039} ).So, the sum of the first 2019 terms is ( frac{2019}{4039} ).Let me just double-check my steps to make sure I didn't make any mistakes.1. Found ( a_n + 1 = 3^n ) by recognizing the recurrence relation leads to a geometric sequence. That seems correct.2. Then, ( b_n = log_3(a_n + 1) = n ). That makes sense because ( log_3(3^n) = n ).3. Expressed ( frac{1}{(2n - 1)(2n + 1)} ) as partial fractions, which gave me ( frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ). That decomposition looks correct.4. Summed the series, recognizing it's telescoping, so most terms cancel, leaving only the first and the last term. That seems right.5. Calculated ( 2*2019 + 1 = 4039 ), correct.6. Then, ( 1 - frac{1}{4039} = frac{4038}{4039} ), correct.7. Finally, multiplied by ( frac{1}{2} ) to get ( frac{2019}{4039} ). That seems correct.I think all the steps check out. So, the final answer should be ( frac{2019}{4039} ).**Final Answer**The sum of the first 2019 terms is boxed{dfrac{2019}{4039}}.

💡Okay, so I have this problem where a rectangular sheet of paper is folded once, and when it's folded, the point (1,4) maps to (5,2). Then, another point (9,3) aligns with some point (m,n). I need to find m + n. The options are A) 5.7, B) 6, C) 6.5, D) 7, E) 7.5.Alright, let me try to visualize this. Folding a paper such that one point maps to another means that the fold is the perpendicular bisector of the segment connecting those two points. So, first, I should probably find the midpoint between (1,4) and (5,2) because the fold line will pass through that midpoint.Calculating the midpoint: The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). So, plugging in the values:Midpoint = ((1 + 5)/2, (4 + 2)/2) = (6/2, 6/2) = (3, 3). Okay, so the midpoint is (3,3).Next, I need the slope of the line connecting (1,4) and (5,2) to find the perpendicular bisector. The slope formula is (y2 - y1)/(x2 - x1). So:Slope = (2 - 4)/(5 - 1) = (-2)/4 = -1/2.Since the fold is the perpendicular bisector, its slope will be the negative reciprocal of -1/2, which is 2. So, the fold line has a slope of 2 and passes through (3,3). Let me write the equation of this fold line.Using point-slope form: y - y1 = m(x - x1). So:y - 3 = 2(x - 3).Simplifying that:y - 3 = 2x - 6y = 2x - 3.Okay, so the fold line is y = 2x - 3.Now, the point (9,3) is being folded over this line to align with (m,n). So, (m,n) is the reflection of (9,3) over the line y = 2x - 3.To find the reflection of a point over a line, I can use the formula for reflection over a line. But I'm a bit rusty on that, so let me recall.The formula for reflecting a point (x, y) over the line ax + by + c = 0 is:x' = x - 2a(ax + by + c)/(a² + b²)y' = y - 2b(ax + by + c)/(a² + b²)But first, I need to write the fold line in the general form ax + by + c = 0.Given y = 2x - 3, subtract y to get 2x - y - 3 = 0. So, a = 2, b = -1, c = -3.So, plugging into the reflection formulas:x' = x - 2a(ax + by + c)/(a² + b²)Similarly for y'.Let me compute ax + by + c for the point (9,3):ax + by + c = 2*9 + (-1)*3 + (-3) = 18 - 3 - 3 = 12.Then, denominator is a² + b² = 4 + 1 = 5.So,x' = 9 - 2*2*(12)/5 = 9 - 4*(12)/5 = 9 - 48/5 = 9 - 9.6 = -0.6Similarly,y' = 3 - 2*(-1)*(12)/5 = 3 + 2*(12)/5 = 3 + 24/5 = 3 + 4.8 = 7.8Wait, so the reflection point is (-0.6, 7.8). Hmm, but that doesn't seem to match any of the answer choices. Did I do something wrong?Wait, maybe I made a mistake in the reflection formula. Let me double-check.Alternatively, another method to find the reflection is:1. Find the perpendicular line from (9,3) to the fold line y = 2x - 3.2. Find the intersection point of this perpendicular line with the fold line.3. The reflection point will be such that the intersection point is the midpoint between (9,3) and (m,n).So, let's try this method.First, find the slope of the fold line, which is 2. So, the slope of the perpendicular line is -1/2.So, the equation of the line perpendicular to the fold line passing through (9,3) is:y - 3 = (-1/2)(x - 9).Simplify:y = (-1/2)x + (9/2) + 3Convert 3 to halves: 3 = 6/2So,y = (-1/2)x + 9/2 + 6/2 = (-1/2)x + 15/2.Now, find the intersection point of this line with the fold line y = 2x - 3.Set them equal:2x - 3 = (-1/2)x + 15/2.Multiply both sides by 2 to eliminate fractions:4x - 6 = -x + 15Bring all terms to one side:4x + x - 6 - 15 = 05x - 21 = 05x = 21x = 21/5 = 4.2Then, plug back into y = 2x - 3:y = 2*(21/5) - 3 = 42/5 - 15/5 = 27/5 = 5.4So, the intersection point is (4.2, 5.4).This point is the midpoint between (9,3) and (m,n). So, using midpoint formula:Midpoint = ((9 + m)/2, (3 + n)/2) = (4.2, 5.4)So,(9 + m)/2 = 4.2Multiply both sides by 2:9 + m = 8.4So,m = 8.4 - 9 = -0.6Similarly,(3 + n)/2 = 5.4Multiply both sides by 2:3 + n = 10.8So,n = 10.8 - 3 = 7.8So, the reflection point is (-0.6, 7.8). Therefore, m = -0.6 and n = 7.8.Adding them together:m + n = -0.6 + 7.8 = 7.2Wait, 7.2 isn't one of the answer choices. The options are 5.7, 6, 6.5, 7, 7.5. Hmm, did I make a mistake somewhere?Let me check my calculations again.First, midpoint between (1,4) and (5,2):((1+5)/2, (4+2)/2) = (3,3). That's correct.Slope between (1,4) and (5,2):(2-4)/(5-1) = (-2)/4 = -1/2. Correct.Perpendicular slope is 2. Correct.Equation of fold line: y = 2x - 3. Correct.Perpendicular line from (9,3):Slope is -1/2. Equation: y - 3 = (-1/2)(x - 9). Simplify: y = (-1/2)x + 9/2 + 3 = (-1/2)x + 15/2. Correct.Intersection with fold line:2x - 3 = (-1/2)x + 15/2Multiply by 2: 4x - 6 = -x + 155x = 21x = 21/5 = 4.2. Correct.y = 2*(21/5) - 3 = 42/5 - 15/5 = 27/5 = 5.4. Correct.Midpoint is (4.2, 5.4). So,(9 + m)/2 = 4.2 => 9 + m = 8.4 => m = -0.6(3 + n)/2 = 5.4 => 3 + n = 10.8 => n = 7.8So, m + n = -0.6 + 7.8 = 7.2Hmm, 7.2 is not an option. The closest is 7 or 7.5. Maybe I made a mistake in interpreting the fold line?Wait, another thought: Maybe the fold is not the perpendicular bisector but something else? Or perhaps I need to consider that the fold could be a horizontal or vertical line?Wait, the fold line is the perpendicular bisector because when you fold a point onto another, the fold is the perpendicular bisector of the segment connecting them. So, that part seems correct.Alternatively, maybe I need to consider that the fold could be a different line? Or perhaps the reflection is not over the entire line but just a segment?Wait, the paper is rectangular, so maybe the fold is within the rectangle, so perhaps the reflection is constrained within the rectangle? But the problem doesn't specify the dimensions of the rectangle, so I think we can assume it's large enough.Alternatively, perhaps I made a calculation error in the reflection.Wait, let me try another approach. Let me use the formula for reflection over a line.Given the line y = 2x - 3, and the point (9,3).The formula for reflection over line ax + by + c = 0 is:x' = x - 2a(ax + by + c)/(a² + b²)y' = y - 2b(ax + by + c)/(a² + b²)We had earlier:ax + by + c = 2*9 + (-1)*3 + (-3) = 18 - 3 - 3 = 12a² + b² = 4 + 1 = 5So,x' = 9 - 2*2*(12)/5 = 9 - 48/5 = 9 - 9.6 = -0.6y' = 3 - 2*(-1)*(12)/5 = 3 + 24/5 = 3 + 4.8 = 7.8Same result as before. So, m = -0.6, n = 7.8, m + n = 7.2But 7.2 is not an option. Wait, maybe the question is not about reflecting over the fold line, but the fold line is such that (1,4) maps to (5,2), and then (9,3) maps to (m,n). Maybe the fold is not the perpendicular bisector but something else?Wait, no, folding over a line would mean that the line is the perpendicular bisector. So, I think my approach is correct.Wait, perhaps the fold is not through the midpoint but somewhere else? Let me think.Alternatively, maybe the fold is such that (1,4) is reflected over the fold line to (5,2), so the fold line is the perpendicular bisector. So, my previous approach is correct.Wait, maybe the problem is that the fold is not the entire line but a segment? But the problem says the paper is folded once, so it's a straight fold line.Wait, another thought: Maybe the fold is such that the paper is folded along the line y = 2x - 3, but the reflection is only on one side of the fold line? But no, reflection should be across the entire line.Wait, perhaps the answer is 7.2, but since it's not an option, maybe I need to round it? 7.2 is closer to 7 than to 7.5, so maybe the answer is 7.But let me check the options again: A) 5.7, B) 6, C) 6.5, D) 7, E) 7.5.So, 7 is an option. Maybe the answer is 7. But why is my calculation giving 7.2?Wait, perhaps I made a mistake in the reflection formula. Let me double-check.Wait, the formula is:x' = x - 2a(ax + by + c)/(a² + b²)y' = y - 2b(ax + by + c)/(a² + b²)But wait, the general form is ax + by + c = 0. In our case, the fold line is y = 2x - 3, which is 2x - y - 3 = 0. So, a = 2, b = -1, c = -3.So, plugging into the formula:x' = x - 2a(ax + by + c)/(a² + b²)= 9 - 2*2*(2*9 + (-1)*3 + (-3))/(4 + 1)= 9 - 4*(18 - 3 - 3)/5= 9 - 4*(12)/5= 9 - 48/5= 9 - 9.6= -0.6Similarly,y' = 3 - 2*(-1)*(2*9 + (-1)*3 + (-3))/(4 + 1)= 3 - (-2)*(12)/5= 3 + 24/5= 3 + 4.8= 7.8So, same result.Wait, maybe the problem is that the fold is not the perpendicular bisector but a different line? Or perhaps the fold is such that the paper is folded in half, but not necessarily the perpendicular bisector.Wait, no, when you fold a point onto another, the fold line is the perpendicular bisector.Wait, maybe the fold is along the line y = 2x - 3, but the reflection is only on one side, so the image is within the paper. But the problem doesn't specify the size of the paper, so I think we can assume it's large enough.Alternatively, maybe I need to consider that the fold is such that (1,4) maps to (5,2), so the fold line is the perpendicular bisector, but then (9,3) is reflected over that line to get (m,n). So, my calculation is correct, but the answer is 7.2, which is not an option. Hmm.Wait, maybe I made a mistake in the calculation of the intersection point.Let me recalculate the intersection of y = 2x - 3 and y = (-1/2)x + 15/2.Set 2x - 3 = (-1/2)x + 15/2Multiply both sides by 2: 4x - 6 = -x + 15Bring x to left and constants to right: 4x + x = 15 + 65x = 21x = 21/5 = 4.2Then y = 2*(21/5) - 3 = 42/5 - 15/5 = 27/5 = 5.4So, intersection point is (4.2, 5.4). Correct.Midpoint between (9,3) and (m,n) is (4.2,5.4). So,(9 + m)/2 = 4.2 => 9 + m = 8.4 => m = -0.6(3 + n)/2 = 5.4 => 3 + n = 10.8 => n = 7.8So, m + n = -0.6 + 7.8 = 7.2Hmm, 7.2 is not an option. Maybe the answer is 7, as it's the closest integer. But I'm not sure. Alternatively, maybe I made a mistake in the initial assumption.Wait, another thought: Maybe the fold is such that (1,4) maps to (5,2), so the fold line is the perpendicular bisector, but then (9,3) is reflected over that line to get (m,n). So, my calculation is correct, but the answer is 7.2, which is not an option. Maybe the answer is 7, as it's the closest.Alternatively, maybe the fold line is not y = 2x - 3, but another line. Wait, let me check the slope again.The slope between (1,4) and (5,2) is (2-4)/(5-1) = -2/4 = -1/2. So, the perpendicular slope is 2. Correct.Equation of fold line: y - 3 = 2(x - 3) => y = 2x - 3. Correct.Wait, maybe the fold is such that the paper is folded along y = 2x - 3, but the reflection is only on one side, so the image is within the paper. But the problem doesn't specify the size of the paper, so I think we can assume it's large enough.Alternatively, maybe the fold is such that the paper is folded in half, but not necessarily the perpendicular bisector. But no, folding such that a point maps to another requires the fold to be the perpendicular bisector.Wait, maybe the problem is that the fold is not the perpendicular bisector but a different line. Wait, no, that's the definition of reflection.Wait, another thought: Maybe the fold is such that the paper is folded along the line y = 2x - 3, but the reflection is only on one side, so the image is within the paper. But the problem doesn't specify the size of the paper, so I think we can assume it's large enough.Alternatively, maybe I need to consider that the fold is such that the paper is folded along the line y = 2x - 3, but the reflection is only on one side, so the image is within the paper. But the problem doesn't specify the size of the paper, so I think we can assume it's large enough.Wait, maybe the answer is 7, as it's the closest to 7.2, and perhaps the options are approximate. So, I'll go with 7.

💡First, identify the total number of blocks Allen has, which is 49.Next, determine how many blocks are painted with each color, which is 7 blocks per color.To find the number of colors Shiela used, divide the total number of blocks by the number of blocks per color: 49 ÷ 7 = 7.Therefore, Shiela used 7 different colors to repaint Allen's blocks.

💡Okay, so I have this problem here: There's a three-digit number, and if I subtract 7 from it, the result is divisible by 7. If I subtract 8, the result is divisible by 8. And if I subtract 9, the result is divisible by 9. I need to find the smallest such number.Hmm, let's break this down. First, let's denote the three-digit number as ( x ). The problem gives me three conditions:1. ( x - 7 ) is divisible by 7.2. ( x - 8 ) is divisible by 8.3. ( x - 9 ) is divisible by 9.So, in mathematical terms, these conditions can be written as:1. ( x equiv 7 pmod{7} )2. ( x equiv 8 pmod{8} )3. ( x equiv 9 pmod{9} )Wait a second, if ( x equiv 7 pmod{7} ), that means ( x ) leaves a remainder of 7 when divided by 7. But 7 divided by 7 is exactly 1 with no remainder. So, actually, ( x equiv 0 pmod{7} ). Similarly, ( x equiv 8 pmod{8} ) implies ( x equiv 0 pmod{8} ), and ( x equiv 9 pmod{9} ) implies ( x equiv 0 pmod{9} ).So, putting it all together, ( x ) must be divisible by 7, 8, and 9. That means ( x ) is a common multiple of these three numbers. To find the smallest such number, I need to find the least common multiple (LCM) of 7, 8, and 9.Let me recall how to find the LCM. The LCM of multiple numbers is the smallest number that is a multiple of each of them. To find the LCM, I can break each number down into its prime factors:- 7 is a prime number, so its prime factorization is just 7.- 8 is ( 2^3 ).- 9 is ( 3^2 ).The LCM is the product of the highest powers of all prime numbers that appear in the factorizations. So, for 7, 8, and 9, the primes involved are 2, 3, and 7.- The highest power of 2 is ( 2^3 ) from 8.- The highest power of 3 is ( 3^2 ) from 9.- The highest power of 7 is ( 7^1 ) from 7.So, the LCM is ( 2^3 times 3^2 times 7 ).Calculating that:First, ( 2^3 = 8 ).Then, ( 3^2 = 9 ).Multiplying these together: ( 8 times 9 = 72 ).Now, multiply by 7: ( 72 times 7 = 504 ).So, the LCM of 7, 8, and 9 is 504. That means 504 is the smallest number that is divisible by 7, 8, and 9.But wait, the problem specifies that it's a three-digit number. 504 is a three-digit number, so that fits. Let me double-check the conditions to make sure:1. ( 504 - 7 = 497 ). Is 497 divisible by 7? Let's see: ( 7 times 71 = 497 ). Yes, it is.2. ( 504 - 8 = 496 ). Is 496 divisible by 8? ( 8 times 62 = 496 ). Yes, it is.3. ( 504 - 9 = 495 ). Is 495 divisible by 9? ( 9 times 55 = 495 ). Yes, it is.So, all the conditions are satisfied. Therefore, 504 is indeed the smallest three-digit number that meets all the given criteria.I think that's it. It makes sense because 504 is the LCM of 7, 8, and 9, which means it's the smallest number divisible by all three, and subtracting 7, 8, or 9 from it gives a number that's still divisible by those respective numbers. So, 504 is the answer.