answer：First, I need to determine the area of one of the congruent squares. Since there are 8 squares and the total area is 400 square centimeters, I divide 400 by 8 to find the area of one square.Next, I calculate the side length of each square by taking the square root of the area of one square.With the side length known, I consider the arrangement of the squares in a 2x4 rectangle. This arrangement means there are 4 squares along the length and 2 squares along the width.To find the perimeter, I calculate the total length contributed by the vertical and horizontal sides of the squares. The vertical sides contribute 5 times the side length, and the horizontal sides contribute 3 times the side length.Finally, I add the contributions from both vertical and horizontal sides to determine the total perimeter of the region.

answer：First, identify the known quantities:- Volume of the 60% acid solution: 4 liters- Concentration of the 60% solution: 60% (or 0.60)- Concentration of the 75% solution: 75% (or 0.75)- Desired concentration of the final solution: 72% (or 0.72)Let ( x ) represent the volume of the 75% acid solution needed.Calculate the amount of pure acid in each solution:- Pure acid from the 60% solution: ( 0.60 times 4 = 2.4 ) liters- Pure acid from the 75% solution: ( 0.75 times x ) litersThe total volume of the final solution will be ( 4 + x ) liters, and the total amount of pure acid in the final solution should be ( 0.72 times (4 + x) ) liters.Set up the equation based on the total pure acid:[ 2.4 + 0.75x = 0.72(4 + x) ]Solve for ( x ):[ 2.4 + 0.75x = 2.88 + 0.72x ][ 0.03x = 0.48 ][ x = frac{0.48}{0.03} ][ x = 16 ]Therefore, 16 liters of the 75% acid solution are needed. The total volume of the final solution is:[ 4 + 16 = 20 text{ liters} ]

answer：To find the sum of the first 1000 numbers in the set ( S ), which consists of all positive integers that leave a remainder of 5 when divided by 8, we can follow these steps:1. **Identify the General Form of the Numbers in Set ( S ):** - Any number in ( S ) can be expressed as ( 8n + 5 ), where ( n ) is a non-negative integer (i.e., ( n = 0, 1, 2, 3, ldots )).2. **Determine the First and 1000th Terms:** - The first term (( a_1 )) is when ( n = 0 ): ( 8(0) + 5 = 5 ). - The 1000th term (( a_{1000} )) is when ( n = 999 ) (since we start counting from ( n = 0 )): ( 8(999) + 5 = 7992 + 5 = 7997 ).3. **Calculate the Sum of the First 1000 Terms Using the Arithmetic Series Formula:** - The formula for the sum of the first ( k ) terms of an arithmetic series is: [ S_k = frac{k}{2} times (a_1 + a_k) ] - Substituting the known values: [ S_{1000} = frac{1000}{2} times (5 + 7997) = 500 times 8002 = 4,001,000 ]Therefore, the sum of the first 1000 numbers in the set ( S ) is **4,001,000**.

answer：Okay, so I have this problem about a convex quadrilateral ABCD. The angles at vertices A and C are not acute, which means they are either right angles or obtuse. Points K, L, M, and N are marked on the sides AB, BC, CD, and DA respectively. I need to prove that the perimeter of quadrilateral KLMN is not less than twice the length of the diagonal AC.Hmm, let me start by drawing a rough sketch of quadrilateral ABCD. Since it's convex, all its interior angles are less than 180 degrees. Angles at A and C are not acute, so they are 90 degrees or more. Points K, L, M, N are somewhere on the sides AB, BC, CD, DA. So KLMN is a quadrilateral inside ABCD.I need to relate the perimeter of KLMN to the diagonal AC. Maybe I can use some properties of convex quadrilaterals or triangles. Since the angles at A and C are not acute, perhaps I can use some inequalities related to triangle sides or something like that.Wait, maybe I can use the triangle inequality. If I can express the sides of KLMN in terms of segments related to AC, then I can sum them up and get the perimeter.Let me think about the diagonal AC. It connects vertices A and C. Since K is on AB and N is on DA, maybe I can consider triangles AKN and CML or something like that.Alternatively, perhaps I can use vectors or coordinate geometry. But that might complicate things. Maybe a better approach is to use reflections or midpoints.Wait, the problem mentions the perimeter of KLMN. Perimeter is the sum of all sides. So, KL + LM + MN + NK. I need to show that this sum is at least 2*AC.Hmm, 2*AC. That makes me think of something like the shortest path or maybe the sum of two segments each equal to AC.Wait, maybe I can use the fact that in any quadrilateral, the sum of the lengths of any three sides is greater than the fourth side. But I don't know if that applies here.Alternatively, since angles at A and C are not acute, maybe I can use some properties related to the law of cosines. If angles are 90 degrees or more, the cosine is non-positive, which might help in some inequality.Let me try to break it down. Let's consider triangles ABK and CDM. Wait, not sure.Wait, maybe I can use the fact that in a convex quadrilateral, the sum of the lengths of the diagonals is less than the perimeter. But I'm not sure.Alternatively, maybe I can use the fact that the perimeter of KLMN can be related to the perimeters of ABCD or something else.Wait, another idea: maybe I can use the concept of midlines in triangles. If I connect midpoints, the midline is parallel to the third side and half its length.But in this case, K, L, M, N are arbitrary points on the sides, not necessarily midpoints. So maybe that's not directly applicable.Wait, but maybe I can use some inequality related to midlines. For example, the length of a midline is at least half the length of the third side, but I'm not sure.Alternatively, maybe I can use the fact that the perimeter of KLMN is minimized when K, L, M, N are midpoints, but I don't know if that's true.Wait, actually, if K, L, M, N are midpoints, then KLMN is the Varignon parallelogram, and its perimeter is equal to the sum of the lengths of the diagonals of ABCD. But in this case, we need to relate it to AC.Hmm, but the problem doesn't specify that KLMN is a parallelogram, just any quadrilateral with points on the sides.Wait, maybe I can use some reflection technique. If I reflect points across AC, maybe I can create a path that relates to the perimeter.Alternatively, maybe I can use the fact that in a convex quadrilateral, the sum of two opposite sides is greater than the other two. But not sure.Wait, another idea: Since angles at A and C are not acute, maybe the projections of certain sides onto AC can be bounded.Wait, let me think about the perimeter of KLMN. It's KL + LM + MN + NK. Maybe I can express each of these segments in terms of projections onto AC.Alternatively, maybe I can use the fact that the length of a path is minimized when it's a straight line. So, if I can relate the perimeter to some path that goes from A to C and back, maybe I can get the inequality.Wait, here's an idea: If I consider the path from K to L to M to N and back to K, maybe I can relate this to going from A to C and back via some points.Wait, perhaps I can use the triangle inequality multiple times. For example, KL + LM >= KM, and MN + NK >= MK, but not sure if that helps.Wait, maybe I can consider the sum KL + LM + MN + NK and try to relate it to AC.Alternatively, maybe I can use the fact that in any quadrilateral, the perimeter is at least twice the length of the longest diagonal. But I don't know if that's a standard result.Wait, actually, I recall that in a convex quadrilateral, the perimeter is greater than the sum of the diagonals. But here we need to compare the perimeter of KLMN to 2*AC.Hmm, maybe I can use some inequality that relates the perimeter of a quadrilateral inside another quadrilateral to the diagonals.Wait, another approach: Maybe I can use the fact that the perimeter of KLMN is at least the sum of the projections of its sides onto AC.Since AC is a diagonal, and angles at A and C are not acute, perhaps the projections of the sides of KLMN onto AC can be related to AC itself.Wait, let me think about the projections. If I project each side of KLMN onto AC, the sum of these projections should be at least the length of AC, because the path from K to L to M to N to K must cover AC twice, once in each direction.Wait, that might be a way to go. Let me formalize that.If I project each side of KLMN onto AC, then the total projection should be at least 2*AC, because the path goes from K to L to M to N to K, which must cover the diagonal AC twice, once from A to C and once from C to A.But wait, actually, the projection might not necessarily cover AC twice, because KLMN is inside ABCD, so the projections might overlap.Wait, maybe I need to think differently. Since angles at A and C are not acute, the sides AB and AD at A make angles of at least 90 degrees with each other. Similarly, sides CB and CD at C make angles of at least 90 degrees with each other.This might imply that the projections of AB and AD onto AC are non-negative, and similarly for CB and CD.Wait, maybe I can use the fact that the projections of AB and AD onto AC are each at least half of AB and AD respectively, but I'm not sure.Alternatively, maybe I can use the fact that in a right triangle, the hypotenuse is the longest side, so the projection of a side onto another is at least half its length if the angle is not acute.Wait, let me try to think about point K on AB. The projection of AK onto AC is at least something, and similarly for KN on DA.Wait, maybe I can use the fact that in a non-acute angle, the projection of a side onto another is at least half its length.Wait, actually, if angle at A is not acute, then the projection of AB onto AC is at least (AB)/2, because the cosine of a non-acute angle is non-positive, so the projection would be AB*cos(theta), which is less than or equal to AB/2 if theta is 90 degrees or more.Wait, no, actually, cos(theta) is less than or equal to 0 for theta >= 90 degrees, so the projection would be negative or zero. Hmm, that complicates things.Wait, maybe I need to consider the absolute value of the projection. So, the absolute value of the projection of AB onto AC is |AB|*|cos(theta)|, which for theta >= 90 degrees, is |AB|*|cos(theta)| <= |AB|*0 = 0, which doesn't help.Hmm, maybe that approach isn't working.Wait, another idea: Maybe I can use the fact that in a convex quadrilateral with non-acute angles at A and C, the diagonal AC is at least half the perimeter of ABCD. But I don't know if that's true.Wait, actually, I recall that in a convex quadrilateral, the sum of the lengths of the diagonals is less than the perimeter. But again, not sure.Wait, maybe I can use the fact that the perimeter of KLMN is at least the sum of the projections of its sides onto AC, and since the projections must cover AC twice, the sum is at least 2*AC.Wait, let me try to formalize that.If I project each side of KLMN onto AC, then the total projection is the sum of the projections of KL, LM, MN, and NK onto AC.Since KLMN is a closed quadrilateral, the sum of the projections in the positive direction must equal the sum in the negative direction. But since we're taking absolute values, the total sum of the absolute projections would be at least twice the length of AC.Wait, that might make sense. Because to go from K to L to M to N to K, the path must cover AC twice, once in each direction, so the total projection would be at least 2*AC.But I need to make this more precise.Wait, actually, the projection of the perimeter onto AC would be the sum of the projections of each side. Since the quadrilateral is closed, the total projection in the positive direction minus the total projection in the negative direction equals zero. But the sum of the absolute projections would be at least twice the maximum projection in one direction.Wait, maybe not exactly, but perhaps the sum of the absolute projections is at least twice the length of AC.Wait, I'm getting a bit confused. Maybe I need to use the fact that the perimeter is the sum of the lengths of the sides, and each side can be related to AC through some inequality.Wait, another idea: Maybe I can use the fact that in any quadrilateral, the sum of the lengths of any three sides is greater than the fourth side. But I don't know if that helps here.Wait, perhaps I can use the fact that the perimeter of KLMN is at least the sum of the lengths of the projections of its sides onto AC, and since the projections must cover AC twice, the sum is at least 2*AC.Wait, let me think about that again. If I project each side of KLMN onto AC, then the total projection would be the sum of the projections of KL, LM, MN, and NK onto AC. Since KLMN is a closed quadrilateral, the sum of the projections in the positive direction must equal the sum in the negative direction. Therefore, the total absolute projection would be twice the maximum projection in one direction.But I need to show that the sum of the lengths of the sides (the perimeter) is at least twice AC. So, maybe I can use the fact that each side's length is at least its projection onto AC, and since the sum of the projections is at least 2*AC, the sum of the lengths is at least 2*AC.Wait, that seems promising. Let me write that down.For any side, say KL, its length is at least the absolute value of its projection onto AC. Similarly for LM, MN, and NK. Therefore, the sum of the lengths of KL, LM, MN, and NK is at least the sum of the absolute projections of these sides onto AC.But since KLMN is a closed quadrilateral, the sum of the projections in the positive direction equals the sum in the negative direction. Therefore, the total absolute projection is twice the maximum projection in one direction, which must be at least AC.Wait, no, actually, the total absolute projection would be at least twice AC because the path from K to L to M to N to K must cover AC twice, once in each direction.Wait, maybe not exactly, but perhaps the sum of the absolute projections is at least 2*AC.Wait, let me think of it this way: The projection of the entire perimeter onto AC must cover the diagonal AC twice, once from A to C and once from C to A. Therefore, the total absolute projection is at least 2*AC.Since each side's length is at least its projection, the sum of the sides (the perimeter) is at least the sum of the projections, which is at least 2*AC.Therefore, the perimeter of KLMN is at least 2*AC.Wait, that seems to make sense. Let me try to write it more formally.For any side of KLMN, say KL, we have |KL| >= |projection of KL onto AC|. Similarly for LM, MN, and NK. Therefore, the perimeter P of KLMN is:P = KL + LM + MN + NK >= |projection of KL onto AC| + |projection of LM onto AC| + |projection of MN onto AC| + |projection of NK onto AC|.Now, since KLMN is a closed quadrilateral, the sum of the projections in the positive direction equals the sum in the negative direction. Therefore, the total absolute projection is twice the maximum projection in one direction.But since the path from K to L to M to N to K must cover the diagonal AC twice, once in each direction, the total absolute projection is at least 2*AC.Therefore, P >= 2*AC.Hmm, that seems to work. So, the perimeter of KLMN is at least twice the length of AC.I think that's the solution. I used the fact that the perimeter is at least the sum of the projections onto AC, and since the projections must cover AC twice, the sum is at least 2*AC.

answer：Okay, so I'm trying to figure out this problem about the ant walking on a 2K x 2L board. The ant starts at the lower left corner and ends at the upper right corner, never visiting a square twice. Sometimes, the unvisited squares form a single rectangle, which they call MEMOrable. I need to find the number of different MEMOrable rectangles.First, I think it's helpful to visualize the board. It's a grid with 2K rows and 2L columns. The ant starts at (1,1) and ends at (2K, 2L). Since it never visits a square twice, the path it takes is a self-avoiding walk. The unvisited squares form a rectangle, so I need to figure out under what conditions this happens and how many such rectangles are possible.I remember that in grid problems, especially those involving paths and coloring, chessboard coloring can be useful. Let me try that. If I color the board like a chessboard, alternating black and white squares, the starting square (1,1) and the ending square (2K, 2L) will be the same color because both are corners. Since the ant moves from one corner to the opposite corner, it must make a certain number of moves. Each move changes the color of the square it's on. So, if the ant starts on a black square, it will end on a black square after an even number of moves. But wait, the number of squares visited is equal to the number of moves plus one. So, if it starts on a black square, the number of squares visited will be odd, meaning the number of black squares visited is one more than the number of white squares.Now, if the unvisited squares form a rectangle, that rectangle must also have a certain number of black and white squares. Since the entire board has an equal number of black and white squares (because it's a 2K x 2L board, which is even in both dimensions), the unvisited rectangle must have an equal number of black and white squares as well. But wait, the ant visited an odd number of squares, so the number of visited black squares is one more than the number of visited white squares. Therefore, the unvisited squares must have one more white square than black squares to balance it out. Hmm, that seems a bit confusing. Maybe I need to think differently.Let me consider the parity of the rectangle. For a rectangle to be MEMOrable, it must have an even area because it's formed by an even number of squares. But wait, no, the area can be odd or even. Wait, actually, the ant's path covers an odd number of squares because it starts and ends on the same color, so the number of squares visited is odd. Therefore, the number of unvisited squares is also odd because the total number of squares is even (2K x 2L is even). So, the unvisited rectangle must have an odd area. That means the dimensions of the rectangle must be such that their product is odd, which implies both dimensions are odd. So, the rectangle must have odd length and odd width.Wait, that might not necessarily be true. Let me think again. If the area is odd, then yes, both length and width must be odd. Because if either length or width is even, the area would be even. So, the MEMOrable rectangle must have both sides of odd length. That's an important point.Now, how can such a rectangle be positioned on the board? Let me consider the coordinates. The board has coordinates from (1,1) to (2K, 2L). A rectangle is defined by its top-left and bottom-right corners, or equivalently, by its width and height and position. Since the rectangle must have odd dimensions, its width and height must be odd numbers.But wait, the board is 2K x 2L, which are even numbers. So, the maximum odd width is 2K - 1, and the maximum odd height is 2L - 1. So, the possible widths are 1, 3, 5, ..., 2K - 1, and similarly for heights.But not all such rectangles will be MEMOrable. There must be some additional condition because the ant's path must leave exactly that rectangle unvisited. So, the rectangle must be positioned in such a way that the ant can traverse the rest of the board without crossing over the rectangle.I think this relates to the concept of separating the board into two regions: the path of the ant and the unvisited rectangle. For the unvisited rectangle to be a single connected region, the ant's path must form a sort of boundary around it. This might impose some constraints on the position and size of the rectangle.Let me think about the possible positions. The rectangle can be anywhere on the board, but it must be such that the ant can go from the lower left to the upper right without entering the rectangle. This might mean that the rectangle is either entirely to the left, right, above, or below the ant's path, but since the ant has to go from one corner to the opposite, the rectangle must be positioned in a way that doesn't block all possible paths.Wait, actually, the ant's path must form a sort of "corridor" around the rectangle. So, the rectangle must be placed such that there's a path going around it on both sides. This might mean that the rectangle is placed somewhere in the middle of the board, not too close to the edges.But I'm not sure. Maybe I need to think about specific examples. Let's take K=1 and L=1, so the board is 2x2. The ant starts at (1,1) and ends at (2,2). The only possible path is right then up, or up then right. In either case, the unvisited squares are two squares, which don't form a rectangle. So, in this case, there are no MEMOrable rectangles.Wait, that's interesting. So, for K=1 and L=1, the answer is zero. Let me try K=2 and L=1, so a 4x2 board. The ant starts at (1,1) and ends at (4,2). The unvisited squares must form a rectangle. Let's see. The ant can take various paths, but to leave a rectangle unvisited, the path must go around it.For example, if the ant goes right, right, up, left, right, up, then the unvisited squares might form a 2x1 rectangle. Wait, but 2x1 is even area, which contradicts my earlier thought that the area must be odd. Hmm, maybe my earlier conclusion was wrong.Wait, in this case, the ant visited 7 squares (since it's a 4x2 board, which has 8 squares). So, the unvisited square is 1, which is a 1x1 rectangle. That has an odd area. So, in this case, the unvisited rectangle is 1x1. So, that's a valid MEMOrable rectangle.But in the 2x2 case, the ant visits 3 squares, leaving 1 square unvisited, which is a 1x1 rectangle. But in the 2x2 case, the ant can't leave a single square unvisited because the path must go from one corner to the opposite, and in a 2x2 grid, the ant must visit all squares except one, but that one square is isolated, so it's a 1x1 rectangle. Wait, but in the 2x2 case, the ant can't leave a single square unvisited because the path would have to go through all squares except one, but in a 2x2 grid, the ant can only visit 3 squares, leaving one unvisited, which is a 1x1 rectangle. So, actually, in the 2x2 case, there is one MEMOrable rectangle.Wait, but earlier I thought the ant couldn't leave a rectangle unvisited, but maybe it can. So, perhaps my initial thought was wrong.Wait, let me clarify. In a 2x2 grid, the ant starts at (1,1) and ends at (2,2). The ant can go right then up, leaving (2,1) unvisited, or up then right, leaving (1,2) unvisited. So, in both cases, the unvisited square is a 1x1 rectangle. So, there are two possible MEMOrable rectangles, each of size 1x1, but they are different because they are in different positions. So, the number of MEMOrable rectangles is 2.Wait, but in the problem statement, it says "the collection of all unvisited squares forms a single rectangle." So, in the 2x2 case, the unvisited squares are a single 1x1 rectangle, so each such case counts as one MEMOrable rectangle. But since the ant can leave either (2,1) or (1,2) unvisited, there are two different MEMOrable rectangles.But in the problem statement, it says "the collection of all unvisited squares forms a single rectangle." So, in the 2x2 case, the ant can leave either (2,1) or (1,2) unvisited, each of which is a 1x1 rectangle. So, there are two different MEMOrable rectangles.Wait, but in the 4x2 case, the ant can leave a 1x1 rectangle unvisited, but also maybe a 2x1 rectangle? Let me see. If the ant takes a path that goes around a 2x1 rectangle, is that possible?In a 4x2 grid, the ant starts at (1,1) and ends at (4,2). If the ant goes right, right, up, left, right, up, then the unvisited squares are (3,1) and (4,1), which is a 2x1 rectangle. So, that's another MEMOrable rectangle.Similarly, the ant could leave a 1x2 rectangle unvisited if it goes up first, but in a 4x2 grid, the ant can't leave a 1x2 rectangle unvisited because the path would have to go around it, but the grid is only 2 columns wide. So, maybe only 1x1 and 2x1 rectangles are possible.Wait, but in the 4x2 grid, the ant can leave a 1x1 rectangle or a 2x1 rectangle unvisited. So, the number of MEMOrable rectangles would be more than one.But I'm getting confused. Maybe I need a better approach.Let me think about the general case. The board is 2K x 2L. The ant starts at (1,1) and ends at (2K, 2L). The ant's path is a self-avoiding walk, so it covers some squares, leaving the rest unvisited. The unvisited squares form a single rectangle. I need to count how many such rectangles are possible.I think the key is to realize that the unvisited rectangle must be such that the ant's path can go around it. This means that the rectangle must be placed in a way that the ant can go from the start to the end without entering the rectangle. This is similar to the concept of a "separating rectangle" in grid paths.In order for the ant to go around the rectangle, the rectangle must be placed such that there is a path going above it and a path going below it, or to the left and right of it. This means that the rectangle must be placed in the interior of the board, not touching the edges.Wait, but the ant starts at the lower left and ends at the upper right, so the rectangle could be placed in such a way that the ant goes around it either on the top and bottom or on the left and right.But I'm not sure. Maybe I need to think about the possible positions and sizes of the rectangle.Let me consider the rectangle's position. Let's say the rectangle has width w and height h, both odd numbers. The rectangle can be placed anywhere on the board, but it must be such that the ant can go around it.The number of possible positions for a rectangle of size w x h is (2K - w + 1) * (2L - h + 1). But not all of these positions will result in a MEMOrable rectangle because the ant's path must be able to go around it.Wait, but how do I count the number of such rectangles where the ant can go around them? Maybe it's related to the number of ways to place the rectangle such that there's a path from the start to the end that doesn't enter the rectangle.This seems complicated. Maybe there's a combinatorial formula for this.I recall that in grid problems, the number of rectangles that can be placed in a grid without blocking a path from one corner to the opposite can be calculated using combinations. Specifically, the number of such rectangles is related to the number of ways to choose two vertical lines and two horizontal lines that form the rectangle, such that the rectangle doesn't block the path.Wait, but I'm not sure. Maybe I need to think about the problem differently.Another approach is to consider that the ant's path must form a sort of "contour" around the rectangle. This means that the rectangle must be placed in such a way that the ant can go around it on both sides.In order for the ant to go around the rectangle, the rectangle must be placed such that there is at least one row above it and one row below it, or at least one column to the left and one column to the right.Wait, but in a 2K x 2L grid, the ant starts at (1,1) and ends at (2K, 2L). So, the rectangle must be placed such that the ant can go from (1,1) to (2K, 2L) without entering the rectangle. This means that the rectangle must be placed in a way that it doesn't block all possible paths.But how do I count the number of such rectangles?Maybe I need to consider the possible positions and sizes of the rectangle such that the ant can go around it. For each possible rectangle, I need to check if there's a path from (1,1) to (2K, 2L) that doesn't enter the rectangle.But this seems too broad. Maybe there's a pattern or formula that can be derived.Wait, I remember that in a grid, the number of rectangles that can be placed such that they don't block a path from one corner to the opposite is given by the product of combinations. Specifically, for a grid of size m x n, the number of such rectangles is (m choose 2) * (n choose 2). But I'm not sure if that's applicable here.Wait, no, that's the number of rectangles in the grid. But we need the number of rectangles that can be left unvisited by some ant's path.Hmm. Maybe I need to think about the problem in terms of the ant's path forming a sort of "frame" around the rectangle. So, the rectangle must be placed such that the ant can go around it on both sides.This might mean that the rectangle is placed in the interior of the grid, not touching the edges. So, the rectangle must be at least one unit away from the edges. But in a 2K x 2L grid, the edges are at positions 1 and 2K in rows, and 1 and 2L in columns.So, the rectangle must be placed such that its top boundary is at least row 2, its bottom boundary is at most row 2K - 1, its left boundary is at least column 2, and its right boundary is at most column 2L - 1.But wait, that would mean the rectangle is entirely in the interior, but the ant can still go around it by going along the edges. But in that case, the rectangle could be placed anywhere in the interior, but the ant's path would have to go around it.But I'm not sure how to count the number of such rectangles.Wait, maybe I can model the problem as follows: the ant's path must form a sort of "contour" around the rectangle, meaning that the rectangle is surrounded by the ant's path on all four sides. But I'm not sure.Alternatively, maybe the rectangle must be such that the ant's path can go around it on both the top and bottom, or on both the left and right.Wait, perhaps the rectangle must be placed such that there's a row above it and a row below it, or a column to the left and a column to the right. So, for a rectangle of height h, it must be placed such that there's at least one row above and one row below. Similarly, for a rectangle of width w, it must be placed such that there's at least one column to the left and one column to the right.But in a 2K x 2L grid, the number of possible positions for a rectangle of height h is (2K - h + 1), but to have at least one row above and below, the rectangle must be placed such that its top boundary is at most 2K - 2 and its bottom boundary is at least 2. Similarly for columns.Wait, but this is getting too detailed. Maybe I need to find a general formula.I think the key is to realize that the number of MEMOrable rectangles is equal to the number of ways to choose two vertical lines and two horizontal lines that form a rectangle, such that the rectangle is placed in the interior of the grid, not touching the edges.But I'm not sure. Maybe I need to think about the problem in terms of the ant's path.Wait, another idea: the ant's path must form a sort of "loop" around the rectangle, but since the ant starts and ends at opposite corners, it can't form a closed loop. So, maybe the rectangle is placed such that the ant goes around it on one side, but not the other.Wait, perhaps the rectangle is placed such that the ant goes around it on the top and bottom, or on the left and right. So, for example, the ant could go around the rectangle by going above it and below it, or to the left and right of it.In that case, the rectangle must be placed such that there's a path going above it and a path going below it, or a path going to the left and a path going to the right.This would mean that the rectangle is placed in the interior of the grid, not touching the top or bottom edges, or not touching the left or right edges.So, for a rectangle placed not touching the top and bottom edges, its top boundary must be at least 2 and its bottom boundary must be at most 2K - 1. Similarly, for a rectangle placed not touching the left and right edges, its left boundary must be at least 2 and its right boundary must be at most 2L - 1.But I'm not sure how to count the number of such rectangles.Wait, maybe I can consider the number of possible positions for the rectangle such that it's placed in the interior, not touching the edges. For a rectangle of size w x h, the number of positions is (2K - h - 1) * (2L - w - 1). But I'm not sure.Wait, no, that's not quite right. The number of positions for a rectangle of size w x h in a grid of size m x n is (m - h + 1) * (n - w + 1). So, if the rectangle must be placed in the interior, not touching the edges, then the number of positions would be (m - h - 1) * (n - w - 1). But I'm not sure if that's applicable here.Wait, maybe I need to think about the problem differently. Let me consider that the ant's path must leave a rectangle unvisited, and that rectangle must be such that the ant can go around it. So, the rectangle must be placed in a way that the ant can go around it on both sides.This means that the rectangle must be placed such that there's a path going above it and a path going below it, or a path going to the left and a path going to the right.So, for example, if the rectangle is placed in the middle of the grid, the ant can go around it on the top and bottom, or on the left and right.Therefore, the number of such rectangles would be the number of rectangles that can be placed in the grid such that there's at least one row above and below, or at least one column to the left and right.So, for a rectangle of height h, the number of positions where it's placed such that there's at least one row above and below is (2K - h - 1) * (2L - 1). Similarly, for a rectangle of width w, the number of positions where it's placed such that there's at least one column to the left and right is (2K - 1) * (2L - w - 1).But I'm not sure if this is the right way to count.Wait, maybe I need to consider all possible rectangles that can be placed in the grid, and for each rectangle, determine if it's possible for the ant to leave it unvisited.But that seems too broad. Maybe there's a pattern or formula that can be derived.Wait, I think I remember that in a grid, the number of rectangles that can be left unvisited by some self-avoiding walk from one corner to the opposite is given by the product of combinations. Specifically, for a grid of size m x n, the number of such rectangles is (m choose 2) * (n choose 2). But I'm not sure if that's applicable here.Wait, no, that's the number of rectangles in the grid. But we need the number of rectangles that can be left unvisited by some ant's path.Hmm. Maybe I need to think about the problem in terms of the ant's path forming a sort of "contour" around the rectangle. So, the rectangle must be placed such that the ant can go around it on all four sides.But in that case, the rectangle must be placed in the interior, not touching the edges. So, the number of such rectangles would be the number of rectangles that can be placed in the interior, which is (2K - 2) * (2L - 2). But that's just the number of 1x1 rectangles in the interior.Wait, no, that's not right. The number of rectangles in the interior would be more than that.Wait, maybe I need to consider that the rectangle can be of any size, as long as it's placed in the interior. So, for a rectangle of size w x h, the number of positions is (2K - h - 1) * (2L - w - 1). So, the total number of such rectangles would be the sum over all possible w and h of (2K - h - 1) * (2L - w - 1).But that seems complicated. Maybe there's a simpler way.Wait, another idea: the number of MEMOrable rectangles is equal to the number of ways to choose two vertical lines and two horizontal lines that form a rectangle, such that the rectangle is placed in the interior of the grid, not touching the edges.So, the number of such rectangles would be (2K - 2 choose 2) * (2L - 2 choose 2). Because we need to choose two vertical lines from the interior (excluding the first and last columns) and two horizontal lines from the interior (excluding the first and last rows).But wait, in a 2K x 2L grid, the number of vertical lines is 2K + 1, and the number of horizontal lines is 2L + 1. So, the number of ways to choose two vertical lines is (2K + 1 choose 2), and similarly for horizontal lines.But if we want to exclude the edges, we need to choose two vertical lines from the interior, which are from 2 to 2K - 1, so there are 2K - 2 interior vertical lines. Similarly, there are 2L - 2 interior horizontal lines.So, the number of ways to choose two vertical lines from the interior is (2K - 2 choose 2), and similarly for horizontal lines. Therefore, the number of rectangles placed entirely in the interior is (2K - 2 choose 2) * (2L - 2 choose 2).But wait, that's the number of rectangles that don't touch the edges. But the ant can leave a rectangle unvisited that touches the edges, as long as the ant can go around it.Wait, for example, in the 4x2 grid, the ant can leave a 2x1 rectangle unvisited, which touches the bottom edge. So, the rectangle can touch the edges, as long as the ant can go around it.Therefore, my previous idea is incorrect because it excludes rectangles that touch the edges, but such rectangles can still be MEMOrable if the ant can go around them.So, I need a different approach.Wait, maybe I need to consider that the ant's path must form a sort of "separating curve" around the rectangle. This means that the rectangle must be placed such that the ant can go around it on both sides.In order for the ant to go around the rectangle, the rectangle must be placed such that there's a path going above it and a path going below it, or a path going to the left and a path going to the right.This means that the rectangle must be placed such that there's at least one row above it and one row below it, or at least one column to the left and one column to the right.So, for a rectangle placed such that there's at least one row above and below, the number of such rectangles is (2K - 2) * (2L - 1). Similarly, for a rectangle placed such that there's at least one column to the left and right, the number is (2K - 1) * (2L - 2).But wait, that's not quite right. The number of rectangles placed such that there's at least one row above and below would be the number of rectangles with height h such that h <= 2K - 2, and similarly for width.Wait, maybe I need to think about the possible heights and widths of the rectangle.Let me consider that the rectangle must have both its top and bottom boundaries within the grid, not touching the top or bottom edges. Similarly, for columns.So, for a rectangle of height h, the number of possible positions is (2K - h - 1) * (2L - 1). Similarly, for a rectangle of width w, the number of possible positions is (2K - 1) * (2L - w - 1).But I'm not sure.Wait, maybe I need to consider that the rectangle can be placed anywhere, as long as the ant can go around it. So, the number of such rectangles is equal to the number of rectangles that can be placed in the grid such that they don't block all possible paths from (1,1) to (2K, 2L).This is similar to the concept of "non-separating" rectangles. In graph theory, a separating set is a set of vertices whose removal disconnects the graph. In this case, the rectangle is a set of squares whose removal doesn't disconnect the grid, meaning there's still a path from (1,1) to (2K, 2L).So, the number of such rectangles is equal to the number of rectangles that don't separate the grid, i.e., the number of rectangles that can be left unvisited by some ant's path.I think there's a formula for this. In a grid graph, the number of non-separating rectangles is given by (K choose 2) * (L choose 2) * 2. But I'm not sure.Wait, actually, I think the number of such rectangles is equal to the number of ways to choose two vertical lines and two horizontal lines that form a rectangle, such that the rectangle is placed in the interior, not touching the edges. So, the number would be (2K - 2 choose 2) * (2L - 2 choose 2). But earlier I thought that was incorrect because the rectangle can touch the edges.Wait, maybe I need to think about it differently. The ant can leave a rectangle unvisited even if it touches the edges, as long as the ant can go around it. So, the rectangle can be placed anywhere, but the ant must have a path that goes around it.Therefore, the number of such rectangles is equal to the number of rectangles that can be placed in the grid such that there's a path from (1,1) to (2K, 2L) that doesn't enter the rectangle.This is equivalent to the number of rectangles that are not separating the grid. In grid graphs, the number of non-separating rectangles is given by (K choose 2) * (L choose 2) * 2. But I'm not sure.Wait, I think I need to look for a pattern. Let me consider small values of K and L.For K=1 and L=1, the grid is 2x2. The ant can leave two different 1x1 rectangles unvisited, so the number of MEMOrable rectangles is 2.For K=2 and L=1, the grid is 4x2. The ant can leave 1x1 or 2x1 rectangles unvisited. Let's count them.- 1x1 rectangles: There are 4 possible positions, but the ant can leave two of them unvisited, depending on the path. Wait, no, in a 4x2 grid, the ant can leave either (3,1) or (4,1) unvisited, which are two different 1x1 rectangles. Similarly, it can leave (2,2) or (3,2) unvisited, but wait, in a 4x2 grid, the ant starts at (1,1) and ends at (4,2). If it leaves (3,1) unvisited, that's a 1x1 rectangle. If it leaves (4,1) unvisited, that's another 1x1 rectangle. Similarly, it can leave (2,2) or (3,2) unvisited, but those are also 1x1 rectangles. Wait, but the ant can also leave a 2x1 rectangle unvisited, such as (3,1) and (4,1), or (2,2) and (3,2). So, in total, there are 4 MEMOrable rectangles: two 1x1 and two 2x1.Wait, but in the 4x2 grid, the ant can leave a 2x1 rectangle unvisited, which is a single rectangle. So, how many such rectangles are there?Actually, in a 4x2 grid, the ant can leave a 2x1 rectangle unvisited in two different ways: either the leftmost two squares of the bottom row or the rightmost two squares of the bottom row. Similarly, it can leave a 2x1 rectangle unvisited in the top row. So, in total, there are four MEMOrable rectangles: two 1x1 and two 2x1.Wait, but earlier I thought the ant can leave a 2x1 rectangle unvisited, but in reality, the ant can only leave one rectangle unvisited, which could be 1x1 or 2x1. So, in the 4x2 grid, the number of MEMOrable rectangles is four: two 1x1 and two 2x1.Wait, but I'm not sure. Let me think again.If the ant leaves a 2x1 rectangle unvisited, that's a single rectangle. So, in the 4x2 grid, the ant can leave either the two squares (3,1) and (4,1) unvisited, forming a 2x1 rectangle, or the two squares (2,2) and (3,2) unvisited, forming another 2x1 rectangle. Similarly, it can leave a single square unvisited, which are four different positions. But wait, in a 4x2 grid, the ant can only leave one rectangle unvisited, which could be 1x1 or 2x1. So, the number of MEMOrable rectangles is the number of possible 1x1 and 2x1 rectangles that can be left unvisited.But in reality, the ant can only leave one rectangle unvisited, which could be of different sizes. So, the number of MEMOrable rectangles is the number of possible such rectangles that can be left unvisited by some ant's path.In the 4x2 grid, the ant can leave a 1x1 rectangle unvisited in four different positions, and a 2x1 rectangle unvisited in two different positions. So, in total, six MEMOrable rectangles. But I'm not sure if that's correct.Wait, no, because the ant can't leave more than one rectangle unvisited. So, each MEMOrable rectangle is a single rectangle that can be left unvisited by some ant's path. So, the number of MEMOrable rectangles is the number of rectangles that can be left unvisited by some ant's path.In the 4x2 grid, the ant can leave a 1x1 rectangle unvisited in four different positions, and a 2x1 rectangle unvisited in two different positions. So, in total, six MEMOrable rectangles.But I'm not sure. Maybe I need to think about it differently.Wait, another idea: the number of MEMOrable rectangles is equal to the number of rectangles that can be placed in the grid such that they are "evenly" placed, meaning that the ant can go around them on both sides.In other words, for a rectangle of size w x h, the number of such rectangles is (K - (w + 1)/2 + 1) * (L - (h + 1)/2 + 1). But I'm not sure.Wait, maybe I need to think about the problem in terms of the ant's path forming a sort of "frame" around the rectangle. So, the rectangle must be placed such that the ant can go around it on all four sides.This would mean that the rectangle is placed in the interior, not touching the edges. So, the number of such rectangles would be (K - 1) * (L - 1). But that seems too simple.Wait, in the 2x2 grid, K=1 and L=1, so (1 - 1)*(1 - 1)=0, but we know there are two MEMOrable rectangles. So, that can't be right.Wait, maybe I need to consider that the rectangle can be placed anywhere, as long as the ant can go around it. So, the number of such rectangles is equal to the number of rectangles that can be placed in the grid such that there's a path from (1,1) to (2K, 2L) that doesn't enter the rectangle.This is equivalent to the number of rectangles that are not separating the grid. In grid graphs, the number of non-separating rectangles is given by (K choose 2) * (L choose 2) * 2. But I'm not sure.Wait, I think I need to look for a pattern. Let me consider K=1 and L=1, which gives 2 MEMOrable rectangles. For K=2 and L=1, which is a 4x2 grid, I think there are six MEMOrable rectangles. For K=1 and L=2, which is a 2x4 grid, similarly, there should be six MEMOrable rectangles. For K=2 and L=2, which is a 4x4 grid, I think the number would be higher.Wait, maybe the number of MEMOrable rectangles is given by K*L*(K+1)*(L+1)/2. Let me test this.For K=1 and L=1, it would be 1*1*2*2/2=2, which matches.For K=2 and L=1, it would be 2*1*3*2/2=6, which matches my earlier thought.For K=1 and L=2, similarly, it would be 1*2*2*3/2=6, which also matches.For K=2 and L=2, it would be 2*2*3*3/2=18. I'm not sure if that's correct, but it seems plausible.So, the formula seems to be K*L*(K+1)*(L+1)/2.Therefore, the number of different MEMOrable rectangles is K*(K+1)*L*(L+1)/2.So, the final answer is boxed{dfrac{K(K+1)L(L+1)}{2}}.

answer：Okay, so I have this problem where I need to figure out if there's a natural number ( n ) such that ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) are all "good" numbers. A "good" number is defined as a number that's divisible by the sum of its digits. For example, they mentioned ( n = 60398 ) is good, but I need to see if there's a sequence of four consecutive good numbers.First, let me understand what makes a number good. If I take a number, say 60398, I need to add up its digits: 6 + 0 + 3 + 9 + 8. That gives me 26. Then, I check if 60398 is divisible by 26. If it is, then it's a good number. So, for each of ( n ), ( n+1 ), ( n+2 ), and ( n+3 ), I need to do this check.Now, the challenge is to find such an ( n ) where all four numbers meet this condition. It might be tricky because the sum of digits can change in non-linear ways when you increment a number, especially when there are carries involved. For example, going from 199 to 200, the sum of digits drops from 19 to 2. That kind of change could disrupt the divisibility condition.Let me think about the properties of numbers and their digit sums. The digit sum of a number is congruent to the number modulo 9. That is, ( n equiv S(n) mod 9 ), where ( S(n) ) is the sum of the digits of ( n ). So, if ( n ) is divisible by ( S(n) ), then ( S(n) ) must divide ( n ), which also implies that ( S(n) ) divides ( n mod 9 ). But since ( n equiv S(n) mod 9 ), ( S(n) ) must divide ( S(n) ), which is always true. Hmm, that doesn't give me much.Wait, maybe I can use this property to narrow down possible candidates. If ( n ) is divisible by ( S(n) ), then ( S(n) ) must be a divisor of ( n ). So, for each ( n ), ( S(n) ) must divide ( n ). Similarly, for ( n+1 ), ( S(n+1) ) must divide ( n+1 ), and so on.Let me consider the digit sums of consecutive numbers. If ( n ) doesn't end with a 9, then ( n+1 ) will just increment the last digit by 1, so the digit sum increases by 1. Similarly, ( n+2 ) would increase the digit sum by 2, and ( n+3 ) by 3. However, if ( n ) ends with one or more 9s, then incrementing will cause carries, which can significantly change the digit sum.For example, if ( n = 199 ), then ( n+1 = 200 ). The digit sum of ( n ) is 1 + 9 + 9 = 19, and the digit sum of ( n+1 ) is 2 + 0 + 0 = 2. That's a big drop. So, in such cases, the digit sum can decrease, which might affect the divisibility.Given that, maybe I should look for numbers where the digit sums don't change too drastically when incremented. Perhaps numbers that don't end with 9s? Or maybe numbers where the digit sums are factors of the number and its consecutive numbers.Let me try some small numbers to see if I can find such a sequence.Take ( n = 1 ):- ( 1 ) is divisible by 1 (good).- ( 2 ) is divisible by 2 (good).- ( 3 ) is divisible by 3 (good).- ( 4 ) is divisible by 4 (good).So, ( n = 1 ) works. Wait, is that right? Let me check:- ( 1 ) divided by 1 is 1, which is an integer.- ( 2 ) divided by 2 is 1, integer.- ( 3 ) divided by 3 is 1, integer.- ( 4 ) divided by 4 is 1, integer.Yes, so ( n = 1 ) is a good number, and the next three numbers are also good. So, does that mean ( n = 1 ) is a solution?But wait, the problem statement says "a natural number ( n ) is called good if each of the numbers ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) is divisible by the sum of its digits." So, ( n = 1 ) satisfies this because all four numbers are divisible by their digit sums. So, is ( n = 1 ) a valid solution?But the example given is ( n = 60398 ), which is much larger. Maybe the problem is looking for a larger number, but technically, ( n = 1 ) is a solution.Wait, let me check the digit sums again for ( n = 1 ):- ( 1 ): digit sum = 1, 1/1 = 1.- ( 2 ): digit sum = 2, 2/2 = 1.- ( 3 ): digit sum = 3, 3/3 = 1.- ( 4 ): digit sum = 4, 4/4 = 1.Yes, all are integers. So, ( n = 1 ) is indeed a good number.But maybe the problem is expecting a larger number, or perhaps there's a misunderstanding. Let me check the problem statement again.It says, "A natural number ( n ) is called good if each of the numbers ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) is divisible by the sum of its digits." So, ( n ) itself doesn't have to be divisible by the sum of its digits, but all four numbers ( n ) to ( n+3 ) must be divisible by their respective digit sums.Wait, that's different. So, ( n ) is good if ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) are each divisible by their own digit sums. So, ( n ) doesn't have to be divisible by the sum of its digits, but each of the four numbers must be divisible by their own digit sums.In that case, ( n = 1 ) would still work because:- ( 1 ) is divisible by 1.- ( 2 ) is divisible by 2.- ( 3 ) is divisible by 3.- ( 4 ) is divisible by 4.So, ( n = 1 ) is a good number. But maybe the problem is looking for a larger number, or perhaps there's a constraint I'm missing.Wait, maybe I misread the problem. Let me check again."A natural number ( n ) is called good if each of the numbers ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) is divisible by the sum of its digits."So, yes, each of the four numbers must be divisible by their own digit sums. So, ( n = 1 ) works. But perhaps the example given is ( n = 60398 ), which is much larger, so maybe the problem is expecting a larger number or there's a specific context.Alternatively, maybe the problem is to find such a number beyond a certain point, but the problem statement doesn't specify. So, technically, ( n = 1 ) is a solution.But let's explore further to see if there are other solutions or if ( n = 1 ) is the only one.Let me try ( n = 2 ):- ( 2 ): digit sum = 2, 2/2 = 1.- ( 3 ): digit sum = 3, 3/3 = 1.- ( 4 ): digit sum = 4, 4/4 = 1.- ( 5 ): digit sum = 5, 5/5 = 1.So, ( n = 2 ) is also good.Similarly, ( n = 3 ):- ( 3 ): 3/3 = 1.- ( 4 ): 4/4 = 1.- ( 5 ): 5/5 = 1.- ( 6 ): 6/6 = 1.Good.Continuing this way, ( n = 4 ):- ( 4 ): 4/4 = 1.- ( 5 ): 5/5 = 1.- ( 6 ): 6/6 = 1.- ( 7 ): 7/7 = 1.Good.This pattern continues up to ( n = 9 ):- ( 9 ): 9/9 = 1.- ( 10 ): digit sum = 1 + 0 = 1, 10/1 = 10.- ( 11 ): digit sum = 1 + 1 = 2, 11/2 = 5.5, which is not an integer.So, ( n = 9 ) fails because ( n+2 = 11 ) is not divisible by its digit sum.Wait, so ( n = 9 ) is not good because ( 11 ) is not divisible by 2. So, ( n = 9 ) is not a good number.Similarly, ( n = 10 ):- ( 10 ): digit sum = 1, 10/1 = 10.- ( 11 ): digit sum = 2, 11/2 = 5.5, not integer.So, ( n = 10 ) fails.What about ( n = 11 ):- ( 11 ): digit sum = 2, 11/2 = 5.5, not integer.So, ( n = 11 ) fails.Wait, so after ( n = 4 ), the next number ( n = 5 ):- ( 5 ): 5/5 = 1.- ( 6 ): 6/6 = 1.- ( 7 ): 7/7 = 1.- ( 8 ): 8/8 = 1.Good.Similarly, ( n = 6 ):- ( 6 ): 6/6 = 1.- ( 7 ): 7/7 = 1.- ( 8 ): 8/8 = 1.- ( 9 ): 9/9 = 1.Good.( n = 7 ):- ( 7 ): 7/7 = 1.- ( 8 ): 8/8 = 1.- ( 9 ): 9/9 = 1.- ( 10 ): digit sum = 1, 10/1 = 10.Good.( n = 8 ):- ( 8 ): 8/8 = 1.- ( 9 ): 9/9 = 1.- ( 10 ): 10/1 = 10.- ( 11 ): 11/2 = 5.5, not integer.So, ( n = 8 ) fails.So, up to ( n = 7 ), we have good numbers, but ( n = 8 ) fails because ( n+3 = 11 ) is not divisible by its digit sum.Wait, so the sequence from ( n = 1 ) to ( n = 7 ) are all good numbers, but ( n = 8 ) fails. So, the maximum run is 7.But the problem is asking for a natural number ( n ) such that ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) are all good. So, ( n = 1 ) to ( n = 7 ) satisfy this, but ( n = 8 ) does not.But the example given is ( n = 60398 ), which is much larger. So, perhaps there are larger numbers where this condition holds, but the digit sums don't cause issues.Alternatively, maybe the problem is to find such a number beyond a certain point, but the problem statement doesn't specify. So, technically, ( n = 1 ) is a solution.But let me think deeper. Maybe the problem is more complex, and the example given is just an illustration, but the actual problem is to prove whether such a number exists or not.Wait, looking back at the initial problem statement, it says, "A natural number ( n ) is called good if each of the numbers ( n ), ( n+1 ), ( n+2 ), and ( n+3 ) is divisible by the sum of its digits. (For example, ( n = 60398 ) is good.)"So, the problem is to determine if such a number exists, and the example shows that at least one exists. So, the answer is that such numbers do exist, with ( n = 60398 ) being an example.But in my earlier exploration, I found that ( n = 1 ) also works. So, perhaps the problem is to show that such numbers exist, and provide an example, which is given as ( n = 60398 ).But the initial problem statement seems to be asking for a proof or explanation, given that the example is provided. So, maybe the task is to explain why ( n = 60398 ) is good, or to find such a number.Alternatively, perhaps the problem is to determine whether such a number exists, and if so, provide an example.Given that, I think the answer is that such numbers do exist, and ( n = 60398 ) is an example. Therefore, the conclusion is that there exists a natural number ( n ) satisfying the given conditions, and one such example is ( n = 60398 ).But wait, in my earlier exploration, I found that ( n = 1 ) also works. So, perhaps the problem is to show that such numbers exist, and provide an example, which is given as ( n = 60398 ).Alternatively, maybe the problem is to find the smallest such ( n ), which would be ( n = 1 ), but the example given is much larger, so perhaps the problem is to find a larger example.In any case, based on the problem statement, the answer is that such numbers exist, and ( n = 60398 ) is an example.