answer:Proposition p: From the problem, we have (m−a)(m−3a)leqslant 0. Proposition q: (m+2)(m+1) < 0, solving for m gives -2 < m < -1. (1) If a=-1, the truth of proposition p implies -3leqslant mleqslant -1. When "p or q" is true, we have three scenarios: p is true and q is false, p is false and q is true, or both p and q are true. Thus, the union of the ranges -3leqslant mleqslant -1 and -2 < m < -1 gives -3leqslant mleqslant -1. Therefore, the range of values for m is left{-3leqslant mleqslant -1right}. (2) If p is a necessary but not sufficient condition for neg q, then q is a necessary but not sufficient condition for neg p. From (1), let's denote the set corresponding to condition q as A=left{m|-2 < m < -1right}, and the set corresponding to condition neg q as B=left{m|(m-a)(m-3a) > 0right}. By the problem statement, Asubset B. When a=0, B=left{m|mneq 0right}; When a > 0, B=left{m|m > 3a text{ or } m < aright}, which satisfies Asubset B; When a < 0, B=left{m|m > a text{ or } m < 3aright}. To ensure Asubset B, we need 3ageqslant -1 or aleqslant -2. Thus, - dfrac{1}{3}leqslant a < 0 or aleqslant -2. Therefore, the range of values for the real number a is boxed{- dfrac{1}{3}leqslant a < 0 text{ or } aleqslant -2}.

answer:# Problem: 求最小的自然数 (a) 和 (b(b>1))，使得 (sqrt{a sqrt{a sqrt{a}}}=b)。 1. From the given expression, we start by simplifying: [ sqrt{a sqrt{a sqrt{a}}}=b ] Raising both sides to the power of 8, we get: [ (b)^{8}=(sqrt{a sqrt{a sqrt{a}}})^{8} ] Since, (sqrt{a sqrt{a sqrt{a}}} = a^{(1/2)*(1/2)*(1/2)} = a^{(1/2)^3} = a^{1/8}), we get: [ b^{8}=a^{7} ] 2. Suppose (b) is divisible by a prime (p). Since (b > 1), such a prime (p) exists. Hence, (a) must also be divisible by (p). We express (a) and (b) in terms of (p) as follows: [ a = p^{alpha} cdot d quad text{and} quad b = p^{beta} cdot c ] where (c) and (d) are natural numbers co-prime with (p), with (alpha geq 1) and (beta geq 1). Substituting these into the equation (b^8 = a^7), we get: [ (p^{beta} cdot c)^{8} = (p^{alpha} cdot d)^{7} ] [ p^{8beta} cdot c^{8} = p^{7alpha} cdot d^{7} ] 3. Since ((c^{8}, p) = 1) and ((d^{7}, p) = 1), it follows that: [ 8beta = 7alpha ] Given that (gcd(7,8)=1), we conclude that 7 must divide (beta), and 8 must divide (alpha). Therefore, we have: [ alpha = 8k quad text{and} quad beta = 7k ] for some natural number (k). Since we want the smallest (a) and (b), set (k = 1): [ alpha = 8 quad text{and} quad beta = 7 ] 4. With (alpha=8) and (beta=7), the smallest prime (p=2) so: [ a = 2^{8} cdot d quad text{and} quad b = 2^{7} cdot c ] Given that (c geq 1), (d geq 1), and (p geq 2), the smallest values for (a) and (b) occur when (d = 1) and (c = 1): [ a = 2^{8} times 1 = 256 quad text{and} quad b = 2^{7} times 1 = 128 ] # Conclusion: Thus, the smallest natural numbers (a) and (b(b>1)) that satisfy (sqrt{a sqrt{a sqrt{a}}}=b) are: [ boxed{a=256, b=128} ]

answer:We are given a rectangular grid with dimensions (6 times 5), from which a (2 times 1) rectangle is cut out from the center. We need to determine if this figure can be divided into 6 triangles. Let's consider the given dimensions and placement of the cut-out: 1. **Initial Figure:** - A (6 times 5) rectangle. The total number of unit squares in this initial rectangle is: [ 6 times 5 = 30 text{ unit squares} ] 2. **Subtracted Area:** - A (2 times 1) rectangle is cut out from the center. The area of this cut-out rectangle is: [ 2 times 1 = 2 text{ unit squares} ] 3. **Remaining Area:** - After the cut-out, the remaining area of the rectangle: [ 30 - 2 = 28 text{ unit squares} ] 4. **Dividing into Triangles:** - To determine if we can divide the remaining figure into 6 triangles, we need to visualize the problem or use a diagram similar to the one given in the reference image. From the reference image, we can see how a (6 times 5) rectangle with a (2 times 1) portion cut out can be split into 6 triangles. A helpful approach is to: - Divide the large rectangle into multiple smaller geometric shapes such that when rearranged or connected, they form 6 distinct triangles. - Ensure that each triangle adheres to the properties that their combined area indeed matches the remaining area of 28 unit squares. Now, we can draw some lines to visualize and achieve this specific division: 5. **Example Cutting Lines:** - One possible method of cutting (as per the reference diagram):  In this approach, we can see the figure can be divided along specific diagonal cuts into smaller triangles. We could verify: - Calculate the area of each triangle separately and ensure that the sum of these areas is equal to the total area of the original rectangle minus the cut-out section: Conduct this verification visually or geometrically. # Conclusion: Through visual representation and proper cutting, it is possible to divide the given (6 times 5) grid with the (2 times 1) cut-out in the center into 6 triangles. boxed{text{Yes, it is possible to divide the figure into 6 triangles.}}

answer:1. Let ( N ) be the initial integer chosen by Mary, where ( N > 2017 ). 2. On Mary's turn, she transforms the number ( x ) into ( 2017x + 2 ). 3. On Pat's turn, he transforms the number ( y ) into ( y + 2019 ). We need to determine the smallest ( N ) such that none of the numbers obtained during the game is divisible by ( 2018 ). 4. First, consider the number after Mary's first turn: [ N_1 = 2017N + 2 ] We need ( N_1 notequiv 0 pmod{2018} ). 5. Next, consider the number after Pat's first turn: [ N_2 = 2017N + 2 + 2019 = 2017N + 2021 ] We need ( N_2 notequiv 0 pmod{2018} ). 6. Simplify ( N_2 mod 2018 ): [ N_2 equiv 2017N + 2021 pmod{2018} ] Since ( 2017 equiv -1 pmod{2018} ), we have: [ N_2 equiv -N + 2021 pmod{2018} ] We need ( -N + 2021 notequiv 0 pmod{2018} ), which simplifies to: [ N notequiv 2021 pmod{2018} ] 7. Consider the number after Mary's second turn: [ N_3 = 2017(2017N + 2 + 2019) + 2 = 2017^2N + 2017 cdot 2021 + 2 ] Simplify ( N_3 mod 2018 ): [ N_3 equiv (-1)^2N + (-1) cdot 2021 + 2 pmod{2018} ] [ N_3 equiv N - 2021 + 2 pmod{2018} ] [ N_3 equiv N - 2019 pmod{2018} ] We need ( N - 2019 notequiv 0 pmod{2018} ), which simplifies to: [ N notequiv 2019 pmod{2018} ] 8. Consider the number after Pat's second turn: [ N_4 = N_3 + 2019 = N - 2019 + 2019 = N ] We are back to the original number ( N ). 9. To ensure that no number is divisible by ( 2018 ), we need: [ N notequiv 0, 2, 3, 1 pmod{2018} ] 10. The smallest ( N ) that satisfies these conditions is ( 4 ). Therefore, the smallest initial integer Mary could choose is: [ N = 2018 + 4 = 2022 ] The final answer is ( boxed{2022} ).