💡First, I recognize that (101) can be expressed as (100 + 1). This allows me to rewrite the expression (101^{37}) in a form that is easier to work with modulo (100).Next, I apply the concept of modular arithmetic. Since (101 equiv 1 pmod{100}), raising both sides of this congruence to the power of (37) preserves the equivalence:[101^{37} equiv 1^{37} pmod{100}]Calculating (1^{37}) is straightforward, as any power of (1) remains (1):[1^{37} = 1]Therefore, the remainder when (101^{37}) is divided by (100) is:[boxed{1}]

💡Okay, so I have this problem here: The sum of the x-coordinates of the vertices of a triangle in the Cartesian plane equals 15. I need to find the sum of the x-coordinates of the midpoints of the sides of this triangle. Hmm, let me try to figure this out step by step.First, let's break down what the problem is saying. We have a triangle with three vertices, each having an x-coordinate. The sum of these x-coordinates is 15. So if I denote the x-coordinates of the vertices as a, b, and c, then according to the problem, a + b + c = 15. Got that part.Now, the question is about the midpoints of the sides of this triangle. Each side of the triangle connects two vertices, so each midpoint will be the average of the x-coordinates of those two vertices. That makes sense because the midpoint formula is just the average of the coordinates of the two endpoints.So, let's think about the midpoints. There are three sides to a triangle, so there will be three midpoints. Each midpoint will have an x-coordinate that's the average of the x-coordinates of the two vertices it connects. Let me write that out. The first midpoint connects vertices with x-coordinates a and b, so its x-coordinate is (a + b)/2. The second midpoint connects vertices with x-coordinates a and c, so its x-coordinate is (a + c)/2. The third midpoint connects vertices with x-coordinates b and c, so its x-coordinate is (b + c)/2.Okay, so now I need to find the sum of these three x-coordinates of the midpoints. That would be:(a + b)/2 + (a + c)/2 + (b + c)/2Hmm, let's simplify this expression. If I add them all together, I can combine the numerators since they all have the same denominator:[(a + b) + (a + c) + (b + c)] / 2Let's compute the numerator:(a + b) + (a + c) + (b + c) = a + b + a + c + b + cCombine like terms:= 2a + 2b + 2cSo now the expression becomes:(2a + 2b + 2c) / 2I can factor out a 2 from the numerator:2(a + b + c) / 2The 2 in the numerator and the 2 in the denominator cancel each other out, leaving:a + b + cBut wait, from the problem statement, we know that a + b + c = 15. So substituting that in, the sum of the x-coordinates of the midpoints is 15.Wait, that seems straightforward. So the sum of the x-coordinates of the midpoints is equal to the sum of the x-coordinates of the vertices. Is that always true? Let me think about it.If I consider the midpoints, each midpoint's x-coordinate is an average of two vertices. When I sum all three midpoints, I'm effectively summing each pair of vertices and then dividing by 2. But when I add all three midpoints together, it's like adding each vertex's x-coordinate twice and then dividing by 2, which brings me back to the original sum of the vertices. So yes, it makes sense that the sum of the midpoints' x-coordinates equals the sum of the vertices' x-coordinates.Let me test this with a simple example to make sure I didn't make a mistake. Suppose I have a triangle with vertices at (1, 0), (2, 0), and (3, 0). The sum of the x-coordinates is 1 + 2 + 3 = 6. Now, the midpoints of the sides would be at ((1+2)/2, 0) = (1.5, 0), ((1+3)/2, 0) = (2, 0), and ((2+3)/2, 0) = (2.5, 0). The sum of the x-coordinates of these midpoints is 1.5 + 2 + 2.5 = 6, which is the same as the sum of the vertices' x-coordinates. Okay, that checks out.Another example: Let's take vertices at (0, 0), (0, 0), and (15, 0). The sum of the x-coordinates is 0 + 0 + 15 = 15. The midpoints would be at ((0+0)/2, 0) = (0, 0), ((0+15)/2, 0) = (7.5, 0), and ((0+15)/2, 0) = (7.5, 0). The sum of the x-coordinates of the midpoints is 0 + 7.5 + 7.5 = 15, which again matches the sum of the vertices' x-coordinates.So it seems consistent that the sum of the midpoints' x-coordinates equals the sum of the vertices' x-coordinates. Therefore, in the given problem, since the sum of the vertices' x-coordinates is 15, the sum of the midpoints' x-coordinates should also be 15.I think I've got it. The key insight is recognizing that when you take the midpoints, each vertex's x-coordinate is used twice in the sum of the midpoints, so when you divide by 2 for each midpoint, it effectively cancels out, leaving the original sum. This pattern holds regardless of the specific coordinates, so it's a general property of triangles in the Cartesian plane.Just to recap:1. Let the x-coordinates of the vertices be a, b, c.2. Sum of vertices' x-coordinates: a + b + c = 15.3. Midpoints' x-coordinates: (a + b)/2, (a + c)/2, (b + c)/2.4. Sum of midpoints' x-coordinates: [(a + b) + (a + c) + (b + c)] / 2 = (2a + 2b + 2c)/2 = a + b + c = 15.Yep, that all adds up. I don't see any flaws in this reasoning, and the examples I tried support this conclusion. So I'm confident that the sum of the x-coordinates of the midpoints is 15.

💡Okay, so I have this problem here where I need to find the number of ordered pairs (b, c) of positive integers such that both quadratic equations x² + bx + c = 0 and x² + cx + b = 0 have no real solutions. The options given are A) 0, B) 1, C) 2, D) 3, E) 4. Hmm, interesting. I remember that for a quadratic equation ax² + bx + c = 0 to have no real solutions, the discriminant must be negative. The discriminant is b² - 4ac. So, in this case, since both quadratics need to have no real solutions, their discriminants must both be negative.Let me write that down. For the first equation, x² + bx + c = 0, the discriminant is b² - 4c. For the second equation, x² + cx + b = 0, the discriminant is c² - 4b. Both of these need to be less than zero. So, we have two inequalities:1. b² - 4c < 02. c² - 4b < 0Which can be rewritten as:1. b² < 4c2. c² < 4bSo, now I need to find positive integers b and c such that both of these inequalities hold. Since b and c are positive integers, I can start by testing small values and see if they satisfy both conditions.Let me start with b = 1. Then, the first inequality becomes 1² < 4c, which simplifies to 1 < 4c. Since c is a positive integer, the smallest c can be is 1. So, c = 1. Let's check the second inequality: c² < 4b, which becomes 1 < 4*1, so 1 < 4, which is true. But wait, we also need to check if both quadratics have no real solutions. For b = 1 and c = 1, the discriminants are 1 - 4 = -3 and 1 - 4 = -3, both of which are negative. So, (1,1) is a valid pair. Hmm, but wait, the options don't include 1 as the answer, but let me check more values.Wait, actually, the problem is asking for ordered pairs, so (1,1) is one pair. But let me check if there are more.Next, let's try b = 2. Then, the first inequality becomes 4 < 4c, so c > 1. The smallest c can be is 2. Let's check c = 2. The second inequality is c² < 4b, which becomes 4 < 8, which is true. So, (2,2) is another pair. Let me check the discriminants: For x² + 2x + 2 = 0, discriminant is 4 - 8 = -4, which is negative. For x² + 2x + 2 = 0, same thing. So, (2,2) is valid.Wait, but let me check higher values. Maybe there are more pairs.Let's try b = 3. The first inequality becomes 9 < 4c, so c > 9/4, which is 2.25. So, c must be at least 3. Let's check c = 3. The second inequality is c² < 4b, which becomes 9 < 12, which is true. So, (3,3) is another pair. Checking discriminants: For x² + 3x + 3 = 0, discriminant is 9 - 12 = -3, which is negative. Similarly for the other equation. So, (3,3) is valid.Wait, but I'm getting more than two pairs, but the options only go up to 4. Maybe I'm missing something. Let me check b = 4. The first inequality becomes 16 < 4c, so c > 4. So, c must be at least 5. Let's check c = 5. The second inequality is c² < 4b, which becomes 25 < 16, which is false. So, c = 5 doesn't work. What about c = 4? Wait, c must be greater than 4, so c = 5 is the smallest. But 25 < 16 is false, so no solution for b = 4.Wait, so maybe (4,5) doesn't work. Let me check (3,4). For b = 3, c = 4. First inequality: 9 < 16, which is true. Second inequality: 16 < 12, which is false. So, that doesn't work. How about (4,3)? For b = 4, c = 3. First inequality: 16 < 12, which is false. So, that doesn't work either.Wait, maybe I should try b = 2 and c = 3. Let's see. First inequality: 4 < 12, which is true. Second inequality: 9 < 8, which is false. So, that doesn't work. How about b = 3 and c = 2? First inequality: 9 < 8, which is false. So, that doesn't work either.Hmm, so far, I have (1,1), (2,2), and (3,3) as valid pairs. But the options only go up to 4, and the answer choices are 0,1,2,3,4. Wait, but maybe I'm missing some pairs where b ≠ c.Let me think. Maybe there are pairs where b ≠ c that satisfy both inequalities. Let's try b = 2 and c = 3. As I checked earlier, the first inequality is 4 < 12, which is true, but the second inequality is 9 < 8, which is false. So, that doesn't work.How about b = 3 and c = 2? First inequality: 9 < 8, which is false. So, that doesn't work either.What about b = 2 and c = 4? First inequality: 4 < 16, which is true. Second inequality: 16 < 8, which is false. So, no.b = 4 and c = 2: First inequality: 16 < 8, which is false.b = 1 and c = 2: First inequality: 1 < 8, true. Second inequality: 4 < 4, which is false because 4 is not less than 4. So, that doesn't work.b = 2 and c = 1: First inequality: 4 < 4, which is false.b = 3 and c = 4: First inequality: 9 < 16, true. Second inequality: 16 < 12, false.b = 4 and c = 3: First inequality: 16 < 12, false.b = 1 and c = 3: First inequality: 1 < 12, true. Second inequality: 9 < 4, false.b = 3 and c = 1: First inequality: 9 < 4, false.b = 1 and c = 4: First inequality: 1 < 16, true. Second inequality: 16 < 4, false.b = 4 and c = 1: First inequality: 16 < 4, false.b = 2 and c = 5: First inequality: 4 < 20, true. Second inequality: 25 < 8, false.b = 5 and c = 2: First inequality: 25 < 8, false.Hmm, seems like the only valid pairs are where b = c, specifically (1,1), (2,2), and (3,3). But wait, let me check if (3,3) actually works. For b = 3 and c = 3, the discriminants are 9 - 12 = -3 for both equations, which is negative, so yes, they have no real solutions. So, that's valid.But wait, the options don't include 3 as the answer, but let me double-check. The options are A)0, B)1, C)2, D)3, E)4. So, D)3 is an option. But I thought the answer was 2. Maybe I made a mistake.Wait, let me go back. The problem says "ordered pairs (b,c)", so (1,1), (2,2), and (3,3) are three distinct ordered pairs. But earlier, I thought (2,3) and (3,2) might work, but they don't because one of the inequalities fails. So, maybe the answer is 3. But in the initial thought process, I thought only (2,3) and (3,2) worked, but that was incorrect.Wait, no, actually, when I first thought about it, I considered (2,3) and (3,2), but they don't satisfy both inequalities. So, maybe the only valid pairs are where b = c, which are (1,1), (2,2), and (3,3). So, that's three pairs. But let me check if (3,3) is valid.For b = 3 and c = 3, the discriminants are 9 - 12 = -3 for both equations, which is negative, so yes, they have no real solutions. So, (3,3) is valid. Similarly, (2,2) gives discriminants of 4 - 8 = -4, which is negative. (1,1) gives discriminants of 1 - 4 = -3, which is negative. So, all three are valid.But wait, the options include up to 4, so maybe I missed some pairs where b ≠ c. Let me think again. Maybe b = 4 and c = 5? Let's check. First inequality: 16 < 20, which is true. Second inequality: 25 < 16, which is false. So, no.How about b = 5 and c = 4? First inequality: 25 < 16, which is false.b = 3 and c = 4: First inequality: 9 < 16, true. Second inequality: 16 < 12, false.b = 4 and c = 3: First inequality: 16 < 12, false.b = 2 and c = 4: First inequality: 4 < 16, true. Second inequality: 16 < 8, false.b = 4 and c = 2: First inequality: 16 < 8, false.b = 1 and c = 2: First inequality: 1 < 8, true. Second inequality: 4 < 4, which is false because 4 is not less than 4.b = 2 and c = 1: First inequality: 4 < 4, which is false.b = 1 and c = 3: First inequality: 1 < 12, true. Second inequality: 9 < 4, false.b = 3 and c = 1: First inequality: 9 < 4, false.b = 1 and c = 4: First inequality: 1 < 16, true. Second inequality: 16 < 4, false.b = 4 and c = 1: First inequality: 16 < 4, false.b = 2 and c = 5: First inequality: 4 < 20, true. Second inequality: 25 < 8, false.b = 5 and c = 2: First inequality: 25 < 8, false.Hmm, so it seems like the only valid pairs are where b = c, specifically (1,1), (2,2), and (3,3). So, that's three pairs. But wait, the answer choices include up to 4, so maybe I'm missing something.Wait, let me check b = 4 and c = 5 again. First inequality: 16 < 20, true. Second inequality: 25 < 16, false. So, no.How about b = 5 and c = 6? First inequality: 25 < 24, which is false.Wait, maybe I should try higher values, but it seems like as b and c increase, the inequalities become harder to satisfy because c² grows faster than 4b and b² grows faster than 4c.Wait, let me think about the inequalities again. We have b² < 4c and c² < 4b. If I substitute c from the first inequality into the second, I get c > b²/4. Then, substituting into the second inequality: (b²/4)² < 4b, which simplifies to b⁴/16 < 4b, or b⁴ < 64b, which simplifies to b³ < 64, so b < 4. So, b can be 1, 2, or 3. That means c must be greater than b²/4, but also c² < 4b.So, for b = 1: c > 1/4, so c ≥ 1. Then c² < 4*1 = 4, so c can be 1 or 2 (since 2²=4 is not less than 4). Wait, but c must be a positive integer, so c = 1. So, only (1,1).For b = 2: c > 4/4 = 1, so c ≥ 2. Then c² < 8, so c can be 2 or possibly 3 (since 3²=9 which is greater than 8). So, c = 2. So, (2,2).For b = 3: c > 9/4 = 2.25, so c ≥ 3. Then c² < 12, so c can be 3 (since 3²=9 < 12) or 4 (16 > 12). So, c = 3. So, (3,3).So, according to this, only (1,1), (2,2), and (3,3) are valid. So, that's three pairs. But wait, the answer choices include up to 4, so maybe I'm missing something.Wait, but earlier I thought (2,3) and (3,2) might work, but they don't because one of the inequalities fails. So, maybe the answer is 3, which is option D.But wait, let me double-check. For b = 3 and c = 3, discriminants are 9 - 12 = -3, which is negative. So, yes, valid.For b = 2 and c = 2, discriminants are 4 - 8 = -4, which is negative.For b = 1 and c = 1, discriminants are 1 - 4 = -3, which is negative.So, three pairs. Therefore, the answer should be D) 3.But wait, in the initial thought process, I thought only (2,3) and (3,2) worked, but that was incorrect. So, the correct answer is 3, which is option D.

💡Okay, so I have this problem about a cube that's 13 units on each side. The entire surface of this cube is painted red, and then it's cut into 1x1x1 smaller cubes. After that, all the faces of these small cubes that weren't painted red are painted blue. I need to find out by what factor the total blue area is greater than the total red area.Hmm, let me break this down step by step. First, the original cube is 13x13x13. So, each face of this cube is a 13x13 square. Since a cube has 6 faces, the total red painted area would be 6 times the area of one face. Let me write that down:Total red area = 6 * (13 * 13) = 6 * 169 = 1014 square units.Okay, so that's the total red area. Now, when the cube is cut into 1x1x1 cubes, how many small cubes do we get? Well, it's 13 * 13 * 13, which is 2197 small cubes.Each of these small cubes has 6 faces, so the total number of faces across all small cubes is 2197 * 6. Let me calculate that:Total faces = 2197 * 6 = 13182 square units.Wait, so the total surface area of all the small cubes combined is 13182 square units. But not all of these are painted blue. Some of them were originally painted red. So, the blue area would be the total faces minus the red area.So, blue area = total faces - red area = 13182 - 1014 = 12168 square units.Now, I need to find by what factor the blue area is greater than the red area. That means I need to divide the blue area by the red area.Factor = blue area / red area = 12168 / 1014.Let me compute that. Hmm, 1014 times 12 is 12168, right? Because 1000*12=12000 and 14*12=168, so 12000+168=12168. So, 1014*12=12168.Therefore, the factor is 12. So, the blue area is 12 times greater than the red area.Wait, let me just verify that. So, the total red area is 1014, and the blue area is 12168. Dividing 12168 by 1014 gives exactly 12. That makes sense because each face of the original cube contributes red paint, but when you cut the cube into smaller cubes, each internal face (which was not on the surface) gets painted blue. Since the cube is 13 units on each side, the number of internal layers is 11 on each side, but maybe that's complicating things.Alternatively, thinking in terms of the total surface area: the original cube has 6 faces, each 13x13. When you cut it into small cubes, each small cube can have 0 to 3 red faces, depending on whether it's on a corner, edge, face, or inside. But instead of counting each small cube's red faces, it's easier to consider the total red area as just the original surface area, which is 6*13^2, and the total blue area as the total surface area of all small cubes minus the red area.Total surface area of all small cubes is 6*(13^3), since each small cube has 6 faces and there are 13^3 small cubes. So, 6*(13^3) - 6*(13^2) = 6*13^2*(13 - 1) = 6*13^2*12. Then, dividing by the red area, which is 6*13^2, gives 12. So, yeah, that's consistent.I think that makes sense. So, the blue area is 12 times the red area.

💡First, I need to determine the total number of shirts produced daily by all employees. With 20 employees each making 20 shirts per day, the total production is 400 shirts.Next, I'll calculate the total revenue from selling these shirts. At 35 per shirt, the revenue amounts to 14,000 per day.Then, I'll compute the total wages paid to the employees. Each employee works an 8-hour shift at 12 per hour, resulting in 96 per day per employee. For 20 employees, this totals 1,920 per day.Additionally, each employee receives a 5 bonus per shirt they make. For 20 shirts, this amounts to 100 per employee per day. For all 20 employees, this adds up to 2,000 per day.Adding the base wages and the bonuses gives a total daily wage expense of 3,920.Including the nonemployee expenses of 1,000 per day, the total daily expenses are 4,920.Finally, by subtracting the total expenses from the total revenue, I find that the company's daily profit is 9,080.

💡First, I need to evaluate the function ( f(x) = 2x - 1 ) at ( x = 3 ).I'll substitute ( x = 3 ) into the function:[f(3) = 2(3) - 1]Next, I'll perform the multiplication:[2 times 3 = 6]Then, I'll subtract 1 from the result:[6 - 1 = 5]So, the value of ( f(3) ) is 5.