💡Alright, so I'm trying to figure out this problem about the city with 10 streets parallel to each other and 10 others intersecting them at right angles. The question is asking for the minimum number of turns required for a closed route that passes through all intersections.Okay, let's break this down. First, there are 10 streets in one direction and 10 in the perpendicular direction. That means the city grid is like a 10x10 grid, right? So, there are 100 intersections in total because each of the 10 streets intersects with the other 10.Now, the goal is to find a closed route that goes through all these intersections with the least number of turns. A closed route means that the path starts and ends at the same point, forming a loop.Hmm, so I need to think about how to traverse all these intersections without making too many turns. In a grid like this, every time you change direction from horizontal to vertical or vice versa, that's a turn. So, minimizing the number of turns would mean trying to go as straight as possible for as long as possible.Let me visualize this. If I start at one corner, say the bottom-left corner, and move along the bottom row to the right, that's one direction. Then, at the end, I have to turn up to go along the next column. That's one turn. Then, at the top of that column, I have to turn left to go back along the top row. That's another turn. And then, at the left end, I have to turn down to go back to the starting point. That's a third turn.Wait, but that's just covering a small part of the grid. I need to cover all 100 intersections. So, maybe I need a more efficient way of covering the grid with fewer turns.I remember something about the "traveling salesman problem," but I'm not sure if that applies here. Maybe it's more about graph theory, where intersections are nodes and streets are edges. So, we're looking for a closed walk that visits every node exactly once with the minimum number of turns.But I'm not sure about that either. Maybe I should think about it differently. If I consider each row and each column, I need to traverse all of them. So, for each row, I need to move horizontally, and for each column, I need to move vertically.But to switch from a row to a column, I need to make a turn. So, if I have 10 rows and 10 columns, how many turns would that require?Let's see. If I start at the bottom-left corner, go all the way to the right, that's one row done. Then, I need to turn up to go along the next column. That's one turn. Then, I go up to the top, turn left, and go back along the top row. That's another turn. Then, I go left to the end, turn down, and go back down the column. That's a third turn.Wait, so for each pair of rows and columns, I'm making two turns? Hmm, maybe not. Let's think step by step.If I have 10 rows, I need to traverse each row once. Similarly, I have 10 columns, each of which I need to traverse once. To switch from a row to a column, I need to make a turn. So, for each row, after traversing it, I need to turn to go into a column. Similarly, after traversing a column, I need to turn to go into a row.But since it's a closed route, I need to end up where I started. So, the number of turns should be even because each turn out requires a turn back.Wait, maybe I'm overcomplicating it. Let's think about the number of direction changes needed.In a grid, to cover all intersections, you need to move horizontally and vertically. Each time you switch from horizontal to vertical or vertical to horizontal, that's a turn.So, if I have 10 rows and 10 columns, how many times do I need to switch directions?Well, for each row, I need to enter and exit, which would require two turns per row. Similarly, for each column, I need to enter and exit, which would require two turns per column.But wait, that would be 2 turns per row times 10 rows, which is 20 turns, and similarly 20 turns for columns. But that seems like too many.Wait, no, because when you enter a row, you're exiting a column, and vice versa. So, maybe the total number of turns is 2 times the number of rows or columns.So, 2 times 10 is 20 turns. That seems plausible.But let me verify. If I have 10 rows, and for each row, I need to make two turns: one to enter and one to exit. Similarly, for columns. But since the route is closed, the last exit from the last row would lead back to the starting point, which might not require a turn.Hmm, maybe it's 2 turns per row minus one because of the closed loop.So, 2 times 10 minus 2, which is 18 turns. Hmm, but I'm not sure.Alternatively, think about it as a graph. Each intersection is a node, and each street segment is an edge. To traverse all edges, you need to have a Eulerian circuit, which requires that all nodes have even degrees.But in this case, we're not traversing all edges, just visiting all nodes. So, it's a Hamiltonian circuit.But Hamiltonian circuits don't necessarily have a direct formula for the number of turns. So, maybe that's not helpful.Alternatively, think about the grid as a chessboard. A knight's tour covers all squares with a certain number of moves, but that's different.Wait, maybe think about it as a rook's tour. A rook can move any number of squares horizontally or vertically. So, to cover all squares, the rook needs to make moves that change direction.But in this case, we're not moving like a rook; we're moving along the streets, making turns at intersections.So, perhaps it's similar to a rook's tour but with the constraint of turning at intersections.In a rook's tour, the minimum number of moves to cover all squares is 2n-2 for an n x n board. But that's for moves, not turns.Wait, maybe not directly applicable.Alternatively, think about the number of direction changes needed to cover all rows and columns.If I have 10 rows, I need to traverse each row once. To do that, I need to move horizontally across each row, but to get from one row to the next, I need to move vertically, which requires a turn.Similarly, after moving vertically, I need to turn again to move horizontally in the next row.So, for each row after the first, I need two turns: one to enter the column and one to exit into the next row.Wait, so for 10 rows, I have 9 transitions between rows, each requiring two turns. So, 9 times 2 is 18 turns.Plus, the initial turn to start moving vertically from the first row.Wait, so total turns would be 18 plus 1, which is 19.But that doesn't seem right because it's not a closed route yet.Wait, to make it a closed route, I need to return to the starting point, which would require another turn.So, 19 plus 1 is 20 turns.Okay, that seems to make sense.So, the minimum number of turns required is 20.But let me think again to make sure.If I have 10 rows, I need to traverse each row once. To get from one row to the next, I need to move vertically, which requires a turn. Then, to start moving horizontally in the next row, I need another turn.So, for each transition between rows, I need two turns.There are 9 transitions between 10 rows, so 9 times 2 is 18 turns.Plus, the initial turn to start moving vertically from the first row, and the final turn to return to the starting point.So, 18 plus 2 is 20 turns.Yes, that seems consistent.Therefore, the minimum number of turns required for a closed route that passes through all intersections is 20.

💡First, I need to add the weights of the three marble pieces purchased by Paco's Countertop Company. The weights are 0.3333333333333333 tons, 0.3333333333333333 tons, and 0.08333333333333333 tons.Adding the first two weights together:0.3333333333333333 + 0.3333333333333333 = 0.6666666666666666 tonsNext, I add the third weight to the result:0.6666666666666666 + 0.08333333333333333 = 0.75 tonsTherefore, the total weight of marble purchased by Paco's Countertop Company is 0.75 tons.

💡First, I need to determine the greatest length of equal pieces that Ioana can cut from all three ropes without any waste. This requires finding the greatest common divisor (GCD) of the three rope lengths: 39 inches, 52 inches, and 65 inches.To find the GCD, I'll start by listing the prime factors of each length:- The prime factors of 39 are 3 and 13.- The prime factors of 52 are 2 squared and 13.- The prime factors of 65 are 5 and 13.Next, I'll identify the common prime factor among all three lengths. The only common prime factor is 13.Therefore, the greatest length of each piece that can be cut from all ropes without wasting any material is 13 inches.

💡First, I need to determine the number of male workers the contractor employs. Let's denote the number of male workers as ( M ).The contractor employs:- ( M ) male workers- 15 female workers- 5 child workersThe total number of workers is ( M + 15 + 5 = M + 20 ).Next, I'll calculate the total daily wages paid to each category of workers:- Male workers: ( 25M ) rupees- Female workers: ( 15 times 20 = 300 ) rupees- Child workers: ( 5 times 8 = 40 ) rupeesThe total daily wage paid to all workers is ( 25M + 300 + 40 = 25M + 340 ) rupees.Given that the average wage per day is 21 rupees, I can set up the equation:[frac{25M + 340}{M + 20} = 21]To solve for ( M ), I'll multiply both sides by ( M + 20 ):[25M + 340 = 21(M + 20)]Expanding the right side:[25M + 340 = 21M + 420]Subtracting ( 21M ) from both sides:[4M + 340 = 420]Subtracting 340 from both sides:[4M = 80]Finally, dividing both sides by 4:[M = 20]Therefore, the contractor employs 20 male workers.

💡Okay, so I have this problem where I need to show that there exists an infinite set of positive integers such that the sum of any finite subset is never a perfect square. And then, I also need to consider what happens if instead of avoiding perfect squares, we want the sums to avoid being perfect powers. Hmm, interesting.First, let me make sure I understand the problem correctly. A perfect square is a number like 1, 4, 9, 16, etc., which are squares of integers. A perfect power is a number like 1, 4, 8, 9, 16, 25, etc., which are powers of integers, such as squares, cubes, fourth powers, etc. So, avoiding perfect squares is a bit more specific than avoiding perfect powers.Now, the task is to construct an infinite set of positive integers where any finite sum of its elements is not a perfect square. And then, see if the same or a similar construction works when avoiding perfect powers instead.I think I need to construct such a set step by step, ensuring at each step that adding a new element doesn't create a sum that's a perfect square. Maybe I can use some kind of inductive process where I build the set one element at a time, making sure each new element doesn't interfere with the previous sums.Let me think about starting with the first element. If I start with 1, then the sum of the subset containing just 1 is 1, which is a perfect square. Oh, that's bad because I don't want any finite subset sum to be a perfect square. So maybe I shouldn't start with 1. Let's try starting with 2. Then, the sum of the subset with 2 is 2, which is not a perfect square. Good.Now, moving on to the second element. I need to choose a number such that when added to 2, the sum isn't a perfect square, and also, the number itself isn't a perfect square. Let's try 3. The sum of 2 and 3 is 5, which isn't a perfect square. Also, 3 isn't a perfect square. So, 3 is a good choice.Next, the third element. I need a number that, when added to 2, 3, or both, doesn't result in a perfect square. Let's try 4. But 4 is a perfect square, so I can't include it. How about 5? The sum of 2 and 5 is 7, not a perfect square. The sum of 3 and 5 is 8, not a perfect square. The sum of 2, 3, and 5 is 10, not a perfect square. Also, 5 itself isn't a perfect square. So, 5 works.Continuing this way, I need to find numbers that, when added to any combination of the previous numbers, don't result in a perfect square. This seems tedious, but maybe there's a pattern or a systematic way to choose these numbers.Perhaps I can use modular arithmetic to ensure that the sums avoid certain residues that correspond to perfect squares. For example, perfect squares modulo some number have specific properties. If I can choose numbers such that their sums modulo that number never match the residues of perfect squares, that might work.Let me think about modulo 4. Perfect squares modulo 4 are either 0 or 1. So, if I can ensure that all subset sums are congruent to 2 or 3 modulo 4, then they can't be perfect squares. How can I achieve that?If I start with a number that is 2 modulo 4, like 2, then adding another number that is 2 modulo 4 will result in 0 modulo 4, which is a perfect square. That's bad. So, maybe I need to alternate or use different residues.Wait, if I choose all numbers to be 2 modulo 4, then any single number is 2 modulo 4, which is not a perfect square. The sum of two numbers would be 0 modulo 4, which is a perfect square. That's not good. So, that approach doesn't work.Maybe instead of modulo 4, I can use a larger modulus where the number of residues is more, and I can avoid the residues corresponding to perfect squares.Alternatively, maybe I can use prime numbers. If I choose numbers that are prime, their sums might have properties that make them avoid perfect squares. But I'm not sure if that's the case.Wait, another idea: if I can make sure that each new number is larger than the sum of all previous numbers, then the subset sums will be unique and can be controlled more easily. For example, if I choose the first number as 1, the next as 2, then the next as 4, then 8, and so on, doubling each time. Then, each number is larger than the sum of all previous numbers, which ensures that each subset sum is unique and corresponds to a binary representation.But in this case, the subset sums would be all numbers from 1 up to the total sum, which includes many perfect squares. So that doesn't help.Hmm, maybe I need a different approach. What if I construct the set in such a way that every new element is chosen to avoid creating any perfect square when added to any combination of the previous elements.This sounds like an inductive process where at each step, I ensure that the new element doesn't form a perfect square with any subset of the existing elements.Let me try to formalize this. Suppose I have already chosen elements x1, x2, ..., xn. I need to choose xn+1 such that for any subset S of {x1, x2, ..., xn}, the sum S + xn+1 is not a perfect square.To do this, I need to ensure that xn+1 is not equal to k^2 - S for any integer k and any subset S of the existing elements.Since there are only finitely many subsets S, and for each S, there are only finitely many k such that k^2 - S is positive, I can choose xn+1 to be larger than all these possible k^2 - S.Therefore, by choosing xn+1 sufficiently large, I can avoid all these potential perfect squares.This seems promising. So, starting with x1, I can choose x2 to be larger than x1 + x1^2, or something like that, to ensure that x2 isn't forming a perfect square with x1.Wait, let me think carefully. If I have x1, then to choose x2, I need to ensure that x2 is not equal to k^2 - x1 for any k, and also x2 itself is not a perfect square.Similarly, when choosing x3, I need to ensure that x3 is not equal to k^2 - x1, k^2 - x2, or k^2 - (x1 + x2), and also x3 itself is not a perfect square.So, at each step, the number of conditions I have to satisfy is finite, because the number of subsets is finite, and for each subset, the number of k's such that k^2 - sum(S) is positive is also finite.Therefore, by choosing each new element to be larger than all these potential k^2 - sum(S), I can ensure that the new element doesn't form a perfect square with any subset of the existing elements.This seems like a viable strategy. So, starting with x1, I can choose x2 to be larger than x1 + x1^2, x3 larger than x1 + x2 + (x1 + x2)^2, and so on.But I need to make sure that this process can continue indefinitely, ensuring that the set remains infinite.Yes, because at each step, I'm only excluding finitely many numbers, so there are still infinitely many numbers left to choose from.Therefore, by this inductive process, I can construct an infinite set of positive integers where the sum of any finite subset is never a perfect square.Now, what happens if we change the condition from not being a perfect square to not being a perfect power?A perfect power is a number of the form k^m where m >= 2. So, perfect squares are a subset of perfect powers.Therefore, if I can construct a set where subset sums are not perfect squares, it automatically satisfies that they are not perfect powers. But maybe the construction needs to be adjusted because perfect powers include more numbers.Wait, actually, the previous construction might not necessarily avoid all perfect powers. For example, a subset sum could be a perfect cube or a higher power, which wasn't considered before.So, I need to ensure that the subset sums are not only not perfect squares but also not perfect cubes, perfect fourth powers, etc.This seems more complicated because there are more conditions to satisfy. However, the number of perfect powers is still countable, and for each subset sum, there are only finitely many k and m such that k^m equals the sum.Therefore, using a similar inductive approach, at each step, I can choose the next element to be larger than all k^m - sum(S) for any subset S and any m >= 2.Since for each subset S, the number of k and m such that k^m - sum(S) is positive is still finite, I can choose the next element to be larger than all these values.Thus, by choosing each new element sufficiently large, I can ensure that no finite subset sum is a perfect power.Therefore, the same construction method can be applied to avoid perfect powers as well.In summary, by using an inductive process where each new element is chosen to be larger than all potential perfect squares (or perfect powers) minus the sums of all previous subsets, we can construct an infinite set of positive integers where the sum of any finite subset is never a perfect square (or a perfect power).I think this approach works, but I should double-check if there are any gaps or if I'm missing something. For instance, ensuring that the choice of each new element doesn't inadvertently create a perfect square or power when combined with multiple previous elements. But since I'm considering all subsets at each step, I think it covers all possible combinations.Another thing to consider is whether the set remains infinite. Since at each step, I'm only excluding finitely many numbers, there are still infinitely many numbers left to choose from, so the set can indeed be infinite.Overall, I believe this method successfully constructs the desired sets for both perfect squares and perfect powers.

💡Okay, so I have this problem here with two sets, A and B. Set A is defined as all real numbers x where x is less than 2. That means A includes numbers like 1, 0, -1, and so on, but not 2 itself. Set B is simpler; it's just the numbers 1 and 2. So B is like a small set with only two elements.Now, the question is asking about the union of A and B, which is written as A ∪ B. I remember that the union of two sets includes all the elements that are in either of the sets. So, if I take everything that's in A and everything that's in B, and put them together, that should give me A ∪ B.Let me think about set A first. Since A is all numbers less than 2, it's like an interval from negative infinity up to, but not including, 2. So in interval notation, that's (-∞, 2). Now, set B has the numbers 1 and 2. So, when I take the union of A and B, I'm adding the elements of B to A. But wait, 1 is already in A because 1 is less than 2. So, the only new element I'm adding is 2.Therefore, A ∪ B would include all numbers less than 2 and also the number 2 itself. In interval notation, that would be from negative infinity up to and including 2, which is written as (-∞, 2]. Looking at the options given:A: (-∞, 2) – This is just set A, not the union.B: (-∞, 2] – This includes all of A and adds 2 from B.C: {1) – This doesn't make sense; it's not a proper interval or set.D: {1,2} – This is just set B, not the union.So, the correct answer should be option B: (-∞, 2].