answer:2 Claim: 19. In fact, the "MakeMyDay" procedure does not change the maximum difference between two numbers on the list. Suppose our list is {a, b, c} with a <b<c. The maximum difference between the largest and the smallest number is c-a. The "MakeMyDay" operation creates {b+c, a+c, a+b}.Since a<b, we know that a+c<b+c. Since b<c, we also know that a+b<a+c.Combining these two inequalities, we have a+b<a+c<b+c.The maximum difference between any number is (b+c)-(a+b) or c-a.So the same as the one we started with. For the initial list of {20,1,8}, the maximum difference will always be 19 .

answer:For n=1, the value of the given expression is 0, so (n-1) can be factored out:(n-1)left(n^{4}-2 n^{2}+1right)The second factor of the product is a perfect square left(n^{2}-1right)^{2}, and if n is a natural number, this is the square of a natural number. Thus, the product is a perfect square precisely when either the first factor is a perfect square, or the second factor is 0.In the latter case, |n|=1, and thus for natural numbers, n=1. In the former case, n=k^{2}+1, where k is an integer. Since choosing k=0 gives us the case n=1, the given expression is a perfect square if and only if n=k^{2}+1, where k is an integer.

answer:(1) Since f(x)= frac {1}{3}x^{3}-4x+4,we have the derivative of f(x) as follows: f'(x) = x^2 - 4. For the function to be monotonically increasing, the derivative should be greater than or equal to zero: f'(x) = x^2 - 4 geq 0. This inequality holds when x geq 2 or x leq -2.Thus, the intervals of monotonic increase for the function f(x) are (-infty, -2] and [2, +infty).(2) To find critical points where the derivative is zero, we solve f'(x) = x^2 - 4 = 0. This gives us x_1 = -2 and x_2 = 2.Calculating the function's values at the boundaries of the interval and at the critical point within the interval, we get:f(0) = 4,f(2) = frac {1}{3} cdot 8 - 4 cdot 2 + 4 = - frac {4}{3},f(4) = frac {1}{3} cdot 64 - 4 cdot 4 + 4 = frac {28}{3}.Therefore, the minimum value of the function f(x) for x in [0, 4] is at x=2 and it is boxed{- frac {4}{3}}.

answer:Answer: 6. Solution: If we take the numbers -1,0,1,2, it is easy to verify that each of the numbers written on the board will be equal to -2, -1, 0, 1, 2, or 3 - a total of 6 different values. We will show that fewer than six different numbers could not appear on the board. Let the taken numbers be ac+d. If ac+d. Variant for completing the solution. Let u and v be the two largest numbers in absolute value, with |u| leq|v|. If |u| geq 2, then |uv| geq 2|v|, which is greater than any sum. If |u| leq 1, then among the original numbers there must be -1,0, 1. If v>0, at least 6 different numbers are written on the board: -1,0,1, v,-v, v+1. The case v<0 is handled similarly.Criteria. Only the answer - 0 points.A correct example for 6 - 2 points.Proved that there are no fewer than five values, with no further substantial progress - 1 point.Proved that there are no fewer than 6 different numbers, but no example - 3 points.A correct example for 6 and proof that there are no fewer than five values, without further substantial progress - 3 points.The same plus substantial, but incomplete progress in proving the existence of a sixth value - 4 points.

answer:AnalysisThis problem tests the ability to use the parity and periodicity of a function to find its value. The key is to determine the period of the function from the given conditions, and then use the parity and periodicity of the function to find its value. This is a basic problem.SolutionSince f(x+4)=f(x−2), we have f(x+6)=f(x),Thus, the period of the function is T=6. Also, the function is even,So, f(919)=f(6times153+1)=f(1)=f(-1)={6}^{-(-1)}=6.Hence, the answer is boxed{6}.

answer:AnalysisFrom the problem, we can see that the sum of the numerator and denominator equals 2 for 1 fraction, equals 3 for 2 fractions, equals 4 for 3 fractions, equals 5 for 4 fractions, equals 6 for 5 fractions, and so on. This problem mainly tests inductive reasoning, with the key being to find the pattern, and it is considered a medium-difficulty question.SolutionSolution: frac{1}{1},           frac{2}{1}, frac{1}{2},           frac{3}{1}, frac{2}{2}, frac{1}{3},           frac{4}{1}, frac{3}{2}, frac{2}{3}, frac{1}{4},           cdots, From observation, we know that the sum of the numerator and denominator in the kth row is k+1, and the denominator gradually increases from 1 to k. Therefore, the total number of terms in the first k rows is n= frac{k(k+1)}{2}. It is easy to know that because frac{9 times (9+1)}{2}=45 < 50 < frac{10 times (10+1)}{2}=55, thus a_{50} must be in the 10th row. When k=9, n=45, a_{45}= frac{1}{9}, so for n=46, a_{46}= frac{10}{1}, thus a_{50}= frac{6}{5}, Therefore, the answer is boxed{frac{6}{5}}.