💡Alright, so I've got this problem here about a cube with an edge length of 1 cm being ground into a spherical part, and I need to find the maximum volume of that spherical part. Hmm, okay. Let me try to visualize this. I have a cube, right? Each edge is 1 cm long. Now, if I'm going to turn this cube into a sphere, I guess the sphere has to fit perfectly inside the cube to maximize its volume. That makes sense because if the sphere is too big, it won't fit, and if it's too small, I'm not using the cube's space efficiently.So, first things first, I need to figure out the relationship between the cube and the sphere. Since the sphere has to fit inside the cube, the sphere's diameter can't be larger than the cube's edge length. The cube has an edge length of 1 cm, so the sphere's diameter must be equal to 1 cm to fit perfectly. That means the radius of the sphere is half of that, which is 0.5 cm. Okay, so radius r = 0.5 cm.Now, I remember the formula for the volume of a sphere is (4/3)πr³. Let me write that down: V = (4/3)πr³. So, plugging in the radius I just found, which is 0.5 cm, I get V = (4/3)π(0.5)³. Let's compute that step by step.First, (0.5)³ is 0.125. So, V = (4/3)π * 0.125. Now, multiplying 4/3 by 0.125, I can think of 0.125 as 1/8. So, 4/3 * 1/8 = 4/24, which simplifies to 1/6. Therefore, V = (1/6)π. So, the volume is π/6 cm³.Wait a second, let me double-check that. If the radius is 0.5 cm, then the volume should be (4/3)π*(0.5)³. Calculating (0.5)³ gives 0.125, and multiplying that by 4/3 gives 0.5π/3, which is the same as π/6. Yeah, that seems right.But just to make sure I'm not missing anything, let's think about the cube's volume. The cube has an edge length of 1 cm, so its volume is 1 cm³. The sphere's volume is π/6 cm³, which is approximately 0.5236 cm³. That seems reasonable because the sphere can't occupy the entire volume of the cube; there will be some wasted space, especially in the corners of the cube that the sphere can't reach.Is there any other way to fit a larger sphere inside the cube? Hmm, if I try to fit a sphere that touches the cube's edges or vertices, would that give a larger volume? Wait, no, because the sphere has to be entirely inside the cube. If I try to make the sphere touch the edges or vertices, it would actually require a larger diameter, which wouldn't fit inside the cube. So, the maximum diameter the sphere can have is equal to the cube's edge length, which is 1 cm.Therefore, the radius is 0.5 cm, and the volume is π/6 cm³. Looking back at the options provided:A: π/6 cm³B: √2π/3 cm³C: √3π/2 cm³D: π/3 cm³So, the answer should be A: π/6 cm³.But just to be thorough, let's consider if there's any other interpretation of the problem. Maybe the sphere isn't inscribed in the cube but somehow uses the cube's space differently? But I don't think so. The maximum volume sphere that can fit inside a cube is the one that is inscribed, meaning it touches the centers of each face of the cube. Any other sphere would either not fit or have a smaller volume.Also, let's recall that the formula for the volume of a sphere is correct. Sometimes, people confuse it with the surface area, but no, I'm pretty sure it's (4/3)πr³. So, with r = 0.5 cm, the volume is indeed π/6 cm³.Alright, I think I've thought this through thoroughly. The maximum volume of the spherical part that can be produced from a cube with an edge length of 1 cm is π/6 cm³.

💡To find the units digit of (2008^{2008}), we focus on the units digit of the base number, which is (8). We need to determine the pattern of the units digits of powers of (8).1. **Identify the units digit pattern of powers of (8):** [ begin{align*} 8^1 & = 8 quad (text{units digit is } 8) 8^2 & = 64 quad (text{units digit is } 4) 8^3 & = 512 quad (text{units digit is } 2) 8^4 & = 4096 quad (text{units digit is } 6) 8^5 & = 32768 quad (text{units digit is } 8) end{align*} ] The units digits repeat every 4 powers: (8, 4, 2, 6).2. **Determine the position of (2008) in the cycle:** Since the units digits repeat every 4 powers, we find the remainder when (2008) is divided by (4): [ 2008 mod 4 = 0 ] This means (2008) is a multiple of (4), corresponding to the 4th position in the cycle.3. **Identify the units digit at the 4th position in the cycle:** From the pattern (8, 4, 2, 6), the 4th position has the units digit (6).Therefore, the units digit of (2008^{2008}) is (boxed{6}).

💡Okay, so I have this problem where I need to figure out the value of ( m ) such that one of the "split numbers" of ( m^3 ) is 333. The problem gives examples of how cubes of natural numbers greater than 1 can be split into sums of consecutive odd numbers. For instance, ( 2^3 = 3 + 5 ), ( 3^3 = 7 + 9 + 11 ), ( 4^3 = 13 + 15 + 17 + 19 ), and so on. First, I need to understand the pattern here. Let me list out the examples given:- ( 2^3 = 3 + 5 ) → 2 numbers- ( 3^3 = 7 + 9 + 11 ) → 3 numbers- ( 4^3 = 13 + 15 + 17 + 19 ) → 4 numbersSo, it seems like for ( m^3 ), the cube is split into ( m ) consecutive odd numbers. That makes sense because the number of terms in the sum equals the base of the cube.Next, I notice that the starting number of each split seems to follow a pattern. Let me calculate the starting number for each example:- For ( m = 2 ): The first number is 3.- For ( m = 3 ): The first number is 7.- For ( m = 4 ): The first number is 13.Hmm, let me see if I can find a formula for the first number in the split. Let's denote the first number as ( a_m ) for ( m^3 ).Calculating the differences:- From ( m = 2 ) to ( m = 3 ): 7 - 3 = 4- From ( m = 3 ) to ( m = 4 ): 13 - 7 = 6Wait, that doesn't seem consistent. Let me think differently. Maybe the first number relates to ( m ) in some multiplicative way.Looking at ( m = 2 ): 3 is ( 2 times 1 + 1 )For ( m = 3 ): 7 is ( 3 times 2 + 1 )For ( m = 4 ): 13 is ( 4 times 3 + 1 )Ah, I see! The first number ( a_m ) is given by ( a_m = m times (m - 1) + 1 ). Let me verify this:- For ( m = 2 ): ( 2 times 1 + 1 = 3 ) ✔️- For ( m = 3 ): ( 3 times 2 + 1 = 7 ) ✔️- For ( m = 4 ): ( 4 times 3 + 1 = 13 ) ✔️Great, that formula works. So, the first number in the split for ( m^3 ) is ( m(m - 1) + 1 ).Now, since each split consists of ( m ) consecutive odd numbers, the sequence will be:( a_m, a_m + 2, a_m + 4, ldots, a_m + 2(m - 1) )So, the numbers are ( a_m + 2k ) where ( k = 0, 1, 2, ldots, m - 1 ).Given that one of these numbers is 333, I need to find ( m ) such that 333 is in the sequence for ( m^3 ).Let me denote the position of 333 in the sequence as ( k ). So,( 333 = a_m + 2k )But ( a_m = m(m - 1) + 1 ), so substituting:( 333 = m(m - 1) + 1 + 2k )Simplify:( 333 = m^2 - m + 1 + 2k )Rearranged:( m^2 - m + 1 + 2k = 333 )Which simplifies to:( m^2 - m + 2k = 332 )Since ( k ) is an integer between 0 and ( m - 1 ), inclusive, I can express ( k ) as:( k = frac{332 - m^2 + m}{2} )But ( k ) must be an integer, so ( 332 - m^2 + m ) must be even. Therefore, ( m^2 - m ) must be even because 332 is even.Looking at ( m^2 - m = m(m - 1) ). Since ( m ) and ( m - 1 ) are consecutive integers, one of them is even, so their product is always even. Therefore, ( k ) will always be an integer, which is good.Now, since ( k ) must satisfy ( 0 leq k leq m - 1 ), let's write that:( 0 leq frac{332 - m^2 + m}{2} leq m - 1 )Multiply all parts by 2:( 0 leq 332 - m^2 + m leq 2(m - 1) )Simplify the right inequality:( 332 - m^2 + m leq 2m - 2 )Bring all terms to the left:( 332 - m^2 + m - 2m + 2 leq 0 )Simplify:( 334 - m^2 - m leq 0 )Which is:( -m^2 - m + 334 leq 0 )Multiply both sides by -1 (remember to reverse the inequality):( m^2 + m - 334 geq 0 )Now, solve the quadratic inequality ( m^2 + m - 334 geq 0 ).First, find the roots of the equation ( m^2 + m - 334 = 0 ).Using the quadratic formula:( m = frac{-1 pm sqrt{1 + 4 times 334}}{2} = frac{-1 pm sqrt{1 + 1336}}{2} = frac{-1 pm sqrt{1337}}{2} )Calculate ( sqrt{1337} ). Let's see, ( 36^2 = 1296 ) and ( 37^2 = 1369 ). So, ( sqrt{1337} ) is between 36 and 37. Let me compute 36.5^2 = 1332.25, which is less than 1337. 36.6^2 = 1339.56, which is more than 1337. So, approximately 36.55.Thus, the positive root is approximately ( frac{-1 + 36.55}{2} = frac{35.55}{2} ≈ 17.775 ).Since ( m ) must be a natural number greater than 1, the inequality ( m^2 + m - 334 geq 0 ) holds when ( m geq 18 ) because 17.775 is approximately 17.78, so the next integer is 18.Now, let's check the left inequality:( 0 leq 332 - m^2 + m )Which is:( m^2 - m - 332 leq 0 )Solve ( m^2 - m - 332 = 0 ).Using the quadratic formula:( m = frac{1 pm sqrt{1 + 4 times 332}}{2} = frac{1 pm sqrt{1 + 1328}}{2} = frac{1 pm sqrt{1329}}{2} )Calculate ( sqrt{1329} ). ( 36^2 = 1296 ), ( 37^2 = 1369 ). So, between 36 and 37. 36.5^2 = 1332.25, which is more than 1329. Let's try 36.4^2 = 1324.96, 36.45^2 ≈ 1329. So, approximately 36.45.Thus, the positive root is approximately ( frac{1 + 36.45}{2} ≈ 18.725 ).So, the inequality ( m^2 - m - 332 leq 0 ) holds when ( m leq 18.725 ). Since ( m ) is an integer, ( m leq 18 ).Putting both inequalities together:From the right inequality: ( m geq 18 )From the left inequality: ( m leq 18 )Therefore, ( m = 18 ).Let me verify this. For ( m = 18 ):First number ( a_{18} = 18 times 17 + 1 = 306 + 1 = 307 ).The split numbers are 307, 309, 311, ..., up to 307 + 2*(18 - 1) = 307 + 34 = 341.So, the numbers are from 307 to 341, stepping by 2.Now, is 333 in this sequence?Let's see: 307, 309, 311, ..., 333, ..., 341.Yes, 333 is in this sequence. Let's find its position.Compute how many steps from 307 to 333:333 - 307 = 26. Since each step is 2, the number of steps is 26 / 2 = 13. So, 333 is the 14th term (since we start counting from 0).Since ( m = 18 ), there are 18 terms, so 14 is within the range (1 to 18). Therefore, 333 is indeed one of the split numbers for ( 18^3 ).Let me check if any other options could work, just to be thorough.Option A: ( m = 16 )First number ( a_{16} = 16 times 15 + 1 = 240 + 1 = 241 )The last number is 241 + 2*(16 - 1) = 241 + 30 = 271333 is greater than 271, so it's not in this sequence.Option B: ( m = 17 )First number ( a_{17} = 17 times 16 + 1 = 272 + 1 = 273 )Last number: 273 + 2*(17 - 1) = 273 + 32 = 305333 is greater than 305, so not in this sequence.Option D: ( m = 19 )First number ( a_{19} = 19 times 18 + 1 = 342 + 1 = 343 )Last number: 343 + 2*(19 - 1) = 343 + 36 = 379333 is less than 343, so it's not in this sequence.Therefore, only ( m = 18 ) works.**Final Answer**boxed{C}

💡Alright, so I have this problem where I need to figure out how many ways I can arrange the numbers 1 to 9 in a 3x3 grid. The condition is that no matter which three numbers I pick, none of which are in the same row or column, their sum should always be 15. Hmm, okay, that sounds a bit tricky, but let me try to break it down.First, I need to understand what it means for three numbers not to be in the same row or column. So, if I pick one number from each row and each column, their sum should always be 15. That reminds me of something called a magic square. I think a magic square is a grid where the sums of numbers in each row, each column, and both main diagonals are the same. Is that right?Let me recall. A 3x3 magic square uses the numbers 1 to 9, and each row, column, and diagonal adds up to 15. So, if I can arrange the numbers in a magic square, then any three numbers picked such that they are not in the same row or column would automatically sum to 15 because they would form a diagonal or something similar. Wait, is that true?Actually, in a magic square, the main diagonals sum to 15, but if I pick three numbers that aren't necessarily on a diagonal, would they still sum to 15? For example, if I pick the top-left, middle-center, and bottom-right, that's a diagonal and sums to 15. But what if I pick top-left, middle-right, and bottom-center? Is that also 15? Let me check.In a standard magic square, the numbers are arranged so that every row, column, and diagonal adds up to 15. So, if I pick one number from each row and each column, it's essentially forming a transversal of the square. In a magic square, all transversals sum to the magic constant, which is 15 in this case. So, yes, any three numbers picked such that they are not in the same row or column will sum to 15.Therefore, the problem reduces to finding the number of 3x3 magic squares that can be formed using the numbers 1 to 9. I remember that there is essentially only one 3x3 magic square, up to rotations and reflections. But how many distinct magic squares are there when considering all possible symmetries?Let me think. A square can be rotated in four different ways: 0°, 90°, 180°, and 270°. It can also be reflected over its vertical, horizontal, and two diagonal axes. So, in total, there are 8 symmetries (4 rotations and 4 reflections). Does that mean there are 8 distinct magic squares?Wait, but the magic square is fixed once you choose the center number. The center number in a 3x3 magic square must be 5 because it's the median of the numbers 1 to 9, and it's part of both diagonals, which must sum to 15. So, if the center is fixed as 5, then the rest of the numbers are determined based on the pairs that add up to 10 (since 15 - 5 = 10). The pairs are (1,9), (2,8), (3,7), and (4,6).These pairs need to be placed symmetrically around the center. Each pair can be placed in two different ways: for example, 1 can be in the top-left and 9 in the bottom-right, or vice versa. Since there are four pairs, each with two possible arrangements, that gives 2^4 = 16 possible arrangements. But wait, that seems too high because I thought there were only 8 symmetries.Maybe I'm overcounting. Let me consider the symmetries again. If I fix the center as 5, then the four pairs can be arranged in different ways, but some arrangements are equivalent under rotation or reflection. So, perhaps the number of distinct magic squares is 8, considering all the symmetries.But I'm not entirely sure. Let me look it up in my mind. I recall that the number of essentially different magic squares of order 3 is 1, but when considering all symmetries, there are 8 distinct magic squares. So, if we count all rotations and reflections as distinct, there are 8 magic squares.But wait, the problem doesn't specify whether rotations or reflections are considered different. It just asks for the number of such tables. So, if we consider all possible arrangements, including rotations and reflections, then there are 8 distinct magic squares.However, I might be missing something. Let me think again. The magic square is determined by the placement of the numbers 1 to 9 such that each row, column, and diagonal sums to 15. The standard magic square is:4 9 23 5 78 1 6But this can be rotated and reflected to get different arrangements. Each rotation (0°, 90°, 180°, 270°) and reflection (over vertical, horizontal, and two diagonals) gives a different magic square. So, that's 4 rotations and 4 reflections, totaling 8 distinct magic squares.Therefore, the number of such tables is 8.Wait, but I'm not sure if all these symmetries result in different magic squares. For example, reflecting the magic square over a vertical axis might give a different arrangement, but is it considered distinct? If the problem counts them as different, then yes, it's 8. If not, it's just 1.But the problem says "how many such tables are there," without specifying whether rotations or reflections are considered the same. So, I think we should count all possible distinct arrangements, considering rotations and reflections as different. Therefore, the answer should be 8.But I'm still a bit confused because I remember hearing that there are 8 magic squares of order 3 when considering all symmetries. So, I think that's the answer.Wait, let me double-check. The standard magic square has 8 symmetries: 4 rotations and 4 reflections. Each symmetry operation produces a distinct magic square. Therefore, there are 8 distinct magic squares.Yes, that makes sense. So, the number of such tables is 8.But hold on, I think I might have made a mistake. I remember that the number of essentially different magic squares of order 3 is 1, but when considering all symmetries, it's 8. So, if we count all symmetries as distinct, then it's 8. If we consider them the same, it's 1. But the problem doesn't specify, so I think we should assume that all symmetries are considered distinct.Therefore, the number of such tables is 8.Wait, but I'm not entirely sure. Let me think of another approach. Maybe the number is higher because there are more ways to arrange the numbers without considering symmetries.No, I think the key point is that the magic square is unique up to symmetries. So, if we count all symmetries, it's 8. If we don't, it's 1. Since the problem doesn't specify, I think the answer is 8.But I'm still a bit unsure. Maybe I should look for another way to calculate it.Let me consider the number of ways to arrange the numbers 1 to 9 in the grid such that every row, column, and diagonal sums to 15. The center number must be 5, as I thought earlier. Then, the pairs (1,9), (2,8), (3,7), and (4,6) must be placed symmetrically around the center.Each pair can be placed in two different ways: for example, 1 can be in the top-left and 9 in the bottom-right, or vice versa. Since there are four pairs, each with two possible arrangements, that gives 2^4 = 16 possible arrangements. However, some of these arrangements are equivalent under rotation or reflection.Specifically, rotating the square 90°, 180°, or 270°, or reflecting it over its axes, can produce equivalent arrangements. So, to find the number of distinct magic squares, we need to divide by the number of symmetries.The number of symmetries in a square is 8 (4 rotations and 4 reflections). Therefore, the number of distinct magic squares is 16 / 8 = 2. Wait, that doesn't make sense because I thought it was 8.Hmm, maybe I'm overcomplicating it. Let me try a different approach.Since the center is fixed as 5, we can consider the four pairs (1,9), (2,8), (3,7), and (4,6). Each pair must be placed in opposite corners or opposite edges.For each pair, there are two choices: which number goes where. So, for four pairs, that's 2^4 = 16 possible arrangements.However, some of these arrangements are equivalent under rotation or reflection. Specifically, rotating the square 90°, 180°, or 270°, or reflecting it over its axes, can produce equivalent arrangements.The number of distinct arrangements is therefore 16 divided by the number of symmetries, which is 8. So, 16 / 8 = 2. But that contradicts what I thought earlier.Wait, maybe I'm not accounting for something. Let me think again.If I fix the center as 5, and then consider the four pairs, each pair has two possible arrangements. So, 2^4 = 16.But these 16 arrangements include all possible rotations and reflections. So, if we consider all symmetries as distinct, then it's 16. But if we consider them the same, it's 1.But I think the correct answer is 8 because the standard magic square has 8 symmetries, and each symmetry gives a distinct magic square.Wait, I'm getting confused. Let me try to visualize.If I have the standard magic square:4 9 23 5 78 1 6Then, rotating it 90° gives:8 3 41 5 96 7 2Which is a different arrangement. Similarly, reflecting it over the vertical axis gives:2 9 47 5 36 1 8Which is another distinct arrangement.So, each of these 8 symmetries gives a different magic square. Therefore, there are 8 distinct magic squares.But earlier, I thought that considering the pairs, it's 16 arrangements, but that's if we don't fix the center. Wait, no, the center is fixed as 5.So, if the center is fixed, and the pairs are arranged around it, then the number of distinct magic squares is 8.Therefore, the number of such tables is 8.But I'm still not entirely confident. Let me see if I can find another way to think about it.Each magic square can be transformed into another by rotating or reflecting it. The group of symmetries of a square is called the dihedral group of order 8, which includes 4 rotations and 4 reflections.Therefore, the number of distinct magic squares, considering all symmetries, is 8.So, I think the answer is 8.But wait, I just found out that the number of essentially different magic squares of order 3 is 1, but when considering all symmetries, it's 8. So, if we count all symmetries as distinct, then it's 8.Therefore, the number of such tables is 8.But I'm still a bit unsure because I remember hearing that the number is 72. Wait, that doesn't make sense because 72 is much larger.Wait, maybe I'm confusing the number of Latin squares or something else. No, a magic square is more restrictive than a Latin square.Wait, let me think again. The number of 3x3 magic squares using numbers 1 to 9 is 8, considering all symmetries. So, the answer should be 8.But I'm not entirely sure. Maybe I should look it up in my mind. I think the standard answer is 8.But wait, I think I might have made a mistake earlier. Let me consider the number of ways to arrange the pairs.If I fix the center as 5, then the four pairs (1,9), (2,8), (3,7), and (4,6) need to be placed in the four corners and the four edges.Each pair can be placed in two ways: for example, 1 can be in the top-left and 9 in the bottom-right, or vice versa.So, for four pairs, that's 2^4 = 16 possible arrangements.However, some of these arrangements are equivalent under rotation or reflection.Specifically, rotating the square 90°, 180°, or 270°, or reflecting it over its axes, can produce equivalent arrangements.The number of distinct arrangements is therefore 16 divided by the number of symmetries, which is 8. So, 16 / 8 = 2.Wait, that suggests there are only 2 distinct magic squares, which contradicts what I thought earlier.Hmm, I'm getting conflicting results. Maybe I need to think differently.Let me consider that once the center is fixed as 5, the four pairs must be placed in the four corners and the four edges. Each pair has two possible arrangements.However, the placement of one pair affects the placement of the others because of the overlapping rows and columns.For example, if I place 1 in the top-left corner, then 9 must be in the bottom-right corner. Then, the pair (2,8) can be placed in the top-right and bottom-left corners, or vice versa.Similarly, the pair (3,7) can be placed in the middle-top and middle-bottom, or vice versa, and the pair (4,6) can be placed in the middle-left and middle-right, or vice versa.So, for each pair, there are two choices, leading to 2^4 = 16 possible arrangements.However, some of these arrangements are equivalent under rotation or reflection.Specifically, rotating the square 90°, 180°, or 270°, or reflecting it over its axes, can produce equivalent arrangements.The number of distinct arrangements is therefore 16 divided by the number of symmetries, which is 8. So, 16 / 8 = 2.But that suggests there are only 2 distinct magic squares, which doesn't seem right.Wait, maybe I'm not accounting for something. Let me think again.If I fix the center as 5, and then place the pairs, each pair has two choices. So, 2^4 = 16.But these 16 arrangements include all possible rotations and reflections. So, if we consider all symmetries as distinct, then it's 16.But if we consider them the same, it's 1.But I think the correct answer is 8 because the standard magic square has 8 symmetries, and each symmetry gives a distinct magic square.Therefore, the number of such tables is 8.But I'm still not entirely confident. Maybe I should think of it as the number of Latin squares or something else, but no, a magic square is more restrictive.Wait, I think I found the mistake. The number of essentially different magic squares of order 3 is 1, but when considering all symmetries, it's 8. So, if we count all symmetries as distinct, then it's 8.Therefore, the number of such tables is 8.But I'm still a bit unsure. Let me try to visualize.If I have the standard magic square:4 9 23 5 78 1 6Then, rotating it 90° gives:8 3 41 5 96 7 2Which is a different arrangement. Similarly, reflecting it over the vertical axis gives:2 9 47 5 36 1 8Which is another distinct arrangement.So, each of these 8 symmetries gives a different magic square. Therefore, there are 8 distinct magic squares.Therefore, the number of such tables is 8.But wait, I think I might have made a mistake earlier when I thought it was 16. I think the correct answer is 8.So, after considering all this, I think the number of such tables is 8.

💡First, determine the number of students who wore colored lipstick. Since half of the 200 students attended class wore colored lipstick:200 / 2 = 100 students.Next, calculate the number of students who wore red lipstick. One quarter of the students who wore lipstick wore red lipstick:100 * 1/4 = 25 students.Finally, find out how many students wore blue lipstick. There were one-fifth as many students wearing blue lipstick as those wearing red lipstick:25 * 1/5 = 5 students.Thus, 5 students wore blue lipstick.

💡First, I'll define variables for the number of each type of item Jane purchased. Let ( x ) be the number of 20-cent items, ( y ) be the number of 1.50 items, and ( z ) be the number of 2.50 items.From the problem, I know two main things:1. The total number of items is 50, so ( x + y + z = 50 ).2. The total cost is 50.00, which is equivalent to 5000 cents. This gives the equation ( 20x + 150y + 250z = 5000 ).To simplify the cost equation, I'll divide everything by 10:[2x + 15y + 25z = 500]Next, I'll substitute ( x = 50 - y - z ) from the first equation into the simplified cost equation:[2(50 - y - z) + 15y + 25z = 500 100 - 2y - 2z + 15y + 25z = 500 13y + 23z = 400]Now, I need to find integer values of ( y ) and ( z ) that satisfy ( 13y + 23z = 400 ) and ensure that ( x, y, z ) are non-negative integers.Testing possible values:- When ( y = 3 ): ( z = frac{400 - 39}{23} = frac{361}{23} = 15.7 ), not an integer.- When ( y = 4 ): ( z = frac{400 - 52}{23} = frac{348}{23} = 15 ), an integer.So, if ( y = 4 ) and ( z = 15 ), then:[x = 50 - 4 - 15 = 31]Therefore, Jane bought 31 items priced at 20 cents.