💡Okay, so I have this problem here: a complex number z satisfies (1 + i)z = 3 + 2i. I need to find the imaginary part of the conjugate of z, which is written as (overline{z}). The options are A: 1/2, B: -1/2i, C: -1/2, D: 1/2i.Hmm, complex numbers can sometimes be tricky, but let me break it down step by step. First, I remember that to solve for z, I need to isolate it. The equation is (1 + i)z = 3 + 2i. So, z must be equal to (3 + 2i) divided by (1 + i). That is, z = (3 + 2i)/(1 + i).But dividing complex numbers isn't straightforward like real numbers. I think I need to multiply the numerator and the denominator by the conjugate of the denominator to make the denominator real. The conjugate of (1 + i) is (1 - i). So, I'll multiply both the top and bottom by (1 - i).Let me write that out:z = (3 + 2i)/(1 + i) * (1 - i)/(1 - i)Multiplying the numerators: (3 + 2i)(1 - i). Let me expand that.3*1 = 33*(-i) = -3i2i*1 = 2i2i*(-i) = -2i²So, combining these terms: 3 - 3i + 2i - 2i².Wait, I remember that i² is equal to -1, so -2i² is the same as -2*(-1) = 2.So, substituting that in, the numerator becomes 3 - 3i + 2i + 2.Combine like terms: 3 + 2 = 5, and -3i + 2i = -i.So, the numerator simplifies to 5 - i.Now, the denominator is (1 + i)(1 - i). I think this is a difference of squares, so (1)^2 - (i)^2 = 1 - (-1) = 2.So, putting it all together, z = (5 - i)/2.Which can be written as z = 5/2 - (1/2)i.Okay, so z is 5/2 minus (1/2)i. Now, the problem asks for the imaginary part of the conjugate of z, which is (overline{z}).I remember that the conjugate of a complex number a + bi is a - bi. So, if z is 5/2 - (1/2)i, then (overline{z}) should be 5/2 + (1/2)i.So, (overline{z}) is 5/2 + (1/2)i. Now, the imaginary part of a complex number is the coefficient of i. In this case, that's 1/2.Looking back at the options, A is 1/2, which should be the answer. But let me double-check my steps to make sure I didn't make any mistakes.First, solving for z: (1 + i)z = 3 + 2i. So, z = (3 + 2i)/(1 + i). Multiplying numerator and denominator by (1 - i):Numerator: (3 + 2i)(1 - i) = 3 - 3i + 2i - 2i² = 3 - i + 2 = 5 - i.Denominator: (1 + i)(1 - i) = 1 - i² = 1 - (-1) = 2.So, z = (5 - i)/2 = 5/2 - (1/2)i. That seems correct.Then, the conjugate of z is 5/2 + (1/2)i. The imaginary part is 1/2, which is option A. So, yeah, I think that's right.Wait, but let me think again. Sometimes when dealing with conjugates, especially in equations, it's easy to mix up signs. Let me verify the conjugate step.If z = a + bi, then (overline{z}) = a - bi. So, if z is 5/2 - (1/2)i, then (overline{z}) is 5/2 + (1/2)i. So, the imaginary part is positive 1/2. That matches option A.I think I'm confident with that. So, the answer should be A: 1/2.**Final Answer**boxed{A}

💡Okay, so I have this problem where I'm given that ( a > 17 ) is an odd number, and ( 3a - 2 ) is a perfect square. I need to show that there are positive integers ( b ) and ( c ) (with ( b neq c )) such that ( a + b ), ( a + c ), ( b + c ), and ( a + b + c ) are all perfect squares.Hmm, let me start by understanding the given information. Since ( a ) is odd and greater than 17, I can represent ( a ) as ( 2k + 1 ) where ( k ) is an integer. That makes sense because any odd number can be written in the form ( 2k + 1 ).Now, the problem states that ( 3a - 2 ) is a square. Let me substitute ( a = 2k + 1 ) into this expression:[3a - 2 = 3(2k + 1) - 2 = 6k + 3 - 2 = 6k + 1]So, ( 6k + 1 ) must be a perfect square. Let me denote this square as ( m^2 ), so:[6k + 1 = m^2]From this equation, I can solve for ( k ):[k = frac{m^2 - 1}{6}]Since ( k ) must be an integer, ( m^2 - 1 ) must be divisible by 6. Let me check for small values of ( m ) to see if this holds.For ( m = 1 ):[k = frac{1 - 1}{6} = 0]But ( k = 0 ) would make ( a = 1 ), which is less than 17. So, this doesn't work.For ( m = 2 ):[k = frac{4 - 1}{6} = frac{3}{6} = 0.5]Not an integer. So, discard.For ( m = 3 ):[k = frac{9 - 1}{6} = frac{8}{6} approx 1.333]Still not an integer.For ( m = 4 ):[k = frac{16 - 1}{6} = frac{15}{6} = 2.5]Nope.For ( m = 5 ):[k = frac{25 - 1}{6} = frac{24}{6} = 4]Ah, this works! So, ( k = 4 ) when ( m = 5 ). Let me check if this satisfies the original condition.Calculating ( a ):[a = 2k + 1 = 2(4) + 1 = 9]But ( a = 9 ) is less than 17. Hmm, not acceptable.Wait, maybe I need a larger ( m ). Let's try ( m = 7 ):[k = frac{49 - 1}{6} = frac{48}{6} = 8]So, ( k = 8 ). Then, ( a = 2(8) + 1 = 17 ). Still, ( a = 17 ) is not greater than 17. Close, but not quite.Next, ( m = 11 ):[k = frac{121 - 1}{6} = frac{120}{6} = 20]So, ( a = 2(20) + 1 = 41 ). Great, ( a = 41 ) is greater than 17 and odd. Let me verify ( 3a - 2 ):[3(41) - 2 = 123 - 2 = 121 = 11^2]Perfect! So, for ( m = 11 ), ( a = 41 ) satisfies the conditions.Now, I need to find ( b ) and ( c ) such that ( a + b ), ( a + c ), ( b + c ), and ( a + b + c ) are all squares.Let me think about how to choose ( b ) and ( c ). Maybe I can express them in terms of ( k ) or ( m ). Since ( k = 20 ) in this case, perhaps I can set ( b ) and ( c ) as functions of ( k ).Looking at the problem, I need four expressions to be squares. Let me denote:1. ( a + b = s^2 )2. ( a + c = t^2 )3. ( b + c = u^2 )4. ( a + b + c = v^2 )I need to find ( b ) and ( c ) such that all these are satisfied. Let me see if I can express ( b ) and ( c ) in terms of ( a ).From equations 1 and 2:( b = s^2 - a )( c = t^2 - a )Substituting into equation 3:( (s^2 - a) + (t^2 - a) = u^2 )Simplify:( s^2 + t^2 - 2a = u^2 )From equation 4:( a + (s^2 - a) + (t^2 - a) = v^2 )Simplify:( s^2 + t^2 - a = v^2 )So now, I have two equations:1. ( s^2 + t^2 - 2a = u^2 )2. ( s^2 + t^2 - a = v^2 )Subtracting equation 1 from equation 2:( (s^2 + t^2 - a) - (s^2 + t^2 - 2a) = v^2 - u^2 )Simplify:( a = v^2 - u^2 )Which can be factored as:( a = (v - u)(v + u) )Since ( a = 41 ) is prime, the only factors are 1 and 41. Therefore:( v - u = 1 ) and ( v + u = 41 )Adding these two equations:( 2v = 42 ) => ( v = 21 )Subtracting:( 2u = 40 ) => ( u = 20 )So, ( v = 21 ) and ( u = 20 ).Now, from equation 2:( s^2 + t^2 - a = v^2 )Substitute ( a = 41 ) and ( v = 21 ):( s^2 + t^2 - 41 = 441 )Thus:( s^2 + t^2 = 482 )From equation 1:( s^2 + t^2 - 2a = u^2 )Substitute ( a = 41 ) and ( u = 20 ):( s^2 + t^2 - 82 = 400 )Thus:( s^2 + t^2 = 482 )So, both equations lead to the same result, which is consistent.Now, I need to find integers ( s ) and ( t ) such that ( s^2 + t^2 = 482 ).Let me list squares less than 482:( 1^2 = 1 )( 2^2 = 4 )( 3^2 = 9 )( 4^2 = 16 )( 5^2 = 25 )( 6^2 = 36 )( 7^2 = 49 )( 8^2 = 64 )( 9^2 = 81 )( 10^2 = 100 )( 11^2 = 121 )( 12^2 = 144 )( 13^2 = 169 )( 14^2 = 196 )( 15^2 = 225 )( 16^2 = 256 )( 17^2 = 289 )( 18^2 = 324 )( 19^2 = 361 )( 20^2 = 400 )( 21^2 = 441 )( 22^2 = 484 ) (Too big)So, looking for two squares that add up to 482.Let me try ( 21^2 = 441 ). Then, ( 482 - 441 = 41 ). Is 41 a square? No.Next, ( 20^2 = 400 ). ( 482 - 400 = 82 ). Not a square.( 19^2 = 361 ). ( 482 - 361 = 121 ). 121 is ( 11^2 ). Perfect!So, ( s = 19 ) and ( t = 11 ) (or vice versa).Thus, ( s = 19 ), ( t = 11 ).Now, let's find ( b ) and ( c ):From earlier:( b = s^2 - a = 19^2 - 41 = 361 - 41 = 320 )( c = t^2 - a = 11^2 - 41 = 121 - 41 = 80 )So, ( b = 320 ) and ( c = 80 ).Let me verify all conditions:1. ( a + b = 41 + 320 = 361 = 19^2 ) ✔️2. ( a + c = 41 + 80 = 121 = 11^2 ) ✔️3. ( b + c = 320 + 80 = 400 = 20^2 ) ✔️4. ( a + b + c = 41 + 320 + 80 = 441 = 21^2 ) ✔️All conditions are satisfied! So, for ( a = 41 ), ( b = 320 ) and ( c = 80 ) work.But wait, the problem says "for any ( a > 17 ) odd with ( 3a - 2 ) square, show that such ( b ) and ( c ) exist." So, I need a general solution, not just for ( a = 41 ).Let me think about how ( b ) and ( c ) relate to ( a ). From the specific case, ( b = 320 ) and ( c = 80 ) when ( a = 41 ). Let me express ( b ) and ( c ) in terms of ( k ) or ( m ).Recall that ( a = 2k + 1 ) and ( 6k + 1 = m^2 ). So, ( k = frac{m^2 - 1}{6} ).In the specific case, ( m = 11 ), so ( k = frac{121 - 1}{6} = 20 ).Then, ( b = 320 = 16 times 20 = 16k ). Wait, 16k is 320. Hmm, but 16 is 4 squared.Wait, ( b = 16k ) and ( c = 4k ). Let me check:( b = 16k = 16 times 20 = 320 )( c = 4k = 4 times 20 = 80 )Yes, that works.So, in general, if I set ( b = 16k ) and ( c = 4k ), then:1. ( a + b = 2k + 1 + 16k = 18k + 1 ). Wait, in the specific case, this was ( 361 = 19^2 ). Let me see:( 18k + 1 = 18 times 20 + 1 = 361 ), which is indeed ( 19^2 ).But wait, ( 18k + 1 ) should be a square. Let me see if this holds in general.Given ( 6k + 1 = m^2 ), can I express ( 18k + 1 ) in terms of ( m )?Yes:( 18k + 1 = 3(6k) + 1 = 3(m^2 - 1) + 1 = 3m^2 - 3 + 1 = 3m^2 - 2 )Wait, but ( 3m^2 - 2 ) needs to be a square. Hmm, that might not necessarily be true.Wait, in the specific case, ( m = 11 ):( 3(11)^2 - 2 = 3(121) - 2 = 363 - 2 = 361 = 19^2 ). So, it worked there.But does ( 3m^2 - 2 ) always result in a square? Probably not for all ( m ), but in our case, since ( a = 2k + 1 ) and ( 6k + 1 = m^2 ), perhaps there's a relationship.Wait, let me think differently. Maybe I can express ( b ) and ( c ) in terms of ( m ).From the specific case, ( b = 320 = 16k = 16 times 20 ), and ( c = 80 = 4k ). Also, ( m = 11 ).Notice that ( 16k = (4k)^2 / k ), but that might not help.Alternatively, maybe ( b = (m - 1)^2 times something ). Wait, ( m = 11 ), ( (m - 1)^2 = 100 ), which is not directly related to ( b = 320 ).Alternatively, perhaps ( b = (m + 1)^2 times something ). ( (m + 1)^2 = 144 ), still not directly.Wait, let me think about the expressions for ( b ) and ( c ) in terms of ( k ):From the specific case, ( b = 16k ) and ( c = 4k ). Let me see if these can be generalized.If I set ( b = 16k ) and ( c = 4k ), then:1. ( a + b = 2k + 1 + 16k = 18k + 1 )2. ( a + c = 2k + 1 + 4k = 6k + 1 = m^2 )3. ( b + c = 16k + 4k = 20k )4. ( a + b + c = 2k + 1 + 16k + 4k = 22k + 1 )Now, I need all these to be squares.From the specific case, ( 18k + 1 = 361 = 19^2 ), ( 6k + 1 = 121 = 11^2 ), ( 20k = 400 = 20^2 ), and ( 22k + 1 = 441 = 21^2 ).So, in general, if I can set:1. ( 18k + 1 = (something)^2 )2. ( 6k + 1 = m^2 )3. ( 20k = (something)^2 )4. ( 22k + 1 = (something)^2 )But this seems too restrictive unless ( k ) is chosen such that all these expressions are squares.Alternatively, maybe I can find a relationship between ( m ) and these expressions.Given ( 6k + 1 = m^2 ), then ( k = frac{m^2 - 1}{6} ).Substituting into ( 18k + 1 ):[18k + 1 = 18 left( frac{m^2 - 1}{6} right) + 1 = 3(m^2 - 1) + 1 = 3m^2 - 3 + 1 = 3m^2 - 2]So, ( 18k + 1 = 3m^2 - 2 ). For this to be a square, say ( n^2 ), we have:[3m^2 - 2 = n^2]This is a Diophantine equation. Let me see if I can find integer solutions for ( m ) and ( n ).Rearranging:[n^2 - 3m^2 = -2]This is a Pell-type equation. The general solution can be found using continued fractions or other methods, but perhaps I can find small solutions.Let me try small values of ( m ):For ( m = 1 ):[n^2 - 3(1)^2 = n^2 - 3 = -2 implies n^2 = 1 implies n = 1]So, ( m = 1 ), ( n = 1 ). But ( k = frac{1 - 1}{6} = 0 ), which is not acceptable as ( a = 1 ).For ( m = 2 ):[n^2 - 12 = -2 implies n^2 = 10 ). Not a square.For ( m = 3 ):[n^2 - 27 = -2 implies n^2 = 25 implies n = 5]So, ( m = 3 ), ( n = 5 ). Then, ( k = frac{9 - 1}{6} = frac{8}{6} approx 1.333 ). Not integer.For ( m = 4 ):[n^2 - 48 = -2 implies n^2 = 46 ). Not a square.For ( m = 5 ):[n^2 - 75 = -2 implies n^2 = 73 ). Not a square.For ( m = 6 ):[n^2 - 108 = -2 implies n^2 = 106 ). Not a square.For ( m = 7 ):[n^2 - 147 = -2 implies n^2 = 145 ). Not a square.For ( m = 8 ):[n^2 - 192 = -2 implies n^2 = 190 ). Not a square.For ( m = 9 ):[n^2 - 243 = -2 implies n^2 = 241 ). Not a square.For ( m = 10 ):[n^2 - 300 = -2 implies n^2 = 298 ). Not a square.For ( m = 11 ):[n^2 - 363 = -2 implies n^2 = 361 implies n = 19]So, ( m = 11 ), ( n = 19 ). Then, ( k = frac{121 - 1}{6} = 20 ), which is integer. This is the case we already saw.Next, ( m = 12 ):[n^2 - 432 = -2 implies n^2 = 430 ). Not a square.( m = 13 ):[n^2 - 507 = -2 implies n^2 = 505 ). Not a square.( m = 14 ):[n^2 - 588 = -2 implies n^2 = 586 ). Not a square.( m = 15 ):[n^2 - 675 = -2 implies n^2 = 673 ). Not a square.( m = 16 ):[n^2 - 768 = -2 implies n^2 = 766 ). Not a square.( m = 17 ):[n^2 - 867 = -2 implies n^2 = 865 ). Not a square.( m = 18 ):[n^2 - 972 = -2 implies n^2 = 970 ). Not a square.( m = 19 ):[n^2 - 1083 = -2 implies n^2 = 1081 ). Not a square.( m = 20 ):[n^2 - 1200 = -2 implies n^2 = 1198 ). Not a square.Hmm, seems like the next solution is not coming up quickly. Maybe Pell's equation has solutions that are spaced apart. The minimal solution is ( m = 11 ), ( n = 19 ). The next solution can be found using the recurrence relations for Pell equations.The general solution for the equation ( n^2 - 3m^2 = -2 ) can be generated from the minimal solution ( (n, m) = (1, 1) ) and ( (19, 11) ). The solutions can be generated using the fundamental solution of the Pell equation ( x^2 - 3y^2 = 1 ), which is ( (2, 1) ).Using the method for Pell equations, the solutions can be generated by:[n + msqrt{3} = (19 + 11sqrt{3})(2 + sqrt{3})^k]for ( k = 0, 1, 2, ldots )Calculating for ( k = 1 ):[(19 + 11sqrt{3})(2 + sqrt{3}) = 19 times 2 + 19 times sqrt{3} + 11sqrt{3} times 2 + 11sqrt{3} times sqrt{3}][= 38 + 19sqrt{3} + 22sqrt{3} + 33][= 38 + 33 + (19 + 22)sqrt{3}][= 71 + 41sqrt{3}]So, the next solution is ( n = 71 ), ( m = 41 ).Let me verify:[71^2 - 3(41)^2 = 5041 - 3(1681) = 5041 - 5043 = -2]Yes, it works.So, ( m = 41 ), ( n = 71 ). Then, ( k = frac{41^2 - 1}{6} = frac{1681 - 1}{6} = frac{1680}{6} = 280 ).Thus, ( a = 2k + 1 = 2(280) + 1 = 561 ).Now, let's find ( b ) and ( c ):From earlier, ( b = 16k = 16 times 280 = 4480 )( c = 4k = 4 times 280 = 1120 )Let me verify all conditions:1. ( a + b = 561 + 4480 = 5041 = 71^2 ) ✔️2. ( a + c = 561 + 1120 = 1681 = 41^2 ) ✔️3. ( b + c = 4480 + 1120 = 5600 ). Wait, is 5600 a square? Let me check:( 74^2 = 5476 )( 75^2 = 5625 )So, 5600 is between ( 74^2 ) and ( 75^2 ). Not a square. Hmm, that's a problem.Wait, what did I do wrong? Earlier, in the specific case, ( b + c = 400 = 20^2 ). But in this case, ( b + c = 5600 ), which is not a square.Wait, maybe my assumption that ( b = 16k ) and ( c = 4k ) is only valid for the first solution ( m = 11 ), ( n = 19 ). For the next solution, perhaps the expressions for ( b ) and ( c ) change.Alternatively, maybe I need a different approach to express ( b ) and ( c ).Let me go back to the general case. We have:1. ( a + b = s^2 )2. ( a + c = t^2 )3. ( b + c = u^2 )4. ( a + b + c = v^2 )From these, we derived:( a = v^2 - u^2 )And since ( a ) is prime (in the specific case), ( v - u = 1 ) and ( v + u = a ). But in general, ( a ) might not be prime. Wait, in the specific case, ( a = 41 ) is prime, but in general, ( a ) could be composite.Wait, actually, ( a = 2k + 1 ), and ( 6k + 1 = m^2 ). So, ( a ) is not necessarily prime. For example, if ( m = 7 ), ( k = 8 ), ( a = 17 ), which is prime. If ( m = 11 ), ( a = 41 ), prime. If ( m = 41 ), ( a = 561 ), which is composite (561 = 3 × 11 × 17).So, in the general case, ( a ) might not be prime, so ( v - u ) and ( v + u ) might not be 1 and ( a ). Instead, they could be other factor pairs of ( a ).But in the specific case where ( a ) is prime, the only factor pairs are 1 and ( a ). So, ( v - u = 1 ) and ( v + u = a ), leading to ( v = frac{a + 1}{2} ) and ( u = frac{a - 1}{2} ).But when ( a ) is composite, there could be other factor pairs. However, for simplicity, let's assume that ( v - u = 1 ) and ( v + u = a ), which would still hold even if ( a ) is composite, as long as ( a ) is odd (which it is). Because ( a ) is odd, ( v ) and ( u ) would be integers.So, proceeding with ( v = frac{a + 1}{2} ) and ( u = frac{a - 1}{2} ).Then, from equation 2:( s^2 + t^2 - a = v^2 = left( frac{a + 1}{2} right)^2 )Similarly, from equation 1:( s^2 + t^2 - 2a = u^2 = left( frac{a - 1}{2} right)^2 )Subtracting equation 1 from equation 2:( (s^2 + t^2 - a) - (s^2 + t^2 - 2a) = left( frac{a + 1}{2} right)^2 - left( frac{a - 1}{2} right)^2 )Simplify:( a = left( frac{a + 1}{2} right)^2 - left( frac{a - 1}{2} right)^2 )Calculating the right-hand side:[left( frac{a + 1}{2} right)^2 - left( frac{a - 1}{2} right)^2 = frac{(a + 1)^2 - (a - 1)^2}{4} = frac{(a^2 + 2a + 1) - (a^2 - 2a + 1)}{4} = frac{4a}{4} = a]So, the equation holds, which is consistent.Now, going back to equation 2:( s^2 + t^2 - a = left( frac{a + 1}{2} right)^2 )Thus:( s^2 + t^2 = a + left( frac{a + 1}{2} right)^2 )Similarly, from equation 1:( s^2 + t^2 = 2a + left( frac{a - 1}{2} right)^2 )But both expressions equal ( s^2 + t^2 ), so:[a + left( frac{a + 1}{2} right)^2 = 2a + left( frac{a - 1}{2} right)^2]Simplify:[a + frac{(a + 1)^2}{4} = 2a + frac{(a - 1)^2}{4}]Multiply both sides by 4 to eliminate denominators:[4a + (a + 1)^2 = 8a + (a - 1)^2]Expand both sides:Left side:[4a + a^2 + 2a + 1 = a^2 + 6a + 1]Right side:[8a + a^2 - 2a + 1 = a^2 + 6a + 1]Both sides are equal, so the equation holds. Thus, our approach is consistent.Now, to find ( s ) and ( t ), we need to solve:( s^2 + t^2 = a + left( frac{a + 1}{2} right)^2 )Let me compute this:[s^2 + t^2 = a + frac{(a + 1)^2}{4} = frac{4a + a^2 + 2a + 1}{4} = frac{a^2 + 6a + 1}{4}]So,[s^2 + t^2 = frac{a^2 + 6a + 1}{4}]But ( a = 2k + 1 ), so substituting:[s^2 + t^2 = frac{(2k + 1)^2 + 6(2k + 1) + 1}{4} = frac{4k^2 + 4k + 1 + 12k + 6 + 1}{4} = frac{4k^2 + 16k + 8}{4} = k^2 + 4k + 2]Wait, that doesn't seem right. Let me recalculate:Wait, ( a = 2k + 1 ), so ( a^2 = (2k + 1)^2 = 4k^2 + 4k + 1 )( 6a = 6(2k + 1) = 12k + 6 )So,[a^2 + 6a + 1 = 4k^2 + 4k + 1 + 12k + 6 + 1 = 4k^2 + 16k + 8]Thus,[s^2 + t^2 = frac{4k^2 + 16k + 8}{4} = k^2 + 4k + 2]Wait, that's not matching the specific case. When ( k = 20 ):( s^2 + t^2 = 20^2 + 4(20) + 2 = 400 + 80 + 2 = 482 ), which matches our earlier result. So, it's correct.So, in general, ( s^2 + t^2 = k^2 + 4k + 2 ).But how do I find ( s ) and ( t ) such that their squares add up to ( k^2 + 4k + 2 )?In the specific case, ( k = 20 ), ( s = 19 ), ( t = 11 ). Let me see if there's a pattern.Notice that ( s = k - 1 ) and ( t = sqrt{k^2 + 4k + 2 - (k - 1)^2} ).Calculating ( t ):[t^2 = k^2 + 4k + 2 - (k^2 - 2k + 1) = k^2 + 4k + 2 - k^2 + 2k - 1 = 6k + 1]But ( 6k + 1 = m^2 ), so ( t = m ).Ah! So, ( s = k - 1 ) and ( t = m ).Thus, in general:( s = k - 1 )( t = m )Therefore, ( b = s^2 - a = (k - 1)^2 - (2k + 1) = k^2 - 2k + 1 - 2k - 1 = k^2 - 4k )Similarly, ( c = t^2 - a = m^2 - (2k + 1) ). But ( m^2 = 6k + 1 ), so:( c = 6k + 1 - 2k - 1 = 4k )So, in general, ( b = k^2 - 4k ) and ( c = 4k ).Let me verify this with the specific case where ( k = 20 ):( b = 20^2 - 4(20) = 400 - 80 = 320 )( c = 4(20) = 80 )Which matches our earlier result.Now, let's check if ( b + c ) is a square:( b + c = (k^2 - 4k) + 4k = k^2 ), which is indeed a square.And ( a + b + c = (2k + 1) + (k^2 - 4k) + 4k = k^2 + 2k + 1 = (k + 1)^2 ), which is also a square.So, in general, setting ( b = k^2 - 4k ) and ( c = 4k ) satisfies all the required conditions.Therefore, for any ( a > 17 ) odd with ( 3a - 2 ) being a square, we can express ( a = 2k + 1 ) and ( 3a - 2 = m^2 ), leading to ( b = k^2 - 4k ) and ( c = 4k ), which are positive integers with ( b neq c ) (since ( k^2 - 4k neq 4k ) for ( k > 4 )).Thus, the required integers ( b ) and ( c ) exist.

💡First, I need to determine how many blocks differ from the 'plastic medium red striped circle' in exactly three attributes. The block has five attributes: material, size, color, pattern, and shape.1. **Material**: There are 3 materials (plastic, wood, metal). Since the original block is plastic, there are 2 other materials it can differ in.2. **Size**: There are 3 sizes (small, medium, large). The original block is medium, so there are 2 other sizes it can differ in.3. **Color**: There are 4 colors (blue, green, red, yellow). The original block is red, so there are 3 other colors it can differ in.4. **Pattern**: There are 2 patterns (striped, plain). The original block is striped, so there is 1 other pattern it can differ in.5. **Shape**: There are 4 shapes (circle, hexagon, square, triangle). The original block is a circle, so there are 3 other shapes it can differ in.To find the number of blocks that differ in exactly three attributes, I need to consider all possible combinations of three attributes changing while the other two remain the same.There are ( binom{5}{3} = 10 ) ways to choose which three attributes will differ. For each combination, I multiply the number of possible changes for each attribute:- **Material, Size, Color**: ( 2 times 2 times 3 = 12 )- **Material, Size, Pattern**: ( 2 times 2 times 1 = 4 )- **Material, Size, Shape**: ( 2 times 2 times 3 = 12 )- **Material, Color, Pattern**: ( 2 times 3 times 1 = 6 )- **Material, Color, Shape**: ( 2 times 3 times 3 = 18 )- **Material, Pattern, Shape**: ( 2 times 1 times 3 = 6 )- **Size, Color, Pattern**: ( 2 times 3 times 1 = 6 )- **Size, Color, Shape**: ( 2 times 3 times 3 = 18 )- **Size, Pattern, Shape**: ( 2 times 1 times 3 = 6 )- **Color, Pattern, Shape**: ( 3 times 1 times 3 = 9 )Adding all these possibilities together:( 12 + 4 + 12 + 6 + 18 + 6 + 6 + 18 + 6 + 9 = 112 )Therefore, there are 112 blocks that differ from the 'plastic medium red striped circle' in exactly three ways.

💡Okay, so I have this problem where I'm given two functions: f(x) = 2 sin(πx + π/3) and g(x) = 2 cos(πx + π/3). Their graphs intersect at points A, B, and C within the interval [-4/3, 7/6]. I need to find the area of triangle ABC. The options are A: √2, B: 3√2/2, C: 5√2/2, D: 2√2. Hmm, okay.First, I remember that to find the points of intersection between two functions, I need to set them equal to each other and solve for x. So, I'll set f(x) equal to g(x):2 sin(πx + π/3) = 2 cos(πx + π/3)I can divide both sides by 2 to simplify:sin(πx + π/3) = cos(πx + π/3)Hmm, when does sin(θ) equal cos(θ)? I recall that sin(θ) = cos(θ) when θ = π/4 + kπ, where k is any integer. So, applying that here:πx + π/3 = π/4 + kπLet me solve for x:πx = π/4 - π/3 + kπFirst, let's compute π/4 - π/3. To subtract these, I need a common denominator, which is 12.π/4 = 3π/12 and π/3 = 4π/12, so:π/4 - π/3 = 3π/12 - 4π/12 = -π/12So, πx = -π/12 + kπDivide both sides by π:x = -1/12 + kSo, the solutions are x = -1/12 + k, where k is any integer.Now, I need to find all x in the interval [-4/3, 7/6]. Let's figure out which values of k will give me x within that interval.Let me compute the possible k values:Start with k = 0: x = -1/12 ≈ -0.0833. Is this within [-4/3, 7/6]? Yes, because -4/3 ≈ -1.333 and 7/6 ≈ 1.1667. So, -0.0833 is within the interval.k = 1: x = -1/12 + 1 = 11/12 ≈ 0.9167. Is this within the interval? Yes, 0.9167 < 1.1667.k = -1: x = -1/12 - 1 = -13/12 ≈ -1.0833. Is this within the interval? -1.0833 is greater than -1.333, so yes.k = 2: x = -1/12 + 2 = 23/12 ≈ 1.9167, which is greater than 7/6 ≈ 1.1667, so that's outside.k = -2: x = -1/12 - 2 = -25/12 ≈ -2.0833, which is less than -4/3 ≈ -1.333, so that's outside.So, the solutions within the interval are x = -13/12, x = -1/12, and x = 11/12.So, the points of intersection are at x = -13/12, -1/12, and 11/12.Now, I need to find the corresponding y-values for each x to get the coordinates of points A, B, and C.Let me compute f(x) at each of these x-values.First, x = -13/12:f(-13/12) = 2 sin(π*(-13/12) + π/3)Let me compute the argument inside the sine:π*(-13/12) + π/3 = (-13π/12) + (4π/12) = (-13π + 4π)/12 = (-9π)/12 = (-3π)/4So, sin(-3π/4) = -sin(3π/4) = -√2/2Therefore, f(-13/12) = 2*(-√2/2) = -√2So, point A is (-13/12, -√2)Next, x = -1/12:f(-1/12) = 2 sin(π*(-1/12) + π/3)Compute the argument:π*(-1/12) + π/3 = (-π/12) + (4π/12) = (3π)/12 = π/4sin(π/4) = √2/2Therefore, f(-1/12) = 2*(√2/2) = √2So, point B is (-1/12, √2)Lastly, x = 11/12:f(11/12) = 2 sin(π*(11/12) + π/3)Compute the argument:π*(11/12) + π/3 = (11π/12) + (4π/12) = 15π/12 = 5π/4sin(5π/4) = -√2/2Therefore, f(11/12) = 2*(-√2/2) = -√2So, point C is (11/12, -√2)So, now I have the three points:A: (-13/12, -√2)B: (-1/12, √2)C: (11/12, -√2)Now, I need to find the area of triangle ABC.To find the area of a triangle given three points, I can use the shoelace formula.The shoelace formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Let me assign:Point A: (x1, y1) = (-13/12, -√2)Point B: (x2, y2) = (-1/12, √2)Point C: (x3, y3) = (11/12, -√2)Plugging into the formula:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Compute each term:First term: x1(y2 - y3) = (-13/12)(√2 - (-√2)) = (-13/12)(2√2) = (-13/12)*2√2 = (-13/6)√2Second term: x2(y3 - y1) = (-1/12)(-√2 - (-√2)) = (-1/12)(0) = 0Third term: x3(y1 - y2) = (11/12)(-√2 - √2) = (11/12)(-2√2) = (-11/6)√2So, adding all three terms:(-13/6)√2 + 0 + (-11/6)√2 = (-24/6)√2 = (-4)√2Take the absolute value and divide by 2:|(-4√2)/2| = | -2√2 | = 2√2So, the area is 2√2.Looking back at the options, D is 2√2, so the answer should be D.Wait, let me double-check in case I made a mistake.Alternatively, maybe I can plot the points or see if the triangle is symmetric.Looking at the points:A: (-13/12, -√2)B: (-1/12, √2)C: (11/12, -√2)So, points A and C are both at y = -√2, but different x-values. Point B is at y = √2.So, the triangle has a base between A and C, which are both on the line y = -√2, and a peak at B on y = √2.So, the vertical distance between the base and the peak is √2 - (-√2) = 2√2.The horizontal distance between A and C is 11/12 - (-13/12) = 24/12 = 2.So, the base length is 2, and the height is 2√2.Wait, but actually, in a triangle, the area is (base * height)/2. So, if the base is 2 and the height is 2√2, then the area would be (2 * 2√2)/2 = 2√2. Which matches what I got earlier.Alternatively, if I consider the base as the distance between A and C, which is 2, and the height as the vertical distance from B to that base, which is 2√2, then yes, area is (2 * 2√2)/2 = 2√2.Alternatively, using coordinates, I can also compute vectors or use determinants, but shoelace seems straightforward.Alternatively, maybe using coordinates to compute the lengths of sides and then Heron's formula, but that might be more complicated.But since both methods gave me 2√2, I think that's correct.So, the answer is D: 2√2.**Final Answer**boxed{D}

💡First, I need to determine how many students traveled by bus. Since 4 students traveled in cars, I subtract this number from the total number of students to find out how many were on the buses.Next, I divide the number of students who traveled by bus by the number of buses available. This will give me the number of students in each bus.Finally, performing the division will provide the final answer.

💡Okay, so I have this problem where there are three boys: Petya, Vasya, and Vitya. Masha knows their names but doesn't know which boy is which. She can ask them yes or no questions, and each question is directed to one boy. The tricky part is that each boy behaves differently when answering questions:- Petya always tells the truth.- Vasya lies on the first question he's asked but tells the truth from the second question onwards.- Vitya lies on the first two questions he's asked but tells the truth from the third question onwards.Masha needs to figure out who is who with just three questions. Hmm, that sounds challenging but also interesting. Let me try to break it down step by step.First, I need to understand the behavior of each boy clearly:1. **Petya**: Always truthful. So, no matter what question you ask, he'll answer honestly.2. **Vasya**: Lies on the first question but tells the truth from the second question onwards. So, if you ask him two questions, the first answer will be a lie, and the second will be truthful.3. **Vitya**: Lies on the first two questions but tells the truth from the third question onwards. So, if you ask him three questions, the first two answers will be lies, and the third will be truthful.Given that Masha can ask three questions, she needs to design them in such a way that she can distinguish between these three boys based on their answers.Let me think about how to approach this. Maybe she can ask each boy a series of questions and analyze their responses. But since she can only ask three questions in total, she needs to be strategic about whom she asks and what she asks.Perhaps she can start by asking one boy a question and then use his response to determine who he might be. If she can identify one boy, she can then use the remaining questions to identify the others.Let's consider asking one boy two questions. If she asks the same boy two questions, she can observe the pattern of his answers:- If the boy is Petya, both answers will be truthful.- If the boy is Vasya, the first answer will be a lie, and the second will be truthful.- If the boy is Vitya, the first two answers will be lies.So, by asking two questions to one boy, Masha can potentially determine who he is based on the pattern of his answers.For example, suppose she asks a boy the following two questions:1. "Are you Vasya?"2. "Are you Vitya?"Depending on who the boy is, the answers will vary:- **If the boy is Petya**: - He will answer truthfully. If he's Petya, he's not Vasya or Vitya, so both answers will be "No."- **If the boy is Vasya**: - He lies on the first question. Since he's Vasya, the truthful answer to "Are you Vasya?" is "Yes," so he will lie and say "No." - On the second question, he tells the truth. Since he's not Vitya, he will answer "No."- **If the boy is Vitya**: - He lies on the first two questions. The truthful answer to "Are you Vasya?" is "No," so he will lie and say "Yes." - The truthful answer to "Are you Vitya?" is "Yes," so he will lie and say "No."Wait, hold on. If Vitya is asked "Are you Vitya?" and he lies, he would say "No," even though he is Vitya. So, for Vitya, both answers would be "Yes" and "No," but that might not be consistent with the previous analysis.Let me correct that:- **Vitya**: - First question: "Are you Vasya?" Truthful answer is "No," so he lies and says "Yes." - Second question: "Are you Vitya?" Truthful answer is "Yes," so he lies and says "No."So, Vitya's answers would be "Yes" and "No."Wait, that's different from what I initially thought. So, let's summarize:- **Petya**: "No," "No"- **Vasya**: "No," "No"- **Vitya**: "Yes," "No"Hmm, that's a problem because both Petya and Vasya would answer "No" to both questions, making it impossible to distinguish between them based on these two questions.So, maybe asking "Are you Vasya?" and "Are you Vitya?" isn't the best approach because it doesn't differentiate between Petya and Vasya.Let me think of a different question to ask. Maybe a question that would give different patterns of answers.What if she asks, "Are you Petya?" to a boy twice?- **If the boy is Petya**: - He will answer "Yes" both times.- **If the boy is Vasya**: - First question: Lie. Since he's not Petya, the truthful answer is "No," so he lies and says "Yes." - Second question: Truth. Since he's not Petya, he answers "No."- **If the boy is Vitya**: - First question: Lie. Since he's not Petya, the truthful answer is "No," so he lies and says "Yes." - Second question: Lie. Since he's not Petya, the truthful answer is "No," so he lies and says "Yes."So, the responses would be:- **Petya**: "Yes," "Yes"- **Vasya**: "Yes," "No"- **Vitya**: "Yes," "Yes"Wait, that's interesting. So, if a boy answers "Yes" to both questions, he could be either Petya or Vitya. If he answers "Yes" and then "No," he's Vasya. But how can we distinguish between Petya and Vitya?Since Vitya lies on the first two questions, if we ask him a third question, he would tell the truth. So, maybe after identifying Vasya, we can use the third question to determine between Petya and Vitya.But Masha only has three questions in total. So, if she uses two questions on one boy, she only has one question left for the other two. That might not be enough.Alternatively, maybe she can ask each boy one question each, but that might not give enough information because the responses depend on the number of questions asked to each boy.Wait, let's think differently. Maybe she can ask a question that can help her identify one boy for sure, and then use the remaining questions to identify the others.For example, if she can identify Vitya, who lies on the first two questions, she can then ask him a third question to get the truth. Similarly, if she can identify Vasya, who lies on the first question, she can ask him a second question to get the truth.But how can she structure the questions to achieve this?Let me try to outline a possible strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" - If the answer is "Yes," then Boy A could be Vitya (lying) or Vasya (lying on the first question). - If the answer is "No," then Boy A could be Petya (truthful) or Vasya (lying on the first question).2. **Second Question**: Depending on the first answer, ask Boy A another question or move to Boy B.But this seems too vague. Maybe a better approach is to ask the same question to two different boys and analyze their responses.Alternatively, maybe she can use a question that is self-referential or about another boy.Wait, here's an idea: ask each boy about another boy's identity. For example, ask Boy A, "Is Boy B Vasya?" This way, the answer depends on who Boy A is and who Boy B is.But this might complicate things because the truthfulness depends on the number of questions asked to each boy.Alternatively, maybe she can ask a question that is independent of their identities, like "Is 2+2=4?" But that might not help because all boys would answer truthfully or lie based on their behavior.Wait, actually, Petya would say "Yes," Vasya would lie on the first question and say "No," and Vitya would lie on the first question and say "No." So, if she asks "Is 2+2=4?" to a boy:- Petya: "Yes"- Vasya: "No"- Vitya: "No"So, if the answer is "Yes," she knows it's Petya. If it's "No," it could be Vasya or Vitya. But she only has three questions, so she might need to use this information.But asking "Is 2+2=4?" might not be the most efficient because it only helps identify Petya if the answer is "Yes," but if it's "No," she still doesn't know between Vasya and Vitya.Alternatively, maybe she can ask a question that is more informative, like "Are you Vasya or Vitya?" or something similar.Wait, let's think about the behavior again:- Petya always tells the truth.- Vasya lies on the first question, then tells the truth.- Vitya lies on the first two questions, then tells the truth.So, if Masha can ask a boy two questions, she can determine his identity based on the pattern of answers.For example, if she asks a boy:1. "Are you Vasya?"2. "Are you Vitya?"Then:- If the boy is Petya: Both answers are "No."- If the boy is Vasya: First answer is "No" (lie), second answer is "No" (truth).- If the boy is Vitya: First answer is "Yes" (lie), second answer is "No" (lie).Wait, that's different from before. Let me re-examine:- **Petya**: - "Are you Vasya?" – "No" - "Are you Vitya?" – "No"- **Vasya**: - First question: Lie. "Are you Vasya?" – "No" (lie) - Second question: Truth. "Are you Vitya?" – "No"- **Vitya**: - First question: Lie. "Are you Vasya?" – "Yes" (lie) - Second question: Lie. "Are you Vitya?" – "No" (lie)Wait, so Vitya would answer "Yes" to the first question and "No" to the second question.So, the responses would be:- Petya: "No," "No"- Vasya: "No," "No"- Vitya: "Yes," "No"So, if a boy answers "Yes," "No," he must be Vitya. If he answers "No," "No," he could be Petya or Vasya.So, if Masha asks two questions to one boy, she can identify Vitya if he answers "Yes," "No." If he answers "No," "No," she can't distinguish between Petya and Vasya with just two questions.But she has a third question left. So, maybe she can use the third question to ask another boy a question that can help her determine their identities.Alternatively, maybe she can ask the third question to the same boy, but since Vitya starts telling the truth from the third question, she can get a truthful answer.Wait, let's outline a possible strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" - If "Yes," then Boy A is either Vitya (lying) or Vasya (lying on first question). - If "No," then Boy A is either Petya (truthful) or Vasya (lying on first question).2. **Second Question**: Ask Boy A, "Are you Vitya?" - Based on the first answer, analyze the second answer. - If first answer was "Yes": - If second answer is "No," then Boy A is Vitya (since Vitya lies on the first two questions). - If second answer is "Yes," this is impossible because Vitya would lie on the second question as well. - If first answer was "No": - If second answer is "No," then Boy A is either Petya or Vasya. - If second answer is "Yes," then Boy A is Vitya (but he would have lied on the first question, which was "No," so this is inconsistent).Wait, this is getting confusing. Maybe a better approach is to consider that after two questions, if the responses are "Yes," "No," then it's Vitya. If it's "No," "No," then it's either Petya or Vasya.Then, for the third question, she can ask Boy A, "Are you Petya?" If Boy A is Petya, he will answer "Yes." If Boy A is Vasya, he will lie on the first question, but since this is the third question, he would have already told the truth on the second question, so he would lie again? Wait, no.Wait, Vasya lies on the first question, then tells the truth from the second question onwards. So, if she asks him a third question, he would tell the truth.Similarly, Vitya lies on the first two questions, then tells the truth from the third question onwards.So, if after two questions, Boy A is identified as Vitya, then she can ask him a third question to confirm.Alternatively, if Boy A is identified as "No," "No," then she can ask him a third question to determine if he's Petya or Vasya.Wait, let's try to formalize this:1. **First Question**: Ask Boy A, "Are you Vasya?" - If "Yes," proceed to second question. - If "No," proceed to second question.2. **Second Question**: Ask Boy A, "Are you Vitya?" - If first answer was "Yes": - If second answer is "No," then Boy A is Vitya. - If second answer is "Yes," this is inconsistent because Vitya would have lied on both questions, so he would have said "No" to the second question. - If first answer was "No": - If second answer is "No," then Boy A is either Petya or Vasya. - If second answer is "Yes," then Boy A is Vitya (but this is inconsistent because Vitya would have lied on the first question, which was "No," so he would have said "Yes" to the second question, but that would mean he's Vitya, which is consistent).Wait, this is getting too tangled. Maybe a better approach is to use the third question to ask a different boy.Alternatively, maybe she can ask each boy one question each, but that might not give enough information.Wait, here's another idea: ask each boy a question about another boy. For example:1. **First Question**: Ask Boy A, "Is Boy B Vasya?"2. **Second Question**: Ask Boy B, "Is Boy C Vitya?"3. **Third Question**: Ask Boy C, "Is Boy A Petya?"But this might not work because the truthfulness depends on the number of questions each boy has been asked.Alternatively, maybe she can ask a question that is independent of their identities, like "Is 2+2=4?" as I thought before.But let's think about it again:- Petya: "Yes"- Vasya: "No" (lying on first question)- Vitya: "No" (lying on first question)So, if she asks "Is 2+2=4?" to a boy:- If the answer is "Yes," it's Petya.- If the answer is "No," it's either Vasya or Vitya.Then, she can use the remaining questions to ask the other boys similar questions.But she only has three questions, so she needs to be efficient.Wait, maybe she can ask each boy one question, each about a different identity.For example:1. **First Question**: Ask Boy A, "Are you Vasya?"2. **Second Question**: Ask Boy B, "Are you Vitya?"3. **Third Question**: Ask Boy C, "Are you Petya?"But this might not work because the answers depend on the number of questions each boy has been asked.Alternatively, maybe she can ask the same question to all three boys, but that might not be efficient.Wait, perhaps she can use the fact that Vitya lies on the first two questions, so if she asks him two questions, the third one will be truthful.Similarly, Vasya lies on the first question, so the second one will be truthful.So, maybe she can structure her questions to exploit this.Here's a possible strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" - If "Yes," then Boy A is either Vitya (lying) or Vasya (lying on first question). - If "No," then Boy A is either Petya (truthful) or Vasya (lying on first question).2. **Second Question**: Ask Boy A, "Are you Vitya?" - If first answer was "Yes": - If second answer is "No," then Boy A is Vitya. - If second answer is "Yes," this is inconsistent because Vitya would have lied on both questions, so he would have said "No" to the second question. - If first answer was "No": - If second answer is "No," then Boy A is either Petya or Vasya. - If second answer is "Yes," then Boy A is Vitya.3. **Third Question**: Depending on the previous answers, ask Boy A or another boy a question to confirm.Wait, this is similar to what I thought before. Let me try to make it more concrete.Suppose Masha asks Boy A:1. "Are you Vasya?" – Answer: A12. "Are you Vitya?" – Answer: A2Based on A1 and A2, she can determine who Boy A is.- If A1 = "Yes" and A2 = "No," then Boy A is Vitya.- If A1 = "No" and A2 = "No," then Boy A is either Petya or Vasya.- If A1 = "No" and A2 = "Yes," then Boy A is Vitya (but this contradicts because Vitya would have lied on both questions, so he would have said "No" to the second question).Wait, so if A1 = "No" and A2 = "Yes," that's inconsistent because Vitya would have lied on both questions, so he would have said "No" to the second question. Therefore, this combination is impossible.So, the only possible consistent answers are:- A1 = "Yes," A2 = "No" → Vitya- A1 = "No," A2 = "No" → Petya or VasyaIf she gets A1 = "No," A2 = "No," she can't distinguish between Petya and Vasya with just two questions. So, she needs to use the third question to ask another boy.Alternatively, she can ask Boy A a third question to determine if he's Petya or Vasya.Since Vasya tells the truth from the second question onwards, if she asks him a third question, he will tell the truth. Similarly, Vitya tells the truth from the third question onwards.So, if she has identified that Boy A is either Petya or Vasya, she can ask him a third question that only Petya would answer truthfully.For example, she can ask him, "Is 2+2=4?" Again:- If Boy A is Petya, he will answer "Yes."- If Boy A is Vasya, since this is his third question, he will tell the truth, so he will answer "Yes" as well.Wait, that doesn't help because both would answer "Yes."Alternatively, she can ask a question that only Petya would answer truthfully, like "Are you Petya?"- If Boy A is Petya, he will answer "Yes."- If Boy A is Vasya, since this is his third question, he will tell the truth, so he will answer "No."Ah, that works!So, here's the revised strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy A, "Are you Petya?" – Answer: A3Now, let's analyze the possible outcomes:- If A1 = "Yes," A2 = "No," then Boy A is Vitya. Then, A3 would be a truthful answer, so if he's Vitya, he would answer "No" to "Are you Petya?" which is consistent.- If A1 = "No," A2 = "No," then Boy A is either Petya or Vasya. Then, A3: - If Boy A is Petya, he will answer "Yes." - If Boy A is Vasya, he will answer "No."So, based on A3, she can determine if Boy A is Petya or Vasya.Once she has identified Boy A, she can deduce the identities of the other two boys by elimination.For example:- If Boy A is Vitya, then the other two boys are Petya and Vasya.- If Boy A is Petya, then the other two boys are Vasya and Vitya.- If Boy A is Vasya, then the other two boys are Petya and Vitya.But wait, she only asked three questions to one boy. She hasn't asked the other two boys anything. So, she needs to make sure that her three questions are directed in a way that she can identify all three boys.Alternatively, maybe she can ask one question to each boy, but that might not give enough information because the answers depend on the number of questions each boy has been asked.Wait, perhaps she can ask the first two questions to one boy and the third question to another boy.For example:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy B, "Are you Petya?" – Answer: B1But then, she can analyze the answers as follows:- If A1 = "Yes," A2 = "No," then Boy A is Vitya. Then, she asks Boy B, "Are you Petya?" If Boy B is Petya, he will answer "Yes." If Boy B is Vasya, he will lie on the first question he's asked, which is this one, so he will answer "No." But since she only asked Boy B one question, if he answers "Yes," he's Petya; if "No," he's Vasya.- If A1 = "No," A2 = "No," then Boy A is either Petya or Vasya. Then, she asks Boy B, "Are you Petya?" If Boy B is Petya, he will answer "Yes." If Boy B is Vasya, he will lie on the first question he's asked, which is this one, so he will answer "No." If Boy B is Vitya, he will lie on the first question, so he will answer "No."Wait, but she doesn't know who Boy B is yet. This might not be the most efficient approach.Alternatively, maybe she can use the third question to ask Boy A a question that reveals his identity if he's Petya or Vasya.As I thought earlier, asking "Are you Petya?" as the third question can help distinguish between Petya and Vasya.So, to summarize, here's a possible strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy A, "Are you Petya?" – Answer: A3Based on the answers:- If A1 = "Yes," A2 = "No," then Boy A is Vitya. Then, A3 would be "No," confirming he's not Petya.- If A1 = "No," A2 = "No," then: - If A3 = "Yes," Boy A is Petya. - If A3 = "No," Boy A is Vasya.Once she identifies Boy A, she can deduce the other two boys by elimination.For example:- If Boy A is Vitya, then the other two boys are Petya and Vasya.- If Boy A is Petya, then the other two boys are Vasya and Vitya.- If Boy A is Vasya, then the other two boys are Petya and Vitya.But wait, she only asked three questions to one boy. She hasn't asked the other two boys anything. So, she might need to adjust her strategy to involve all three boys.Alternatively, maybe she can ask one question to each boy, but that might not give enough information because the answers depend on the number of questions each boy has been asked.Wait, here's another idea: ask each boy one question, but design the questions in such a way that the pattern of answers reveals their identities.For example:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy B, "Are you Vitya?" – Answer: B13. **Third Question**: Ask Boy C, "Are you Petya?" – Answer: C1But this approach doesn't take into account the number of questions each boy has been asked, which affects their truthfulness.Alternatively, maybe she can ask the same question to all three boys, but that might not be efficient.Wait, perhaps she can use the fact that Vitya lies on the first two questions, so if she asks him two questions, the third one will be truthful. Similarly, Vasya lies on the first question, so the second one will be truthful.So, maybe she can structure her questions to exploit this.Here's a possible strategy:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy A, "Are you Petya?" – Answer: A3As before, this can help identify Boy A.Alternatively, she can ask:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy B, "Are you Vitya?" – Answer: B13. **Third Question**: Ask Boy C, "Are you Petya?" – Answer: C1But again, this doesn't account for the number of questions each boy has been asked, which affects their truthfulness.Wait, maybe she can ask each boy a question about another boy, ensuring that the number of questions each boy has been asked is known.For example:1. **First Question**: Ask Boy A, "Is Boy B Vasya?" – Answer: A12. **Second Question**: Ask Boy B, "Is Boy C Vitya?" – Answer: B13. **Third Question**: Ask Boy C, "Is Boy A Petya?" – Answer: C1But this approach might not work because the truthfulness depends on the number of questions each boy has been asked.Alternatively, maybe she can ask each boy a question about themselves, but that might not help.Wait, perhaps she can ask each boy a question that is independent of their identities, like "Is 2+2=4?" as I thought before.But let's think about it again:- Petya: "Yes"- Vasya: "No" (lying on first question)- Vitya: "No" (lying on first question)So, if she asks "Is 2+2=4?" to each boy:1. **First Question**: Ask Boy A, "Is 2+2=4?" – Answer: A12. **Second Question**: Ask Boy B, "Is 2+2=4?" – Answer: B13. **Third Question**: Ask Boy C, "Is 2+2=4?" – Answer: C1Then:- If a boy answers "Yes," he's Petya.- If a boy answers "No," he's either Vasya or Vitya.But this only identifies Petya, and leaves the other two as Vasya or Vitya, which she can't distinguish with just three questions.Alternatively, she can use the fact that Vasya and Vitya have different lying patterns.Wait, here's a better idea: ask each boy a question that reveals their identity based on their lying pattern.For example:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy A, "Are you Petya?" – Answer: A3As before, this can help identify Boy A.But she needs to do this for all three boys, which would require more than three questions.Alternatively, she can ask one boy two questions and another boy one question.For example:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy B, "Are you Petya?" – Answer: B1Then, based on A1 and A2, she can identify Boy A, and based on B1, she can identify Boy B, and then deduce Boy C.But this might not always work because the answers depend on the number of questions each boy has been asked.Alternatively, maybe she can ask the first two questions to one boy and the third question to another boy, using the information from the first two to inform the third.For example:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy B, "Are you Petya?" – Answer: B1Then:- If A1 = "Yes," A2 = "No," then Boy A is Vitya. Then, if B1 = "Yes," Boy B is Petya; if B1 = "No," Boy B is Vasya.- If A1 = "No," A2 = "No," then Boy A is either Petya or Vasya. Then, if B1 = "Yes," Boy B is Petya; if B1 = "No," Boy B is Vasya.But this approach might not always work because if Boy A is Vasya, he would have lied on the first question, so A1 = "No" (lie), and told the truth on the second question, so A2 = "No" (truth). Then, if she asks Boy B, "Are you Petya?" and Boy B is Petya, he will answer "Yes." If Boy B is Vitya, he will lie on the first question he's asked, which is this one, so he will answer "No."Wait, but if Boy A is Vasya, then the other two boys are Petya and Vitya. So, if she asks Boy B, "Are you Petya?" and gets "Yes," then Boy B is Petya, and Boy C is Vitya. If she gets "No," then Boy B is Vitya, and Boy C is Petya.Similarly, if Boy A is Petya, then the other two boys are Vasya and Vitya. If she asks Boy B, "Are you Petya?" and gets "No," then Boy B is either Vasya or Vitya. But since Boy A is Petya, she can deduce that Boy B is Vasya or Vitya based on the answers.Wait, this seems to work. Let me outline it clearly:1. **First Question**: Ask Boy A, "Are you Vasya?" – Answer: A12. **Second Question**: Ask Boy A, "Are you Vitya?" – Answer: A23. **Third Question**: Ask Boy B, "Are you Petya?" – Answer: B1Now, analyze:- If A1 = "Yes," A2 = "No": - Boy A is Vitya. - Then, Boy B and Boy C are Petya and Vasya. - Ask Boy B, "Are you Petya?" – B1 - If B1 = "Yes," Boy B is Petya, Boy C is Vasya. - If B1 = "No," Boy B is Vasya, Boy C is Petya.- If A1 = "No," A2 = "No": - Boy A is either Petya or Vasya. - Ask Boy B, "Are you Petya?" – B1 - If B1 = "Yes," Boy B is Petya, so Boy A must be Vasya, and Boy C is Vitya. - If B1 = "No," Boy B is either Vasya or Vitya. But since Boy A is either Petya or Vasya, and Boy B is not Petya, Boy B must be Vitya, and Boy A must be Petya.Wait, this seems to cover all possibilities.So, to summarize:1. Ask Boy A, "Are you Vasya?" (A1)2. Ask Boy A, "Are you Vitya?" (A2)3. Ask Boy B, "Are you Petya?" (B1)Based on A1 and A2, determine Boy A's identity, then use B1 to determine Boy B's identity, and deduce Boy C's identity by elimination.This strategy should allow Masha to determine the names of all three boys with just three questions.