answer：Alright, so I have this geometry problem here with two tangent circles, O₁ and O₂, touching at point A. There's a line l passing through A that intersects O₁ at C₁ and O₂ at C₂. Then there's another circle O passing through C₁ and C₂, which intersects O₁ again at B₁ and O₂ again at B₂. Circle n is the circumcircle of triangle AB₁B₂, and circle k, which is tangent to circle n at A, intersects O₁ and O₂ at D₁ and D₂ respectively. The problem has two parts: first, I need to prove that points C₁, C₂, D₁, and D₂ are either concyclic or collinear. Second, I have to show that B₁, B₂, D₁, and D₂ are concyclic if and only if AC₁ and AC₂ are the diameters of circles O₁ and O₂.Okay, let me start by drawing a rough sketch of the figure to visualize everything. I have two tangent circles at A, line l through A cutting both circles at C₁ and C₂. Then circle O goes through C₁ and C₂ and intersects the original circles again at B₁ and B₂. Then circle n is the circumcircle of AB₁B₂, and circle k is tangent to n at A and cuts O₁ and O₂ again at D₁ and D₂.For part (1), I need to show that C₁, C₂, D₁, D₂ lie on a circle or are on a straight line. Hmm, maybe I can use some properties of cyclic quadrilaterals or radical axes here. Since all these points lie on circles that are related through tangents and intersections, perhaps inversion could be a useful tool. Inversion often simplifies problems involving tangent circles and radical axes.Let me recall that inversion preserves circles and lines, and tangency. If I invert the figure with respect to point A, which is the point of tangency of O₁ and O₂, then O₁ and O₂, being tangent at A, will invert to two parallel lines. That's because when you invert a circle passing through the center of inversion, it becomes a line. Since O₁ and O₂ are tangent at A, their images after inversion should be parallel lines.So, let's perform an inversion with center at A. Let me denote the inversion as inv(A, r), where r is some radius. The exact radius might not matter, but it's good to keep in mind.After inversion, O₁ becomes a line l₁, and O₂ becomes another line l₂, both parallel since O₁ and O₂ were tangent at A. The line l, which passes through A, will invert to itself because it passes through the center of inversion. So, points C₁ and C₂, which are on l, will invert to points C₁' and C₂' on l.Circle O passes through C₁ and C₂, so its image under inversion, let's call it O', will pass through C₁' and C₂'. Since circle O intersects O₁ again at B₁ and O₂ again at B₂, their images B₁' and B₂' will lie on O' as well. So, O' is the image of circle O and passes through C₁', C₂', B₁', and B₂'.Circle n is the circumcircle of triangle AB₁B₂. After inversion, point A remains the same (since it's the center), and points B₁ and B₂ invert to B₁' and B₂'. So, the image of circle n, let's call it n', is the circumcircle of triangle AB₁'B₂'. Circle k is tangent to circle n at A and intersects O₁ and O₂ at D₁ and D₂. After inversion, circle k becomes a circle k' tangent to n' at A and intersecting l₁ and l₂ at D₁' and D₂'.Now, since l₁ and l₂ are parallel lines, and k' intersects them at D₁' and D₂', the line D₁'D₂' must be a transversal cutting the two parallel lines. In projective geometry, if two lines are parallel, any transversal will create equal corresponding angles. Also, since k' is tangent to n' at A, the tangent condition implies that k' and n' share a common tangent at A, meaning their centers lie along the same line through A. Given that O' passes through C₁', C₂', B₁', and B₂', and n' is the circumcircle of AB₁'B₂', there might be some radical axis properties or power of a point that can be applied here.Wait, perhaps I should think about the radical axis of circles O' and n'. The radical axis is the set of points with equal power concerning both circles. Since both circles pass through A (as n' is the circumcircle of AB₁'B₂' and O' passes through B₁' and B₂'), the radical axis would be the line through A and the other intersection point of O' and n'. But since n' is the circumcircle of AB₁'B₂', and O' passes through B₁' and B₂', their radical axis is line AB₁'B₂', which is the same as the radical axis of O and n before inversion.But I'm not sure if this is directly helpful. Maybe instead, I should consider the power of point D₁' with respect to O'. Since D₁' lies on l₁, which is the image of O₁, and O₁ is tangent to O₂ at A, perhaps there's some power relation here.Alternatively, since O' passes through C₁', C₂', B₁', and B₂', and k' passes through D₁' and D₂', and is tangent to n' at A, maybe I can find some cyclic quadrilateral properties.Wait, let me think about the original points before inversion. If I can show that the power of D₁ and D₂ with respect to circle O is equal, then they would lie on the radical axis of O and another circle, which might be the circle passing through C₁, C₂, D₁, D₂.But I'm getting a bit tangled here. Maybe another approach: since circle k is tangent to circle n at A, and n is the circumcircle of AB₁B₂, perhaps there's some homothety or tangent properties that can link D₁ and D₂ to C₁ and C₂.Alternatively, maybe using the power of point A with respect to circle O. Since A lies on both O₁ and O₂, and line l passes through A, the power of A with respect to O is AC₁ * AC₂. But since A is also on circle n, which is tangent to k at A, the power of A with respect to k is zero, meaning AD₁ * AD₂ = 0, but that doesn't make sense because D₁ and D₂ are distinct points. Wait, no, because k is tangent to n at A, so the power of A with respect to k is equal to the square of the tangent length from A to k, which is zero since A is on k.Hmm, perhaps I need to consider the radical axis of circles O and k. The radical axis is the set of points with equal power concerning both circles. If I can show that points C₁, C₂, D₁, D₂ lie on this radical axis, then they would be concyclic or collinear.Alternatively, maybe using the fact that points C₁, C₂, B₁, B₂ lie on circle O, and points D₁, D₂ lie on circle k, which is tangent to n at A. Maybe there's some angle chasing that can be done here.Let me try angle chasing. Since A is the point of tangency of O₁ and O₂, the tangent at A is common to both circles. So, the tangent line at A is the same for both O₁ and O₂. Now, circle O passes through C₁ and C₂, and intersects O₁ again at B₁ and O₂ again at B₂. So, angles at B₁ and B₂ might have some relation to angles at C₁ and C₂.Also, circle n is the circumcircle of AB₁B₂, so angles subtended by AB₁ and AB₂ at points on n should be equal. Circle k is tangent to n at A, so the tangent at A for k is the same as the tangent at A for n. This implies that the centers of k and n lie along the same line through A, which is the tangent line.Since k intersects O₁ at D₁ and O₂ at D₂, maybe there's a relation between angles at D₁ and D₂ and angles at C₁ and C₂.Wait, perhaps using power of a point. For point D₁, which lies on O₁ and k, the power with respect to O should be equal to the power with respect to k. Similarly for D₂.But I'm not sure. Maybe I need to consider the power of point D₁ with respect to circle O. The power would be D₁B₁ * D₁C₁ = D₁D₂ * D₁ something. Hmm, not sure.Alternatively, since circle O passes through C₁, C₂, B₁, B₂, and circle k passes through D₁, D₂ and is tangent to n at A, maybe there's a radical axis that includes C₁, C₂, D₁, D₂.Wait, if I can show that the power of D₁ and D₂ with respect to circle O is equal, then they would lie on the radical axis of O and another circle, which could be the circle passing through C₁, C₂, D₁, D₂.But I'm not making much progress here. Maybe I should try to use inversion more effectively.Since after inversion, O₁ and O₂ become parallel lines l₁ and l₂, and circle O becomes a circle passing through C₁', C₂', B₁', B₂'. Circle n becomes the circumcircle of AB₁'B₂', and circle k becomes a circle tangent to n' at A and intersecting l₁ and l₂ at D₁' and D₂'.In this inverted figure, since l₁ and l₂ are parallel, and k' intersects them at D₁' and D₂', the line D₁'D₂' is a transversal. Also, since k' is tangent to n' at A, which is a point on both k' and n', the tangent condition implies that the centers of k' and n' lie along the same line through A.Now, in the inverted figure, points C₁', C₂', B₁', B₂' lie on circle O', and points D₁', D₂' lie on circle k'. Since O' and k' are both circles in the inverted plane, their radical axis would be the set of points with equal power concerning both circles.If I can show that points C₁', C₂', D₁', D₂' lie on the same circle or line, then their pre-images C₁, C₂, D₁, D₂ would lie on a circle or line as well.Alternatively, since l₁ and l₂ are parallel, and D₁' and D₂' lie on them, the line D₁'D₂' is a transversal. If I can show that C₁', C₂', D₁', D₂' lie on a circle, then their pre-images would be concyclic or collinear.Wait, maybe in the inverted figure, since O' passes through C₁', C₂', B₁', B₂', and k' passes through D₁', D₂', and is tangent to n' at A, there might be some cyclic quadrilateral properties or similar triangles that can be used.Alternatively, since l₁ and l₂ are parallel, the angles formed by the transversal D₁'D₂' with them are equal. Maybe this can help in establishing some angle relations that translate back to the original figure.I'm still not entirely sure, but perhaps the key is to recognize that after inversion, the problem reduces to showing that four points lie on a circle or line, which might be more straightforward in the inverted figure.For part (2), I need to show that B₁, B₂, D₁, D₂ are concyclic if and only if AC₁ and AC₂ are diameters of O₁ and O₂.Hmm, so this is an if and only if statement, meaning I need to prove both directions: 1. If AC₁ and AC₂ are diameters, then B₁, B₂, D₁, D₂ are concyclic.2. If B₁, B₂, D₁, D₂ are concyclic, then AC₁ and AC₂ are diameters.Let me start with the first direction. Suppose AC₁ and AC₂ are diameters of O₁ and O₂. Then, since AC₁ is a diameter of O₁, angle AB₁C₁ is a right angle because any angle inscribed in a semicircle is a right angle. Similarly, angle AB₂C₂ is a right angle.So, in circle O, points B₁ and B₂ are such that angles AB₁C₁ and AB₂C₂ are right angles. This might imply some properties about circle O and its relation to circle n.Since circle n is the circumcircle of AB₁B₂, and angles at B₁ and B₂ are right angles, maybe circle n has some special properties, like being orthogonal to O₁ and O₂.Now, circle k is tangent to n at A and intersects O₁ and O₂ at D₁ and D₂. If AC₁ and AC₂ are diameters, then perhaps D₁ and D₂ have some symmetric properties with respect to O₁ and O₂.Alternatively, since angles at B₁ and B₂ are right angles, maybe the circle passing through B₁, B₂, D₁, D₂ has some relation to the right angles, making it a circle.For the converse, suppose that B₁, B₂, D₁, D₂ are concyclic. Then, perhaps this implies that angles at B₁ and B₂ are right angles, forcing AC₁ and AC₂ to be diameters.But I'm not entirely sure about this. Maybe I need to use some power of a point or cyclic quadrilateral properties here.Alternatively, using inversion again, if B₁, B₂, D₁, D₂ are concyclic in the original figure, then their images B₁', B₂', D₁', D₂' are concyclic in the inverted figure. Since in the inverted figure, B₁', B₂' lie on O', and D₁', D₂' lie on k', which is tangent to n' at A, maybe this implies some orthogonality or right angles that translate back to AC₁ and AC₂ being diameters.I think I need to work more carefully on the inversion approach for both parts. Let me try to formalize the inversion steps.Let me perform an inversion with center at A and radius r. Let me denote the inversion as inv(A, r). Under this inversion:- Circle O₁, passing through A, inverts to a line l₁.- Circle O₂, passing through A, inverts to another line l₂.- Since O₁ and O₂ are tangent at A, l₁ and l₂ are parallel lines.- Line l, passing through A, inverts to itself.- Points C₁ and C₂ on l invert to points C₁' and C₂' on l.- Circle O, passing through C₁ and C₂, inverts to a circle O' passing through C₁' and C₂'.- Points B₁ and B₂, where O intersects O₁ and O₂ again, invert to points B₁' and B₂' on O'.- Circle n, the circumcircle of AB₁B₂, inverts to the circumcircle n' of AB₁'B₂'.- Circle k, tangent to n at A, inverts to a circle k' tangent to n' at A.- Points D₁ and D₂, where k intersects O₁ and O₂, invert to points D₁' and D₂' on l₁ and l₂.Now, in the inverted figure, l₁ and l₂ are parallel lines, and k' intersects them at D₁' and D₂'. Since k' is tangent to n' at A, the centers of k' and n' lie along the same line through A, which is the tangent line.Since O' passes through C₁', C₂', B₁', and B₂', and n' is the circumcircle of AB₁'B₂', there might be some radical axis properties or cyclic quadrilaterals here.For part (1), I need to show that C₁', C₂', D₁', D₂' are concyclic or collinear. Since l₁ and l₂ are parallel, and D₁' and D₂' lie on them, the line D₁'D₂' is a transversal. If I can show that C₁', C₂', D₁', D₂' lie on a circle, then their pre-images would be concyclic or collinear.Alternatively, since O' passes through C₁', C₂', B₁', B₂', and k' passes through D₁', D₂', maybe there's a radical axis that includes these points.Wait, the radical axis of O' and k' would be the set of points with equal power concerning both circles. If C₁' and C₂' lie on O', their power with respect to k' would be equal to the power of C₁' and C₂' with respect to k'. Similarly, D₁' and D₂' lie on k', so their power with respect to O' would be equal to the power of D₁' and D₂' with respect to O'.If I can show that the power of C₁' and C₂' with respect to k' is equal, then they would lie on the radical axis of O' and k', implying that C₁', C₂', D₁', D₂' lie on a circle or line.Alternatively, since l₁ and l₂ are parallel, and D₁' and D₂' lie on them, the angles formed by D₁'D₂' with l₁ and l₂ are equal. This might imply that angles subtended by D₁'D₂' at C₁' and C₂' are equal, leading to C₁', C₂', D₁', D₂' being concyclic.I think this is the right direction. So, in the inverted figure, since l₁ and l₂ are parallel, the angles at C₁' and C₂' subtended by D₁'D₂' are equal, implying that C₁', C₂', D₁', D₂' lie on a circle. Therefore, their pre-images C₁, C₂, D₁, D₂ lie on a circle or line, proving part (1).For part (2), I need to show that B₁, B₂, D₁, D₂ are concyclic if and only if AC₁ and AC₂ are diameters.Assuming AC₁ and AC₂ are diameters, then angles AB₁C₁ and AB₂C₂ are right angles. In the inverted figure, this would mean that angles AB₁'C₁' and AB₂'C₂' are right angles. Since O' passes through C₁', C₂', B₁', B₂', and angles at B₁' and B₂' are right angles, O' must be orthogonal to l₁ and l₂.Since k' is tangent to n' at A, and n' is the circumcircle of AB₁'B₂', which has right angles at B₁' and B₂', n' must be a circle with diameter AB₁' and AB₂', but since angles at B₁' and B₂' are right angles, n' is actually the circle with diameter AB₁' and AB₂', which are both right angles, so n' is orthogonal to l₁ and l₂.Therefore, k', being tangent to n' at A, must also be orthogonal to l₁ and l₂, implying that D₁' and D₂' are such that angles AD₁'D₂' are right angles. Translating back, this implies that D₁ and D₂ lie on circles orthogonal to O₁ and O₂, which would mean that AC₁ and AC₂ are diameters.Conversely, if B₁, B₂, D₁, D₂ are concyclic, then their images B₁', B₂', D₁', D₂' are concyclic in the inverted figure. Since B₁' and B₂' lie on O', and D₁' and D₂' lie on k', which is tangent to n' at A, this implies that angles at B₁' and B₂' are right angles, forcing AC₁ and AC₂ to be diameters.So, putting it all together, part (1) is proven by showing that after inversion, the points lie on a circle or line, and part (2) is proven by showing the equivalence between the concyclicity and the diameters condition.I think this makes sense, but I might have skipped some steps or made some assumptions that need to be verified. Maybe I should check if the inversion preserves the necessary properties and if the radical axis argument holds.Also, for part (2), I need to ensure that the converse is properly addressed, making sure that the concyclicity indeed implies the diameters condition and not just one direction.Overall, I believe the inversion approach is the key here, simplifying the problem by transforming the tangent circles into parallel lines, making the concyclicity or collinearity more apparent.

answer：Okay, so I have this problem here about random variables X and Y. Let me try to understand what it's asking. First, there's a table that gives the distribution of X. It shows that X can take the values 2, 3, and 6, with probabilities 1/2, a, and 1/6 respectively. So, the probabilities for X=2 is 1/2, for X=3 it's some unknown value 'a', and for X=6 it's 1/6. Then, there's another random variable Y defined as Y = 4X + 3. The question is asking for the variance of Y, denoted as D(Y). Alright, so I remember that variance has some properties, especially when dealing with linear transformations of random variables. Specifically, if Y is a linear function of X, like Y = aX + b, then the variance of Y is a squared times the variance of X. That is, D(Y) = a² * D(X). In this case, a is 4 and b is 3. So, D(Y) should be 4² * D(X), which is 16 * D(X). But wait, I don't know D(X) yet. So, I need to find the variance of X first. To find the variance of X, I need to know the expected value (mean) of X and then use the formula for variance, which is E[X²] - (E[X])².First, let's find the expected value of X, E[X]. The expected value is the sum of each possible value multiplied by its probability. So, E[X] = 2*(1/2) + 3*a + 6*(1/6). Let me calculate that step by step. 2*(1/2) is 1. 3*a is just 3a. 6*(1/6) is 1. So, E[X] = 1 + 3a + 1 = 2 + 3a. Now, I need to find E[X²], which is the expected value of X squared. That would be 2²*(1/2) + 3²*a + 6²*(1/6). Calculating each term:2²*(1/2) = 4*(1/2) = 2.3²*a = 9a.6²*(1/6) = 36*(1/6) = 6.So, E[X²] = 2 + 9a + 6 = 8 + 9a.Now, the variance D(X) is E[X²] - (E[X])². So, D(X) = (8 + 9a) - (2 + 3a)².Let me expand (2 + 3a)². That's 4 + 12a + 9a².So, D(X) = 8 + 9a - (4 + 12a + 9a²) = 8 + 9a - 4 - 12a - 9a² = (8 - 4) + (9a - 12a) - 9a² = 4 - 3a - 9a².Hmm, that seems a bit complicated. Maybe I made a mistake somewhere.Wait, let's double-check the calculations.E[X] = 2*(1/2) + 3*a + 6*(1/6) = 1 + 3a + 1 = 2 + 3a. That seems correct.E[X²] = 4*(1/2) + 9*a + 36*(1/6) = 2 + 9a + 6 = 8 + 9a. That also seems correct.So, D(X) = E[X²] - (E[X])² = (8 + 9a) - (2 + 3a)².Expanding (2 + 3a)²: 4 + 12a + 9a². So, D(X) = 8 + 9a - 4 - 12a - 9a² = 4 - 3a - 9a². Yeah, that's correct.But wait, variance can't be negative, right? So, 4 - 3a - 9a² must be positive. That gives us some constraint on 'a'.Also, since probabilities must sum to 1, the sum of the probabilities for X=2, X=3, and X=6 should be 1. So, 1/2 + a + 1/6 = 1.Let me solve for 'a':1/2 + a + 1/6 = 1Convert to common denominator, which is 6:3/6 + a + 1/6 = 1So, 4/6 + a = 1Simplify 4/6 to 2/3:2/3 + a = 1Therefore, a = 1 - 2/3 = 1/3.So, a is 1/3.Now, let's plug that back into D(X):D(X) = 4 - 3*(1/3) - 9*(1/3)²Calculate each term:3*(1/3) = 19*(1/3)² = 9*(1/9) = 1So, D(X) = 4 - 1 - 1 = 2.Okay, so the variance of X is 2.Now, going back to Y = 4X + 3.As I remembered earlier, the variance of Y is 4² * D(X) = 16 * D(X).So, D(Y) = 16 * 2 = 32.Therefore, the variance of Y is 32.I think that makes sense. Let me just recap to make sure I didn't skip any steps.1. Found the expected value E[X] = 2 + 3a.2. Found E[X²] = 8 + 9a.3. Calculated variance D(X) = 4 - 3a - 9a².4. Found 'a' by ensuring probabilities sum to 1: a = 1/3.5. Plugged 'a' back into D(X) to get D(X) = 2.6. Used the property of variance for linear transformations: D(Y) = 16 * D(X) = 32.Everything seems to check out. I think I'm confident that D(Y) is 32.

answer：Okay, so I have this problem where I need to find the points in common between two graphs on modulo 9 graph paper. The equations given are:y equiv 3x + 5 pmod{9}andy equiv 7x + 3 pmod{9}.I need to find the x-coordinates where these two graphs intersect and then sum them up. Hmm, okay, let me think about how to approach this.First, since both equations are congruent modulo 9, I can set them equal to each other to find the x-values where they intersect. That is, I can set:3x + 5 equiv 7x + 3 pmod{9}.Alright, so now I have this congruence. Let me try to solve for x. I'll subtract 3x from both sides to get:5 equiv 4x + 3 pmod{9}.Wait, that doesn't seem right. Let me double-check. If I subtract 3x from both sides, it should be:5 equiv 4x + 3 pmod{9}.Yes, that's correct. Now, I need to isolate the term with x. So, I'll subtract 3 from both sides:5 - 3 equiv 4x pmod{9}.Which simplifies to:2 equiv 4x pmod{9}.Okay, so now I have 4x ≡ 2 mod 9. I need to solve for x. To do that, I should find the multiplicative inverse of 4 modulo 9. The multiplicative inverse of a number a modulo m is a number b such that:a cdot b equiv 1 pmod{m}.So, I need to find a number b such that 4b ≡ 1 mod 9. Let me think. Trying small integers:4*1 = 4 ≡ 4 mod 94*2 = 8 ≡ 8 mod 94*3 = 12 ≡ 3 mod 94*4 = 16 ≡ 7 mod 94*5 = 20 ≡ 2 mod 94*6 = 24 ≡ 6 mod 94*7 = 28 ≡ 1 mod 9Oh, there it is! 4*7 = 28 ≡ 1 mod 9. So, the multiplicative inverse of 4 modulo 9 is 7.Great, so now I can multiply both sides of the congruence 4x ≡ 2 mod 9 by 7 to solve for x:7*4x ≡ 7*2 mod 9Which simplifies to:28x ≡ 14 mod 9But 28 mod 9 is 1, because 9*3=27, so 28-27=1. Similarly, 14 mod 9 is 5, because 9*1=9, so 14-9=5.So, 1x ≡ 5 mod 9, which simplifies to:x ≡ 5 mod 9So, the solution is x = 5 mod 9. Since we're working modulo 9, the possible x-values are from 0 to 8. Therefore, the only solution in this range is x = 5.Wait, but the problem says "points in common," which implies there might be more than one solution. Did I miss something? Let me check my steps again.Starting from:3x + 5 ≡ 7x + 3 mod 9Subtract 3x: 5 ≡ 4x + 3 mod 9Subtract 3: 2 ≡ 4x mod 9Inverse of 4 is 7, so multiply both sides: 14 ≡ 28x mod 9Which simplifies to 5 ≡ x mod 9, so x = 5.Hmm, seems like only one solution. Maybe because 4 and 9 are coprime, so the equation 4x ≡ 2 mod 9 has a unique solution modulo 9. So, only x = 5 is the solution.Therefore, the only x-coordinate where the two graphs intersect is 5. So, the sum of the x-coordinates is just 5.Wait, but let me confirm by plugging x = 5 back into both equations to make sure I didn't make a mistake.First equation: y ≡ 3x + 5 mod 9Plugging x = 5: y ≡ 3*5 + 5 = 15 + 5 = 20 ≡ 2 mod 9 (since 20 - 18 = 2)Second equation: y ≡ 7x + 3 mod 9Plugging x = 5: y ≡ 7*5 + 3 = 35 + 3 = 38 ≡ 2 mod 9 (since 38 - 36 = 2)Yes, both give y ≡ 2 mod 9, so (5, 2) is indeed a common point.Therefore, the sum of the x-coordinates is 5.I think that's it. Only one point of intersection, so the sum is 5.**Final Answer**The sum of the x-coordinates is boxed{5}.

answer：First, I need to determine the largest square tile size that can be used without cutting any tiles. This involves finding the greatest common divisor (GCD) of the room's length and breadth.I'll start by converting the measurements from meters and centimeters to centimeters for consistency:- Length: 50 meters 00 centimeters = 5000 centimeters- Breadth: 11 meters 25 centimeters = 1125 centimetersNext, I'll use the Euclidean algorithm to find the GCD of 5000 cm and 1125 cm:1. Divide 5000 by 1125, which gives a quotient of 4 and a remainder of 500.2. Divide 1125 by 500, resulting in a quotient of 2 and a remainder of 125.3. Finally, divide 500 by 125, which leaves a remainder of 0.The last non-zero remainder is 125, so the GCD is 125 cm. This means the largest square tile that can be used measures 125 cm by 125 cm.Now, I'll calculate the area of the room and the area of one tile:- Area of the room = Length × Breadth = 5000 cm × 1125 cm = 5,625,000 cm²- Area of one tile = Side × Side = 125 cm × 125 cm = 15,625 cm²To find the number of tiles required, I'll divide the total area of the room by the area of one tile:- Number of tiles = 5,625,000 cm² ÷ 15,625 cm² = 360Therefore, the least number of square tiles needed to pave the floor is 360.

answer：First, I need to determine the number of students expected to attend the game night after a 40% drop in attendance. This is calculated by reducing the original number of students by 40%.Next, I'll calculate the total number of cookies needed. Assuming each student eats an average of two cookies, I'll multiply the number of attending students by two.Finally, to find out how many full recipes Walter and Gretel need to prepare, I'll divide the total number of cookies required by the number of cookies each recipe produces. If the result is a whole number, that's the number of recipes needed. If not, I'll round up to ensure there are enough cookies.

answer：Okay, so I have this problem where I need to find all positive integers ( n ) such that the sum ( 1! + 2! + ldots + n! ) is a perfect square. Then, I have to find the sum of all such ( n ). Hmm, let me try to figure this out step by step.First, I know that a factorial of a number ( k ), written as ( k! ), is the product of all positive integers up to ( k ). So, ( 1! = 1 ), ( 2! = 2 ), ( 3! = 6 ), ( 4! = 24 ), and so on. The problem is asking for the sum of these factorials up to some ( n ), and this sum has to be a perfect square. A perfect square is a number that can be expressed as ( m^2 ) where ( m ) is an integer.Let me start by calculating the sum for small values of ( n ) and checking if the result is a perfect square.For ( n = 1 ):( 1! = 1 ). Is 1 a perfect square? Yes, because ( 1 = 1^2 ). So, ( n = 1 ) is a possible value.For ( n = 2 ):( 1! + 2! = 1 + 2 = 3 ). Is 3 a perfect square? No, because ( 1^2 = 1 ) and ( 2^2 = 4 ), so 3 is not a perfect square. So, ( n = 2 ) doesn't work.For ( n = 3 ):( 1! + 2! + 3! = 1 + 2 + 6 = 9 ). Is 9 a perfect square? Yes, because ( 3^2 = 9 ). So, ( n = 3 ) is another possible value.Okay, so far, ( n = 1 ) and ( n = 3 ) work. Let me check ( n = 4 ):( 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 ). Is 33 a perfect square? No, because ( 5^2 = 25 ) and ( 6^2 = 36 ), so 33 is not a perfect square.Hmm, maybe ( n = 4 ) doesn't work. Let me try ( n = 5 ):( 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 ). Is 153 a perfect square? Let me see, ( 12^2 = 144 ) and ( 13^2 = 169 ), so 153 is between them and not a perfect square.Hmm, not working. Let me try ( n = 6 ):( 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873 ). Is 873 a perfect square? ( 29^2 = 841 ) and ( 30^2 = 900 ), so 873 is not a perfect square.This is getting bigger quickly. Maybe I should look for a pattern or a mathematical reason why the sum can't be a perfect square beyond a certain point.I notice that starting from ( 4! ), the factorials end with a zero because they include the factor 5 and 2, which makes 10. So, ( 4! = 24 ), ( 5! = 120 ), ( 6! = 720 ), etc. So, adding these to the sum will add numbers ending with zero, which means the last digit of the sum will be determined by the sum of the factorials up to ( 3! ).Let me calculate the sum up to ( 3! ):( 1! + 2! + 3! = 1 + 2 + 6 = 9 ). So, the last digit is 9. Now, when I add ( 4! ) which is 24, the sum becomes 33, which ends with 3. Adding ( 5! = 120 ) gives 153, which ends with 3 again. Adding ( 6! = 720 ) gives 873, which still ends with 3. It seems that from ( n = 4 ) onwards, the sum ends with a 3.Wait, but perfect squares can end with certain digits only. Let me recall the possible last digits of perfect squares. They can end with 0, 1, 4, 5, 6, or 9. So, a perfect square cannot end with 2, 3, 7, or 8. Since our sum for ( n geq 4 ) ends with 3, which is not a possible last digit for a perfect square, this means that for ( n geq 4 ), the sum cannot be a perfect square.Therefore, the only possible values of ( n ) are 1 and 3. Adding these together, ( 1 + 3 = 4 ).Wait, let me double-check my reasoning. I concluded that for ( n geq 4 ), the sum ends with 3, which cannot be a perfect square. Is that correct? Let me verify.The sum up to ( 3! ) is 9. Adding ( 4! = 24 ) gives 33, which ends with 3. Adding ( 5! = 120 ) gives 153, which ends with 3. Adding ( 6! = 720 ) gives 873, which ends with 3. So, yes, every time we add a factorial from ( 4! ) onwards, we're adding a number ending with 0, so the last digit remains 3. Since 3 is not a possible last digit for a perfect square, the sum cannot be a perfect square for ( n geq 4 ).Therefore, the only possible values of ( n ) are 1 and 3, and their sum is 4.I think that's solid reasoning. I don't see any mistakes in my calculations or logic. So, I'm confident that the answer is 4.