answer:To solve for the minimum distance |PQ|, we need to determine the standard forms of the line and the curve C and then find the minimum distance between a point on the line and the curve. The curve C is given by: begin{cases} x=1+sqrt{5}costheta y=-2+sqrt{5}sintheta end{cases} Eliminating the parameter theta, we can rewrite the curve C as a standard form of a circular equation: (x-1)^2+(y+2)^2=5 From this equation, we know that the center of the circle is at (1, -2) and the radius is r=sqrt{5}. The line is represented as: begin{cases} x=t y=6-2t end{cases} Eliminating the parameter t, we get the standard form of the line: 2x+y-6=0 We now calculate the distance d from the center of the circle to the line using the formula for the distance from a point to a line: d = frac{|2 cdot 1 + 1 cdot (-2) - 6|}{sqrt{2^2 + 1^2}} = frac{|-6|}{sqrt{5}} = frac{6 sqrt{5}}{5} The minimum distance |PQ| is then the distance from the center to the line minus the radius of the circle: |PQ|_{text{min}} = d - r = frac{6 sqrt{5}}{5} - sqrt{5} = frac{sqrt{5}}{5} Therefore, the final answer is: boxed{frac{sqrt{5}}{5}} The examination of this problem involves converting parametric equations to their standard form and utilizing knowledge of plane geometry to solve for extremal values.

answer:To find the smallest ten-digit natural number with all distinct digits such that removing all even digits leaves ( 97531 ) and removing all odd digits leaves ( 02468 ), we proceed as follows: 1. **Identify the Given Condition and Objective**: - Given ( 97531 ) after eliminating even digits, - Given ( 02468 ) after eliminating odd digits, - We need to construct the smallest ten-digit number ( N ) (hereafter referred to as ( A )) that satisfies these criteria. 2. **Represent the Number Mathematically**: - Let ( N = overline{a_1a_2dots a_{10}} ). 3. **Determine the Odd Digits Placeholders**: - The odd digits ( 97531 ) must occupy specific places such that the remaining places are for even digits. - The smallest permissible combination while adhering to non-increasing order constraint would have odd digits in positions that maintain the sequence being minimal. 4. **Determine the Even Digits Placement**: - This implies positioning even digits ( 02468 ) in the remaining placeholders without altering the order of odd digits ( 97531 ). 5. **Formulate the Exact Composition**: - First digit must be 9 (since the digit sequence maintaining ascending/descending order should start with a non-zero for achieving the smallest integer). - Consequently, regarding placement permutations and maintaining relative positions for smallest possible formation: [ A = 9024675318 ] 6. **Verification**: - Removing all even digits (0,2,4,6,8) from ( 9024675318 ) leaves: [ 97531 ] - Removing all odd digits (9,7,5,3,1) from ( 9024675318 ) leaves: [ 02468 ] 7. **Conclusion:** [ boxed{9024675318} ] Thus, the smallest ten-digit number meeting the criteria is ( 9024675318 ).

answer:(1) The general formula for the sequence {a_n} is a_n = 3 times (-2)^{n-1}. (2) It exists. The set of all such n is boxed{{n | n = 2k + 1, k in mathbb{N}, k geq 5}}. The reason is explained in the analysis.

answer:① Since 2011 div 5 = 402 with a remainder of 1, it follows that 2011 in [1], so statement ① is correct; ② Since -4 = 5 times (-1) + 1, it follows that -4 notin [4], so statement ② is incorrect; ③ Because the set of integers can be and only can be divided into five classes when divided by 5, it follows that Z = [0] cup [1] cup [2] cup [3] cup [4], so statement ③ is correct; ④ Since integers a, b belong to the same "class", the remainder of a, b when divided by 5 is the same, hence the remainder of a-b when divided by 5 is 0. The converse is also true, so the necessary and sufficient condition for "integers a, b belong to the same 'class'" is "a-b in [0]". Therefore, statement ④ is correct. Thus, the answer is: boxed{text{①③④}}