💡First, I need to determine how much Jose invested in the shop. Tom invested Rs. 30,000 for the entire year, while Jose joined 2 months later, so he invested for 10 months.The total profit earned after one year is Rs. 27,000, and Jose's share of the profit is Rs. 15,000. This means Tom's share is Rs. 12,000.The ratio of their profits should be equal to the ratio of their investments multiplied by the time they invested. Therefore, the ratio of Tom's investment to Jose's investment is:[frac{30,000 times 12}{X times 10} = frac{4}{5}]Solving for ( X ):[30,000 times 12 times 5 = 4 times X times 10][1,800,000 times 5 = 40X][9,000,000 = 40X][X = frac{9,000,000}{40} = 225,000]Therefore, Jose invested Rs. 225,000.

💡Okay, so I have this problem about "equation number pairs." It sounds a bit complicated, but I think I can figure it out step by step. Let me start by understanding the definition they gave.They say that two real numbers ( a ) and ( b ) are called an "equation number pair" if they satisfy the fractional equation ( frac{a}{x} + 1 = b ) with the solution ( x = frac{1}{a + b} ). So, basically, if I plug ( a ) and ( b ) into that equation, the solution for ( x ) should be ( frac{1}{a + b} ). Got it.Let me look at part (1) first. They give me two number pairs: ( [3, -5] ) and ( [-2, 4] ). I need to determine which of these are "equation number pairs." Starting with ( [3, -5] ):1. Substitute ( a = 3 ) and ( b = -5 ) into the equation: ( frac{3}{x} + 1 = -5 ).2. Let me solve for ( x ). Subtract 1 from both sides: ( frac{3}{x} = -6 ).3. Then, multiply both sides by ( x ): ( 3 = -6x ).4. Divide both sides by -6: ( x = -frac{3}{6} = -frac{1}{2} ).5. Now, check if this ( x ) is equal to ( frac{1}{a + b} ). Calculate ( a + b = 3 + (-5) = -2 ).6. So, ( frac{1}{a + b} = frac{1}{-2} = -frac{1}{2} ), which matches the solution ( x = -frac{1}{2} ).7. Therefore, ( [3, -5] ) is an "equation number pair."Now, checking ( [-2, 4] ):1. Substitute ( a = -2 ) and ( b = 4 ) into the equation: ( frac{-2}{x} + 1 = 4 ).2. Solve for ( x ). Subtract 1 from both sides: ( frac{-2}{x} = 3 ).3. Multiply both sides by ( x ): ( -2 = 3x ).4. Divide both sides by 3: ( x = -frac{2}{3} ).5. Check if this ( x ) is equal to ( frac{1}{a + b} ). Calculate ( a + b = -2 + 4 = 2 ).6. So, ( frac{1}{a + b} = frac{1}{2} ).7. But our solution was ( x = -frac{2}{3} ), which is not equal to ( frac{1}{2} ).8. Therefore, ( [-2, 4] ) is not an "equation number pair."Alright, so part (1) is done. Only ( [3, -5] ) is an "equation number pair."Moving on to part (2): If the number pair ( [n, 3 - n] ) is an "equation number pair," find the value of ( n ).1. So, ( a = n ) and ( b = 3 - n ).2. Substitute into the equation: ( frac{n}{x} + 1 = 3 - n ).3. Let me solve for ( x ). Subtract 1 from both sides: ( frac{n}{x} = 2 - n ).4. Multiply both sides by ( x ): ( n = (2 - n)x ).5. So, ( x = frac{n}{2 - n} ).6. According to the definition, ( x ) should also be ( frac{1}{a + b} = frac{1}{n + (3 - n)} = frac{1}{3} ).7. Therefore, set ( frac{n}{2 - n} = frac{1}{3} ).8. Cross-multiplying: ( 3n = 2 - n ).9. Bring ( n ) to the left: ( 3n + n = 2 ) => ( 4n = 2 ).10. Divide both sides by 4: ( n = frac{2}{4} = frac{1}{2} ).So, ( n = frac{1}{2} ).Now, part (3): If the number pair ( [m - k, k] ) (with ( m neq -1 ) and ( m neq 0 ), ( k neq 1 )) is an "equation number pair," express ( k ) in terms of ( m ).1. Here, ( a = m - k ) and ( b = k ).2. Substitute into the equation: ( frac{m - k}{x} + 1 = k ).3. Solve for ( x ). Subtract 1 from both sides: ( frac{m - k}{x} = k - 1 ).4. Multiply both sides by ( x ): ( m - k = (k - 1)x ).5. So, ( x = frac{m - k}{k - 1} ).6. According to the definition, ( x ) should also be ( frac{1}{a + b} = frac{1}{(m - k) + k} = frac{1}{m} ).7. Therefore, set ( frac{m - k}{k - 1} = frac{1}{m} ).8. Cross-multiplying: ( m(m - k) = k - 1 ).9. Expand the left side: ( m^2 - mk = k - 1 ).10. Bring all terms to one side: ( m^2 - mk - k + 1 = 0 ).11. Factor terms with ( k ): ( m^2 + 1 - k(m + 1) = 0 ).12. Solve for ( k ): ( k(m + 1) = m^2 + 1 ).13. Therefore, ( k = frac{m^2 + 1}{m + 1} ).Let me double-check this result. If I substitute ( k = frac{m^2 + 1}{m + 1} ) back into the equation, does it satisfy?1. Compute ( a = m - k = m - frac{m^2 + 1}{m + 1} ).2. Let's get a common denominator: ( frac{m(m + 1) - (m^2 + 1)}{m + 1} = frac{m^2 + m - m^2 - 1}{m + 1} = frac{m - 1}{m + 1} ).3. Compute ( b = k = frac{m^2 + 1}{m + 1} ).4. Then, ( a + b = frac{m - 1}{m + 1} + frac{m^2 + 1}{m + 1} = frac{m - 1 + m^2 + 1}{m + 1} = frac{m^2 + m}{m + 1} = frac{m(m + 1)}{m + 1} = m ).5. So, ( frac{1}{a + b} = frac{1}{m} ).6. Now, let's check the original equation: ( frac{a}{x} + 1 = b ).7. Substitute ( a = frac{m - 1}{m + 1} ), ( b = frac{m^2 + 1}{m + 1} ), and ( x = frac{1}{m} ).8. Compute ( frac{a}{x} = frac{frac{m - 1}{m + 1}}{frac{1}{m}} = frac{(m - 1)m}{m + 1} = frac{m^2 - m}{m + 1} ).9. Then, ( frac{a}{x} + 1 = frac{m^2 - m}{m + 1} + 1 = frac{m^2 - m + m + 1}{m + 1} = frac{m^2 + 1}{m + 1} = b ).10. It checks out! So, ( k = frac{m^2 + 1}{m + 1} ) is correct.So, summarizing my findings:1. Only ( [3, -5] ) is an "equation number pair."2. The value of ( n ) is ( frac{1}{2} ).3. ( k ) is expressed as ( frac{m^2 + 1}{m + 1} ).I think I've covered all parts and double-checked my work. Hopefully, I didn't make any mistakes!**Final Answer**(1) boxed{3, -5}(2) boxed{dfrac{1}{2}}(3) boxed{dfrac{m^2 + 1}{m + 1}}

💡Alright, so I have this problem where I need to construct a segment ( x ) using just a compass and straightedge, and it has to satisfy the ratio ( frac{x}{a} = frac{b}{c} ). Okay, let's break this down. I have three given segments: ( a ), ( b ), and ( c ). My goal is to find ( x ) such that when I divide ( x ) by ( a ), it's the same as dividing ( b ) by ( c ).First, I need to recall some basic principles of compass and straightedge constructions. I remember that one common method to create proportional segments is by using similar triangles. If I can set up two triangles that are similar, then their corresponding sides will be proportional. That seems like a good approach here.So, maybe I can construct two triangles where one triangle has sides ( a ) and ( c ), and the other triangle will have sides ( x ) and ( b ). If these triangles are similar, then the ratios of their corresponding sides should be equal, which would give me ( frac{x}{a} = frac{b}{c} ).But how do I actually construct these triangles? I think I need to start by drawing a base line or an angle. Let me try drawing an angle with vertex ( O ). I'll label one side of the angle as ( OA ) and the other as ( OB ). Now, I can place points along these sides to represent the segments ( a ), ( b ), and ( c ).Wait, maybe I should use the concept of similar triangles by constructing parallel lines. If I can create a line parallel to one side of a triangle that intersects the other two sides proportionally, then I can use that to find the desired segment ( x ). This is similar to the Basic Proportionality Theorem, also known as Thales' theorem.Okay, let's try that. I'll draw a line segment ( OC ) of length ( c ) on one side of the angle, and from point ( C ), I'll mark another segment ( CB ) of length ( b ). On the other side of the angle, I'll mark a segment ( OA ) of length ( a ). Now, if I can draw a line through point ( B ) that's parallel to segment ( AC ), this line should intersect the extension of ( OA ) at some point ( D ). According to the theorem, the ratio ( frac{AD}{OA} ) should be equal to ( frac{BC}{OC} ), which is ( frac{b}{c} ).So, substituting the known values, ( frac{AD}{a} = frac{b}{c} ), which means ( AD = a cdot frac{b}{c} ). Therefore, the segment ( OD ) should be the desired ( x ).But wait, I need to make sure that I can actually perform all these steps with just a compass and straightedge. Drawing an angle is straightforward, and marking segments of given lengths is also something I can do with a compass. Drawing a parallel line might be a bit tricky, but I remember there's a method to construct a line parallel to a given line through a specific point using a compass and straightedge.Let me recall how to construct a parallel line. I think it involves creating congruent angles. If I can replicate the angle that segment ( AC ) makes with one side of the angle ( AOB ) at point ( B ), then the line I draw through ( B ) with that angle should be parallel to ( AC ).So, to do that, I'll first draw a line segment ( AC ). Then, using my compass, I'll measure the angle at ( C ) between ( AC ) and ( CB ). I'll replicate that angle at point ( B ) on the other side of the angle ( AOB ). This should give me the direction for the parallel line.Once I have that parallel line, it should intersect the extension of ( OA ) at point ( D ). Then, measuring the segment ( OD ) should give me the length ( x ) that satisfies the ratio ( frac{x}{a} = frac{b}{c} ).Hmm, let me visualize this. I have angle ( AOB ), with ( OA = a ), ( OC = c ), and ( CB = b ). Drawing ( AC ), then constructing a parallel line through ( B ) to ( AC ), which intersects the extension of ( OA ) at ( D ). Then ( OD = x ).I think this makes sense. By constructing similar triangles ( AOC ) and ( BOD ), the sides should be proportional. So, ( frac{AO}{OC} = frac{BO}{OD} ), which translates to ( frac{a}{c} = frac{b}{x} ), hence ( x = frac{b}{c} cdot a ).But I need to make sure that all these steps are possible with just a compass and straightedge. Drawing the angle, marking the segments, constructing the parallel line—all of these are standard compass and straightedge operations. So, I think this method is valid.Let me summarize the steps to ensure I haven't missed anything:1. Draw an angle ( AOB ) with vertex ( O ).2. On one side, mark point ( C ) such that ( OC = c ).3. From ( C ), mark point ( B ) such that ( CB = b ).4. On the other side, mark point ( A ) such that ( OA = a ).5. Draw segment ( AC ).6. Construct a line through ( B ) parallel to ( AC ).7. Let this parallel line intersect the extension of ( OA ) at point ( D ).8. The segment ( OD ) is the desired ( x ).Yes, that seems comprehensive. I don't see any steps that would require tools beyond a compass and straightedge. Each operation—drawing lines, marking segments, constructing parallel lines—is achievable with those tools.I might also want to verify this construction with an example. Suppose ( a = 2 ), ( b = 3 ), and ( c = 4 ). Then, ( x ) should be ( frac{3}{4} times 2 = 1.5 ). If I follow the steps, I should end up with ( OD = 1.5 ). Let me try to sketch this mentally.Drawing angle ( AOB ), placing ( OC = 4 ), then ( CB = 3 ), and ( OA = 2 ). Drawing ( AC ), then a parallel line through ( B ) intersecting the extension of ( OA ) at ( D ). Measuring ( OD ) should give ( 1.5 ). That seems correct.Another thing to consider is if the construction works for different configurations. What if ( a ), ( b ), and ( c ) are not colinear or form different angles? I think the key here is that the construction relies on the proportionality of similar triangles, which should hold regardless of the specific lengths or angles, as long as the parallel line is correctly constructed.Also, I should remember that when extending ( OA ), it might be necessary to extend it beyond ( A ) if the parallel line doesn't intersect within the original segment. But since we're dealing with ratios, the extension should still maintain the proportionality.Is there another method to achieve this? Perhaps using similar triangles in a different configuration? I could consider constructing a triangle where one side is ( a ) and another side is ( c ), and then scaling it by the ratio ( frac{b}{c} ) to get ( x ). But that might essentially be the same as what I've already outlined.Alternatively, I could use the concept of intercept theorems or use a number line approach, but those might be more complicated with just compass and straightedge. The method I've thought through seems straightforward and relies on fundamental geometric principles.I also need to ensure that I'm not assuming any prior knowledge beyond basic compass and straightedge constructions. Each step should be explainable in terms of those basic operations. For example, constructing a parallel line involves creating congruent angles, which can be done by copying an angle using a compass.In summary, the construction involves setting up an angle, marking the given segments, drawing a key segment, constructing a parallel line, and then using the intersection point to determine the desired segment ( x ). This approach leverages the properties of similar triangles and proportional segments to achieve the required ratio.I think I've covered all the necessary steps and verified that each part is achievable with the given tools. I don't see any gaps in the logic, and the example I considered seems to validate the method. Therefore, I'm confident that this construction will yield the correct segment ( x ) such that ( frac{x}{a} = frac{b}{c} ).

💡Alright, so I have this sequence problem here. Let me try to figure it out step by step. The sequence is defined as {a_n}, with a_1 = 1, and the recursive formula a_{n+1} = (2a_n)/(a_n + 2) for any natural number n. I need to find a_5.Hmm, okay. So starting with a_1 = 1, I can compute a_2, then a_3, and so on up to a_5. That seems straightforward, but maybe there's a pattern or a formula I can find to make it easier.Let me write down the first few terms to see if I can spot a pattern.Given:a_1 = 1Compute a_2:a_2 = (2 * a_1)/(a_1 + 2) = (2 * 1)/(1 + 2) = 2/3Compute a_3:a_3 = (2 * a_2)/(a_2 + 2) = (2 * (2/3))/((2/3) + 2) = (4/3)/(8/3) = (4/3) * (3/8) = 12/24 = 1/2Compute a_4:a_4 = (2 * a_3)/(a_3 + 2) = (2 * (1/2))/((1/2) + 2) = (1)/(5/2) = 1 * (2/5) = 2/5Compute a_5:a_5 = (2 * a_4)/(a_4 + 2) = (2 * (2/5))/((2/5) + 2) = (4/5)/(12/5) = (4/5) * (5/12) = 20/60 = 1/3Wait, so a_5 is 1/3. That's option A. But let me double-check my calculations because sometimes when dealing with fractions, it's easy to make a mistake.Starting again:a_1 = 1a_2 = (2*1)/(1+2) = 2/3. That's correct.a_3 = (2*(2/3))/((2/3)+2) = (4/3)/(8/3) = (4/3)*(3/8) = 12/24 = 1/2. Correct.a_4 = (2*(1/2))/((1/2)+2) = (1)/(5/2) = 2/5. Correct.a_5 = (2*(2/5))/((2/5)+2) = (4/5)/(12/5) = (4/5)*(5/12) = 20/60 = 1/3. Yep, that's correct.But wait, maybe there's a smarter way to do this without computing each term step by step. Sometimes, sequences like this can be transformed into something more manageable, like an arithmetic or geometric sequence.Looking at the recursive formula: a_{n+1} = (2a_n)/(a_n + 2). Hmm, maybe I can take reciprocals to simplify this.Let me define b_n = 1/a_n. Then, let's see what the recursive formula becomes.Starting with a_{n+1} = (2a_n)/(a_n + 2), take reciprocals:1/a_{n+1} = (a_n + 2)/(2a_n) = (a_n)/(2a_n) + 2/(2a_n) = 1/2 + 1/a_n.So, 1/a_{n+1} = 1/a_n + 1/2. That simplifies to b_{n+1} = b_n + 1/2.Ah, so the sequence {b_n} is an arithmetic sequence with common difference 1/2.Given that, since b_n = 1/a_n, and a_1 = 1, then b_1 = 1/a_1 = 1.So, the first term of the arithmetic sequence is 1, and each subsequent term increases by 1/2.Therefore, the general formula for b_n is:b_n = b_1 + (n - 1)*d = 1 + (n - 1)*(1/2) = 1 + (n - 1)/2 = (2 + n - 1)/2 = (n + 1)/2.So, b_n = (n + 1)/2.Therefore, since b_n = 1/a_n, we have a_n = 2/(n + 1).So, a_n = 2/(n + 1). Let me test this formula with the terms I computed earlier.For n = 1: a_1 = 2/(1 + 1) = 2/2 = 1. Correct.For n = 2: a_2 = 2/(2 + 1) = 2/3. Correct.For n = 3: a_3 = 2/(3 + 1) = 2/4 = 1/2. Correct.For n = 4: a_4 = 2/(4 + 1) = 2/5. Correct.For n = 5: a_5 = 2/(5 + 1) = 2/6 = 1/3. Correct.So, the formula works. Therefore, a_5 is indeed 1/3.But just to make sure, let me think if there's another way to approach this problem, maybe using induction or another method.Suppose I try mathematical induction to prove that a_n = 2/(n + 1).Base case: n = 1. a_1 = 2/(1 + 1) = 1. Correct.Inductive step: Assume that for some k ≥ 1, a_k = 2/(k + 1). Then, a_{k+1} = (2a_k)/(a_k + 2) = (2*(2/(k + 1)))/(2/(k + 1) + 2).Simplify numerator: 4/(k + 1).Denominator: 2/(k + 1) + 2 = (2 + 2(k + 1))/(k + 1) = (2 + 2k + 2)/(k + 1) = (2k + 4)/(k + 1) = 2(k + 2)/(k + 1).Therefore, a_{k+1} = (4/(k + 1)) / (2(k + 2)/(k + 1)) = (4/(k + 1)) * ((k + 1)/(2(k + 2))) = 4/(2(k + 2)) = 2/(k + 2).Which is 2/((k + 1) + 1), so the formula holds for k + 1.Thus, by induction, the formula a_n = 2/(n + 1) holds for all natural numbers n.Therefore, a_5 = 2/(5 + 1) = 2/6 = 1/3.So, the answer is A: 1/3.But just to make sure, let me see if I can think of another perspective. Maybe using continued fractions or something else, but I think the reciprocal method is the most straightforward here.Alternatively, I could write out the terms again:a_1 = 1a_2 = 2/3a_3 = 1/2a_4 = 2/5a_5 = 1/3Yes, that's consistent with the formula.Alternatively, if I didn't see the reciprocal trick, I might have tried to compute each term step by step, which I did initially, and that worked too, but it's more time-consuming.So, in conclusion, the fifth term a_5 is 1/3, which is option A.**Final Answer**boxed{A}

💡Okay, so Annie has this test with 6 questions, each having two parts. Each part is graded with one of nine letter grades: A, B, C, D, E, F, G, H, I. Each of these grades corresponds to a unique numerical score. She knows that A is 1, E is 0.5, and I is 0. The rest of the grades (B, C, D, F, G, H) have unique numerical scores that she doesn't know yet, but she can figure them out based on her test results.Annie got two of each of A, E, and I. That means she has 2 A's, 2 E's, and 2 I's across all 12 parts of the test (since there are 6 questions, each with 2 parts). She also knows the total numerical scores for each question, which are the sums of the scores of the two parts. From this information, she can determine the numerical scores corresponding to all 9 letter grades. We need to find the number of ways she can receive these letter grades, and then determine the exponent of 2 in the prime factorization of that number.First, let's break down the problem step by step.1. **Determine the placement of A, E, and I:** - Annie has 6 questions, each with two parts. She has 2 A's, 2 E's, and 2 I's to distribute across these 12 parts. - Each question must have one part with one of these grades (A, E, I). So, for each question, we need to decide which part (first or second) gets which grade. - The number of ways to distribute 2 A's, 2 E's, and 2 I's among the 6 questions is given by the multinomial coefficient: [ frac{6!}{2!2!2!} = 90 ] - This is because we're dividing 6 questions into three groups of 2 each (for A, E, and I).2. **Assigning parts to each grade:** - For each question, once we've decided which grade (A, E, or I) it gets, we need to assign it to either the first part or the second part of the question. - Since each question has two parts, and each grade can be assigned to either part, there are 2 choices per question. - Therefore, for 6 questions, the total number of ways to assign the parts is: [ 2^6 = 64 ]3. **Ordering the remaining grades:** - After placing the A's, E's, and I's, there are 6 remaining parts that need to be filled with the remaining 6 grades (B, C, D, F, G, H). - These remaining grades must be assigned in such a way that each grade is used exactly once across the 6 parts. - The number of ways to order these 6 remaining grades is: [ 6! = 720 ]4. **Calculate the total number of ways:** - The total number of ways to assign the grades is the product of the three quantities calculated above: [ 90 times 64 times 720 ] - Let's compute this: [ 90 times 64 = 5760 ] [ 5760 times 720 = 4,147,200 ]5. **Prime factorization of the total number of ways:** - Now, we need to find the prime factorization of 4,147,200 to determine the exponent of 2. - Let's break it down: [ 4,147,200 = 4147200 ] - Divide by 2 repeatedly: [ 4147200 ÷ 2 = 2073600 ] [ 2073600 ÷ 2 = 1036800 ] [ 1036800 ÷ 2 = 518400 ] [ 518400 ÷ 2 = 259200 ] [ 259200 ÷ 2 = 129600 ] [ 129600 ÷ 2 = 64800 ] [ 64800 ÷ 2 = 32400 ] [ 32400 ÷ 2 = 16200 ] [ 16200 ÷ 2 = 8100 ] [ 8100 ÷ 2 = 4050 ] [ 4050 ÷ 2 = 2025 ] - We divided by 2 a total of 11 times before we got an odd number (2025). So, the exponent of 2 is 11.Therefore, the exponent of 2 in the prime factorization of the number of ways Annie can receive her letter grades is 11.**Final Answer**boxed{11}

💡Okay, so I have this equation: ||x + 3| - 1| = a, and I need to find the value of 'a' that makes this equation have three solutions. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that absolute value equations can sometimes have multiple solutions because the expression inside the absolute value can be positive or negative. So, maybe I need to consider different cases based on the value of 'x'. Let me start by understanding the inner absolute value: |x + 3|. This expression will always be non-negative, right? So, |x + 3| is equal to x + 3 when x + 3 is non-negative, and it's equal to -(x + 3) when x + 3 is negative. That means:- If x + 3 ≥ 0, which is the same as x ≥ -3, then |x + 3| = x + 3.- If x + 3 < 0, which is x < -3, then |x + 3| = -(x + 3) = -x - 3.Okay, so now I can rewrite the original equation ||x + 3| - 1| = a by considering these two cases.**Case 1: x ≥ -3**In this case, |x + 3| = x + 3, so the equation becomes |(x + 3) - 1| = a, which simplifies to |x + 2| = a.Now, solving |x + 2| = a gives two possibilities:1. x + 2 = a, which leads to x = a - 2.2. x + 2 = -a, which leads to x = -a - 2.But wait, since we're in the case where x ≥ -3, we need to make sure that these solutions satisfy x ≥ -3.So, for x = a - 2 to be valid, we need a - 2 ≥ -3. That simplifies to a ≥ -1.Similarly, for x = -a - 2 to be valid, we need -a - 2 ≥ -3. Let's solve that:-a - 2 ≥ -3 Add 2 to both sides: -a ≥ -1 Multiply both sides by -1 (and reverse the inequality): a ≤ 1.So, in this case, the solutions are x = a - 2 and x = -a - 2, provided that a ≥ -1 and a ≤ 1, respectively.**Case 2: x < -3**Here, |x + 3| = -x - 3, so the equation becomes |(-x - 3) - 1| = a, which simplifies to |-x - 4| = a.Again, solving |-x - 4| = a gives two possibilities:1. -x - 4 = a, which leads to x = -a - 4.2. -x - 4 = -a, which leads to x = a - 4.But since we're in the case where x < -3, we need to ensure these solutions satisfy x < -3.For x = -a - 4 to be valid, we need -a - 4 < -3. Let's solve that:-a - 4 < -3 Add 4 to both sides: -a < 1 Multiply both sides by -1 (and reverse the inequality): a > -1.For x = a - 4 to be valid, we need a - 4 < -3. Let's solve that:a - 4 < -3 Add 4 to both sides: a < 1.So, in this case, the solutions are x = -a - 4 and x = a - 4, provided that a > -1 and a < 1, respectively.Now, let's summarize all the possible solutions we've found:- From Case 1 (x ≥ -3): - x = a - 2, valid if a ≥ -1. - x = -a - 2, valid if a ≤ 1.- From Case 2 (x < -3): - x = -a - 4, valid if a > -1. - x = a - 4, valid if a < 1.So, depending on the value of 'a', we can have different numbers of solutions.But the question asks for the value of 'a' that results in three solutions. Hmm, how can that happen?Well, let's think about how many solutions we can get from each case.If 'a' is such that both cases contribute solutions, we might have up to four solutions. But we need only three. That suggests that one of the solutions from either Case 1 or Case 2 must coincide or perhaps be invalid, reducing the total number of solutions by one.Let me consider the conditions for each solution to be valid.From Case 1:- x = a - 2 is valid if a ≥ -1.- x = -a - 2 is valid if a ≤ 1.From Case 2:- x = -a - 4 is valid if a > -1.- x = a - 4 is valid if a < 1.So, if a = -1, let's see what happens.At a = -1:- From Case 1: - x = a - 2 = -1 - 2 = -3. But x = -3 is the boundary between Case 1 and Case 2. So, does this count as a solution in Case 1? Well, x = -3 is included in Case 1 (x ≥ -3), so yes, it's a valid solution. - x = -a - 2 = 1 - 2 = -1. This is also valid since a = -1 ≤ 1.- From Case 2: - x = -a - 4 = 1 - 4 = -3. But x = -3 is not less than -3, so this is invalid. - x = a - 4 = -1 - 4 = -5. This is valid since a = -1 < 1.So, at a = -1, we have solutions at x = -3, x = -1, and x = -5. That's three solutions. Hmm, interesting.But wait, let me check if a = -1 is valid. Because absolute values are always non-negative, so 'a' must be greater than or equal to 0. So, a = -1 would make the equation ||x + 3| - 1| = -1, which is impossible because the left side is always non-negative. Therefore, a cannot be negative. So, a = -1 is invalid.Okay, so a must be non-negative. Let's try a = 1.At a = 1:- From Case 1: - x = a - 2 = 1 - 2 = -1. Valid since a = 1 ≥ -1. - x = -a - 2 = -1 - 2 = -3. Valid since a = 1 ≤ 1.- From Case 2: - x = -a - 4 = -1 - 4 = -5. Valid since a = 1 > -1. - x = a - 4 = 1 - 4 = -3. But x = -3 is not less than -3, so this is invalid.So, at a = 1, we have solutions at x = -1, x = -3, and x = -5. Again, three solutions. But wait, is x = -3 counted twice? Once from Case 1 and once from Case 2? Let me check:From Case 1, x = -3 is a solution because when a = 1, x = -a - 2 = -3.From Case 2, x = a - 4 = -3, but since x must be less than -3, this solution is invalid. So, x = -3 is only counted once.Therefore, at a = 1, we have three distinct solutions: x = -1, x = -3, and x = -5.Wait a second, but when a = 1, let's plug it back into the original equation to verify:||x + 3| - 1| = 1.So, let's solve this:Case 1: x ≥ -3|x + 3| = x + 3So, |x + 3 - 1| = |x + 2| = 1Which gives x + 2 = 1 => x = -1Or x + 2 = -1 => x = -3Case 2: x < -3|x + 3| = -x - 3So, |-x - 3 - 1| = |-x - 4| = 1Which gives -x - 4 = 1 => -x = 5 => x = -5Or -x - 4 = -1 => -x = 3 => x = -3But x = -3 is not less than -3, so only x = -5 is valid.So, the solutions are x = -1, x = -3, and x = -5. That's three solutions.But wait, when a = 1, the equation ||x + 3| - 1| = 1 has three solutions. So, a = 1 is the answer.But earlier, when I thought about a = -1, it gave three solutions, but a can't be negative because absolute values are non-negative. So, a must be 1.Wait, but let me think again. Is there another value of 'a' that could give three solutions? For example, if 'a' is between 0 and 1, would that also give three solutions?Let me try a = 0.5.At a = 0.5:From Case 1:- x = 0.5 - 2 = -1.5- x = -0.5 - 2 = -2.5From Case 2:- x = -0.5 - 4 = -4.5- x = 0.5 - 4 = -3.5So, all four solutions are valid because:- For x = -1.5, x ≥ -3: valid- For x = -2.5, x ≥ -3: valid- For x = -4.5, x < -3: valid- For x = -3.5, x < -3: validSo, at a = 0.5, we have four solutions: x = -1.5, x = -2.5, x = -4.5, x = -3.5.So, when a is between 0 and 1, we have four solutions. When a = 1, one of the solutions from Case 2 becomes invalid (x = -3), so we have three solutions. Similarly, when a = 0, let's see:At a = 0:From Case 1:- x = 0 - 2 = -2- x = -0 - 2 = -2So, x = -2 is a double root.From Case 2:- x = -0 - 4 = -4- x = 0 - 4 = -4So, x = -4 is a double root.But wait, does that mean at a = 0, we have two solutions: x = -2 and x = -4, each with multiplicity two? Or are they considered as two distinct solutions?In terms of distinct solutions, x = -2 and x = -4 are two solutions. So, at a = 0, we have two solutions.Therefore, the number of solutions changes as 'a' varies:- For a < 0: No solutions, because absolute value can't be negative.- For a = 0: Two solutions.- For 0 < a < 1: Four solutions.- For a = 1: Three solutions.- For a > 1: Let's check.At a = 2:From Case 1:- x = 2 - 2 = 0- x = -2 - 2 = -4From Case 2:- x = -2 - 4 = -6- x = 2 - 4 = -2But x = -2 is not less than -3, so invalid.So, solutions are x = 0, x = -4, and x = -6. That's three solutions.Wait, so at a = 2, we also have three solutions. Hmm, that contradicts my earlier thought that only a = 1 gives three solutions.Wait, let me check again.At a = 2:Case 1: x ≥ -3|x + 3| = x + 3So, |x + 3 - 1| = |x + 2| = 2Which gives x + 2 = 2 => x = 0Or x + 2 = -2 => x = -4But x = -4 is less than -3, so in Case 1, x = -4 is not valid because x must be ≥ -3. Wait, no, in Case 1, x ≥ -3, so x = -4 is not in this case. So, only x = 0 is valid from Case 1.From Case 2: x < -3|x + 3| = -x - 3So, |-x - 3 - 1| = |-x - 4| = 2Which gives -x - 4 = 2 => -x = 6 => x = -6Or -x - 4 = -2 => -x = 2 => x = -2But x = -2 is not less than -3, so invalid.So, only x = -6 is valid from Case 2.Therefore, at a = 2, we have two solutions: x = 0 and x = -6.Wait, that's only two solutions, not three. So, my earlier thought was wrong. So, at a = 2, we have two solutions.Wait, but earlier when I thought a = 1 gives three solutions, and a = 2 gives two solutions, but when a = 1, we have three solutions, and when a increases beyond 1, the number of solutions decreases.Wait, let me think again. Maybe I made a mistake in counting.At a = 1:Case 1: x = -1 and x = -3Case 2: x = -5So, three solutions.At a = 2:Case 1: x = 0Case 2: x = -6So, two solutions.At a = 0.5:Case 1: x = -1.5 and x = -2.5Case 2: x = -4.5 and x = -3.5So, four solutions.At a = 0:Case 1: x = -2 (double root)Case 2: x = -4 (double root)So, two distinct solutions.Wait, so the number of solutions is:- 0 solutions for a < 0- 2 solutions for a = 0- 4 solutions for 0 < a < 1- 3 solutions for a = 1- 2 solutions for a > 1So, the only value of 'a' that gives exactly three solutions is a = 1.Therefore, the answer is a = 1.