💡Okay, so I have this polynomial g(x) = ax³ - 8x² + bx - 7, and I need to find the values of a and b. The problem says that when g(x) is divided by x - 2, the remainder is 1, and when it's divided by x + 3, the remainder is -89. Hmm, I remember something about the Remainder Theorem from my algebra class. Let me try to recall how that works.The Remainder Theorem states that if a polynomial f(x) is divided by (x - c), the remainder is f(c). So, in this case, if g(x) is divided by x - 2, the remainder is g(2), and that's equal to 1. Similarly, when g(x) is divided by x + 3, which is the same as x - (-3), the remainder is g(-3), which is -89. Okay, so I can set up two equations using these remainders.Let me write down the first equation. When x = 2, g(2) should be 1. So plugging 2 into g(x):g(2) = a*(2)³ - 8*(2)² + b*(2) - 7 = 1.Calculating each term:2³ is 8, so the first term is 8a.2² is 4, multiplied by 8 gives 32, so the second term is -32.Then, 2*b is 2b.And the constant term is -7.Putting it all together: 8a - 32 + 2b - 7 = 1.Simplify the constants: -32 -7 is -39. So, 8a + 2b - 39 = 1.Let me write that as equation (1):8a + 2b = 1 + 39Which is 8a + 2b = 40.Okay, that's the first equation. Now, the second remainder is when x = -3, so g(-3) = -89.Let me plug x = -3 into the polynomial:g(-3) = a*(-3)³ - 8*(-3)² + b*(-3) - 7 = -89.Calculating each term:(-3)³ is -27, so the first term is -27a.(-3)² is 9, multiplied by 8 gives 72, so the second term is -72.Then, b*(-3) is -3b.And the constant term is -7.Putting it all together: -27a - 72 - 3b - 7 = -89.Simplify the constants: -72 -7 is -79. So, -27a - 3b - 79 = -89.Let me write that as equation (2):-27a - 3b = -89 + 79Which is -27a - 3b = -10.Alright, now I have two equations:1) 8a + 2b = 402) -27a - 3b = -10I need to solve this system of equations for a and b. Let me see how to approach this. Maybe I can use the elimination method. If I can eliminate one variable by making the coefficients equal, that would work.Looking at the coefficients of b in both equations: equation (1) has 2b and equation (2) has -3b. To eliminate b, I can make the coefficients of b to be the same in absolute value. So, I can multiply equation (1) by 3 and equation (2) by 2. Let's try that.Multiply equation (1) by 3:3*(8a + 2b) = 3*40Which gives 24a + 6b = 120. Let's call this equation (3).Multiply equation (2) by 2:2*(-27a - 3b) = 2*(-10)Which gives -54a - 6b = -20. Let's call this equation (4).Now, if I add equation (3) and equation (4), the b terms will cancel out.Adding equation (3): 24a + 6b = 120And equation (4): -54a - 6b = -20Adding them together:24a - 54a + 6b - 6b = 120 - 20Simplify:-30a = 100Now, solving for a:a = 100 / (-30) = -10/3Okay, so a is -10/3. Now, I need to find b. I can substitute this value of a back into one of the original equations. Let's use equation (1) since the numbers seem smaller.Equation (1): 8a + 2b = 40Substitute a = -10/3:8*(-10/3) + 2b = 40Calculate 8*(-10/3):8*(-10) is -80, so -80/3.So, -80/3 + 2b = 40Let me solve for 2b:2b = 40 + 80/3Convert 40 to thirds: 40 = 120/3So, 2b = 120/3 + 80/3 = 200/3Therefore, b = (200/3) / 2 = 100/3So, b is 100/3.Let me double-check these values in equation (2) to make sure I didn't make a mistake.Equation (2): -27a - 3b = -10Substitute a = -10/3 and b = 100/3:-27*(-10/3) - 3*(100/3) = ?Calculate each term:-27*(-10/3) = (27*10)/3 = 270/3 = 90-3*(100/3) = -300/3 = -100So, 90 - 100 = -10, which matches equation (2). Perfect, that checks out.Just to be thorough, let me plug a and b back into the original polynomial and verify the remainders.First, when x = 2:g(2) = a*(8) - 8*(4) + b*(2) -7Substitute a = -10/3 and b = 100/3:g(2) = (-10/3)*8 - 32 + (100/3)*2 -7Calculate each term:(-10/3)*8 = -80/3(100/3)*2 = 200/3So, g(2) = (-80/3) - 32 + (200/3) -7Combine the fractions:(-80/3 + 200/3) = 120/3 = 40Now, the constants: -32 -7 = -39So, 40 - 39 = 1, which is correct.Now, check when x = -3:g(-3) = a*(-27) - 8*(9) + b*(-3) -7Substitute a = -10/3 and b = 100/3:g(-3) = (-10/3)*(-27) - 72 + (100/3)*(-3) -7Calculate each term:(-10/3)*(-27) = (10*27)/3 = 270/3 = 90(100/3)*(-3) = -300/3 = -100So, g(-3) = 90 - 72 - 100 -7Combine the constants:90 -72 = 1818 -100 = -82-82 -7 = -89Which is correct.So, both remainders check out. Therefore, the values of a and b are -10/3 and 100/3 respectively.**Final Answer**The ordered pair is boxed{left(-dfrac{10}{3}, dfrac{100}{3}right)}.

💡First, I need to determine the total increase in weight for the group of 8 persons. Since the average weight increases by 3.5 kg, the total weight increase is 8 multiplied by 3.5 kg, which equals 28 kg.Next, I know that the new person weighs 90 kg. This new person's weight contributed to the total weight increase of 28 kg. To find the weight of the person who was replaced, I subtract the total weight increase from the new person's weight.So, the weight of the replaced person is 90 kg minus 28 kg, which equals 62 kg.

💡Okay, so I have this problem about a right triangle ABC. The legs are lengths a and b, and the hypotenuse is c. They tell me that a + b equals 21, and c is 15. I need to find the area of the triangle. The options are A: 25, B: 54, C: 63, and D: Cannot be determined. Hmm, let me think about how to approach this.First, since it's a right triangle, I remember the Pythagorean theorem, which says that a squared plus b squared equals c squared. So, I can write that as:a² + b² = c²Given that c is 15, that means:a² + b² = 15²a² + b² = 225Okay, so that's one equation. They also told me that a + b = 21. Maybe I can use that somehow. I know that (a + b) squared is equal to a squared plus 2ab plus b squared. Let me write that down:(a + b)² = a² + 2ab + b²Since a + b is 21, then:21² = a² + 2ab + b²441 = a² + 2ab + b²But from the Pythagorean theorem, I know that a² + b² is 225. So I can substitute that into the equation:441 = 225 + 2abNow, I can solve for ab. Let me subtract 225 from both sides:441 - 225 = 2ab216 = 2abThen, divide both sides by 2:ab = 108Wait, but the area of a right triangle is (1/2)*ab, right? So, the area should be:Area = (1/2)*ab = (1/2)*108 = 54So, the area is 54. Looking back at the options, that's option B. Hmm, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. I used the Pythagorean theorem correctly: a² + b² = 225.2. I expanded (a + b)² correctly: 441 = a² + 2ab + b².3. Substituted a² + b² with 225: 441 = 225 + 2ab.4. Subtracted 225 from 441 to get 216: 216 = 2ab.5. Divided by 2 to find ab = 108.6. Calculated the area as (1/2)*108 = 54.Everything seems to check out. I don't think I made any calculation errors. So, yeah, the area should be 54. That's option B.**Final Answer**boxed{B}

💡Alright, so I'm trying to solve this problem about a school surveying students and their preferences for outdoor sports and their test scores. It's a bit confusing, but I'll try to break it down step by step.First, part (1) asks me to complete the 2x2 contingency table. Okay, so I know that a contingency table is a table that shows the frequency distribution of variables. In this case, the variables are "Likes Outdoor Sports" and "Does Not Like Outdoor Sports" versus "Non-Excellent Test Score" and "Excellent Test Score."The table they've given me has some numbers filled in, but it's incomplete. Let me look at the numbers they've provided:- In the "Non-Excellent Test Score" row, they have 10 students who like outdoor sports and 15 who don't. So, the total for that row should be 10 + 15 = 25.- In the "Excellent Test Score" row, the total is 75, but the numbers for "Likes Outdoor Sports" and "Does Not Like Outdoor Sports" are missing.- The total number of students surveyed is 100, and out of those, 60 like outdoor sports. So, the total for "Does Not Like Outdoor Sports" should be 100 - 60 = 40.Alright, let's try to fill in the missing numbers.First, I know that the total number of students with excellent test scores is 75. Since the total number of students is 100, the number of students with non-excellent test scores is 100 - 75 = 25. That matches the total I calculated earlier for the "Non-Excellent Test Score" row.Now, for the "Likes Outdoor Sports" column, we know that 60 students like outdoor sports. Out of these, 10 have non-excellent test scores, so the number of students who like outdoor sports and have excellent test scores should be 60 - 10 = 50.Similarly, for the "Does Not Like Outdoor Sports" column, we know that 40 students don't like outdoor sports. Out of these, 15 have non-excellent test scores, so the number of students who don't like outdoor sports and have excellent test scores should be 40 - 15 = 25.Let me check if these numbers add up correctly:- In the "Excellent Test Score" row, we have 50 who like outdoor sports and 25 who don't, so the total is 50 + 25 = 75, which matches the given total.- In the "Non-Excellent Test Score" row, we have 10 who like outdoor sports and 15 who don't, so the total is 10 + 15 = 25, which also matches.- The total number of students who like outdoor sports is 50 (excellent) + 10 (non-excellent) = 60, which is correct.- The total number of students who don't like outdoor sports is 25 (excellent) + 15 (non-excellent) = 40, which is also correct.Okay, so I think I've filled in the table correctly.Now, moving on to part (2). It asks whether we can be 95% confident that the students' performance in the physical test is related to their preference for outdoor sports. They've given me a formula for K², which looks similar to the chi-squared test statistic.The formula is:[ K^2 = frac{n(ad - bc)^2}{(a + b)(c + d)(a + c)(b + d)} ]Where:- ( n = a + b + c + d )- ( a, b, c, d ) are the counts from the contingency table.They've also provided a table of critical values for K² and the corresponding probabilities. The critical values are 2.706, 3.841, 6.635, 7.879, and 10.828, with probabilities 0.10, 0.05, 0.01, 0.005, and 0.001 respectively.So, to determine if there's a significant relationship between test performance and outdoor sports preference, I need to calculate the K² statistic and compare it to the critical value at the 0.05 significance level (since 95% confidence corresponds to a 5% significance level).First, let's assign the values from our contingency table to ( a, b, c, d ):- ( a = 50 ) (Likes Outdoor Sports and Excellent Test Score)- ( b = 10 ) (Likes Outdoor Sports and Non-Excellent Test Score)- ( c = 25 ) (Does Not Like Outdoor Sports and Excellent Test Score)- ( d = 15 ) (Does Not Like Outdoor Sports and Non-Excellent Test Score)- ( n = 100 ) (Total number of students)Now, plug these values into the formula:[ K^2 = frac{100(50 times 15 - 10 times 25)^2}{(50 + 10)(25 + 15)(50 + 25)(10 + 15)} ]Let's compute the numerator first:Calculate ( ad - bc ):[ 50 times 15 = 750 ][ 10 times 25 = 250 ][ ad - bc = 750 - 250 = 500 ]Now, square this value:[ (500)^2 = 250,000 ]Multiply by ( n ):[ 100 times 250,000 = 25,000,000 ]Now, compute the denominator:First, calculate each part:- ( a + b = 50 + 10 = 60 )- ( c + d = 25 + 15 = 40 )- ( a + c = 50 + 25 = 75 )- ( b + d = 10 + 15 = 25 )Multiply these together:[ 60 times 40 = 2,400 ][ 75 times 25 = 1,875 ][ 2,400 times 1,875 = 4,500,000 ]Now, divide the numerator by the denominator:[ frac{25,000,000}{4,500,000} approx 5.555... ]So, ( K^2 approx 5.56 )Now, compare this value to the critical value at the 0.05 significance level, which is 3.841.Since 5.56 > 3.841, we can reject the null hypothesis that there is no relationship between test performance and outdoor sports preference. Therefore, we can be 95% confident that there is a relationship between the two variables.Wait, but I should double-check my calculations to make sure I didn't make any mistakes.Let me verify the numerator again:- ( ad = 50 times 15 = 750 )- ( bc = 10 times 25 = 250 )- ( ad - bc = 750 - 250 = 500 )- ( (ad - bc)^2 = 500^2 = 250,000 )- ( n times (ad - bc)^2 = 100 times 250,000 = 25,000,000 )Denominator:- ( a + b = 60 )- ( c + d = 40 )- ( a + c = 75 )- ( b + d = 25 )- Multiply all together: 60 × 40 × 75 × 25Wait, I think I might have miscalculated the denominator earlier. Let's compute it step by step.First, 60 × 40 = 2,400Then, 75 × 25 = 1,875Now, 2,400 × 1,875Let me compute that:2,400 × 1,875First, 2,400 × 1,000 = 2,400,0002,400 × 800 = 1,920,0002,400 × 75 = 180,000Adding them together: 2,400,000 + 1,920,000 = 4,320,0004,320,000 + 180,000 = 4,500,000So, the denominator is indeed 4,500,000.Therefore, ( K^2 = 25,000,000 / 4,500,000 ≈ 5.555... ), which is approximately 5.56.Yes, that seems correct.Now, looking back at the critical values table:- 2.706 corresponds to 0.10 probability- 3.841 corresponds to 0.05- 6.635 corresponds to 0.01- 7.879 corresponds to 0.005- 10.828 corresponds to 0.001Our calculated K² value is 5.56, which is greater than 3.841 but less than 6.635.This means that the p-value is between 0.01 and 0.05. Specifically, since 5.56 is closer to 6.635 than to 3.841, the p-value is less than 0.01 but greater than 0.005. Wait, actually, no. The critical value at 0.01 is 6.635, and our value is 5.56, which is less than 6.635. So, the p-value is greater than 0.01 but less than 0.05.Wait, I think I might have confused the interpretation. Let me think again.The critical value at 0.05 is 3.841. Since our calculated K² is 5.56, which is greater than 3.841, we can reject the null hypothesis at the 0.05 significance level. This means that there is a statistically significant relationship between test performance and outdoor sports preference.However, since 5.56 is less than 6.635, which is the critical value for 0.01, we cannot reject the null hypothesis at the 0.01 significance level. Therefore, the p-value is between 0.01 and 0.05.But for the question, we only need to determine if we can be 95% confident, which corresponds to a 0.05 significance level. Since our K² value exceeds 3.841, we can indeed be 95% confident that there is a relationship between the two variables.I think that's it. I've double-checked my calculations, and everything seems to add up correctly. So, the conclusion is that there is a statistically significant relationship between students' preference for outdoor sports and their performance in the physical test at the 95% confidence level.

💡First, I need to determine the total number of digits used when writing the first 1500 positive odd integers.I know that the nth positive odd integer can be represented as (2n - 1). Therefore, the 1500th positive odd integer is (2 times 1500 - 1 = 2999).Next, I'll categorize these odd numbers based on the number of digits they have:1. **One-digit odd numbers**: These range from 1 to 9. There are 5 one-digit odd numbers.2. **Two-digit odd numbers**: These range from 11 to 99. There are 45 two-digit odd numbers.3. **Three-digit odd numbers**: These range from 101 to 999. There are 450 three-digit odd numbers.4. **Four-digit odd numbers**: These range from 1001 to 2999. There are 1000 four-digit odd numbers.Now, I'll calculate the total number of digits for each category:- **One-digit numbers**: (5 times 1 = 5) digits- **Two-digit numbers**: (45 times 2 = 90) digits- **Three-digit numbers**: (450 times 3 = 1350) digits- **Four-digit numbers**: (1000 times 4 = 4000) digitsFinally, I'll sum up all the digits to find the total number of digits used:(5 + 90 + 1350 + 4000 = 5445) digits

💡Alright, so I came across this interesting problem: "Which natural numbers ( n ) can be represented as ( n = [a, b] + [a, c] + [b, c] ) for some natural numbers ( a, b, c )? Here, ([x, y]) denotes the least common multiple of the natural numbers ( x ) and ( y )." At first glance, it seems a bit abstract, but I think I can break it down. Let me start by recalling what the least common multiple (LCM) actually is. The LCM of two numbers is the smallest number that is a multiple of both. So, for example, the LCM of 4 and 6 is 12 because 12 is the smallest number that both 4 and 6 divide into.Now, the problem is asking for which natural numbers ( n ) we can find three natural numbers ( a, b, c ) such that when we take the LCM of each pair and add them up, we get ( n ). So, ( n = [a, b] + [a, c] + [b, c] ).I think a good starting point is to consider small natural numbers and see if they can be expressed in this form. Maybe by examining specific cases, I can spot a pattern or come up with a conjecture.Let's start with ( n = 1 ). Is there a way to choose ( a, b, c ) such that their pairwise LCMs add up to 1? Well, the LCM of any two natural numbers is at least as large as the larger of the two numbers. So, if ( a, b, c ) are all 1, then their LCMs would all be 1, and the sum would be 3. That's too big. If I try to use numbers larger than 1, the LCMs will only get bigger. So, it seems ( n = 1 ) is not possible.Moving on to ( n = 2 ). Again, trying ( a = b = c = 1 ) gives a sum of 3, which is too big. If I try ( a = 1 ), ( b = 1 ), and ( c = 2 ), then the LCMs would be ( [1,1] = 1 ), ( [1,2] = 2 ), and ( [1,2] = 2 ). Adding these up gives ( 1 + 2 + 2 = 5 ), which is still too big. If I try ( a = 1 ), ( b = 2 ), ( c = 2 ), then the LCMs are ( [1,2] = 2 ), ( [1,2] = 2 ), and ( [2,2] = 2 ). The sum is ( 2 + 2 + 2 = 6 ). Hmm, still too big. It seems challenging to get a sum as low as 2. Maybe ( n = 2 ) isn't possible either.Let's try ( n = 3 ). If I set ( a = b = c = 1 ), then the sum is ( 1 + 1 + 1 = 3 ). Perfect! So, ( n = 3 ) is achievable.Next, ( n = 4 ). Let's see. If I take ( a = 1 ), ( b = 1 ), ( c = 1 ), the sum is 3. If I increase one of them to 2, say ( a = 1 ), ( b = 1 ), ( c = 2 ), the sum is ( 1 + 2 + 2 = 5 ). If I set ( a = 1 ), ( b = 2 ), ( c = 2 ), the sum is 6. What if I set ( a = 1 ), ( b = 2 ), ( c = 3 )? Then, ( [1,2] = 2 ), ( [1,3] = 3 ), ( [2,3] = 6 ). The sum is ( 2 + 3 + 6 = 11 ). That's way too big. Maybe ( n = 4 ) isn't possible.Wait, maybe I need to try different combinations. What if ( a = 1 ), ( b = 1 ), ( c = 3 )? Then, ( [1,1] = 1 ), ( [1,3] = 3 ), ( [1,3] = 3 ). The sum is ( 1 + 3 + 3 = 7 ). Still not 4. Maybe ( a = 1 ), ( b = 2 ), ( c = 4 )? Then, ( [1,2] = 2 ), ( [1,4] = 4 ), ( [2,4] = 4 ). The sum is ( 2 + 4 + 4 = 10 ). Not helpful.Perhaps ( n = 4 ) isn't achievable. Let's try ( n = 5 ). If I take ( a = 1 ), ( b = 1 ), ( c = 2 ), the sum is 5. Perfect! So, ( n = 5 ) is possible.Continuing this way, ( n = 6 ) can be achieved by ( a = 1 ), ( b = 2 ), ( c = 2 ), as we saw earlier. The sum is 6.For ( n = 7 ), maybe ( a = 1 ), ( b = 1 ), ( c = 3 ) gives a sum of 7. So, that works.( n = 8 ): Let's see. If I set ( a = 1 ), ( b = 1 ), ( c = 4 ), the sum is ( 1 + 4 + 4 = 9 ). If I set ( a = 1 ), ( b = 2 ), ( c = 4 ), the sum is 10. If I set ( a = 1 ), ( b = 2 ), ( c = 3 ), the sum is 11. Maybe ( a = 2 ), ( b = 2 ), ( c = 2 ), which gives ( [2,2] + [2,2] + [2,2] = 2 + 2 + 2 = 6 ). Not enough. What if ( a = 1 ), ( b = 3 ), ( c = 3 )? Then, ( [1,3] = 3 ), ( [1,3] = 3 ), ( [3,3] = 3 ). The sum is 9. Still not 8.Hmm, maybe ( n = 8 ) isn't possible. Let's try ( n = 9 ). As above, ( a = 1 ), ( b = 1 ), ( c = 4 ) gives 9. So, that works.Wait a second, I'm noticing a pattern here. It seems like some numbers are achievable, and some aren't. Specifically, numbers like 3, 5, 6, 7, 9 are achievable, but 1, 2, 4, 8 are not. Is there a commonality among the achievable numbers? They seem to be numbers that are not powers of 2. Let's check that.Powers of 2 are 1, 2, 4, 8, 16, etc. From our earlier attempts, ( n = 1, 2, 4, 8 ) don't seem achievable, while other numbers like 3, 5, 6, 7, 9 are achievable. Maybe the conjecture is that all natural numbers except powers of 2 can be expressed as the sum of three pairwise LCMs.But before jumping to conclusions, let's test this conjecture further. Let's try ( n = 10 ). If I set ( a = 1 ), ( b = 2 ), ( c = 5 ), then ( [1,2] = 2 ), ( [1,5] = 5 ), ( [2,5] = 10 ). The sum is ( 2 + 5 + 10 = 17 ). That's too big. Maybe ( a = 1 ), ( b = 3 ), ( c = 3 ). Then, the sum is 9, which is less than 10. What if ( a = 2 ), ( b = 2 ), ( c = 3 )? Then, ( [2,2] = 2 ), ( [2,3] = 6 ), ( [2,3] = 6 ). The sum is ( 2 + 6 + 6 = 14 ). Still too big. Maybe ( a = 1 ), ( b = 2 ), ( c = 4 ). The sum is 10. Perfect! So, ( n = 10 ) is achievable.Similarly, ( n = 11 ) can be achieved by ( a = 1 ), ( b = 2 ), ( c = 3 ), as we saw earlier, giving a sum of 11.What about ( n = 12 )? Let's try ( a = 1 ), ( b = 3 ), ( c = 4 ). Then, ( [1,3] = 3 ), ( [1,4] = 4 ), ( [3,4] = 12 ). The sum is ( 3 + 4 + 12 = 19 ). Too big. Maybe ( a = 2 ), ( b = 2 ), ( c = 4 ). Then, ( [2,2] = 2 ), ( [2,4] = 4 ), ( [2,4] = 4 ). The sum is ( 2 + 4 + 4 = 10 ). Not enough. What if ( a = 2 ), ( b = 3 ), ( c = 4 )? Then, ( [2,3] = 6 ), ( [2,4] = 4 ), ( [3,4] = 12 ). The sum is ( 6 + 4 + 12 = 22 ). Still too big. Maybe ( a = 1 ), ( b = 4 ), ( c = 4 ). Then, ( [1,4] = 4 ), ( [1,4] = 4 ), ( [4,4] = 4 ). The sum is 12. Perfect!So, ( n = 12 ) is achievable. It seems like my initial conjecture holds so far. Powers of 2 like 1, 2, 4, 8, 16, etc., are not achievable, while other numbers are.But to be thorough, let's check ( n = 16 ). If I set ( a = 1 ), ( b = 1 ), ( c = 16 ), the sum is ( 1 + 16 + 16 = 33 ). If I set ( a = 1 ), ( b = 2 ), ( c = 16 ), the sum is ( 2 + 16 + 16 = 34 ). If I set ( a = 2 ), ( b = 2 ), ( c = 8 ), the sum is ( 2 + 8 + 8 = 18 ). If I set ( a = 4 ), ( b = 4 ), ( c = 4 ), the sum is ( 4 + 4 + 4 = 12 ). If I set ( a = 2 ), ( b = 4 ), ( c = 8 ), the sum is ( 4 + 8 + 8 = 20 ). It seems challenging to get exactly 16. Maybe ( a = 1 ), ( b = 8 ), ( c = 8 ). Then, ( [1,8] = 8 ), ( [1,8] = 8 ), ( [8,8] = 8 ). The sum is 24. Still too big.Perhaps ( n = 16 ) isn't achievable. Let's try ( n = 17 ). If I set ( a = 1 ), ( b = 2 ), ( c = 16 ), the sum is 34. If I set ( a = 1 ), ( b = 4 ), ( c = 16 ), the sum is ( 4 + 16 + 16 = 36 ). If I set ( a = 1 ), ( b = 8 ), ( c = 16 ), the sum is ( 8 + 16 + 16 = 40 ). Maybe ( a = 2 ), ( b = 4 ), ( c = 8 ). The sum is 20. If I set ( a = 1 ), ( b = 3 ), ( c = 16 ), the sum is ( 3 + 16 + 48 = 67 ). That's way too big. It seems like ( n = 16 ) might not be achievable, supporting the conjecture.Now, to ensure that all non-powers of 2 are achievable, let's try a larger number, say ( n = 20 ). If I set ( a = 1 ), ( b = 4 ), ( c = 5 ), then ( [1,4] = 4 ), ( [1,5] = 5 ), ( [4,5] = 20 ). The sum is ( 4 + 5 + 20 = 29 ). If I set ( a = 2 ), ( b = 4 ), ( c = 5 ), then ( [2,4] = 4 ), ( [2,5] = 10 ), ( [4,5] = 20 ). The sum is ( 4 + 10 + 20 = 34 ). If I set ( a = 1 ), ( b = 5 ), ( c = 5 ), the sum is ( 5 + 5 + 5 = 15 ). If I set ( a = 1 ), ( b = 2 ), ( c = 10 ), the sum is ( 2 + 10 + 10 = 22 ). If I set ( a = 1 ), ( b = 4 ), ( c = 10 ), the sum is ( 4 + 10 + 20 = 34 ). Maybe ( a = 2 ), ( b = 2 ), ( c = 10 ). Then, ( [2,2] = 2 ), ( [2,10] = 10 ), ( [2,10] = 10 ). The sum is ( 2 + 10 + 10 = 22 ). Still not 20.Wait, maybe ( a = 1 ), ( b = 5 ), ( c = 10 ). Then, ( [1,5] = 5 ), ( [1,10] = 10 ), ( [5,10] = 10 ). The sum is ( 5 + 10 + 10 = 25 ). Close, but not 20. What if ( a = 2 ), ( b = 5 ), ( c = 10 )? Then, ( [2,5] = 10 ), ( [2,10] = 10 ), ( [5,10] = 10 ). The sum is ( 10 + 10 + 10 = 30 ). Hmm.Maybe I need a different approach. Let's think about the structure of the problem. The sum of three LCMs. Since LCMs are multiplicative, maybe I can choose ( a, b, c ) such that their LCMs contribute to the sum in a way that avoids powers of 2.Wait, earlier I noticed that when I set ( a = 1 ), ( b = 2^s ), ( c = 2^s cdot t ), where ( t ) is an odd number, the sum becomes ( 2^s (1 + 2t) ). This suggests that any number that can be written as ( 2^s (2t + 1) ) can be expressed as the sum of three LCMs. Since ( 2t + 1 ) is odd, this covers all numbers that are not pure powers of 2.So, if ( n ) is not a power of 2, it can be expressed as ( 2^s (2t + 1) ), and thus can be written as the sum of three LCMs. On the other hand, if ( n ) is a power of 2, say ( n = 2^k ), then it cannot be expressed in the form ( 2^s (2t + 1) ) unless ( s = k ) and ( t = 0 ), but ( t ) must be a natural number, so ( t = 0 ) is not allowed. Therefore, powers of 2 cannot be expressed as such sums.This seems to confirm my earlier conjecture. To solidify this, let's try to formalize it.Suppose ( n ) is a natural number that is not a power of 2. Then, ( n ) can be written as ( n = 2^s cdot m ), where ( m ) is an odd number greater than 1. Let's set ( a = 1 ), ( b = 2^s ), and ( c = 2^s cdot m ). Then:- ( [a, b] = [1, 2^s] = 2^s )- ( [a, c] = [1, 2^s cdot m] = 2^s cdot m )- ( [b, c] = [2^s, 2^s cdot m] = 2^s cdot m )Adding these up: ( 2^s + 2^s cdot m + 2^s cdot m = 2^s (1 + 2m) ). Since ( m ) is odd, ( 1 + 2m ) is also odd, and thus ( n = 2^s (1 + 2m) ) is achievable.Conversely, if ( n ) is a power of 2, say ( n = 2^k ), then any representation ( n = [a, b] + [a, c] + [b, c] ) would require that each LCM is a power of 2 as well. However, since LCMs are multiplicative, the sum of three powers of 2 cannot itself be a single power of 2 unless two of them are equal and the third is zero, which is not possible with natural numbers. Therefore, powers of 2 cannot be expressed in this form.This reasoning seems solid, but I should check if there are any exceptions or special cases I might have missed. For instance, what if ( a, b, c ) are not of the form I assumed? Could there be another way to express a power of 2 as the sum of three LCMs?Suppose ( n = 8 ). Let's try different combinations:- ( a = 1 ), ( b = 2 ), ( c = 4 ): Sum is ( 2 + 4 + 4 = 10 )- ( a = 2 ), ( b = 2 ), ( c = 4 ): Sum is ( 2 + 4 + 4 = 10 )- ( a = 1 ), ( b = 1 ), ( c = 8 ): Sum is ( 1 + 8 + 8 = 17 )- ( a = 1 ), ( b = 4 ), ( c = 4 ): Sum is ( 4 + 4 + 4 = 12 )- ( a = 2 ), ( b = 4 ), ( c = 4 ): Sum is ( 4 + 4 + 4 = 12 )- ( a = 1 ), ( b = 2 ), ( c = 8 ): Sum is ( 2 + 8 + 8 = 18 )None of these combinations give a sum of 8. It seems impossible. Therefore, my conclusion holds.In summary, through testing specific cases and then generalizing, I've arrived at the conclusion that all natural numbers except powers of 2 can be expressed as the sum of three pairwise LCMs. Powers of 2 cannot be expressed in this form because their representation would require an impossible combination of LCMs.