answer:Firstly, we determine the total number of elements, N, in the list. Each integer n from 1 to 100 appears n^2 times. Therefore, N is the sum of squares of the first 100 positive integers: [ N = 1^2 + 2^2 + 3^2 + ldots + 100^2 = frac{100 times 101 times 201}{6} = 338350 ] Because N is even, the median is the average of the elements at positions frac{N}{2} = 169175 and frac{N}{2} + 1 = 169176. We now determine which number n occurs at these positions by calculating cumulative counts. The cumulative count up to a number n is given by: [ text{Cumulative count up to } n = sum_{k=1}^n k^2 = frac{n times (n+1) times (2n+1)}{6} ] We need to find the smallest n such that this cumulative count reaches 169175. Solving the equation: [ frac{n times (n+1) times (2n+1)}{6} geq 169175 ] This is a cubic inequality in n, which is less straightforward to solve directly, but trial, simplification, or numerical methods can be used to find: [ n approx 72 ] Checking cumulative counts for n=71 and n=72: [ text{Cumulative count for } 71 = frac{71 times 72 times 143}{6} = 162906 ] [ text{Cumulative count for } 72 = frac{72 times 73 times 145}{6} = 171288 ] Since both 169175 and 169176 fall between 162906 and 171288, the 169175-th and 169176-th numbers are 72. Thus, the median is: [ 72 ] Conclusion: The median is valid and the calculation aligns with the hypothesis, making the problem solvable and coherent. The final answer is boxed{C} (72)

answer:We follow a similar procedure as before, arranging the numbers in a circle but using the numbers 1, 2, 3, 4, 6. Place 6 at the top and denote the others as ( a, b, c, d ) around it. The sum to maximize is ( 6a + ab + bc + cd + 6d ). After examining permutations and their circle arrangements, the assumption is that bigger numbers next to 6 will create a higher product. We start with the highest numbers next to 6, which are 4 and 5. Suppose ( a = 4 ) and ( d = 5 ), and now arrange ( b ) and ( c ). Let ( b = 3 ) and ( c = 2 ), then: [ 6 times 4 + 4 times 3 + 3 times 2 + 2 times 1 + 6 times 1 = 24 + 12 + 6 + 2 + 6 = 50. ] Trying different permutations, one can observe that higher sums generally involve placing the bigger numbers next to 6 and arranging the remaining in descending order around the circle. Testing all valid permutations gives a maximal sum of 50 in the cases: 1. ( (6, 4, 3, 2, 1) ) 2. ( (6, 1, 2, 3, 4) ) (reverse order) 3. and their cyclic permutations. Thus, ( M ) is 50, and ( N ) (number of such permutations including cyclic rotations and the reverse arrangement) is 10. Thus, ( M + N = 50 + 10 = boxed{60} ).

answer:Let lambda be the Lebesgue measure on mathbb{R}. We aim to show that, even though the Lebesgue measure is countably additive, there are scenarios where it fails to be continuous from above when dealing with measures that can take the value +infty. 1. Consider the family of sets [n, infty) for n in mathbb{N}. 2. The Lebesgue measure lambda([n, infty)) is given by: [ lambda([n, infty)) = infty ] for any n in mathbb{N} because the length (measure) of [n, infty) is unbounded in mathbb{R}. 3. The intersection of all such sets [n, infty) is: [ bigcap_{n in mathbb{N}} [n, infty) = varnothing ] Indeed, there is no real number that lies in every interval [n, infty) for all n in mathbb{N} because for any real number x, there exists n in mathbb{N} such that x < n. 4. Continuity from above states that if {A_n}_{n in mathbb{N}} is a decreasing sequence of measurable sets with A_1 supseteq A_2 supseteq A_3 supseteq cdots and bigcap_{n in mathbb{N}} A_n = varnothing, then: [ lim_{n to infty} lambda(A_n) = lambdaleft(bigcap_{n in mathbb{N}} A_nright) ] 5. However, in this case: [ lim_{n to infty} lambda([n, infty)) = infty ] while: [ lambdaleft(bigcap_{n in mathbb{N}} [n, infty)right) = lambda(varnothing) = 0 ] 6. Therefore, we observe that in the context of measures that can take the value +infty, countable additivity does not imply continuity from above. Specifically, when dealing with sets whose measures are +infty, their infinite measure does not translate to the measure of their intersection, which can be zero. This example demonstrates that for measures which can take the value +infty, countable additivity does not generally ensure continuity at the empty set varnothing. blacksquare

answer:A. Since b < a < 0, we have -b > -a > 0, which implies a^2 < b^2, so option A is correct. B. Since b < a < 0, and both are negative, multiplying both sides by b (which is negative), we get ab > b^2, so option B is incorrect because we expected ab < b^2. The comparison with b^2 should be with a^2, as in option A. C. Since 0 < frac{1}{2} < 1 and b < a, raising (frac{1}{2}) to a higher power makes it smaller, so (frac{1}{2})^b > (frac{1}{2})^a, and thus option C is incorrect. D. Since b < a < 0, both frac{a}{b} and frac{b}{a} are positive. Using the arithmetic mean-geometric mean inequality (AM-GM inequality), we have frac{a}{b} + frac{b}{a} > 2sqrt{frac{a}{b} cdot frac{b}{a}} = 2, which confirms that D is correct. Therefore, the only incorrect option is C. [ boxed{text{The answer is C.}} ]