answer:Applying Simon’s Favorite Factoring Trick as in the original problem, we will add 1 to both sides of the modified equation: [ xy + x + y + 1 = 153. ] We rewrite this as: [ (x + 1)(y + 1) = 153. ] The factors of 153 that satisfy (x > y > 0) (keeping in mind (x+1) and (y+1) since (x) and (y) themselves must be positive) are: - (153 = 1 times 153), - (153 = 3 times 51), - (153 = 9 times 17). Testing these, the only pair where (x > y) is (9 times 17), giving us: [ x+1 = 17, quad y+1 = 9. ] Thus, [ x = 17 - 1 = 16, quad y = 9 - 1 = 8. ] So (x = boxed{16}).

answer:For (1), since f(x)=ax^2-e^x has two zeros in (0,+infty), therefore the equation a= frac{e^x}{x^2} has two roots, which is equivalent to the curve y=a intersecting with y= frac{e^x}{x^2} at two points. Let h(x)= frac{e^x}{x^2}, then h'(x)= frac{e^x(x-2)}{x^3}, thus, when xin(0,2), h'(x) < 0, meaning h(x) is monotonically decreasing in (0,2); when xin(2,+infty), h'(x) > 0, meaning h(x) is monotonically increasing in (2,+infty), therefore h_{min}(x)=h(2)= frac{e^2}{4}, therefore the range of a is boxed{(frac{e^2}{4} ,+infty)}. For (2), since x_1,x_2 (x_1 < x_2) are the zeros of f(x)=ax^2-e^x in (0,+infty), therefore ax_1^2=e^{x_1}, ax_2^2=e^{x_2}, dividing the two equations, we get left( frac{x_2}{x_1}right)^2=e^{x_2-x_1}. Let frac{x_2}{x_1}=t(t > 1), (1) the equation becomes t^2=e^{x_2-x_1}, i.e., x_2-x_1=ln t^2=2ln t, (2) solving (1) and (2) together, we get: x_1= frac{2ln t}{t-1}, x_2= frac{2tln t}{t-1}. To prove x_1+x_2 > 4, we need to prove frac{2ln t}{t-1}+ frac{2tln t}{t-1} > 4, which is to prove ln t+tln t > 2t-2. Let h(t)=ln t+tln t-2t+2, then h'(t)= frac{1}{t}+ln t-1. Let varphi(t)= frac{1}{t}+ln t-1, varphi'(t)= frac{t-1}{t^2} > 0, thus, varphi(t) is monotonically increasing in (1,+infty), hence varphi(t) > varphi(1)=0, i.e., h'(t) > 0, thus, h(t) is monotonically increasing in (1,+infty), hence h(t) > h(1)=0, i.e., ln t+tln t > 2t-2, proved; therefore x_1+x_2 > 4, boxed{x_1+x_2 > 4}.

answer:Let's calculate the amount the man saves before the increase in expenses: He saves 20% of his monthly salary, which is Rs. 6250. So, the amount saved = 20% of Rs. 6250 = (20/100) * 6250 = 0.20 * 6250 = Rs. 1250 Now, let's calculate his expenses before the increase: His expenses would be the remaining 80% of his salary (since he saves 20%). So, the expenses = 80% of Rs. 6250 = (80/100) * 6250 = 0.80 * 6250 = Rs. 5000 After the increase in expenses by 20%, his new expenses will be: New expenses = 120% of the original expenses = (120/100) * 5000 = 1.20 * 5000 = Rs. 6000 Now, let's calculate how much he is able to save after the increase in expenses: His salary remains the same, which is Rs. 6250. So, the amount he is able to save after the increase in expenses = Salary - New expenses = Rs. 6250 - Rs. 6000 = Rs. 250 Therefore, after the increase in expenses, he is able to save Rs. boxed{250} per month.

answer:Given a set of three numbers with both the mean and median being 6: 1. The numbers can be represented as (a leq 6 leq b), with (a) being the smallest. 2. The smallest number (a) is given as 2, hence the set can be represented as (2, 6, b). 3. The mean of the set is given as 6. Therefore, the sum of the numbers (2 + 6 + b = 3 times 6 = 18). 4. Solving for (b), the largest number, we have: [ 2 + 6 + b = 18 ] [ b = 18 - 8 ] [ b = 10 ] Conclusion: The largest number in the set is ( boxed{10} ).