💡Okay, so I have this problem where I need to compute the value of ( F_{1020} F_{1022} - F_{1021}^2 ). Hmm, Fibonacci numbers, right? I remember that Fibonacci numbers have some interesting properties and identities, so maybe I can use one of those to solve this without having to compute such large Fibonacci numbers directly.The problem also mentions a matrix identity for Fibonacci numbers. It says that for any positive integer ( n ),[begin{pmatrix} 1 & 1 1 & 0 end{pmatrix}^n = begin{pmatrix} F_{n + 1} & F_n F_n & F_{n - 1} end{pmatrix}.]I think this is related to how powers of that matrix can generate Fibonacci numbers. Maybe I can use the properties of determinants here because determinants have multiplicative properties, and that might help me relate the Fibonacci numbers.So, if I take the determinant of both sides of that matrix equation, I should get something useful. The determinant of a product of matrices is the product of their determinants, right? So,[det left( begin{pmatrix} 1 & 1 1 & 0 end{pmatrix}^n right) = det left( begin{pmatrix} F_{n + 1} & F_n F_n & F_{n - 1} end{pmatrix} right).]Calculating the determinant on the left side first. The determinant of the matrix ( begin{pmatrix} 1 & 1 1 & 0 end{pmatrix} ) is ( (1)(0) - (1)(1) = -1 ). So, raising that to the power ( n ), the determinant becomes ( (-1)^n ).On the right side, the determinant of ( begin{pmatrix} F_{n + 1} & F_n F_n & F_{n - 1} end{pmatrix} ) is ( F_{n + 1}F_{n - 1} - F_n^2 ). So, putting it all together, we have:[F_{n + 1}F_{n - 1} - F_n^2 = (-1)^n.]That's a nice identity. It relates the product of two Fibonacci numbers separated by one index minus the square of the middle Fibonacci number to ( (-1)^n ).Now, looking back at the problem, I need to compute ( F_{1020}F_{1022} - F_{1021}^2 ). Let me see how this fits into the identity I just found. If I set ( n = 1021 ), then according to the identity:[F_{1021 + 1}F_{1021 - 1} - F_{1021}^2 = (-1)^{1021}.]Simplifying that, it becomes:[F_{1022}F_{1020} - F_{1021}^2 = (-1)^{1021}.]Which is exactly the expression we have in the problem. So, ( F_{1020}F_{1022} - F_{1021}^2 = (-1)^{1021} ).Now, ( (-1)^{1021} ) is just ( -1 ) because 1021 is an odd number. So, the value of ( F_{1020}F_{1022} - F_{1021}^2 ) is ( -1 ).Wait, let me double-check to make sure I didn't make a mistake. So, the identity is ( F_{n+1}F_{n-1} - F_n^2 = (-1)^n ). Plugging in ( n = 1021 ), we get ( F_{1022}F_{1020} - F_{1021}^2 = (-1)^{1021} ). Since 1021 is odd, ( (-1)^{1021} = -1 ). Yeah, that seems right.I don't think I made any errors in the reasoning. The key was recognizing the matrix identity and using the determinant property to derive the Fibonacci identity. Once I had that, it was just a matter of plugging in the correct value of ( n ).So, I'm confident that the answer is ( -1 ).**Final Answer**boxed{-1}

💡First, I need to determine the side lengths of both the square and the original equilateral triangle. For the square, the perimeter is 40 inches. Since a square has four equal sides, I divide the perimeter by 4 to find the length of one side:40 inches ÷ 4 = 10 inches.Next, for the equilateral triangle with a perimeter of 45 inches, I divide the perimeter by 3 to find the length of one side:45 inches ÷ 3 = 15 inches.Now, I need to halve the side length of the square:10 inches ÷ 2 = 5 inches.I then add this halved side length to the side length of the original triangle:15 inches + 5 inches = 20 inches.This new side length of 20 inches forms a new equilateral triangle. To find the area of this new triangle, I use the formula for the area of an equilateral triangle:Area = (√3 / 4) × (side length)².Plugging in the new side length:Area = (√3 / 4) × (20 inches)² = (√3 / 4) × 400 square inches = 100√3 square inches.Therefore, the total area of the new triangle is 100√3 square inches.

💡I need to find the terminating decimal expansion of the fraction (frac{13}{200}).First, I'll look at the denominator, which is 200. I know that terminating decimals occur when the denominator can be expressed as a product of powers of 2 and 5. So, I'll factorize 200 into its prime components.200 can be broken down into (2^3 times 5^2). This means the denominator is already in a form that allows for a terminating decimal.To convert the fraction into a decimal, I can adjust the denominator to be a power of 10. Since the denominator is (2^3 times 5^2), I'll multiply both the numerator and the denominator by (2^2) to make the denominator (10^2).Multiplying the numerator 13 by (2^2) gives me 52, and the denominator becomes (10^2), which is 100. So, the fraction simplifies to (frac{52}{100}).Finally, dividing 52 by 100 gives me the decimal 0.52. Therefore, the terminating decimal expansion of (frac{13}{200}) is 0.52.

💡Alright, so I have this problem here that I need to prove. It says that if ( ab + bc + ac = 1 ), then[frac{a}{1 - a^2} + frac{b}{1 - b^2} + frac{c}{1 - c^2} = frac{4abc}{(1 - a^2)(1 - b^2)(1 - c^2)}]given that ( a neq pm 1 ), ( b neq pm 1 ), and ( c neq pm 1 ).Hmm, okay. Let me try to understand what's going on here. The condition is ( ab + bc + ac = 1 ), which is symmetric in ( a ), ( b ), and ( c ). The equation we need to prove is also symmetric, so maybe there's some symmetry we can exploit here.First, let me look at the left-hand side (LHS) and the right-hand side (RHS) of the equation we need to prove. The LHS is the sum of three fractions, each involving one variable, while the RHS is a single fraction involving the product of all three variables in the numerator and the product of the denominators from the LHS in the denominator.Maybe if I can manipulate the LHS to look more like the RHS, that would help. Let me try to combine the fractions on the LHS into a single fraction. To do that, I need a common denominator, which would be ( (1 - a^2)(1 - b^2)(1 - c^2) ).So, let's write each term with this common denominator:[frac{a}{1 - a^2} = frac{a(1 - b^2)(1 - c^2)}{(1 - a^2)(1 - b^2)(1 - c^2)}][frac{b}{1 - b^2} = frac{b(1 - a^2)(1 - c^2)}{(1 - a^2)(1 - b^2)(1 - c^2)}][frac{c}{1 - c^2} = frac{c(1 - a^2)(1 - b^2)}{(1 - a^2)(1 - b^2)(1 - c^2)}]Adding these together, the LHS becomes:[frac{a(1 - b^2)(1 - c^2) + b(1 - a^2)(1 - c^2) + c(1 - a^2)(1 - b^2)}{(1 - a^2)(1 - b^2)(1 - c^2)}]So, now the LHS is a single fraction with the same denominator as the RHS. That means if I can show that the numerator equals ( 4abc ), then the equality will hold.Let me expand the numerator:First, expand each term:1. ( a(1 - b^2)(1 - c^2) = a(1 - b^2 - c^2 + b^2c^2) = a - ab^2 - ac^2 + ab^2c^2 )2. ( b(1 - a^2)(1 - c^2) = b(1 - a^2 - c^2 + a^2c^2) = b - ba^2 - bc^2 + ba^2c^2 )3. ( c(1 - a^2)(1 - b^2) = c(1 - a^2 - b^2 + a^2b^2) = c - ca^2 - cb^2 + ca^2b^2 )Now, let's add these three expanded terms together:[(a - ab^2 - ac^2 + ab^2c^2) + (b - ba^2 - bc^2 + ba^2c^2) + (c - ca^2 - cb^2 + ca^2b^2)]Combine like terms:- Constants: ( a + b + c )- Terms with ( a^2 ): ( -ab^2 - ba^2 - ca^2 )- Terms with ( b^2 ): ( -ac^2 - bc^2 - cb^2 )- Terms with ( c^2 ): ( -ab^2 - bc^2 - cb^2 ) (Wait, I think I messed up here. Let me check again.)Wait, actually, let's list all the terms properly:1. Constants: ( a + b + c )2. Terms with ( a^2 ): ( -ab^2 - ba^2 - ca^2 )3. Terms with ( b^2 ): ( -ac^2 - bc^2 - cb^2 )4. Terms with ( c^2 ): ( -ab^2 - bc^2 - cb^2 )5. Terms with ( a^2b^2c^2 ): ( ab^2c^2 + ba^2c^2 + ca^2b^2 )Wait, that seems a bit messy. Maybe I should group them differently.Looking back, after expanding, the numerator is:[a - ab^2 - ac^2 + ab^2c^2 + b - ba^2 - bc^2 + ba^2c^2 + c - ca^2 - cb^2 + ca^2b^2]Let me rearrange the terms:- Constants: ( a + b + c )- Terms with ( a^2 ): ( -ab^2 - ba^2 - ca^2 )- Terms with ( b^2 ): ( -ac^2 - bc^2 - cb^2 )- Terms with ( c^2 ): ( -ab^2 - bc^2 - cb^2 )- Terms with ( a^2b^2c^2 ): ( ab^2c^2 + ba^2c^2 + ca^2b^2 )Wait, I think I'm overcomplicating this. Maybe I should factor out common terms.Looking at the numerator:[a(1 - b^2 - c^2 + b^2c^2) + b(1 - a^2 - c^2 + a^2c^2) + c(1 - a^2 - b^2 + a^2b^2)]Let me factor out ( a ), ( b ), and ( c ) respectively:[a[1 - b^2 - c^2 + b^2c^2] + b[1 - a^2 - c^2 + a^2c^2] + c[1 - a^2 - b^2 + a^2b^2]]Hmm, maybe I can factor these expressions inside the brackets.Notice that ( 1 - b^2 - c^2 + b^2c^2 = (1 - b^2)(1 - c^2) ). Wait, is that true?Let me check:( (1 - b^2)(1 - c^2) = 1 - b^2 - c^2 + b^2c^2 ). Yes, exactly. So, that simplifies each bracket.So, the numerator becomes:[a(1 - b^2)(1 - c^2) + b(1 - a^2)(1 - c^2) + c(1 - a^2)(1 - b^2)]Wait, but that's just the original numerator before expanding. So, maybe this approach isn't helping.Let me think differently. Since the condition is ( ab + bc + ac = 1 ), maybe I can use this to simplify the numerator.Let me denote ( ab + bc + ac = 1 ) as equation (1).Looking back at the numerator:[a(1 - b^2)(1 - c^2) + b(1 - a^2)(1 - c^2) + c(1 - a^2)(1 - b^2)]Let me expand each term:First term: ( a(1 - b^2 - c^2 + b^2c^2) )Second term: ( b(1 - a^2 - c^2 + a^2c^2) )Third term: ( c(1 - a^2 - b^2 + a^2b^2) )So, adding them all together:[a - ab^2 - ac^2 + ab^2c^2 + b - ba^2 - bc^2 + ba^2c^2 + c - ca^2 - cb^2 + ca^2b^2]Now, let's collect like terms:- Constants: ( a + b + c )- Terms with ( a^2 ): ( -ab^2 - ba^2 - ca^2 )- Terms with ( b^2 ): ( -ac^2 - bc^2 - cb^2 )- Terms with ( c^2 ): ( -ab^2 - bc^2 - cb^2 )- Terms with ( a^2b^2c^2 ): ( ab^2c^2 + ba^2c^2 + ca^2b^2 )Wait, this seems repetitive. Maybe I can factor out some terms.Looking at the terms with ( a^2 ):( -ab^2 - ba^2 - ca^2 = -a^2b - ab^2 - a^2c )Similarly, terms with ( b^2 ):( -ac^2 - bc^2 - cb^2 = -b^2c - bc^2 - b^2a )And terms with ( c^2 ):( -ab^2 - bc^2 - cb^2 = -c^2a - ac^2 - c^2b )Hmm, not sure if that helps.Wait, maybe I can factor out ( a ), ( b ), and ( c ) from each group.For the ( a^2 ) terms:( -a^2b - ab^2 - a^2c = -a(ab + b^2 + ac) )Similarly, for the ( b^2 ) terms:( -b^2c - bc^2 - b^2a = -b(bc + c^2 + ab) )And for the ( c^2 ) terms:( -c^2a - ac^2 - c^2b = -c(ac + c^2 + bc) )But from equation (1), ( ab + bc + ac = 1 ), so maybe we can substitute that in.Wait, let's see:For the ( a^2 ) terms:( -a(ab + b^2 + ac) = -a(ab + ac + b^2) )But ( ab + ac = 1 - bc ) from equation (1). So,( -a(1 - bc + b^2) )Similarly, for the ( b^2 ) terms:( -b(bc + c^2 + ab) = -b(ab + bc + c^2) )Again, ( ab + bc = 1 - ac ), so:( -b(1 - ac + c^2) )And for the ( c^2 ) terms:( -c(ac + c^2 + bc) = -c(ac + bc + c^2) )Again, ( ac + bc = 1 - ab ), so:( -c(1 - ab + c^2) )Hmm, not sure if this is helpful. Maybe I need a different approach.Let me consider the denominators. The LHS has denominators ( 1 - a^2 ), ( 1 - b^2 ), ( 1 - c^2 ). The RHS has the product of these denominators in the denominator and ( 4abc ) in the numerator.Maybe if I can write the LHS in terms of the product of denominators, I can relate it to the RHS.Alternatively, perhaps I can use substitution. Let me set ( x = a ), ( y = b ), ( z = c ), just to make it easier to write.So, the condition is ( xy + yz + xz = 1 ), and we need to prove:[frac{x}{1 - x^2} + frac{y}{1 - y^2} + frac{z}{1 - z^2} = frac{4xyz}{(1 - x^2)(1 - y^2)(1 - z^2)}]Hmm, maybe I can express each fraction in terms of ( xy + yz + xz ).Wait, another idea: perhaps I can use the identity that if ( ab + bc + ac = 1 ), then ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) = a^2 + b^2 + c^2 + 2 ). So, ( a^2 + b^2 + c^2 = (a + b + c)^2 - 2 ).But I'm not sure if that helps directly.Alternatively, maybe I can consider the reciprocals. Let me denote ( u = frac{a}{1 - a^2} ), ( v = frac{b}{1 - b^2} ), ( w = frac{c}{1 - c^2} ). Then, the LHS is ( u + v + w ), and the RHS is ( 4abc / [(1 - a^2)(1 - b^2)(1 - c^2)] ).But I don't see an immediate relationship between ( u ), ( v ), ( w ) and ( abc ).Wait, maybe I can express ( a ) in terms of ( u ). Let's solve for ( a ):( u = frac{a}{1 - a^2} Rightarrow u(1 - a^2) = a Rightarrow u - u a^2 = a Rightarrow u a^2 + a - u = 0 )This is a quadratic in ( a ):( u a^2 + a - u = 0 )Solving for ( a ):( a = frac{-1 pm sqrt{1 + 4u^2}}{2u} )Hmm, that seems complicated. Maybe not the best approach.Let me think again about the numerator. Maybe I can factor it differently.Looking back at the expanded numerator:[a + b + c - ab^2 - ac^2 - ba^2 - bc^2 - ca^2 - cb^2 + ab^2c^2 + ba^2c^2 + ca^2b^2]Wait, maybe I can group terms in a way that uses the condition ( ab + bc + ac = 1 ).Let me see:First, notice that ( ab + bc + ac = 1 ). Maybe I can express some terms in terms of this.Looking at the numerator:[a + b + c - ab^2 - ac^2 - ba^2 - bc^2 - ca^2 - cb^2 + ab^2c^2 + ba^2c^2 + ca^2b^2]Let me factor out ( a ), ( b ), and ( c ) from some terms:- ( -ab^2 - ac^2 = -a(b^2 + c^2) )- ( -ba^2 - bc^2 = -b(a^2 + c^2) )- ( -ca^2 - cb^2 = -c(a^2 + b^2) )So, the numerator becomes:[a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + ab^2c^2 + ba^2c^2 + ca^2b^2]Now, let's see if we can express ( b^2 + c^2 ), ( a^2 + c^2 ), and ( a^2 + b^2 ) in terms of the given condition.From the condition ( ab + bc + ac = 1 ), we can square both sides:( (ab + bc + ac)^2 = 1 )Expanding the left side:( a^2b^2 + b^2c^2 + a^2c^2 + 2a b^2c + 2ab c^2 + 2a^2bc = 1 )Hmm, not sure if that helps directly.Wait, maybe I can express ( a^2 + b^2 + c^2 ) in terms of ( (a + b + c)^2 ):( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) = a^2 + b^2 + c^2 + 2 )So, ( a^2 + b^2 + c^2 = (a + b + c)^2 - 2 )But I'm not sure if that helps with ( b^2 + c^2 ), etc.Alternatively, maybe I can use the condition to express ( a ), ( b ), or ( c ) in terms of the others.For example, from ( ab + bc + ac = 1 ), we can solve for ( c ):( c(a + b) + ab = 1 Rightarrow c = frac{1 - ab}{a + b} )But this might complicate things further.Wait, another idea: perhaps I can consider the product ( (1 - a^2)(1 - b^2)(1 - c^2) ) and see if it relates to the numerator.Let me compute ( (1 - a^2)(1 - b^2)(1 - c^2) ):First, expand two terms:( (1 - a^2)(1 - b^2) = 1 - a^2 - b^2 + a^2b^2 )Then multiply by ( (1 - c^2) ):[(1 - a^2 - b^2 + a^2b^2)(1 - c^2) = 1 - c^2 - a^2 + a^2c^2 - b^2 + b^2c^2 + a^2b^2 - a^2b^2c^2]So,[(1 - a^2)(1 - b^2)(1 - c^2) = 1 - a^2 - b^2 - c^2 + a^2c^2 + b^2c^2 + a^2b^2 - a^2b^2c^2]Hmm, interesting. Now, looking back at the numerator:[a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + ab^2c^2 + ba^2c^2 + ca^2b^2]I notice that the numerator has terms like ( -a(b^2 + c^2) ), which could relate to the expansion of ( (1 - a^2)(1 - b^2)(1 - c^2) ).Wait, let me see:From the expansion of ( (1 - a^2)(1 - b^2)(1 - c^2) ), we have:[1 - a^2 - b^2 - c^2 + a^2c^2 + b^2c^2 + a^2b^2 - a^2b^2c^2]Let me denote this as ( D = (1 - a^2)(1 - b^2)(1 - c^2) ).Now, looking back at the numerator ( N ):[N = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + ab^2c^2 + ba^2c^2 + ca^2b^2]Let me see if I can express ( N ) in terms of ( D ).Notice that ( D ) has terms like ( -a^2 ), ( -b^2 ), ( -c^2 ), and positive terms like ( a^2c^2 ), etc.In ( N ), we have terms like ( -a(b^2 + c^2) ), which is ( -ab^2 - ac^2 ).Wait, maybe I can factor ( N ) as follows:[N = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + abc^2(b + a + a) + text{something}]Wait, that seems unclear. Maybe another approach.Let me consider the numerator ( N ) and the denominator ( D ). If I can write ( N ) as ( 4abc ) times something, maybe that something is related to ( D ).Wait, let me try to factor ( N ).Looking at ( N ):[N = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + ab^2c^2 + ba^2c^2 + ca^2b^2]Let me factor out ( a ), ( b ), and ( c ) from the negative terms:[N = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + abc^2(b + a + a)]Wait, no, that's not accurate. Let me try again.Looking at the positive terms at the end:( ab^2c^2 + ba^2c^2 + ca^2b^2 = abc^2(b + a + a) ). Wait, no, that's not correct.Actually, ( ab^2c^2 + ba^2c^2 + ca^2b^2 = abc^2(b) + abc^2(a) + abc^2(a) ). Hmm, not quite.Wait, let's factor ( abc ) from these terms:( ab^2c^2 + ba^2c^2 + ca^2b^2 = abc(bc + ac + ab) = abc(1) = abc )Because ( ab + bc + ac = 1 ). Oh, that's a key insight!So, the last three terms in ( N ):( ab^2c^2 + ba^2c^2 + ca^2b^2 = abc(ab + bc + ac) = abc(1) = abc )So, now ( N ) simplifies to:[N = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2) + abc]Now, let's look at the remaining terms:[a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2)]Let me factor out ( a ), ( b ), and ( c ):[= a(1 - b^2 - c^2) + b(1 - a^2 - c^2) + c(1 - a^2 - b^2)]Now, recall that ( ab + bc + ac = 1 ). Let me see if I can express ( 1 - b^2 - c^2 ) in terms of ( a ).From the condition ( ab + bc + ac = 1 ), we can write:( 1 = ab + bc + ac )Let me solve for ( 1 - b^2 - c^2 ):Hmm, not directly obvious. Maybe I can use the identity ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) ).From this, we have:( a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) = (a + b + c)^2 - 2 )So,( 1 - b^2 - c^2 = 1 - (b^2 + c^2) = 1 - [(a^2 + b^2 + c^2) - a^2] = 1 - [( (a + b + c)^2 - 2 ) - a^2] )This seems complicated. Maybe another approach.Wait, let's consider ( 1 - b^2 - c^2 = (1 - b^2) - c^2 = (1 - b^2) - c^2 ). Not helpful.Alternatively, perhaps I can express ( 1 - b^2 - c^2 ) in terms of ( a ).From ( ab + bc + ac = 1 ), we can write ( a(b + c) + bc = 1 ). Maybe solve for ( b + c ) or something.But I'm not sure. Let me think differently.Looking back at the expression for ( N ):[N = a(1 - b^2 - c^2) + b(1 - a^2 - c^2) + c(1 - a^2 - b^2) + abc]Let me denote ( S = a + b + c ). Then, from the identity ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) ), we have:( a^2 + b^2 + c^2 = S^2 - 2 )So, ( 1 - b^2 - c^2 = 1 - (b^2 + c^2) = 1 - (S^2 - 2 - a^2) = 1 - S^2 + 2 + a^2 = 3 - S^2 + a^2 )Wait, that seems messy. Maybe not helpful.Alternatively, let's consider ( 1 - b^2 - c^2 = (1 - b^2) - c^2 ). From the condition ( ab + bc + ac = 1 ), maybe I can express ( 1 - b^2 ) in terms of ( a ) and ( c ).Wait, ( 1 - b^2 = ab + bc + ac - b^2 ). Hmm, not sure.Alternatively, maybe I can use the fact that ( 1 - b^2 = (1 - b)(1 + b) ), but I don't see how that helps.Wait, another idea: perhaps I can write ( 1 - b^2 - c^2 = (1 - b^2) - c^2 = (1 - b^2) - c^2 ). Hmm, not helpful.Wait, maybe I can consider the terms ( a(1 - b^2 - c^2) ), ( b(1 - a^2 - c^2) ), and ( c(1 - a^2 - b^2) ) and see if they can be expressed in terms of ( D ), the denominator.Recall that ( D = (1 - a^2)(1 - b^2)(1 - c^2) ). Maybe I can relate ( N ) to ( D ).Wait, let me think about the relationship between ( N ) and ( D ). If I can express ( N ) as ( 4abc ), then ( N = 4abc ), and since the denominator is ( D ), the LHS would equal ( 4abc / D ), which is the RHS.But how can I show that ( N = 4abc )?Wait, earlier I found that the last three terms in ( N ) sum to ( abc ). So, ( N = [a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2)] + abc ).If I can show that the expression in the brackets equals ( 3abc ), then ( N = 3abc + abc = 4abc ), which would complete the proof.So, let me focus on the expression in the brackets:[E = a + b + c - a(b^2 + c^2) - b(a^2 + c^2) - c(a^2 + b^2)]I need to show that ( E = 3abc ).Let me try to manipulate ( E ):First, expand ( E ):[E = a + b + c - ab^2 - ac^2 - ba^2 - bc^2 - ca^2 - cb^2]Now, let me group the terms:[E = (a - ab^2 - ac^2) + (b - ba^2 - bc^2) + (c - ca^2 - cb^2)]Factor out ( a ), ( b ), and ( c ):[E = a(1 - b^2 - c^2) + b(1 - a^2 - c^2) + c(1 - a^2 - b^2)]Wait, this is the same as before. Maybe I can express ( 1 - b^2 - c^2 ) in terms of ( a ).From the condition ( ab + bc + ac = 1 ), let me solve for ( 1 - b^2 - c^2 ):Hmm, not directly. Maybe I can use the identity ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2 ), so ( a^2 + b^2 + c^2 = S^2 - 2 ), where ( S = a + b + c ).But I'm not sure if that helps.Wait, another idea: perhaps I can express ( 1 - b^2 - c^2 ) as ( (1 - b^2) - c^2 ). From the condition ( ab + bc + ac = 1 ), maybe I can relate ( 1 - b^2 ) to ( a ) and ( c ).Wait, let me consider ( 1 - b^2 = ab + bc + ac - b^2 ). Hmm, not helpful.Alternatively, maybe I can write ( 1 - b^2 = (1 - b)(1 + b) ), but I don't see how that helps.Wait, perhaps I can use the fact that ( ab + bc + ac = 1 ) to express ( a ), ( b ), or ( c ) in terms of the others and substitute back into ( E ).For example, solve for ( c ):( c = frac{1 - ab}{a + b} )Then, substitute this into ( E ):But this might get too complicated. Let me try.Let ( c = frac{1 - ab}{a + b} ). Then,( E = a + b + frac{1 - ab}{a + b} - a(b^2 + (frac{1 - ab}{a + b})^2) - b(a^2 + (frac{1 - ab}{a + b})^2) - frac{1 - ab}{a + b}(a^2 + b^2) )This seems very messy. Maybe not the best approach.Wait, another idea: perhaps I can consider the expression ( E ) and see if it can be written as ( 3abc ).Let me assume ( E = 3abc ) and see if that leads to a contradiction or not.So, ( E = 3abc )Thus,[a + b + c - ab^2 - ac^2 - ba^2 - bc^2 - ca^2 - cb^2 = 3abc]Let me rearrange:[a + b + c = ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 + 3abc]Hmm, not sure if that's helpful.Wait, maybe I can factor the right-hand side.Looking at the right-hand side:[ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 + 3abc]Let me factor terms:- ( ab^2 + ba^2 = ab(a + b) )- ( ac^2 + ca^2 = ac(a + c) )- ( bc^2 + cb^2 = bc(b + c) )- Then, we have ( 3abc )So,[ab(a + b) + ac(a + c) + bc(b + c) + 3abc]Hmm, not sure.Wait, another idea: perhaps I can factor ( a + b + c ) out of the right-hand side.But I don't see an obvious way.Wait, maybe I can write the right-hand side as:[ab(a + b) + ac(a + c) + bc(b + c) + 3abc = ab(a + b + c) + ac(a + b + c) + bc(a + b + c) - abc(a + b + c)]Wait, let me check:( ab(a + b + c) = ab(a + b) + abc )Similarly,( ac(a + b + c) = ac(a + c) + abc )( bc(a + b + c) = bc(b + c) + abc )So,[ab(a + b + c) + ac(a + b + c) + bc(a + b + c) = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]Which matches the right-hand side. Therefore,[ab(a + b + c) + ac(a + b + c) + bc(a + b + c) = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]So, the right-hand side can be written as:[(ab + ac + bc)(a + b + c) = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]But from the condition ( ab + bc + ac = 1 ), we have:[1 cdot (a + b + c) = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]Thus,[a + b + c = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]But from earlier, we have:[a + b + c = ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 + 3abc]Which is the same as:[a + b + c = ab(a + b) + ac(a + c) + bc(b + c) + 3abc]Therefore, this shows that ( E = 3abc ), which means:[N = E + abc = 3abc + abc = 4abc]Thus, the numerator ( N = 4abc ), and the denominator is ( D = (1 - a^2)(1 - b^2)(1 - c^2) ).Therefore, the LHS:[frac{a}{1 - a^2} + frac{b}{1 - b^2} + frac{c}{1 - c^2} = frac{N}{D} = frac{4abc}{D} = frac{4abc}{(1 - a^2)(1 - b^2)(1 - c^2)}]Which is exactly the RHS. Therefore, the equality holds.So, after a lot of manipulation and using the given condition ( ab + bc + ac = 1 ), we were able to show that the numerator simplifies to ( 4abc ), making the LHS equal to the RHS.**Final Answer**boxed{frac{a}{1 - a^2} + frac{b}{1 - b^2} + frac{c}{1 - c^2} = frac{4abc}{(1 - a^2)(1 - b^2)(1 - c^2)}}

💡Alright, so I've got this problem here: I need to invent a ten-digit number with no zeros such that when I add the product of its digits to the number itself, the resulting number has the same product of digits as the original number. Hmm, that sounds a bit tricky, but let's break it down step by step.First off, let's understand what the problem is asking. I need a ten-digit number, so it has to be between 1,000,000,000 and 9,999,999,999. Also, none of the digits can be zero. That means each digit has to be from 1 to 9. Got it.Now, the main condition is that if I take the product of all the digits of this number and add it to the number itself, the resulting number should have the same product of digits as the original number. So, if my number is N, and P is the product of its digits, then N + P should be a number where the product of its digits is also P.Let me write that down:Let N be the ten-digit number, and P = product of digits of N.Then, N + P should be a number where the product of its digits is also P.So, P = product of digits of (N + P).Okay, that makes sense.Now, how do I approach finding such a number? Maybe I can start by thinking about smaller numbers and see if I can find a pattern or a rule that might help me extend it to ten digits.Let's consider a smaller example first. Suppose I have a two-digit number, say 13. The product of its digits is 1*3 = 3. If I add that to 13, I get 13 + 3 = 16. Now, the product of the digits of 16 is 1*6 = 6, which is not equal to 3. So, that doesn't work.Wait, but if I take 28, the product is 2*8 = 16. Adding that to 28 gives 28 + 16 = 44. The product of the digits of 44 is 4*4 = 16, which is the same as the original product. So, 28 is a number that satisfies this condition for a two-digit number.Interesting. So, 28 is a two-digit number where adding the product of its digits to itself results in another number with the same product of digits.Maybe I can use this idea to construct a ten-digit number. But how?One thought is that if I have a number where adding the product of its digits just rearranges the digits or changes some digits in a way that the overall product remains the same, then that might work.Looking back at the two-digit example, 28 becomes 44 when we add the product. The product remains 16 because 4*4 is still 16. So, in a way, the digits are transformed but the product stays the same.Perhaps I can create a ten-digit number where adding the product of its digits only affects certain digits, keeping the overall product the same.But how do I ensure that the product remains the same after adding it to the number? It seems like a delicate balance because adding the product could change the digits in unpredictable ways.Maybe I can look for numbers where the product of digits is a relatively small number, so that when I add it to the original number, it doesn't drastically change the digits. For example, if the product is small, adding it might only affect the last few digits, leaving the rest of the digits unchanged.Let's test this idea with a three-digit number. Suppose I have 132. The product is 1*3*2 = 6. Adding that to 132 gives 138. The product of digits of 138 is 1*3*8 = 24, which is different from 6. So, that doesn't work.What about 144? The product is 1*4*4 = 16. Adding that to 144 gives 160. But 160 has a zero, which is not allowed. So, that's out.How about 129? The product is 1*2*9 = 18. Adding that to 129 gives 147. The product of digits of 147 is 1*4*7 = 28, which is different from 18. Not good.Hmm, maybe I need a different approach. Perhaps instead of trying random numbers, I can think about the properties that such a number must have.First, the product of the digits must be such that when added to the number, it doesn't introduce any zeros. So, adding P (the product) to N must not result in any digit becoming zero.Also, the product of the digits of N + P must be equal to P. So, the way the digits change when adding P must preserve the product.This seems complicated. Maybe I can think about the digits in terms of their prime factors. Since the product of digits is P, and we want the product of digits of N + P to also be P, the digits must somehow rearrange or change in a way that the prime factors remain the same.But I'm not sure how to apply that directly.Another idea: perhaps if the product P is a multiple of 9, then adding P to N might carry over in a way that preserves the digit product. But I'm not sure.Wait, let's think about the two-digit example again: 28. The product is 16, and adding 16 gives 44, which also has a product of 16. So, in this case, the digits changed from 2 and 8 to 4 and 4, but the product stayed the same.So, maybe if I can find a way to have the digits of N + P be a rearrangement or a transformation of the digits of N that preserves the product.But how?Perhaps if I have digits that can be transformed into other digits that multiply to the same product. For example, 2 and 8 can be transformed into 4 and 4, since 2*8 = 4*4 = 16.So, maybe I can design the number N such that when I add P, certain digits change in a way that their product remains the same.But how can I ensure that adding P will only affect certain digits and not disrupt the entire number?Maybe I can have most of the digits as 1s, since multiplying by 1 doesn't change the product. Then, the significant digits can be arranged in a way that adding P only affects those significant digits.For example, if I have a number like 1111111613, which is a ten-digit number with mostly 1s and ending with 613. The product of its digits is 1*1*1*1*1*1*1*6*1*3 = 18. Adding 18 to 1111111613 gives 1111111631. The product of the digits of 1111111631 is 1*1*1*1*1*1*1*6*3*1 = 18, which is the same as before.So, in this case, adding the product of digits (18) to the number only changes the last two digits from 613 to 631, but the product remains the same because 6*1*3 = 6*3*1.Ah, so this seems to work! By having most digits as 1s, the product is dominated by the non-1 digits, and adding the product only affects the non-1 digits in a way that preserves the product.So, the key idea is to have a number with mostly 1s and a few digits that can be transformed when adding the product, keeping the product the same.Let me verify this with the example:N = 1111111613P = product of digits = 1*1*1*1*1*1*1*6*1*3 = 18N + P = 1111111613 + 18 = 1111111631Product of digits of N + P = 1*1*1*1*1*1*1*6*3*1 = 18Yes, it works!So, the strategy is to create a ten-digit number with mostly 1s and a few digits that can be adjusted when adding the product, ensuring the product remains the same.Another example could be 111111128, but wait, that's only nine digits. To make it ten digits, I can add another 1 at the beginning: 1111111281.Let's check:N = 1111111281P = 1*1*1*1*1*1*1*2*8*1 = 16N + P = 1111111281 + 16 = 1111111297Product of digits of N + P = 1*1*1*1*1*1*1*2*9*7 = 126Wait, that's not equal to 16. So, this doesn't work.Hmm, why didn't this work? Because adding 16 changed the digits in a way that increased the product. So, maybe I need to be careful about which digits I choose.In the previous example, adding 18 to 1111111613 changed 613 to 631, which kept the product the same because 6*1*3 = 6*3*1.But in this case, adding 16 to 1111111281 changed 281 to 297, and 2*8*1 = 16, but 2*9*7 = 126, which is different.So, the key is to have the digits that change when adding P to multiply to the same product.Therefore, I need to choose the non-1 digits such that when I add P, their product remains the same.In the first example, 613 becomes 631, and 6*1*3 = 6*3*1 = 18.So, the digits rearrange but keep the same product.In the second example, 281 becomes 297, but 2*8*1 ≠ 2*9*7.Therefore, to make it work, the digits that change should rearrange or transform in a way that their product remains the same.So, perhaps I can choose the non-1 digits such that their product is a number that, when added to the number, only affects those digits in a way that preserves the product.Another idea: if the non-1 digits are such that their product is a number that, when added, only increments certain digits without changing the overall product.But this seems too vague.Wait, let's think about the first example again. The number is 1111111613.The non-1 digits are 6, 1, 3. Their product is 18.Adding 18 to the number changes the last three digits from 613 to 631.So, 613 + 18 = 631.Now, 6*1*3 = 18, and 6*3*1 = 18.So, essentially, adding 18 to 613 just rearranges the digits, keeping the product the same.Therefore, if I can find a set of digits where adding their product just rearranges them, then the overall number will satisfy the condition.So, perhaps I can choose the non-1 digits such that their product is a number that, when added to the number formed by those digits, results in a rearrangement of those digits.In the case of 613, adding 18 gives 631, which is a rearrangement.Similarly, if I have digits 2, 8, adding their product 16 gives 28 + 16 = 44, which is a rearrangement in a way (though not a direct rearrangement, but the product remains the same).So, maybe I can use this idea to construct a ten-digit number.Let me try to create another example.Suppose I have the digits 3, 2, 6. Their product is 3*2*6 = 36.If I add 36 to the number formed by these digits, say 326, I get 326 + 36 = 362.Now, the product of digits of 362 is 3*6*2 = 36, which is the same as before.So, 326 + 36 = 362, and both have the same product of digits.Therefore, if I create a ten-digit number with mostly 1s and ending with 326, then adding the product 36 would change the last three digits to 362, preserving the product.So, let's try that.N = 1111111326P = product of digits = 1*1*1*1*1*1*1*3*2*6 = 36N + P = 1111111326 + 36 = 1111111362Product of digits of N + P = 1*1*1*1*1*1*1*3*6*2 = 36Yes, it works!So, this seems to be a valid approach. By having a ten-digit number with mostly 1s and ending with a set of digits whose product, when added, rearranges those digits to maintain the same product.Another example could be using the digits 8, 1, 9. Their product is 8*1*9 = 72.Adding 72 to 819 gives 819 + 72 = 891.The product of digits of 891 is 8*9*1 = 72, which is the same.So, if I create a ten-digit number ending with 819, adding 72 would change it to 891, preserving the product.Let's test it:N = 1111111819P = product of digits = 1*1*1*1*1*1*1*8*1*9 = 72N + P = 1111111819 + 72 = 1111111891Product of digits of N + P = 1*1*1*1*1*1*1*8*9*1 = 72Yes, it works!So, it seems like a reliable method to construct such numbers. The key is to have a ten-digit number with mostly 1s and ending with a set of digits whose product, when added, rearranges those digits to maintain the same product.Therefore, to answer the original problem, I can construct a ten-digit number using this approach.Let me choose another set of digits to ensure variety. Suppose I take the digits 4, 4. Their product is 4*4 = 16.Adding 16 to 44 gives 44 + 16 = 60, but 60 has a zero, which is not allowed. So, that doesn't work.Wait, but if I have more digits, maybe it can work. Let's try with three digits: 4, 4, 1. Their product is 4*4*1 = 16.Adding 16 to 441 gives 441 + 16 = 457.The product of digits of 457 is 4*5*7 = 140, which is different from 16. So, that doesn't work.Hmm, maybe I need to choose digits where adding their product doesn't introduce zeros and keeps the product the same.Another idea: use digits that are all the same, like 3, 3, 3. Their product is 27.Adding 27 to 333 gives 333 + 27 = 360, which has a zero. Not good.Alternatively, 2, 2, 2. Product is 8.Adding 8 to 222 gives 230, which has a zero. Again, not good.Hmm, seems like using multiple same digits can lead to zeros when adding the product.Maybe I need to stick with digits that when added, don't produce a zero.Going back to the previous examples, using digits like 6, 1, 3 or 3, 2, 6 or 8, 1, 9 seems to work because adding their product doesn't introduce zeros and keeps the product the same.So, perhaps I can use these sets of digits to construct my ten-digit number.Let me try one more example.Suppose I have the digits 5, 5, 2. Their product is 5*5*2 = 50.Adding 50 to 552 gives 552 + 50 = 602, which has a zero. Not good.Alternatively, 5, 5, 5. Product is 125.Adding 125 to 555 gives 555 + 125 = 680, which has a zero. Not good.Hmm, tricky.Wait, maybe I can use four digits instead of three. Let's try 2, 2, 2, 2. Product is 16.Adding 16 to 2222 gives 2222 + 16 = 2238.Product of digits of 2238 is 2*2*3*8 = 96, which is different from 16.Not good.Alternatively, 2, 2, 3, 3. Product is 2*2*3*3 = 36.Adding 36 to 2233 gives 2233 + 36 = 2269.Product of digits of 2269 is 2*2*6*9 = 216, which is different from 36.Hmm.Maybe I need to stick with three digits for the non-1 part.Let me try 7, 1, 1. Product is 7*1*1 = 7.Adding 7 to 711 gives 711 + 7 = 718.Product of digits of 718 is 7*1*8 = 56, which is different from 7.Not good.Alternatively, 7, 7, 1. Product is 7*7*1 = 49.Adding 49 to 771 gives 771 + 49 = 820, which has a zero. Not good.Hmm, seems challenging.Wait, maybe I can use digits that when added, the carryover doesn't introduce zeros.For example, if I have digits 9, 9, 9. Product is 729.Adding 729 to 999 gives 999 + 729 = 1728.Product of digits of 1728 is 1*7*2*8 = 112, which is different from 729.Not good.Alternatively, 9, 9, 1. Product is 81.Adding 81 to 991 gives 991 + 81 = 1072, which has a zero. Not good.Hmm.Maybe I need to think differently. Instead of trying to find a set of digits where adding their product rearranges them, perhaps I can find a number where adding the product doesn't change the digits at all.Wait, that would mean P = 0, but the number has no zeros, so P can't be zero. So, that's not possible.Alternatively, maybe adding P doesn't change the digits except for some that don't affect the product.But I'm not sure.Wait, going back to the first example, 1111111613 + 18 = 1111111631.Here, the digits that changed were the last three digits: 613 became 631.So, the product of those digits remained the same because 6*1*3 = 6*3*1.Therefore, the key was that adding P only affected the last few digits in a way that preserved the product.So, perhaps I can generalize this.If I have a ten-digit number where the last few digits are such that adding their product only rearranges them, keeping the product the same, and the rest of the digits are 1s, then the overall product remains the same.Therefore, to construct such a number, I need to:1. Choose a set of digits (let's say k digits) such that their product P, when added to the number formed by these digits, results in a rearrangement of these digits, keeping the product P.2. The rest of the digits in the ten-digit number should be 1s, so that they don't affect the product.3. The total number of digits should be ten.So, for example, in the first example, k = 3, and the digits are 6, 1, 3. Their product is 18, and adding 18 to 613 gives 631, which is a rearrangement.Therefore, the ten-digit number is 1111111613.Similarly, for the set 3, 2, 6, their product is 36, and adding 36 to 326 gives 362, which is a rearrangement.Therefore, the ten-digit number is 1111111326.Another example with digits 8, 1, 9: product is 72, adding 72 to 819 gives 891, which is a rearrangement.Therefore, the ten-digit number is 1111111819.So, the strategy is solid. Now, to ensure that the number is ten digits, I need to have seven 1s followed by the three digits whose product, when added, rearranges them.But wait, in the examples above, it's seven 1s followed by three digits, making a total of ten digits.Yes, that's correct.So, to generalize, the number would be:1111111XYZWhere XYZ is a three-digit number such that XYZ + (X*Y*Z) is a rearrangement of XYZ, keeping the product X*Y*Z the same.Therefore, to find such numbers, I need to find all three-digit numbers XYZ where XYZ + (X*Y*Z) is a permutation of XYZ.This seems like a specific condition, but there are known numbers that satisfy this.For example:- 132 + (1*3*2) = 132 + 6 = 138. But 138 is not a permutation of 132.- 144 + (1*4*4) = 144 + 16 = 160. Not a permutation and has a zero.- 222 + (2*2*2) = 222 + 8 = 230. Not a permutation and has a zero.- 28 + (2*8) = 28 + 16 = 44. This is a two-digit number, but 44 is a permutation of 28 in a way (though not directly, but the product remains the same).Wait, in the two-digit case, 28 + 16 = 44, and 4*4 = 16, which is the same as 2*8 = 16.So, even though it's not a direct permutation, the product remains the same.Therefore, perhaps in the three-digit case, we can have numbers where adding the product results in a number with the same product, even if it's not a direct permutation.But in the examples above, like 613 + 18 = 631, it is a permutation.Similarly, 326 + 36 = 362, which is a permutation.And 819 + 72 = 891, which is a permutation.So, these are valid.Therefore, to find such numbers, I can look for three-digit numbers where adding their digit product results in a permutation of themselves.These are known as "digit product additive permutations."I think there are only a few such numbers, but for the purpose of this problem, I can use one of them to construct the ten-digit number.So, to answer the original question, I can choose one of these three-digit numbers and prepend seven 1s to make it ten digits.For example, using 613:N = 1111111613P = 1*1*1*1*1*1*1*6*1*3 = 18N + P = 1111111613 + 18 = 1111111631Product of digits of N + P = 1*1*1*1*1*1*1*6*3*1 = 18Therefore, N = 1111111613 satisfies the condition.Similarly, using 326:N = 1111111326P = 1*1*1*1*1*1*1*3*2*6 = 36N + P = 1111111326 + 36 = 1111111362Product of digits of N + P = 1*1*1*1*1*1*1*3*6*2 = 36Therefore, N = 1111111326 also satisfies the condition.Another example using 819:N = 1111111819P = 1*1*1*1*1*1*1*8*1*9 = 72N + P = 1111111819 + 72 = 1111111891Product of digits of N + P = 1*1*1*1*1*1*1*8*9*1 = 72Therefore, N = 1111111819 also works.So, there are multiple such numbers, but the key is to use a three-digit number where adding the product of its digits results in a permutation of itself, and then prepend seven 1s to make it ten digits.Therefore, the answer to the problem is any ten-digit number constructed in this way. For simplicity, I can choose the first example:N = 1111111613But to ensure variety, I can also present another example like 1111111326 or 1111111819.However, since the problem asks to "invent" such a number, I can choose any one of them. For definiteness, I'll go with 1111111613.So, the final answer is:**Final Answer**boxed{1111111613}

💡To determine the percentage of employees at Newton Inc. who have been employed for 7 years or more, we need to analyze the given bar graph. However, the vertical axis lacks a scale, which means we need to assign a variable to represent the number of employees each count on the graph stands for.Let's denote the number of employees each count represents as ( y ).From the problem, we have the following distribution of employees based on their years of employment:- Less than 1 year: ( 4y )- 1 to 2 years: ( 6y )- 2 to 3 years: ( 5y )- 3 to 4 years: ( 2y )- 4 to 5 years: ( 3y )- 5 to 6 years: ( 1y )- 6 to 7 years: ( 1y )- 7 to 8 years: ( 2y )- 8 to 9 years: ( 2y )- 9 to 10 years: ( 1y )- 10+ years: ( 1y )First, we calculate the total number of employees by summing all these values:[text{Total employees} = 4y + 6y + 5y + 2y + 3y + 1y + 1y + 2y + 2y + 1y + 1y = 28y]Next, we determine the number of employees who have been employed for 7 years or more. This includes the employees in the following categories:- 7 to 8 years: ( 2y )- 8 to 9 years: ( 2y )- 9 to 10 years: ( 1y )- 10+ years: ( 1y )Adding these together:[text{Employees with 7+ years} = 2y + 2y + 1y + 1y = 6y]To find the percentage of these employees relative to the total number of employees, we use the formula:[text{Percentage} = left( frac{text{Number of employees with 7+ years}}{text{Total number of employees}} right) times 100 = left( frac{6y}{28y} right) times 100 = frac{6}{28} times 100 = 21.43%]Therefore, the percentage of employees at Newton Inc. who have been employed for 7 years or more is ( 21.43% ).