answer:Step 1: Understanding the problem We need to express the number 1000 as a sum of the maximum possible number of natural numbers such that the sums of their digits are pairwise distinct. Step 2: Identifying the approach To achieve this, we first identify the smallest natural number corresponding to any given sum of digits. 1. For a sum of digits (A), the minimal number corresponds to the form (a99 ldots 99), where the number consists of digit 'a' followed by as many '9's as necessary. Here, 'a' is the remainder when (A) is divided by 9, and the number of '9's exactly equals the quotient of (A) divided by 9. Step 3: Listing numbers with unique sums of digits Let's consider the smallest natural number representation for each possible sum of digits from 1 to 20: [ begin{aligned} &1 implies 1 &2 implies 2 &3 implies 3 &4 implies 4 &5 implies 5 &6 implies 6 &7 implies 7 &8 implies 8 &9 implies 9 &10 implies 19 : left(1 + 9right) &11 implies 29 : left(2 + 9right) &12 implies 39 : left(3 + 9right) &13 implies 49 : left(4 + 9right) &14 implies 59 : left(5 + 9right) &15 implies 69 : left(6 + 9right) &16 implies 79 : left(7 + 9right) &17 implies 89 : left(8 + 9right) &18 implies 99 : left(9 + 9right) &19 implies 199 : left(1 + 9 + 9 + 9right) &20 implies 299 : left(2 + 9 + 9 + 9right) end{aligned} ] Step 4: Calculating the partial sum Sum of the first 19 different unique digit sums (excluding 299 for sum=20 as needed): [ begin{aligned} &1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 19 + 29 + 39 + 49 + 59 + 69 + 79 + 89 + 99 + 199 &=775 end{aligned} ] Step 5: Utilizing the sum constraint Checking the residue modulo 9 of the problem input (1000): [ 1000 div 9 = 111 text{ remainder } 1 ] The smallest additional number replacing one of the sums must maintain distinct sums while adjusting for the deficit: Adding 1 more to the sum (775 + 225 = 1000), could come from adjusting 99 to 234 (since the sum of digits 2+3+4=9 remains same). Thus, we modify this making sum as necessary: [ 1, 2, 3, 4, 5, 6, 7, 8, 19, 29, 39, 49, 59, 69, 79, 89, 199, 234 ] Conclusion: Finally summing various 19 entries covering every incremental unique sum condition: [ boxed{19} ]

answer:1. When a = -1, f(x) = |x - 1| + |x + 2| = begin{cases} 2x + 1, & x > 1 3, & -2 leq x leq 1 -2x - 1, & x < -2 end{cases} For x < -2, f(x) ≥ x + 5 implies -2x - 1 ≥ x + 5, which gives x < -2. For -2 ≤ x ≤ 1, f(x) ≥ x + 5 implies 3 ≥ x + 5, which gives x = -2. For x > 1, f(x) ≥ x + 5 implies 2x + 1 ≥ x + 5, which gives x ≥ 4. Therefore, the solution set for the original inequality is {x | x ≤ -2 or x ≥ 4}. 2. If a < 2, when x ∈ (-5, -3), f(x) = -2x - a - 2. Since f(x) > x^2 + 2x - 5 holds true for x ∈ (-5, -3), We have -2x - a - 2 > x^2 + 2x - 5, This implies a < -x^2 - 4x + 3, which is always true for x ∈ (-5, -3). Let g(x) = -x^2 - 4x + 3, where x ∈ (-5, -3). Thus, g(x) ∈ (-2, 6), so a ≤ -2. The final answers are: 1. Solution set: boxed{x leq -2 text{ or } x geq 4} 2. Range of 'a': boxed{a leq -2}

answer:1. First, we determine the range of numbers for which the square root lies between 30 and 30.5: - The square of 30 is 30^2 = 900. - The square of 30.5 is (30.5)^2 = 930.25. - Therefore, the number must be between 900 and 930.25. 2. Next, we find the multiples of 18 within this range: - Start checking from 900: 900 div 18 = 50, thus 900 is a multiple of 18. - 18 times 51 = 918 is the next multiple. - 18 times 52 = 936 is outside the required range. 3. The multiples of 18 within our range are 900 and 918. 4. We need to check if there is any other condition to refine our choice: - Both numbers fit the initial condition: the square root is between 30 and 30.5. - As there are no further constraints, both numbers are solutions. Conclusion: There are two integers that meet the criteria: boxed{900 text{ and } 918}.

answer:Let's denote the amount of 10% butterfat milk that needs to be added as x gallons. The amount of butterfat in the 8 gallons of 40% butterfat milk is: 40% of 8 gallons = 0.40 * 8 = 3.2 gallons of butterfat. The amount of butterfat in the x gallons of 10% butterfat milk is: 10% of x gallons = 0.10 * x = 0.1x gallons of butterfat. When we mix the two types of milk, we want to have a mixture that is 20% butterfat. The total volume of the mixture will be (8 + x) gallons, and the total amount of butterfat will be (3.2 + 0.1x) gallons. The equation for the final mixture to be 20% butterfat is: 20% of (8 + x) gallons = (3.2 + 0.1x) gallons of butterfat. Now we can set up the equation and solve for x: 0.20 * (8 + x) = 3.2 + 0.1x 0.20 * 8 + 0.20x = 3.2 + 0.1x 1.6 + 0.20x = 3.2 + 0.1x 0.20x - 0.1x = 3.2 - 1.6 0.10x = 1.6 x = 1.6 / 0.10 x = 16 gallons. So, 16 gallons of 10% butterfat milk must be added to 8 gallons of 40% butterfat milk to obtain milk that is boxed{20%} butterfat.