answer:1. **Identify the variables and initial weights:** - Let ( x ) be the difference in weight between any two consecutive gnomes. - The weight of the heaviest gnome, which is the first gnome, is ( 5 text{ kg} ). - Thus, the weights follow the pattern where the ( n )-th gnome weighs ( 5 - (n-1)x text{ kg} ). 2. **Calculate the total weight of the 76th to 80th gnomes:** - The weights of these gnomes are: [ begin{aligned} text{76th gnome:} & quad 5 - 75x, text{77th gnome:} & quad 5 - 76x, text{78th gnome:} & quad 5 - 77x, text{79th gnome:} & quad 5 - 78x, text{80th gnome:} & quad 5 - 79x. end{aligned} ] - The sum of these weights is: [ (5 - 75x) + (5 - 76x) + (5 - 77x) + (5 - 78x) + (5 - 79x) = 25 - 385x ] 3. **Calculate the total weight of the 96th to 101st gnomes:** - The weights include: [ begin{aligned} text{96th gnome:} & quad 5 - 95x, text{97th gnome:} & quad 5 - 96x, text{98th gnome:} & quad 5 - 97x, text{99th gnome:} & quad 5 - 98x, text{100th gnome:} & quad 5 - 99x, text{101st gnome:} & quad 5 - 100x. end{aligned} ] - The sum of these weights is: [ (5 - 95x) + (5 - 96x) + (5 - 97x) + (5 - 98x) + (5 - 99x) + (5 - 100x) = 30 - 585x ] 4. **Set up the equation:** - According to the problem, the total weights of the 76th to 80th gnomes equal the total weight of the 96th to 101st gnomes. Thus: [ 25 - 385x = 30 - 585x ] 5. **Solve for ( x ):** [ begin{aligned} 25 - 385x & = 30 - 585x, 200x & = 5, x & = frac{5}{200} = 0.025 text{ kg}. end{aligned} ] 6. **Calculate the weight of the lightest (101st) gnome:** - The lightest gnome is the 101st gnome, where ( n = 101 ): [ text{Weight of the 101st gnome} = 5 - 100 cdot 0.025 = 5 - 2.5 = 2.5 text{ kg}. ] # Conclusion: The weight of the lightest gnome is ( boxed{2.5 text{ kg}} ).

answer:1. **Interpreting the Given Condition:** We need to find all real-coefficient polynomials ( P(x) ) such that if ( P(a) in mathbb{Z} ) for some ( a in mathbb{R} ), then ( a in mathbb{Z} ). 2. **Applying the Intermediate Value Theorem:** If ( P(a) in mathbb{Z} ) for ( a in mathbb{R} ) implies ( a in mathbb{Z} ), then consider ( P(n) ) where ( n in mathbb{Z} ). By the Intermediate Value Theorem, the difference in the values of ( P ) at consecutive integers must be relatively small: [ |P(n+1) - P(n)| leq 1 ] This is because ( P(x) ) must transition from one integer value to another without crossing any non-integer values. 3. **Analyzing the Polynomial Degree:** Assume ( P(x) ) is a polynomial of degree ( d ): [ P(x) = a_d x^d + a_{d-1} x^{d-1} + cdots + a_1 x + a_0 ] Evaluate the change ( P(n+1) - P(n) ) where ( n ) is an integer: [ |P(n+1) - P(n)| = |f(n)| ] for some polynomial ( f(n) ) with degree ( d-1 ). 4. **Considering Constant and Linear Polynomials:** - If ( d geq 2 ), ( P(n+1) - P(n) ) will typically involve more significant changes than ±1 as ( x to infty ). Therefore, higher-degree terms must be zero to satisfy the given condition. - This restricts ( P(x) ) either to be a constant polynomial or a linear polynomial (degree ( leq 1 )): [ P(x) = ax + b ] 5. **Conditions on Linear Polynomials:** For ( P(x) = ax + b ): - ( P(n) in mathbb{Z} ) implies ( an + b in mathbb{Z} ) for any integer ( n ). - Consider ( a = 0 ). In this case, ( P(x) = b ) where ( b ) must be an integer for it to hold that ( b in mathbb{Z} ). If ( a neq 0 ): - For ( P(n) = an + b ) to map integers ( n ) to integers, ( a ) must be such that the term ( an ) (where ( a = 1/k ) for some integer ( k )) and ( b in k mathbb{Z} ). - Simplifying, ( a = 1 ) (the coefficient term will properly account for integer transitions, preserving the conditions of ( a ) being integer multiplicative). 6. **Final Form of the Polynomial:** The polynomials that satisfy the given conditions are: - Constant polynomials where the constant is an integer. - Linear polynomials where ( a in mathbb{Z} ) and ( b in mathbb{Z} ). 7. **Conclusion:** [ boxed{P(x) = kx + b text{ where } k, b in mathbb{Z}} ] or [ boxed{P(x) = c text{ where } c in mathbb{Z}} ]

answer:To solve the problem, we start by simplifying the given complex number 1+frac{1}{{1+i}}. Step 1: Simplify the fraction frac{1}{{1+i}}. We multiply the numerator and the denominator by the conjugate of the denominator to get rid of the complex number in the denominator: frac{1}{{1+i}} cdot frac{1-i}{1-i} = frac{1-i}{(1+i)(1-i)} Step 2: Simplify the denominator (1+i)(1-i). We use the difference of squares formula, a^2 - b^2 = (a+b)(a-b), where a=1 and b=i: (1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 2 Step 3: Continue the simplification. Now, we have: 1+frac{1-i}{2} = 1 + frac{1}{2} - frac{i}{2} = frac{3}{2} - frac{1}{2}i Step 4: Determine the quadrant of the complex number. The complex number frac{3}{2} - frac{1}{2}i corresponds to the point (frac{3}{2}, -frac{1}{2}) in the complex plane. Since the real part is positive and the imaginary part is negative, this point is in the fourth quadrant. Therefore, the correct answer is: boxed{D}

answer:Analyzing the given points, we can see which point could be the center by checking which one is equidistant from the other two. Points (1, 5) and (11, 5) are clearly endpoints of the same axis (horizontal) because they share the same y-coordinate. The midpoint of this segment, which is the center of the ellipse, is calculated as: [ left(frac{1+11}{2}, frac{5+5}{2}right) = (6, 5). ] This means point (4, -3) is on the vertical axis, and the missing point must also lie on this vertical line, symmetrically opposite about the center (6, 5). To find this point: [ 5 - (-3) = 8 quad text{implies} quad 5 + 8 = 13. ] So, the missing point is (6, 13). Now we can find the lengths of the semi-major and semi-minor axes: - Semi-major axis: Distance between (6, -3) and (6, 13) is 13 - (-3) = 16, so the full length is 16 (semi-major axis is 8). - Semi-minor axis: Distance between (1, 5) and (11, 5) is 11 - 1 = 10, so the full length is 10 (semi-minor axis is 5). The distance between the foci c of the ellipse is: [ c = 2sqrt{8^2 - 5^2} = 2sqrt{64 - 25} = 2sqrt{39}. ] Thus, the distance between the foci is boxed{2sqrt{39}}.