💡Okay, so I have this function f(x) that's defined in two parts. It's a piecewise function, right? For x between 0 and 3, inclusive, f(x) is equal to 2x minus x squared. And for x between -2 and 0, not including 0, f(x) is equal to x squared plus 6x. I need to find the range of this function. Hmm, the range is all the possible output values, or y-values, that f(x) can take. So, I guess I need to figure out the range for each piece separately and then combine them.Let me start with the first piece: f(x) = 2x - x² for 0 ≤ x ≤ 3. This looks like a quadratic function. Quadratic functions have either a maximum or a minimum value, depending on whether the coefficient of x² is negative or positive. In this case, the coefficient is -1, which is negative, so the parabola opens downward, meaning it has a maximum point. That maximum point is the vertex of the parabola.To find the vertex, I can use the formula for the x-coordinate of the vertex, which is -b/(2a). In this quadratic, a is -1 and b is 2. So, plugging in, x = -2/(2*(-1)) = -2/(-2) = 1. So, the vertex is at x = 1. Now, let me find the corresponding y-value by plugging x = 1 into f(x). That gives f(1) = 2(1) - (1)² = 2 - 1 = 1. So, the maximum value of this piece is 1.Now, I should check the endpoints of the interval, x = 0 and x = 3, to see what the function values are there. At x = 0, f(0) = 2(0) - (0)² = 0. At x = 3, f(3) = 2(3) - (3)² = 6 - 9 = -3. So, for this first piece, the function ranges from -3 to 1.Alright, moving on to the second piece: f(x) = x² + 6x for -2 ≤ x < 0. Again, this is a quadratic function. The coefficient of x² is positive (1), so the parabola opens upward, meaning it has a minimum point at its vertex.Let me find the vertex of this quadratic. Using the vertex formula again, x = -b/(2a). Here, a is 1 and b is 6, so x = -6/(2*1) = -3. Wait, but our interval is from -2 to 0. So, the vertex is at x = -3, which is outside the interval we're considering. That means the minimum value isn't within our interval, so the function is either increasing or decreasing throughout the interval.Since the parabola opens upward and the vertex is at x = -3, which is to the left of our interval, the function will be increasing on the interval from -2 to 0. So, the minimum value on this interval will be at the left endpoint, x = -2, and the maximum will be at the right endpoint, x approaching 0 from the left.Let me compute f(-2): f(-2) = (-2)² + 6*(-2) = 4 - 12 = -8. Wait, but hold on, that's not right. Let me double-check. (-2) squared is 4, and 6 times -2 is -12. So, 4 - 12 is indeed -8. Hmm, but earlier, I thought the function was increasing on this interval, so the minimum should be at x = -2 and the maximum as x approaches 0.But when I plug in x = 0, f(0) would be 0 + 0 = 0, but since x approaches 0 from the left, the function approaches 0. So, the range for this second piece is from -8 up to, but not including, 0. Wait, but when I calculated f(-2), I got -8, but earlier I thought the vertex was at x = -3, which is outside the interval. So, maybe I made a mistake there.Wait, let me think again. The quadratic f(x) = x² + 6x can be rewritten by completing the square. Let's try that. x² + 6x = (x + 3)^2 - 9. So, the vertex is at (-3, -9). So, the minimum value is -9 at x = -3, which is outside our interval. So, on the interval from -2 to 0, the function is increasing because the vertex is to the left of this interval. Therefore, the minimum value on this interval is at x = -2, which is f(-2) = (-2)^2 + 6*(-2) = 4 - 12 = -8, and as x approaches 0, f(x) approaches 0.Wait, but if the vertex is at x = -3, which is less than -2, then from x = -2 to x = 0, the function is increasing. So, the lowest point on this interval is at x = -2, which is -8, and it goes up to 0 as x approaches 0. So, the range for this piece is [-8, 0).Wait, but earlier, when I calculated f(-2), I got -8, and f(0) is 0, but since x approaches 0 from the left, it's approaching 0 but doesn't include 0. So, the range is from -8 to 0, including -8 but not including 0.Wait, but earlier, I thought the function was increasing, so the minimum is at x = -2, which is -8, and the maximum approaches 0 as x approaches 0. So, the range for this piece is [-8, 0).But wait, when I first thought about the vertex, I thought the function was increasing on the interval, so the minimum is at x = -2, which is -8, and the maximum is approaching 0. So, the range is [-8, 0).But hold on, when I completed the square, I saw that f(x) = (x + 3)^2 - 9. So, at x = -2, f(-2) = (-2 + 3)^2 - 9 = (1)^2 - 9 = 1 - 9 = -8, which matches. And as x approaches 0, f(x) approaches (0 + 3)^2 - 9 = 9 - 9 = 0, but since x is approaching 0 from the left, it's approaching 0 from below, right? Because x is negative, so (x + 3) is positive, but x is approaching 0, so (x + 3) approaches 3, so (x + 3)^2 approaches 9, so f(x) approaches 0 from below? Wait, no, because f(x) is (x + 3)^2 - 9, so as x approaches 0, (x + 3)^2 approaches 9, so f(x) approaches 0. But since x is approaching 0 from the left, does that affect the value? Hmm, no, because (x + 3)^2 is always positive, regardless of x being negative. So, as x approaches 0 from the left, f(x) approaches 0 from below? Wait, no, because (x + 3)^2 is 9 when x is 0, so f(x) is 0. But when x is just less than 0, say x = -0.1, then (x + 3)^2 is (2.9)^2 = 8.41, so f(x) = 8.41 - 9 = -0.59. Wait, that's negative. So, as x approaches 0 from the left, f(x) approaches 0 from below, meaning it approaches 0 from the negative side. So, the function approaches 0 but doesn't reach it on this interval.So, the range for the second piece is from -8 up to, but not including, 0. So, [-8, 0).Wait, but earlier, when I calculated f(-2), I got -8, and as x increases from -2 to 0, f(x) increases from -8 to 0. So, the range is [-8, 0).But wait, in the first piece, the range was from -3 to 1. So, combining both pieces, the overall range would be from the lowest value, which is -8, up to the highest value, which is 1. So, the range of f(x) is [-8, 1].Wait, but earlier, I thought the first piece had a range of [-3, 1], and the second piece had a range of [-8, 0). So, combining these, the overall range is from -8 to 1, including both endpoints because in the first piece, the function reaches -3 and 1, and in the second piece, it reaches -8 and approaches 0.Wait, but in the second piece, the function doesn't actually reach 0, because x approaches 0 from the left. So, 0 is not included in the second piece, but in the first piece, when x = 0, f(0) = 0. So, 0 is included in the overall range because it's achieved at x = 0 in the first piece.So, putting it all together, the overall range is from -8 to 1, including both -8 and 1, and including 0 because it's achieved at x = 0.Wait, but in the second piece, the function approaches 0 but doesn't reach it, but in the first piece, at x = 0, f(x) = 0. So, 0 is included in the overall range.So, the range is from -8 to 1, including both endpoints. So, the range is [-8, 1].Wait, but earlier, I thought the first piece had a range of [-3, 1], and the second piece had a range of [-8, 0). So, combining these, the overall range is [-8, 1].Yes, that makes sense. So, the function can take any value from -8 up to 1.Wait, but let me double-check my calculations to make sure I didn't make any mistakes.For the first piece, f(x) = 2x - x² for 0 ≤ x ≤ 3.We found the vertex at x = 1, f(1) = 1. Then, at x = 0, f(0) = 0, and at x = 3, f(3) = -3. So, the range is from -3 to 1.For the second piece, f(x) = x² + 6x for -2 ≤ x < 0.We completed the square to get f(x) = (x + 3)^2 - 9. The vertex is at x = -3, which is outside our interval. So, on the interval from -2 to 0, the function is increasing because the vertex is to the left. So, at x = -2, f(-2) = (-2 + 3)^2 - 9 = 1 - 9 = -8. As x approaches 0 from the left, f(x) approaches (0 + 3)^2 - 9 = 0. So, the range is from -8 to 0, not including 0.But since in the first piece, f(0) = 0, 0 is included in the overall range. So, combining both pieces, the overall range is from -8 to 1, including both endpoints.Wait, but in the second piece, the function doesn't reach 0, but in the first piece, it does. So, 0 is included. So, the overall range is [-8, 1].Wait, but earlier, I thought the first piece had a range of [-3, 1], and the second piece had a range of [-8, 0). So, combining these, the overall range is [-8, 1].Yes, that seems correct.But let me just plot these functions mentally to make sure.For the first piece, f(x) = 2x - x² from x = 0 to x = 3. It's a downward opening parabola with vertex at (1, 1). So, it starts at (0, 0), goes up to (1, 1), then down to (3, -3). So, the y-values go from 0 up to 1, then down to -3.For the second piece, f(x) = x² + 6x from x = -2 to x = 0. It's an upward opening parabola with vertex at (-3, -9). So, on the interval from -2 to 0, it's increasing because the vertex is to the left. So, at x = -2, f(-2) = (-2)^2 + 6*(-2) = 4 - 12 = -8. At x = 0, f(0) = 0 + 0 = 0, but since x approaches 0 from the left, the function approaches 0 from below. So, the y-values go from -8 up to just below 0.So, combining both pieces, the function can take any value from -8 up to 1. So, the range is [-8, 1].Wait, but in the first piece, the function goes down to -3, which is higher than -8. So, the lowest point overall is -8, and the highest point is 1. So, the range is [-8, 1].Yes, that makes sense.I think I'm confident now that the range of the function is from -8 to 1, including both endpoints. So, the range is [-8, 1].

💡Okay, so I have this problem where there are five points marked on a grid plane, and I need to prove that among these five points, there are at least two such that the midpoint of the segment connecting them is also a grid point. Hmm, interesting. Let me try to break this down step by step.First, I need to understand what it means for a midpoint to be a grid point. A grid point has integer coordinates, right? So, if I have two points, say (x1, y1) and (x2, y2), the midpoint would be ((x1 + x2)/2, (y1 + y2)/2). For this midpoint to also be a grid point, both (x1 + x2)/2 and (y1 + y2)/2 need to be integers. That means x1 + x2 and y1 + y2 must both be even numbers because dividing an even number by 2 gives an integer.So, for the sum of two numbers to be even, both numbers must be either even or both must be odd. That is, x1 and x2 must have the same parity, and y1 and y2 must have the same parity. Parity means whether a number is even or odd. So, if x1 is even, x2 must also be even, and if x1 is odd, x2 must also be odd. The same goes for y1 and y2.Now, each point on the grid can be classified based on the parity of its coordinates. There are four possible parity combinations for a point:1. (Even, Even)2. (Even, Odd)3. (Odd, Even)4. (Odd, Odd)So, each point falls into one of these four categories based on whether its x-coordinate is even or odd and whether its y-coordinate is even or odd.Now, we have five points. Since there are only four categories, by the pigeonhole principle, at least two of these five points must fall into the same category. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least two pigeons must share a pigeonhole. In this case, the pigeons are the five points, and the pigeonholes are the four parity categories.So, if two points fall into the same parity category, that means their x-coordinates have the same parity, and their y-coordinates have the same parity. Therefore, when we take the midpoint of these two points, both the x and y coordinates of the midpoint will be integers, making the midpoint a grid point.Let me double-check this reasoning. Suppose we have two points in the (Even, Even) category. Then, both x1 and x2 are even, so x1 + x2 is even, and y1 and y2 are even, so y1 + y2 is even. Therefore, the midpoint ((x1 + x2)/2, (y1 + y2)/2) will have integer coordinates. The same logic applies if both points are in the (Even, Odd), (Odd, Even), or (Odd, Odd) categories.Wait, what if the two points are in different categories? For example, one is (Even, Even) and the other is (Even, Odd). Then, their x-coordinates are both even, so x1 + x2 is even, but their y-coordinates are even and odd, so y1 + y2 is odd. Therefore, the midpoint would have an integer x-coordinate but a non-integer y-coordinate, which means it's not a grid point. Similarly, if one point is (Even, Even) and the other is (Odd, Even), their x-coordinates are even and odd, so x1 + x2 is odd, and their y-coordinates are both even, so y1 + y2 is even. Thus, the midpoint would have a non-integer x-coordinate and an integer y-coordinate, again not a grid point.So, only when two points are in the same parity category do we get both coordinates of the midpoint as integers. Therefore, the key is that among five points, at least two must share the same parity category, ensuring that their midpoint is a grid point.Let me think if there's any other way this could go wrong. Suppose someone tries to place five points such that no two share the same parity category. But since there are only four categories, it's impossible to place five points without having at least two in the same category. Hence, the pigeonhole principle guarantees that at least two points must share a category, ensuring the existence of such a midpoint.I think that covers it. The pigeonhole principle is the key here, along with understanding the parity of coordinates and how it affects the midpoint. It's a neat application of combinatorial principles to a geometric problem.

💡Okay, so I've got this problem here: Triangle ABC is congruent to triangle DEF, and they tell me that AB is 3 units long. I need to find the length of DE. Hmm, congruent triangles... I remember that congruent means they're exactly the same shape and size, right? So, all their corresponding sides and angles are equal.First, let me recall what congruent triangles mean. If two triangles are congruent, it means that all their corresponding sides are equal in length, and all their corresponding angles are equal in measure. So, if triangle ABC is congruent to triangle DEF, then side AB in triangle ABC corresponds to side DE in triangle DEF. That makes sense because the order of the letters matters in congruence; the first letter of one triangle corresponds to the first letter of the other triangle.So, AB corresponds to DE. Since AB is given as 3, DE should also be 3. Wait, is that right? Let me make sure I'm not mixing up the sides. Sometimes, the order can be confusing. Let me write down the correspondence: A corresponds to D, B corresponds to E, and C corresponds to F. So, side AB in triangle ABC is between points A and B, which correspond to points D and E in triangle DEF. Therefore, side DE in triangle DEF is the same as side AB in triangle ABC.Just to double-check, I'll think about the properties of congruent triangles. One of the main properties is that corresponding parts of congruent triangles are congruent, often abbreviated as CPCTC. So, if two triangles are congruent, then their corresponding parts (sides and angles) are equal. Therefore, since AB is a side in triangle ABC and DE is the corresponding side in triangle DEF, they must be equal in length.I don't think there's any need for more complicated calculations here because the problem is straightforward. It's just about understanding the correspondence between the sides of the two triangles. I might be overcomplicating it, but it's always good to verify.Let me visualize the triangles. If I imagine triangle ABC with points A, B, and C, and triangle DEF with points D, E, and F, and they are congruent, then each side in ABC has an equal counterpart in DEF. So, AB corresponds to DE, BC corresponds to EF, and AC corresponds to DF. Similarly, the angles at each vertex are equal: angle A equals angle D, angle B equals angle E, and angle C equals angle F.Since AB is given as 3, DE must also be 3 because they are corresponding sides of congruent triangles. I don't think I need to use the Pythagorean theorem or any other formulas here because the problem doesn't provide any other information that would require that. It's purely about the definition of congruent triangles.Wait, just to make sure, let me think about the different congruence criteria. There are several ways to prove that two triangles are congruent, like SSS (side-side-side), SAS (side-angle-side), ASA (angle-side-angle), and AAS (angle-angle-side). But in this case, we're already told that the triangles are congruent, so we don't need to prove it. We just need to use the fact that their corresponding sides are equal.So, if I were to write this out step by step, it would be:1. Given that triangle ABC is congruent to triangle DEF.2. Therefore, corresponding sides are equal.3. AB corresponds to DE.4. Since AB is 3, DE is also 3.That seems pretty solid. I don't see any gaps in this reasoning. It's a direct application of the definition of congruent triangles.Just to cover all bases, let me think about if there's any possibility that DE could be a different length. For example, if the triangles weren't congruent, then DE could be different, but since they are congruent, DE has to be the same as AB. There's no other option here.Also, considering the notation, when we write triangle ABC congruent to triangle DEF, the order of the letters is important. It tells us which sides correspond. So, A corresponds to D, B to E, and C to F. Therefore, side AB corresponds to side DE, side BC corresponds to side EF, and side AC corresponds to side DF.If I were to label the triangles, I could draw triangle ABC with points A, B, and C, and triangle DEF with points D, E, and F. Then, connecting the corresponding points would show that AB is equal to DE, BC is equal to EF, and AC is equal to DF.Another way to think about it is through transformations. If triangle ABC is congruent to triangle DEF, then one can be transformed into the other through a series of rigid transformations (translations, rotations, reflections). These transformations preserve the lengths of sides and the measures of angles. So, after any such transformation, the length of AB would remain 3, and thus DE, being the image of AB, would also be 3.I think I've covered all the angles here. It's a pretty straightforward problem once you understand the basics of congruent triangles. The key takeaway is that corresponding parts of congruent triangles are congruent, so knowing one side allows you to find the corresponding side in the other triangle.Just to recap:- Congruent triangles have all corresponding sides equal.- The order of the letters in the congruence statement tells us which sides correspond.- Therefore, AB corresponds to DE, and since AB is 3, DE must also be 3.I don't think I've missed anything here. It's a fundamental concept in geometry, and understanding it is essential for solving more complex problems involving congruent triangles.

💡First, I need to determine the original salary before the increase. The new salary is 25,000, and the increase was 5,000. By subtracting the increase from the new salary, I can find the original salary.Next, I'll use the percent increase formula to calculate the percentage increase. The formula is:[text{Percent Increase} = left( frac{text{Increase}}{text{Original Salary}} right) times 100]Plugging in the values:[text{Percent Increase} = left( frac{5,000}{20,000} right) times 100 = 25%]

💡Okay, so I need to find the greatest integer ( y ) such that ( frac{8}{11} > frac{y}{15} ). Hmm, let's see. I think I should start by understanding what this inequality means. It's saying that the fraction ( frac{8}{11} ) is larger than ( frac{y}{15} ). So, I need to find the largest whole number ( y ) that makes this true.Maybe I can solve the inequality for ( y ) to find its maximum value. To do that, I can cross-multiply to get rid of the fractions. If I multiply both sides by 15, that should help. Let me write that down:[frac{8}{11} > frac{y}{15}]Multiplying both sides by 15:[frac{8}{11} times 15 > y]Calculating ( frac{8 times 15}{11} ):[frac{120}{11} approx 10.909]So, ( y ) has to be less than approximately 10.909. Since ( y ) has to be an integer, the largest integer less than 10.909 is 10. Therefore, the greatest integer ( y ) that satisfies the inequality is 10.Wait, let me double-check to make sure I didn't make a mistake. If ( y = 10 ), then ( frac{10}{15} = frac{2}{3} approx 0.6667 ). And ( frac{8}{11} approx 0.7273 ). Yes, 0.7273 is greater than 0.6667, so that works.What if I try ( y = 11 )? Then ( frac{11}{15} approx 0.7333 ). Comparing that to ( frac{8}{11} approx 0.7273 ), 0.7333 is actually greater than 0.7273. So, ( y = 11 ) doesn't satisfy the inequality because ( frac{8}{11} ) is not greater than ( frac{11}{15} ).Therefore, 10 is indeed the largest integer that works. I think that's the correct answer.

💡Okay, so I have this problem here where I need to find the largest possible value for a positive integer ( a ) given some inequalities involving ( a ), ( b ), ( c ), and ( e ). The inequalities are ( a < 2b ), ( b < 3c ), and ( c < 5e ). Additionally, it's given that ( e < 100 ). The options provided are A) 2960, B) 2961, C) 2962, D) 2963, and E) 2964. I need to figure out which one is the correct answer.First, let me try to understand the relationships between these variables. The inequalities are all less than inequalities, which means each variable is strictly less than some multiple of another variable. So, ( a ) is less than twice ( b ), ( b ) is less than three times ( c ), and ( c ) is less than five times ( e ). Since all these are positive integers, I can think of them as being as large as possible without violating the inequalities.Given that ( e < 100 ), the maximum integer value ( e ) can take is 99. That seems straightforward. So, ( e = 99 ) is the largest possible value for ( e ).Next, since ( c < 5e ), and ( e = 99 ), substituting that in, we get ( c < 5 times 99 ). Calculating that, ( 5 times 99 = 495 ). So, ( c ) must be less than 495. Since ( c ) is a positive integer, the largest possible value for ( c ) is 494.Moving on to ( b ), we have ( b < 3c ). Substituting the maximum value of ( c ) which is 494, we get ( b < 3 times 494 ). Calculating that, ( 3 times 494 = 1482 ). Therefore, ( b ) must be less than 1482, meaning the largest integer value ( b ) can take is 1481.Now, finally, looking at ( a < 2b ). Substituting the maximum value of ( b ) which is 1481, we get ( a < 2 times 1481 ). Calculating that, ( 2 times 1481 = 2962 ). So, ( a ) must be less than 2962. Since ( a ) is a positive integer, the largest possible value ( a ) can take is 2961.Wait, but let me double-check my steps to make sure I didn't make a mistake. Starting from ( e = 99 ), ( c < 5e ) gives ( c < 495 ), so ( c = 494 ). Then ( b < 3c ) gives ( b < 1482 ), so ( b = 1481 ). Then ( a < 2b ) gives ( a < 2962 ), so ( a = 2961 ). That seems consistent.But let me think again. Is there a way to make ( a ) larger? Maybe if ( e ) is smaller, but then ( c ) would be smaller, which would make ( b ) smaller, and consequently ( a ) would be smaller. So, to maximize ( a ), I need to maximize ( e ), which is 99, then maximize ( c ), ( b ), and so on. So, I think my initial conclusion is correct.Alternatively, maybe I can express all variables in terms of ( e ) and see if that gives me a different perspective. Let's try that.Starting from ( c < 5e ), so ( c ) can be at most ( 5e - 1 ). Then ( b < 3c ), so substituting ( c ), ( b < 3(5e - 1) = 15e - 3 ). Therefore, ( b ) can be at most ( 15e - 4 ). Then ( a < 2b ), so substituting ( b ), ( a < 2(15e - 4) = 30e - 8 ). Therefore, ( a ) can be at most ( 30e - 9 ).Given that ( e < 100 ), the maximum ( e ) is 99. Substituting ( e = 99 ), we get ( a < 30 times 99 - 8 ). Calculating that, ( 30 times 99 = 2970 ), so ( 2970 - 8 = 2962 ). Therefore, ( a ) must be less than 2962, so the maximum ( a ) is 2961. That matches my previous result.Wait, so in this case, is it 2961? But the options include 2960, 2961, 2962, 2963, 2964. So, 2961 is an option, which is B.But let me check if I made a mistake in the substitution. When I expressed ( a ) in terms of ( e ), I had ( a < 30e - 8 ). So, for ( e = 99 ), ( 30 times 99 = 2970 ), minus 8 is 2962. So, ( a < 2962 ), so the maximum integer ( a ) is 2961. That seems correct.Alternatively, maybe I should express all the inequalities together. Let me see:Given ( a < 2b ), ( b < 3c ), ( c < 5e ).So, combining these, ( a < 2b < 2 times 3c = 6c ).Then, ( 6c < 6 times 5e = 30e ).Therefore, ( a < 30e ).But since ( a ) is less than 30e, and ( e < 100 ), so ( a < 30 times 100 = 3000 ). But that's a rough upper bound. However, since ( e ) is less than 100, the maximum ( e ) is 99, so ( a < 30 times 99 = 2970 ). But from the previous step, we saw that ( a ) is actually less than 2962, which is tighter.Wait, so combining all the inequalities, ( a < 2b < 6c < 30e ). So, ( a < 30e ). But since ( a ) must be less than 30e, and ( e ) is 99, ( a < 2970 ). But in reality, because each inequality is strict, we have to subtract 1 at each step.Wait, let me think again. If ( c < 5e ), then ( c leq 5e - 1 ). Similarly, ( b < 3c ) implies ( b leq 3c - 1 ), and ( a < 2b ) implies ( a leq 2b - 1 ).So, starting from ( e = 99 ):( c leq 5 times 99 - 1 = 495 - 1 = 494 ).Then, ( b leq 3 times 494 - 1 = 1482 - 1 = 1481 ).Then, ( a leq 2 times 1481 - 1 = 2962 - 1 = 2961 ).So, that confirms it again. Therefore, the maximum value of ( a ) is 2961, which is option B.But just to be thorough, let me check if ( a = 2961 ) actually satisfies all the inequalities with some integers ( b ), ( c ), and ( e ).Starting from ( a = 2961 ), we need ( b ) such that ( 2961 < 2b ). So, ( b > 2961 / 2 = 1480.5 ). Since ( b ) is an integer, ( b geq 1481 ).Next, ( b = 1481 ) must satisfy ( b < 3c ). So, ( 1481 < 3c ), which implies ( c > 1481 / 3 approx 493.666 ). Since ( c ) is an integer, ( c geq 494 ).Then, ( c = 494 ) must satisfy ( c < 5e ). So, ( 494 < 5e ), which implies ( e > 494 / 5 = 98.8 ). Since ( e ) is an integer, ( e geq 99 ).Given that ( e < 100 ), ( e = 99 ) is acceptable.So, putting it all together:( e = 99 ), ( c = 494 ), ( b = 1481 ), and ( a = 2961 ).Checking the inequalities:1. ( a = 2961 < 2b = 2 times 1481 = 2962 ). True.2. ( b = 1481 < 3c = 3 times 494 = 1482 ). True.3. ( c = 494 < 5e = 5 times 99 = 495 ). True.4. ( e = 99 < 100 ). True.All inequalities are satisfied. Therefore, ( a = 2961 ) is indeed a valid solution.Is there a possibility of a larger ( a )? Let's see. If I try ( a = 2962 ), then ( 2962 < 2b ) implies ( b > 2962 / 2 = 1481 ). So, ( b geq 1482 ).Then, ( b = 1482 ) must satisfy ( 1482 < 3c ), so ( c > 1482 / 3 = 494 ). Therefore, ( c geq 495 ).But ( c = 495 ) must satisfy ( c < 5e ), so ( 495 < 5e ) implies ( e > 495 / 5 = 99 ). Since ( e ) must be less than 100, ( e ) would have to be 100, but ( e < 100 ), so ( e ) cannot be 100. Therefore, ( c ) cannot be 495 because that would require ( e geq 100 ), which is not allowed.Thus, ( c ) cannot be 495, so ( c ) must be at most 494. But if ( c = 494 ), then ( b < 3c = 1482 ), so ( b leq 1481 ). But earlier, for ( a = 2962 ), we needed ( b geq 1482 ), which conflicts with ( b leq 1481 ). Therefore, ( a = 2962 ) is not possible because it would require ( b ) to be both greater than or equal to 1482 and less than or equal to 1481, which is impossible.Therefore, ( a = 2962 ) is not a valid solution. Hence, the maximum possible value for ( a ) is indeed 2961.I think that's thorough enough. I've checked each step, made sure that each inequality is satisfied, and even tested the next possible value to confirm that it doesn't work. So, I'm confident that 2961 is the correct answer.