answer：Alright, so I've got this problem about vectors and forming a triangle. Let me try to figure it out step by step. First, the problem gives me three vectors: OA, OB, and OC. Their coordinates are OA = (2, -1), OB = (3, 2), and OC = (m, 2m+1). I need to find two things: 1. The condition that the real number m must satisfy so that points A, B, and C can form a triangle.2. If triangle ABC is a right-angled triangle, find the value of m.Okay, starting with the first part. To form a triangle, the three points A, B, and C must not be colinear. That means the vectors AB and AC shouldn't be scalar multiples of each other. So, I need to find the condition on m such that points A, B, and C are not colinear.Let me recall that vectors AB and AC can be found by subtracting the coordinates of the points. So, vector AB is OB - OA, and vector AC is OC - OA.Calculating AB:OA is (2, -1), OB is (3, 2). So, AB = OB - OA = (3-2, 2 - (-1)) = (1, 3).Calculating AC:OC is (m, 2m+1), OA is (2, -1). So, AC = OC - OA = (m - 2, 2m + 1 - (-1)) = (m - 2, 2m + 2).Now, for vectors AB and AC to not be colinear, the determinant of the matrix formed by these vectors should not be zero. The determinant is calculated as:|AB_x * AC_y - AB_y * AC_x| ≠ 0Plugging in the values:AB_x = 1, AB_y = 3AC_x = m - 2, AC_y = 2m + 2So, determinant = (1)(2m + 2) - (3)(m - 2) = 2m + 2 - 3m + 6 = (-m) + 8For the determinant to not be zero, we have:(-m) + 8 ≠ 0=> -m ≠ -8=> m ≠ 8So, the condition is that m should not be equal to 8.Alright, that seems straightforward. Now, moving on to the second part: if triangle ABC is a right-angled triangle, find the value of m.Hmm, right-angled triangle. So, one of the angles at the vertices A, B, or C is 90 degrees. That means the dot product of the vectors forming that angle should be zero.So, I need to check for each vertex whether the dot product of the vectors is zero.First, let me write down the vectors AB, BC, and AC.We already have AB = (1, 3) and AC = (m - 2, 2m + 2). Let me find BC as well.Vector BC = OC - OB = (m - 3, 2m + 1 - 2) = (m - 3, 2m - 1)So, vectors AB = (1, 3), AC = (m - 2, 2m + 2), BC = (m - 3, 2m - 1)Now, for triangle ABC to be right-angled, one of the following must hold:1. AB · AC = 0 (right angle at A)2. AB · BC = 0 (right angle at B)3. AC · BC = 0 (right angle at C)Let me compute each dot product and solve for m.First, AB · AC:AB = (1, 3), AC = (m - 2, 2m + 2)Dot product = (1)(m - 2) + (3)(2m + 2) = m - 2 + 6m + 6 = 7m + 4Set this equal to zero:7m + 4 = 0 => m = -4/7Okay, so if m = -4/7, then angle at A is 90 degrees.Second, AB · BC:AB = (1, 3), BC = (m - 3, 2m - 1)Dot product = (1)(m - 3) + (3)(2m - 1) = m - 3 + 6m - 3 = 7m - 6Set this equal to zero:7m - 6 = 0 => m = 6/7So, if m = 6/7, then angle at B is 90 degrees.Third, AC · BC:AC = (m - 2, 2m + 2), BC = (m - 3, 2m - 1)Dot product = (m - 2)(m - 3) + (2m + 2)(2m - 1)Let me compute each term:First term: (m - 2)(m - 3) = m^2 - 3m - 2m + 6 = m^2 - 5m + 6Second term: (2m + 2)(2m - 1) = 4m^2 - 2m + 4m - 2 = 4m^2 + 2m - 2Adding both terms:m^2 - 5m + 6 + 4m^2 + 2m - 2 = 5m^2 - 3m + 4Set this equal to zero:5m^2 - 3m + 4 = 0Let me solve this quadratic equation:Discriminant D = (-3)^2 - 4*5*4 = 9 - 80 = -71Since the discriminant is negative, there are no real solutions. So, AC · BC cannot be zero for any real m.Therefore, the only possible values of m are -4/7 and 6/7.Wait, but I need to make sure that these values of m don't violate the condition from part (1), which was m ≠ 8. Well, -4/7 and 6/7 are both not equal to 8, so that's fine.So, summarizing:1. The condition is m ≠ 8.2. The values of m that make triangle ABC right-angled are m = -4/7 and m = 6/7.I think that's it. Let me just double-check my calculations to make sure I didn't make any mistakes.For part (1), determinant was (-m + 8) ≠ 0, so m ≠ 8. That seems correct.For part (2):- AB · AC: 7m + 4 = 0 => m = -4/7. Correct.- AB · BC: 7m - 6 = 0 => m = 6/7. Correct.- AC · BC: Quadratic with no real roots. Correct.Yes, everything checks out.**Final Answer**(1) The condition is boxed{m neq 8}.(2) The values of ( m ) are boxed{-dfrac{4}{7}} and boxed{dfrac{6}{7}}.

answer：First, I need to determine the dimensions of the rectangular room. The area of the floor is given as 360 square feet, and the length is twice the width.Let’s denote the width of the room as ( W ) feet. Therefore, the length ( L ) is ( 2W ) feet.Using the formula for the area of a rectangle:[ L times W = 360 ]Substituting ( L = 2W ):[ 2W times W = 360 ][ 2W^2 = 360 ][ W^2 = 180 ][ W = sqrt{180} ][ W = 6sqrt{5} text{ feet} ]Now, the length ( L ) is:[ L = 2 times 6sqrt{5} = 12sqrt{5} text{ feet} ]Next, I need to convert the length from feet to inches because the tile size is given in inches. There are 12 inches in a foot:[ L = 12sqrt{5} times 12 = 144sqrt{5} text{ inches} ]Each tile is 8 inches long. To find the number of tiles that fit along the length:[ text{Number of tiles} = frac{144sqrt{5}}{8} = 18sqrt{5} ]Therefore, the number of tiles in each row along the length of the room is ( 18sqrt{5} ).

answer：Alright, so I have this geometry problem here. Let me try to figure it out step by step. The problem says:"Draw two tangent lines ( l_1 ) and ( l_2 ) from a point on the line ( y = x ) to the circle ( (x - 5)^2 + (y - 1)^2 = 2 ). When the lines ( l_1 ) and ( l_2 ) are symmetric about ( y = x ), the angle between them is (　　). The options are 30°, 45°, 60°, and 90°."Okay, so first, I need to visualize this. There's a circle with center at (5, 1) and radius ( sqrt{2} ). Then, there's a point on the line ( y = x ) from which we draw two tangent lines to this circle. These tangent lines are symmetric about the line ( y = x ), and we need to find the angle between them.Let me recall some concepts. The tangent lines from a point outside a circle have equal lengths. Also, the line connecting the center of the circle to the external point bisects the angle between the two tangent lines. Since the tangent lines are symmetric about ( y = x ), the external point must lie on the line ( y = x ), which is given.So, the external point is somewhere on ( y = x ). Let me denote this point as ( P(a, a) ). From this point, we draw two tangent lines to the circle. The circle is at (5,1), so the distance from point P to the center of the circle is important here.The distance between P(a, a) and the center (5,1) can be calculated using the distance formula:( d = sqrt{(a - 5)^2 + (a - 1)^2} )Since the radius of the circle is ( sqrt{2} ), the length of the tangent from P to the circle is given by:( sqrt{d^2 - r^2} = sqrt{(a - 5)^2 + (a - 1)^2 - 2} )But maybe I don't need to compute this directly. Instead, since the tangent lines are symmetric about ( y = x ), the point P must be such that the line connecting P to the center of the circle is perpendicular to ( y = x ). Wait, is that true?Let me think. If the tangent lines are symmetric about ( y = x ), then the line connecting P to the center should be such that it makes equal angles with the tangent lines. Hmm, perhaps it's better to find the point P where the tangents are symmetric.Alternatively, since the tangents are symmetric about ( y = x ), the point P must lie on the perpendicular bisector of the line segment joining the center of the circle and its reflection over ( y = x ). Wait, reflecting the center over ( y = x ) gives (1,5). So, the line connecting (5,1) and (1,5) is perpendicular to ( y = x ), right?Let me calculate the midpoint of (5,1) and (1,5). The midpoint is ( left( frac{5+1}{2}, frac{1+5}{2} right) = (3,3) ). So, the point (3,3) is on the line ( y = x ). Therefore, this must be the point P from which the tangents are drawn.So, P is (3,3). Now, let's find the angle between the two tangent lines from (3,3) to the circle.First, let's compute the distance from P(3,3) to the center (5,1):( d = sqrt{(5 - 3)^2 + (1 - 3)^2} = sqrt{4 + 4} = sqrt{8} = 2sqrt{2} )The radius of the circle is ( sqrt{2} ). So, the length of the tangent from P to the circle is:( sqrt{d^2 - r^2} = sqrt{(2sqrt{2})^2 - (sqrt{2})^2} = sqrt{8 - 2} = sqrt{6} )Now, the angle between the two tangent lines can be found using the formula:( theta = 2 arcsin left( frac{r}{d} right) )Plugging in the values:( theta = 2 arcsin left( frac{sqrt{2}}{2sqrt{2}} right) = 2 arcsin left( frac{1}{2} right) )Since ( arcsin left( frac{1}{2} right) = 30° ), so:( theta = 2 times 30° = 60° )Therefore, the angle between the two tangent lines is 60°, which corresponds to option C.Wait, let me double-check my reasoning. I assumed that the point P is (3,3) because it's the midpoint of (5,1) and its reflection over ( y = x ). Is that correct?Yes, because reflecting the center over ( y = x ) gives (1,5), and the midpoint between (5,1) and (1,5) is indeed (3,3). Since the tangents are symmetric about ( y = x ), the point P must lie on this perpendicular bisector, which is ( y = x ) in this case. So, (3,3) is the correct point.Calculating the distance from (3,3) to the center (5,1) as ( 2sqrt{2} ) seems right. Then, the tangent length is ( sqrt{6} ). Using the formula for the angle between two tangents, which is ( 2 arcsin(r/d) ), gives 60°, which seems correct.Alternatively, I can think about the triangle formed by the center, point P, and one of the tangent points. This triangle is a right-angled triangle with sides ( sqrt{2} ), ( sqrt{6} ), and hypotenuse ( 2sqrt{2} ). The angle at point P is half the angle between the two tangents. Using trigonometry, ( sin(theta/2) = r/d = sqrt{2}/2sqrt{2} = 1/2 ), so ( theta/2 = 30° ), hence ( theta = 60° ).Yes, that all checks out. So, the angle between the two tangent lines is 60°, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this problem where I need to prove that any natural number ( n ) can be uniquely represented in the form ( n = a_{1} cdot 1! + a_{2} cdot 2! + ldots + a_{k} cdot k! ), where each ( a_{i} ) is an integer satisfying ( 0 leqslant a_{i} leqslant i ), and ( a_{k} neq 0 ). Hmm, interesting. I think this is related to something called factorial number system or something like that. Let me try to figure this out step by step.First, I need to understand what the problem is asking. It says that for any natural number ( n ), there's a unique way to express it as a sum of multiples of factorials, where each multiple is bounded by the index of the factorial. So, for example, the coefficient ( a_1 ) can be 0 or 1, ( a_2 ) can be 0, 1, or 2, and so on. Also, the highest non-zero coefficient ( a_k ) must be at least 1.I think the key here is to use a process similar to converting a number into a different base, but instead of powers of a base, we're using factorials. In the decimal system, we use powers of 10, but here, each "digit" is multiplied by a factorial instead. So, maybe we can use a division algorithm approach, dividing by the largest factorial less than or equal to ( n ), then proceeding with the remainder.Let me try to formalize this. Suppose I have a natural number ( n ). I need to find the largest ( k ) such that ( k! leq n ). Once I have that ( k ), I can determine how many times ( k! ) fits into ( n ), which would give me ( a_k ). Then, I subtract ( a_k cdot k! ) from ( n ) and repeat the process with ( (k-1)! ), and so on, until I reach 1!.But wait, I need to ensure that each ( a_i ) satisfies ( 0 leq a_i leq i ). So, when I divide ( n ) by ( k! ), the quotient ( a_k ) must be less than or equal to ( k ). Is that always possible?Let me think about the inequality ( 1 cdot 1! + 2 cdot 2! + ldots + k cdot k! < (k+1)! ). If this holds, then the sum of all possible terms up to ( k ) is less than ( (k+1)! ), which means that any number less than ( (k+1)! ) can be represented with coefficients up to ( k ). This seems to be a crucial point.Let me test this inequality for small ( k ). For ( k = 1 ), the left side is ( 1 cdot 1! = 1 ), and the right side is ( 2! = 2 ). So, ( 1 < 2 ), which is true. For ( k = 2 ), the left side is ( 1 cdot 1! + 2 cdot 2! = 1 + 4 = 5 ), and the right side is ( 3! = 6 ). So, ( 5 < 6 ), which is also true. For ( k = 3 ), left side is ( 1 + 4 + 18 = 23 ), and right side is ( 4! = 24 ). Again, ( 23 < 24 ). Hmm, seems like it's holding.So, if this inequality is true for all ( k ), then it means that the maximum number representable with coefficients up to ( k ) is ( (k+1)! - 1 ). Therefore, any number less than ( (k+1)! ) can be represented with coefficients up to ( k ), each ( a_i ) satisfying ( 0 leq a_i leq i ).This suggests that for any ( n ), there exists a unique ( k ) such that ( k! leq n < (k+1)! ). Once we have this ( k ), we can find ( a_k ) as the quotient when ( n ) is divided by ( k! ), ensuring ( 0 leq a_k leq k ). Then, subtract ( a_k cdot k! ) from ( n ) and repeat the process for ( (k-1)! ), and so on, until we reach ( 1! ).Let me try an example to see if this works. Let's take ( n = 10 ). First, find the largest ( k ) such that ( k! leq 10 ). ( 3! = 6 leq 10 ), and ( 4! = 24 > 10 ), so ( k = 3 ). Now, divide 10 by 3! (which is 6). The quotient is 1, so ( a_3 = 1 ). Subtract ( 1 cdot 6 = 6 ) from 10, leaving a remainder of 4.Next, consider ( 2! = 2 ). Divide 4 by 2, which gives a quotient of 2. So, ( a_2 = 2 ). Subtract ( 2 cdot 2 = 4 ), leaving a remainder of 0. Finally, ( a_1 = 0 ) since there's no remainder left. So, the representation is ( 10 = 1 cdot 3! + 2 cdot 2! + 0 cdot 1! ). Let me check: ( 6 + 4 + 0 = 10 ). Yep, that works.Another example: ( n = 7 ). Largest ( k ) with ( k! leq 7 ) is 3 (since ( 3! = 6 leq 7 )). Divide 7 by 6, quotient is 1, so ( a_3 = 1 ). Subtract 6, remainder is 1. Next, ( 2! = 2 ). 1 divided by 2 is 0, so ( a_2 = 0 ). Remainder is still 1. Then, ( a_1 = 1 ). So, ( 7 = 1 cdot 3! + 0 cdot 2! + 1 cdot 1! ). Checking: ( 6 + 0 + 1 = 7 ). Correct.Wait, but what if the remainder is larger than the next factorial? Let me try ( n = 15 ). Largest ( k ) with ( k! leq 15 ) is 3 (since ( 3! = 6 leq 15 ), ( 4! = 24 > 15 )). Divide 15 by 6, quotient is 2 (since ( 2 cdot 6 = 12 leq 15 )), so ( a_3 = 2 ). Subtract 12, remainder is 3. Next, ( 2! = 2 ). Divide 3 by 2, quotient is 1, so ( a_2 = 1 ). Subtract 2, remainder is 1. Then, ( a_1 = 1 ). So, ( 15 = 2 cdot 3! + 1 cdot 2! + 1 cdot 1! ). Checking: ( 12 + 2 + 1 = 15 ). Correct.But wait, what if ( a_i ) exceeds ( i )? For example, if I have ( n = 14 ). Largest ( k ) is 3. ( 14 / 6 = 2 ), so ( a_3 = 2 ). Remainder is 2. Then, ( 2 / 2 = 1 ), so ( a_2 = 1 ). Remainder is 0. So, ( a_1 = 0 ). Thus, ( 14 = 2 cdot 3! + 1 cdot 2! + 0 cdot 1! ). That's fine because ( a_2 = 1 leq 2 ). Wait, but what if I had a remainder that required ( a_i > i )?Let me think. Suppose ( n = 8 ). Largest ( k ) is 3. ( 8 / 6 = 1 ), so ( a_3 = 1 ). Remainder is 2. Then, ( 2 / 2 = 1 ), so ( a_2 = 1 ). Remainder is 0. So, ( a_1 = 0 ). So, ( 8 = 1 cdot 3! + 1 cdot 2! + 0 cdot 1! ). That's fine.Wait, another example: ( n = 5 ). Largest ( k ) is 3 (since ( 3! = 6 > 5 ), so actually ( k = 2 ) because ( 2! = 2 leq 5 ). Wait, no, ( 3! = 6 > 5 ), so ( k = 2 ). So, divide 5 by 2, quotient is 2 (since ( 2 cdot 2 = 4 leq 5 )), so ( a_2 = 2 ). Subtract 4, remainder is 1. Then, ( a_1 = 1 ). So, ( 5 = 2 cdot 2! + 1 cdot 1! ). Checking: ( 4 + 1 = 5 ). Correct.But wait, ( a_2 = 2 ) which is equal to ( i = 2 ). So, that's allowed because ( 0 leq a_i leq i ). So, that's okay.Wait, let me try ( n = 4 ). Largest ( k ) is 2. ( 4 / 2 = 2 ), so ( a_2 = 2 ). Remainder is 0. So, ( a_1 = 0 ). Thus, ( 4 = 2 cdot 2! + 0 cdot 1! ). Correct.Another example: ( n = 3 ). Largest ( k ) is 2. ( 3 / 2 = 1 ), so ( a_2 = 1 ). Remainder is 1. Then, ( a_1 = 1 ). So, ( 3 = 1 cdot 2! + 1 cdot 1! ). Correct.Wait, what about ( n = 2 ). Largest ( k ) is 2. ( 2 / 2 = 1 ), so ( a_2 = 1 ). Remainder is 0. So, ( a_1 = 0 ). Thus, ( 2 = 1 cdot 2! + 0 cdot 1! ). Correct.And ( n = 1 ). Largest ( k ) is 1. ( 1 / 1 = 1 ), so ( a_1 = 1 ). So, ( 1 = 1 cdot 1! ). Correct.Okay, so far, all these examples are working. It seems like the process is:1. Find the largest ( k ) such that ( k! leq n ).2. Divide ( n ) by ( k! ) to get ( a_k ), ensuring ( 0 leq a_k leq k ).3. Subtract ( a_k cdot k! ) from ( n ) to get the remainder.4. Repeat the process with ( (k-1)! ) until all coefficients are found.This seems similar to the greedy algorithm, where at each step, we take the largest possible factorial and subtract as many as possible, then proceed to the next smaller factorial.Now, to prove this works for any ( n ), I need to show two things: existence and uniqueness.**Existence**: For any ( n ), we can find such coefficients ( a_1, a_2, ldots, a_k ) satisfying the given conditions.**Uniqueness**: There is only one such representation.Let me tackle existence first.**Existence Proof**:We can use induction on ( n ).**Base Case**: ( n = 1 ). As shown earlier, ( 1 = 1 cdot 1! ). So, the representation exists.**Inductive Step**: Assume that for all natural numbers ( m < n ), the representation exists. We need to show it exists for ( n ).Find the largest ( k ) such that ( k! leq n ). Let ( a_k = leftlfloor frac{n}{k!} rightrfloor ). Since ( k! leq n < (k+1)! ), we have ( a_k leq frac{n}{k!} < frac{(k+1)!}{k!} = k+1 ). Therefore, ( a_k leq k ). Also, ( a_k geq 1 ) because ( k! leq n ).Let ( n' = n - a_k cdot k! ). Then, ( n' = n - a_k cdot k! < n ). By the inductive hypothesis, ( n' ) can be represented as ( n' = a_1 cdot 1! + a_2 cdot 2! + ldots + a_{k-1} cdot (k-1)! ).Thus, combining these, ( n = a_k cdot k! + a_1 cdot 1! + a_2 cdot 2! + ldots + a_{k-1} cdot (k-1)! ), which is the desired representation.Therefore, by induction, every natural number ( n ) can be represented in the given form.**Uniqueness Proof**:Suppose, for contradiction, that there are two different representations for ( n ):1. ( n = a_1 cdot 1! + a_2 cdot 2! + ldots + a_k cdot k! )2. ( n = b_1 cdot 1! + b_2 cdot 2! + ldots + b_m cdot m! )where ( a_i ) and ( b_i ) satisfy ( 0 leq a_i, b_i leq i ), and ( a_k, b_m neq 0 ).Without loss of generality, assume ( k geq m ). If ( k > m ), then in the second representation, the highest factorial is ( m! ), but in the first, it's ( k! ). Since ( k! > m! ), the coefficients for ( k! ) in the second representation must be zero, but in the first, it's non-zero. However, since both equal ( n ), this would imply that ( a_k cdot k! = 0 ), which contradicts ( a_k neq 0 ). Therefore, ( k = m ).Now, consider the highest factorial ( k! ). Since both representations have ( k! ), we can equate the coefficients:( a_k cdot k! = b_k cdot k! )Thus, ( a_k = b_k ).Subtracting ( a_k cdot k! ) from both sides, we get:( n' = a_1 cdot 1! + a_2 cdot 2! + ldots + a_{k-1} cdot (k-1)! )and( n' = b_1 cdot 1! + b_2 cdot 2! + ldots + b_{k-1} cdot (k-1)! )By the same reasoning, we can show that ( a_{k-1} = b_{k-1} ), and so on, down to ( a_1 = b_1 ).Therefore, both representations are identical, proving uniqueness.**Conclusion**:By induction, we've shown that every natural number ( n ) can be expressed in the form ( n = a_1 cdot 1! + a_2 cdot 2! + ldots + a_k cdot k! ) with ( 0 leq a_i leq i ) and ( a_k neq 0 ). Furthermore, this representation is unique because any two representations must coincide at each coefficient step by step.So, the proof is complete.

answer：Okay, so I have this problem here where I need to find the intersection of two sets, M and N. Let me read it again to make sure I understand.Set M is given as {-1, 0, 1}. That's straightforward—those are just the elements in M. Then, set N is defined as {x | x² = x}. Hmm, okay, so N is the set of all x such that when you square x, you get x back. I think that means x squared equals x, so x² = x. I need to figure out what numbers satisfy that equation.Alright, let's solve the equation x² = x. If I subtract x from both sides, I get x² - x = 0. Then, I can factor out an x, which gives me x(x - 1) = 0. So, according to the zero product property, either x = 0 or x - 1 = 0. That means x = 0 or x = 1. So, set N is {0, 1}.Now, I need to find the intersection of M and N. The intersection of two sets is the set of elements that are common to both sets. So, let's look at M and N.Set M is {-1, 0, 1}, and set N is {0, 1}. The elements in both sets are 0 and 1. So, the intersection should be {0, 1}.Wait, let me double-check to make sure I didn't miss anything. Set M has three elements: -1, 0, and 1. Set N has two elements: 0 and 1. The common elements are definitely 0 and 1. So, yeah, the intersection is {0, 1}.I think that's it. I don't see any other elements that are in both sets. So, M ∩ N is {0, 1}.

answer：Okay, let's try to figure out this problem step by step. So, we have a bag with counters. Initially, there are 10 blue counters and Y yellow counters. Then, we add 2 more yellow counters and some blue counters, let's call that number X. After adding these counters, the proportion of yellow counters in the bag remains the same as it was before adding the counters. We need to find all possible values of Y.First, let's understand what the proportion of yellow counters means. Initially, the total number of counters is Y (yellow) + 10 (blue). So, the proportion of yellow counters is Y divided by (Y + 10). After adding 2 yellow counters and X blue counters, the number of yellow counters becomes Y + 2, and the number of blue counters becomes 10 + X. Therefore, the total number of counters is (Y + 2) + (10 + X) = Y + X + 12. The proportion of yellow counters now is (Y + 2) divided by (Y + X + 12).According to the problem, these two proportions are equal. So, we can set up the equation:Y / (Y + 10) = (Y + 2) / (Y + X + 12)Now, we need to solve this equation for Y. Let's cross-multiply to eliminate the fractions:Y * (Y + X + 12) = (Y + 2) * (Y + 10)Expanding both sides:Y^2 + YX + 12Y = Y^2 + 10Y + 2Y + 20Simplify the right side:Y^2 + YX + 12Y = Y^2 + 12Y + 20Now, subtract Y^2 and 12Y from both sides:YX = 20So, we have Y multiplied by X equals 20. Since Y and X are both positive integers (you can't have a negative or fractional number of counters), we need to find all pairs of positive integers (Y, X) such that their product is 20.Let's list the positive integer factors of 20:1, 2, 4, 5, 10, 20So, the possible pairs (Y, X) are:(1, 20), (2, 10), (4, 5), (5, 4), (10, 2), (20, 1)Therefore, the possible values of Y are 1, 2, 4, 5, 10, and 20.Wait, let me double-check if these values make sense in the context of the problem. For each Y, we need to ensure that when we add 2 yellow counters and X blue counters, the proportion remains the same.Let's test Y = 1:Initial proportion: 1 / (1 + 10) = 1/11After adding 2 yellow and 20 blue: (1 + 2) / (1 + 20 + 10) = 3/33 = 1/11Okay, that works.Y = 2:Initial proportion: 2/12 = 1/6After adding 2 yellow and 10 blue: 4/22 = 2/11Wait, that's not equal to 1/6. Hmm, did I make a mistake?Wait, 2/12 is 1/6, and 4/22 simplifies to 2/11, which is approximately 0.1818, whereas 1/6 is approximately 0.1667. These are not equal. So, Y = 2 doesn't work? But according to our earlier equation, YX = 20, so Y = 2 and X = 10 should satisfy the equation.Wait, let's plug back into the equation:Y / (Y + 10) = (Y + 2) / (Y + X + 12)For Y = 2, X = 10:Left side: 2 / 12 = 1/6Right side: 4 / (2 + 10 + 12) = 4 / 24 = 1/6Oh, wait, I miscalculated earlier. 2 + 10 + 12 is 24, not 22. So, 4/24 is indeed 1/6. So, it does work. My mistake earlier was adding incorrectly.Similarly, let's check Y = 4:Initial proportion: 4 / 14 ≈ 0.2857After adding 2 yellow and 5 blue: 6 / (4 + 5 + 12) = 6 / 21 ≈ 0.2857That works.Y = 5:Initial proportion: 5 / 15 = 1/3 ≈ 0.3333After adding 2 yellow and 4 blue: 7 / (5 + 4 + 12) = 7 / 21 = 1/3 ≈ 0.3333Good.Y = 10:Initial proportion: 10 / 20 = 1/2After adding 2 yellow and 2 blue: 12 / (10 + 2 + 12) = 12 / 24 = 1/2Perfect.Y = 20:Initial proportion: 20 / 30 = 2/3 ≈ 0.6667After adding 2 yellow and 1 blue: 22 / (20 + 1 + 12) = 22 / 33 = 2/3 ≈ 0.6667That works too.So, all the values Y = 1, 2, 4, 5, 10, 20 satisfy the condition that the proportion of yellow counters remains unchanged after adding 2 yellow and some blue counters.I think that's all. So, the possible values of Y are 1, 2, 4, 5, 10, and 20.