answer:Let's denote the initial amount of 4% saltwater as x grams, and the amount of water evaporated as y grams. According to the problem, we have:- The equation for the salt content before and after evaporation: 4% times x = (x - y) times 10%.- The equation for the salt content after adding 300 grams of 4% saltwater: frac{(x - y) times 10% + 300 times 4%}{x - y + 300} = 6.4%.Solving these equations, we find x = 500 grams.Therefore, the answer is boxed{500}.

answer:【Answer】Solution: Since the least common multiple of 4,5,7 is 4 times 5 times 7=140, the three-digit numbers that meet the criteria are 140,280,420,560,700,840,980, so the middle number is 560. Therefore, the answer is: 560.

answer:AnalysisThis problem examines the application of the cosine rule, using the cosine rule to establish an equation, and solving the equation to reach a conclusion.SolutionSince a= sqrt{5}, c=2, and cos A= frac{2}{3}, we have 5=b^2+4-4btimes frac{2}{3}, which simplifies to 3b^2-8b-3=0.Solving this, we find b=3 or b=-frac{1}{3} (the latter is discarded).Therefore, the correct choice is boxed{D}.

answer:20. Answer. 95004Solution. We shall prove that for any positive integer a, if f(a) denotes the sum of all nonnegative integer solutions to leftlfloorfrac{n}{a}rightrfloor=leftlfloorfrac{n}{a+1}rightrfloor, thenf(a)=frac{1}{6} aleft(a^{2}-1right)(a+2) .Thus f(27)=95004.Let n be a solution to leftlfloorfrac{n}{a}rightrfloor=leftlfloorfrac{n}{a+1}rightrfloor. Write n=a q+r, where 0 leq r<a. Thus leftlfloorfrac{n}{a}rightrfloor=q. Also n=(a+1) q+r-q. Since leftlfloorfrac{n}{a+1}rightrfloor=q, we have 0 leq r-q, that is, q leq r<a. Therefore for each q=0,1, ldots, a-1, r can be anyone of the values q, q+1, ldots, a-1. Thusbegin{aligned}A & =sum_{q=0}^{a-1} sum_{r=q}^{a-1}(q a+r) & =sum_{q=0}^{a-1}(a-q) q a+sum_{q=0}^{a-1} sum_{r=q}^{a-1} r & =a^{2} sum_{q=0}^{a-1} q-a sum_{q=0}^{a-1} q^{2}+sum_{r=0}^{a-1} sum_{q=0}^{r} r & =a^{2} sum_{q=0}^{a-1} q-a sum_{q=0}^{a-1} q^{2}+sum_{r=0}^{a-1} r(r+1) & =a^{2} sum_{q=0}^{a-1} q-a sum_{q=0}^{a-1} q^{2}+sum_{r=0}^{a-1} r^{2}+sum_{r=0}^{a-1} r & =left(a^{2}+1right) cdot frac{1}{2} a(a-1)+(1-a) cdot frac{1}{6} a(2 a-1)(a-1) & =frac{1}{6} aleft(a^{2}-1right)(a+2) .end{aligned}

answer:square Answer: 400.Evaluation. If there is an unpicked cell, then on the rows (vertical and horizontal) passing through it, there are no rooks. To attack all cells of one color on such a row, rooks must be placed on at least 20 perpendicular rows. Thus, there are no more than 20 rows in each direction that are free from rooks. Unattacked cells are located at the intersections of such rows, so there are no more than 20^{2} of them.Example. We can assume that the bottom-left cell is black. Place 20 rooks on white cells along the left and bottom edges. Then, at the intersections of the unattacked rows, all cells are black, totaling 400, and all white cells are attacked.(A. Shapovalov)## Fourth Round

answer:Solution 1The sum of the digits is at most 9+9+9=27. Therefore the number is at most 6cdot 27 = 162. Out of the numbers 1 to 162 the one with the largest sum of digits is 99, and the sum is 9+9=18. Hence the sum of digits will be at most 18.Also, each number with this property is divisible by 6, therefore it is divisible by 3, and thus also its sum of digits is divisible by 3. Thus, the number is divisible by 18.We only have six possibilities left for the sum of the digits: 3, 6, 9, 12, 15, and 18, but since the number is divisible by 18, the digits can only add to 9 or 18. This leads to the integers 18, 36, 54, 72, 90, and 108 being possibilities. We can check to see that boxed{1} solution: the number 54 is the only solution that satisfies the conditions in the problem.Solution 2We can write each integer between 1 and 999 inclusive as overline{abc}=100a+10b+c where a,b,cin{0,1,dots,9} and a+b+c>0.The sum of digits of this number is a+b+c, hence we get the equation 100a+10b+c = 6(a+b+c). This simplifies to 94a + 4b - 5c = 0. Clearly for a>0 there are no solutions, hence a=0 and we get the equation 4b=5c. This obviously has only one valid solution (b,c)=(5,4), hence the only solution is the number 54.Solution 3The sum of the digits is at most 9+9+9=27. Therefore the number is at most 6cdot 27 = 162. Since the number is 6 times the sum of its digits, it must be divisible by 6, therefore also by 3, therefore the sum of its digits must be divisible by 3. With this in mind we can conclude that the number must be divisible by 18, not just by 6. Since the number is divisible by 18, it is also divisible by 9, therefore the sum of its digits is divisible by 9, therefore the number is divisible by 54, which leaves us with 54, 108 and 162. Only 54 is 6 times its digits, hence the answer is boxed{1}.