🧠:[<think>Okay, so the problem here is about Allen flipping a fair two-sided coin and rolling a fair six-sided die. We need to find the probability that the coin lands on heads and he rolls a number that's a multiple of 5. The answer choices are given from A to E. Let me think through this step by step.First, let's break down the problem into parts. There are two independent events here: flipping a coin and rolling a die. Since these are independent, the outcome of one doesn't affect the outcome of the other. So, the probability of both events happening together should be the product of their individual probabilities. That makes sense because for independent events, P(A and B) = P(A) * P(B).Alright, so let's start with the coin flip. The coin is fair and two-sided, so the probability of getting heads is 1/2. That seems straightforward. There are two possible outcomes, heads or tails, each equally likely. So, P(Heads) = 1/2.Next, the die roll. It's a standard six-sided die with faces numbered 1 through 6. We need the probability that the number rolled is a multiple of 5. Let me think, the multiples of 5 on a die... Well, 5 is the only multiple of 5 on a die numbered 1 to 6. Because 5*1=5 and 5*2=10, which is beyond 6. So, the only number that's a multiple of 5 here is 5 itself. Therefore, there's only one favorable outcome out of six possible outcomes. So, the probability of rolling a multiple of 5 is 1/6. Got that down. So, P(Multiple of 5) = 1/6.Now, since these two events are independent, we multiply their probabilities to find the probability that both occur. So, that would be 1/2 * 1/6. Let me do the multiplication: 1 times 1 is 1, and 2 times 6 is 12. So, the probability is 1/12. Hmm, looking at the answer choices, option B is 1/12. So, that seems like the answer.Wait, let me double-check to make sure I didn't make any mistakes here. First, the coin flip: two sides, so 1/2. Correct. The die: numbers 1-6. Multiples of 5. Only 5. So, 1 out of 6. Yes. Then multiply them because they are independent. 1/2 * 1/6 = 1/12. That's right. Answer B.But wait a second, just to be thorough, could there be another multiple of 5 that I'm missing? Let's see. 5*1=5, 5*2=10, but 10 is higher than 6, so no. So, only 5. So, 1/6. So, yeah, that seems right. So, the combined probability is 1/12. So, answer B.I don't see any mistakes here. The key points are recognizing the independence of the events and correctly identifying the number of favorable outcomes for the die roll. Since both steps check out, I think this is correct. Let me just make sure that the question is asking for both events happening, which it is: the coin lands on heads AND he rolls a multiple of 5. So, yes, that's the intersection of the two independent events, so multiplication applies. Therefore, 1/2 * 1/6 is 1/12, which is option B. So, confident that's the right answer.Final Answerboxed{B}</think>Allen flips a fair two-sided coin and rolls a fair six-sided die. We need to find the probability that the coin lands on heads and he rolls a number that is a multiple of 5.1. Probability of flipping heads: - The coin has two sides, so the probability of landing on heads is (frac{1}{2}).2. Probability of rolling a multiple of 5: - The die has faces numbered 1 through 6. The only multiple of 5 in this range is 5. Therefore, the probability is (frac{1}{6}).3. Combined probability: - Since the coin flip and die roll are independent events, we multiply their probabilities: [ frac{1}{2} times frac{1}{6} = frac{1}{12} ]Thus, the probability that the coin lands on heads and Allen rolls a multiple of 5 is (boxed{B}).,<think>Okay, let's see. The problem is asking for the probability that Allen gets heads on a coin flip and rolls a multiple of 5 on a 6-sided die. Hmm. Alright, so first, I need to recall how probability works when dealing with two independent events. Since the coin flip and the die roll are separate actions, their outcomes shouldn't affect each other, right? So, maybe I need to use the multiplication rule for independent events. Let me start by breaking down each event separately. The first event is flipping a fair two-sided coin. The probability of getting heads on a coin flip is 1/2 because there are two possible outcomes, heads or tails, and each is equally likely. That seems straightforward.The second event is rolling a fair 6-sided die. The question asks for the probability of rolling a number that is a multiple of 5. Hmm, let's think. The die has numbers 1 through 6. Which of these numbers are multiples of 5? Well, 5 is a multiple of 5 because 5 times 1 is 5. 10 is also a multiple of 5, but the die only goes up to 6. So, the only multiple of 5 on the die is 5 itself. Therefore, there's only one favorable outcome out of six possible outcomes. So, the probability of rolling a multiple of 5 is 1/6.Now, since these two events are independent, the probability that both occur is the product of their individual probabilities. That would be 1/2 (for the coin) multiplied by 1/6 (for the die). Let me calculate that: 1/2 * 1/6 = 1/12. So, is the answer 1/12? Let me check the options. Option B is 1/12. So, that seems right.Wait a second, let me double-check. Maybe I missed something. The die has numbers 1-6. Multiples of 5 in that range are just 5. So, that's definitely 1/6. The coin is heads, which is 1/2. Multiplying them gives 1/12. Yeah, that makes sense. Since the two events are independent, the combined probability is the product. Alternatively, if I wanted to think in terms of total possible outcomes, there are 2 outcomes for the coin and 6 outcomes for the die. So, total number of possible combined outcomes is 2*6=12. The favorable outcome is getting heads and rolling a 5. There's only one such outcome: (Heads, 5). So, 1 favorable out of 12 possible. Therefore, 1/12. That matches the previous result. I don't see any issues with this reasoning. The answer must be 1/12, which is option B.Final Answerboxed{B}</think>Allen flips a fair two-sided coin and rolls a fair 6-sided die. We need to find the probability that the coin lands on heads and he rolls a number that is a multiple of 5.1. Probability of flipping heads: The coin has two sides, so the probability of getting heads is (frac{1}{2}).2. Probability of rolling a multiple of 5: The die has numbers 1 through 6. The only multiple of 5 in this range is 5. Therefore, the probability is (frac{1}{6}).3. Combined probability: Since the coin flip and die roll are independent events, we multiply their probabilities: [ frac{1}{2} times frac{1}{6} = frac{1}{12} ]Thus, the probability that the coin lands on heads and Allen rolls a multiple of 5 is (boxed{B}).]

🔑:1. Identify the events: - Let ( A ) be the event that the coin lands on heads. - Let ( B ) be the event that the die shows a number that is a multiple of 5.2. Determine the probability of each event: - The probability of the coin landing on heads, ( P(A) ), is ( frac{1}{2} ) because the coin is fair. - The die has numbers 1 through 6, and the only number that is a multiple of 5 is 5. Therefore, the probability of rolling a 5, ( P(B) ), is ( frac{1}{6} ).3. Calculate the combined probability: - Since the coin flip and the die roll are independent events, the probability of both events occurring is the product of their individual probabilities: [ P(A cap B) = P(A) cdot P(B) = frac{1}{2} cdot frac{1}{6} = frac{1}{12} ]4. Conclusion: - The probability that the coin lands on heads and the die shows a number that is a multiple of 5 is ( boxed{frac{1}{12}} ).

🧠:[<think>Okay, let me try to figure out how to solve this problem. The question is asking for the sum of the inverses of the squares of the roots of the polynomial P = X³ - 3X - 1. So, if the roots are r1, r2, r3, then we need to find (1/r1²) + (1/r2²) + (1/r3²). First, I remember that for polynomials, there are relationships between the roots and the coefficients called Vieta's formulas. These might be useful here. Vieta's formulas for a cubic equation like X³ + aX² + bX + c = 0 tell us that the sum of the roots is -a, the sum of the product of roots two at a time is b, and the product of the roots is -c. In our case, the polynomial is X³ - 3X - 1. Comparing this to the general form, we see that a = 0 (since there's no X² term), b = -3, and c = -1. So applying Vieta's formulas:1. r1 + r2 + r3 = -a = 02. r1r2 + r1r3 + r2r3 = b = -33. r1r2r3 = -c = 1So we have the sum of roots is 0, the sum of products two at a time is -3, and the product of the roots is 1.Now, we need to compute the sum 1/r1² + 1/r2² + 1/r3². Let me think about how to express this in terms of the roots. I know that 1/r1² + 1/r2² + 1/r3² can be written as (1/r1 + 1/r2 + 1/r3)² - 2(1/(r1r2) + 1/(r1r3) + 1/(r2r3)). Maybe this is a way to break it down using Vieta's formulas. Let me verify that. Yes, because for any numbers x, y, z, we have (x + y + z)² = x² + y² + z² + 2(xy + xz + yz). So rearranging, x² + y² + z² = (x + y + z)² - 2(xy + xz + yz). So if x = 1/r1, y = 1/r2, z = 1/r3, then x² + y² + z² is exactly what we want, and we can compute it using (x + y + z)² - 2(xy + xz + yz). So, first we need to compute x + y + z = 1/r1 + 1/r2 + 1/r3. Then, compute xy + xz + yz = 1/(r1r2) + 1/(r1r3) + 1/(r2r3). Let's compute x + y + z first. 1/r1 + 1/r2 + 1/r3. This can be written as (r2r3 + r1r3 + r1r2)/(r1r2r3). From Vieta's formulas, the numerator is r1r2 + r1r3 + r2r3 = -3, and the denominator is r1r2r3 = 1. Therefore, x + y + z = (-3)/1 = -3. Next, compute xy + xz + yz = [1/(r1r2) + 1/(r1r3) + 1/(r2r3)]. Similarly, this can be written as (r3 + r2 + r1)/(r1r2r3). Wait, let's check that. Let me compute:1/(r1r2) + 1/(r1r3) + 1/(r2r3) = (r3 + r2 + r1)/(r1r2r3). Because:If we take 1/(r1r2) = r3/(r1r2r3), similarly for the others. So adding them up gives (r1 + r2 + r3)/(r1r2r3). But from Vieta, r1 + r2 + r3 = 0. Therefore, xy + xz + yz = 0 / 1 = 0. Therefore, going back to the expression for the sum of squares:x² + y² + z² = (x + y + z)^2 - 2(xy + xz + yz) = (-3)^2 - 2*0 = 9 - 0 = 9. Wait, that seems straightforward, but let me verify if I did everything correctly. Let me check each step.First, the sum of 1/r_i is (sum of products of two roots)/product of all roots) = (-3)/1 = -3. Correct.Then, the sum of 1/(r_i r_j) is (sum of roots)/product of all roots. Since the sum of roots is 0, this is 0. Therefore, the cross terms in the square are 0. Then the sum of squares is (-3)^2 - 0 = 9. Therefore, the sum of the inverses of the squares of the roots is 9. Is that possible? Let me check with an example.Alternatively, maybe there's another way to approach this. Let's denote the roots as r1, r2, r3. Then each root satisfies the equation r_i³ - 3r_i - 1 = 0. So, for each root, r_i³ = 3r_i + 1. Maybe we can use this relation to find expressions involving 1/r_i².For example, let's take the equation r_i³ = 3r_i + 1. If we divide both sides by r_i³, we get 1 = 3/r_i² + 1/r_i³. Hmm, maybe that's useful. Rearranging, 1/r_i³ = 1 - 3/r_i². But I'm not sure if that helps directly. Alternatively, if we want to compute 1/r_i², maybe we can express this in terms of lower powers. Let me see. Starting from r_i³ = 3r_i + 1, divide both sides by r_i^4: 1/r_i = 3/r_i³ + 1/r_i^4. Hmm, not sure. Alternatively, maybe take reciprocal of both sides? Let's see:From r_i³ = 3r_i + 1, take reciprocal: 1/(r_i³) = 1/(3r_i + 1). Not sure if helpful.Alternatively, let's express 1/r_i² in terms of something. Let's solve for 1/r_i². Let me think. Let's take the original equation r_i³ - 3r_i - 1 = 0. Let's solve for 1/r_i². Let's divide both sides by r_i^3: 1 - 3/r_i² - 1/r_i³ = 0. Rearranged, 1 - 3/r_i² = 1/r_i³. Hmm, not helpful yet.Alternatively, let's denote s = 1/r_i. Then s_i = 1/r_i. Then since r_i satisfies r_i³ - 3r_i - 1 = 0, substituting s_i = 1/r_i gives (1/s_i)^3 - 3*(1/s_i) -1 = 0. Multiplying through by s_i^3: 1 - 3s_i² - s_i³ = 0. So s_i³ + 3s_i² -1 = 0. Therefore, the inverses of the roots satisfy the equation s³ + 3s² -1 = 0. Therefore, if we let Q(s) = s³ + 3s² -1, then the inverses of the roots of P are the roots of Q. Therefore, the sum we want is the sum of the squares of the roots of Q. Let's denote the roots of Q as s1, s2, s3. Then we need to compute s1² + s2² + s3². Using Vieta's formulas again for Q(s) = s³ + 3s² -1. Here, the coefficients are a = 3 (coefficient of s²), b = 0 (coefficient of s, since there is no s term), and c = -1 (constant term). So applying Vieta's formulas:1. s1 + s2 + s3 = -a = -32. s1s2 + s1s3 + s2s3 = b = 03. s1s2s3 = -c = 1Therefore, the sum of the roots of Q is -3, the sum of the products two at a time is 0, and the product is 1. Then, the sum of the squares of the roots of Q is (s1 + s2 + s3)^2 - 2(s1s2 + s1s3 + s2s3) = (-3)^2 - 2*0 = 9 - 0 = 9. Therefore, s1² + s2² + s3² = 9, which is exactly the same result as before. Therefore, the sum of the inverses of the squares of the roots of P is equal to 9. Wait, this seems to confirm the initial calculation. Both methods give 9. Let me check once again for errors. First method:Sum of inverses squared = (sum of inverses)^2 - 2*(sum of inverses of products two at a time). Sum of inverses: (r1r2 + r1r3 + r2r3)/(r1r2r3) = (-3)/1 = -3. Sum of inverses of products two at a time: (r1 + r2 + r3)/(r1r2r3) = 0/1 = 0. Therefore, sum of inverses squared = (-3)^2 - 2*0 = 9. Second method:Inverting the roots gives a new polynomial whose roots are 1/r1, 1/r2, 1/r3. Then, sum of squares of these roots is computed via Vieta's, same result. So both methods give 9. So probably the answer is 9. But let me verify with another approach. Maybe using Newton's identities. Newton's identities relate the power sums of roots to the coefficients. The power sum S(k) = r1^k + r2^k + r3^k. But we need S(-2) = 1/r1² + 1/r2² + 1/r3². Newton's identities can handle negative exponents if we use generating functions or reciprocal polynomials, but I think it's more straightforward to use the previous methods. But let's see. Alternatively, consider generating the power sums for the reciprocal roots. Let me define s_i = 1/r_i. Then we can use Newton's identities on the polynomial Q(s) = s³ + 3s² -1, which is the polynomial we derived earlier. For Q(s) = s³ + 3s² -1, we can compute the power sums S'(1) = s1 + s2 + s3 = -3, S'(2) = s1² + s2² + s3² = 9, which we already found. Alternatively, using the original polynomial P(x), perhaps we can find a recurrence relation for the power sums. Given that each root r_i satisfies r_i³ = 3r_i + 1, then for each root, we can write r_i^n = 3r_i^{n-2} + r_i^{n-3} for n >= 3. But since we need S(-2) = sum 1/r_i² = sum s_i² where s_i = 1/r_i. Since s_i satisfies s_i³ + 3s_i² -1 =0, as we found before. Then, for the power sums of s_i, we can set up a recurrence. But since the polynomial for s_i is Q(s) = s³ + 3s² -1, the recurrence for power sums S'(k) = s1^k + s2^k + s3^k would be S'(k) = -3S'(k-1) + 0*S'(k-2) + 1*S'(k-3). Wait, Newton's identities for monic polynomials relate the power sums to the coefficients. Let me recall. For a monic polynomial with roots s1, s2, s3, so Q(s) = (s - s1)(s - s2)(s - s3) = s³ - (s1 + s2 + s3)s² + (s1s2 + s1s3 + s2s3)s - s1s2s3. But in our case, Q(s) = s³ + 3s² -1. So comparing coefficients:Q(s) = s³ + 3s² -1 = s³ - (-3)s² + 0s - (-1). Therefore, the coefficients are:a1 = -(s1 + s2 + s3) = 3 ⇒ s1 + s2 + s3 = -3a2 = s1s2 + s1s3 + s2s3 = 0a3 = -s1s2s3 = -1 ⇒ s1s2s3 = 1So the power sums can be calculated via Newton's identities:For k >= 1,S'(k) + a1 S'(k-1) + a2 S'(k-2) + a3 S'(k-3) = 0But since Q(s) is a monic polynomial, the Newton identities formula for power sums is:For k >= 1,S'(k) + a1 S'(k-1) + a2 S'(k-2) + a3 S'(k-3) = 0But in our case, a1 = 3, a2 = 0, a3 = -1. So,For k >= 1,S'(k) + 3 S'(k-1) + 0*S'(k-2) -1*S'(k-3) = 0Therefore,S'(k) = -3 S'(k-1) + S'(k-3)But we need S'(2). Let's compute S'(0), S'(1), S'(2):S'(0) = s1^0 + s2^0 + s3^0 = 1 + 1 + 1 = 3S'(1) = s1 + s2 + s3 = -3 (from Vieta)S'(2) = ?Using the recurrence for k = 2:S'(2) = -3 S'(1) + S'(-1)Wait, but k starts at 1. Wait, maybe I need to adjust. Wait, the standard Newton's identities apply for k from 1 to n. For k > n, the recurrence holds, but for k < 0, maybe it's not defined. Hmm, perhaps this approach isn't the best here. Alternatively, we can compute S'(2) using the formula we already used earlier: S'(2) = (S'(1))² - 2(s1s2 + s1s3 + s2s3) = (-3)^2 - 2*0 = 9, which matches.Alternatively, if we use the recurrence for k=3:Wait, perhaps for k=3:S'(3) + 3 S'(2) + 0*S'(1) -1*S'(0) = 0But S'(3) = sum s_i³. However, since each s_i satisfies s_i³ = -3s_i² +1 (from Q(s) = s³ +3s² -1 =0 ⇒ s_i³ = -3s_i² +1). Therefore, S'(3) = sum (-3s_i² +1) = -3 sum s_i² + sum 1 = -3 S'(2) + 3. But from the recurrence:S'(3) + 3 S'(2) -1*S'(0) = 0Substitute S'(3) = -3 S'(2) +3:(-3 S'(2) + 3) + 3 S'(2) - 3 = 0 ⇒ 0 = 0. Which is a tautology, so no help. Alternatively, if we already know S'(1) and want S'(2), perhaps we can use the relation:S'(2) = (S'(1))² - 2(s1s2 + s1s3 + s2s3) = (-3)^2 - 2*0 =9.Yes, that's straightforward. Therefore, S'(2) = 9, which is the sum we want. Therefore, all methods point to the answer being 9. But just to be thorough, let me compute an example. Suppose the roots are r1, r2, r3. Let me assume some specific roots and compute the sum. Wait, but the polynomial X³ -3X -1 might have complex roots, but regardless, the sum should be 9. Alternatively, let's suppose we have a cubic equation where Vieta's formulas give sum of roots 0, sum of products -3, product 1. For example, take roots 1, 1, -2. Wait, sum is 0, product is 1*1*(-2)= -2, which is not 1. Not our case. Alternatively, suppose we have roots a, b, c such that a + b + c = 0, ab + ac + bc = -3, abc =1. Let's compute 1/a² +1/b² +1/c². Using the formula, (sum of 1/a)^2 - 2(sum of 1/(ab)). sum of 1/a = (ab + ac + bc)/(abc) = (-3)/1 = -3. sum of 1/(ab) = (a + b + c)/(abc) = 0/1 =0. Thus, sum of squares is (-3)^2 -2*0 =9. Thus, even without knowing the actual roots, the relationships hold. Therefore, regardless of the specific roots, the sum must be 9. Therefore, the answer is 9, which in boxed form is boxed{9}.Final Answerboxed{9}</think>To find the sum of the inverses of the squares of the roots of the polynomial ( P = X^3 - 3X - 1 ), we start by noting the roots ( r_1, r_2, r_3 ) satisfy Vieta's formulas:1. ( r_1 + r_2 + r_3 = 0 )2. ( r_1r_2 + r_1r_3 + r_2r_3 = -3 )3. ( r_1r_2r_3 = 1 )We need to calculate ( frac{1}{r_1^2} + frac{1}{r_2^2} + frac{1}{r_3^2} ). This can be expressed as:[left( frac{1}{r_1} + frac{1}{r_2} + frac{1}{r_3} right)^2 - 2 left( frac{1}{r_1r_2} + frac{1}{r_1r_3} + frac{1}{r_2r_3} right)]First, we calculate ( frac{1}{r_1} + frac{1}{r_2} + frac{1}{r_3} ):[frac{r_2r_3 + r_1r_3 + r_1r_2}{r_1r_2r_3} = frac{-3}{1} = -3]Next, we calculate ( frac{1}{r_1r_2} + frac{1}{r_1r_3} + frac{1}{r_2r_3} ):[frac{r_1 + r_2 + r_3}{r_1r_2r_3} = frac{0}{1} = 0]Substituting these into the expression for the sum of the squares of the inverses:[(-3)^2 - 2 cdot 0 = 9 - 0 = 9]Thus, the sum of the inverses of the squares of the roots is (boxed{9}).,<think>Okay, so I have this problem here where I need to calculate the sum of the inverses of the squares of the roots of the polynomial P = X³ - 3X - 1. Hmm, let's see. I remember that for polynomials, there are relationships between the roots and the coefficients, called Vieta's formulas. Maybe I can use those? Let me recall. If a polynomial has roots α, β, γ, then the sum of the roots is - coefficient of X² divided by the coefficient of X³. In this case, the polynomial is monic (coefficient of X³ is 1), and there is no X² term, so the sum of the roots α + β + γ = 0. That's right.Then, the sum of the product of roots two at a time is the coefficient of X divided by the coefficient of X³. Here, the coefficient of X is -3, so αβ + αγ + βγ = -3. And the product of the roots is - the constant term divided by the coefficient of X³, so αβγ = 1, since the constant term is -1. So, Vieta's formulas give us:α + β + γ = 0,αβ + αγ + βγ = -3,αβγ = 1.Now, the problem asks for the sum of the inverses of the squares of the roots, which is (1/α²) + (1/β²) + (1/γ²). I need to express this in terms of the coefficients of the polynomial, using Vieta's formulas. Let me think about how to manipulate this expression.First, note that (1/α²) + (1/β²) + (1/γ²) can be written as (1/α + 1/β + 1/γ)² - 2(1/(αβ) + 1/(αγ) + 1/(βγ)). That's using the identity that (a + b + c)² = a² + b² + c² + 2(ab + ac + bc), so rearranged, a² + b² + c² = (a + b + c)² - 2(ab + ac + bc). So applying this to 1/α, 1/β, 1/γ.Therefore, sum of inverses squared = (sum of inverses)² - 2(sum of inverses products two at a time). Let me compute each part.First, sum of inverses: 1/α + 1/β + 1/γ. That can be written as (βγ + αγ + αβ)/(αβγ). From Vieta, the numerator is αβ + αγ + βγ = -3, and the denominator is αβγ = 1. So sum of inverses = -3/1 = -3.Next, sum of inverses products two at a time: 1/(αβ) + 1/(αγ) + 1/(βγ). Similarly, this can be written as (γ + β + α)/(αβγ). The numerator is α + β + γ = 0, so this sum is 0/1 = 0.Therefore, substituting back into the expression for the sum of inverses squared: (-3)² - 2*0 = 9 - 0 = 9.Wait, but that seems too straightforward. Let me verify. If the sum of inverses is -3 and the sum of inverses products two at a time is 0, then indeed, squaring the sum of inverses gives 9 and subtracting twice the sum of the products gives 9 - 0 = 9. So, is the answer 9?But hold on, let me check this again because sometimes these problems have a twist. Let me consider the roots of the polynomial P = X³ - 3X - 1. Maybe there's a specific property here. Let me actually compute the roots numerically or check for possible roots.First, check if there are any rational roots. By Rational Root Theorem, possible roots are ±1. Let's test x=1: 1 - 3 -1 = -3 ≠ 0. x=-1: -1 +3 -1 = 1 ≠ 0. So no rational roots. Therefore, the polynomial is irreducible over the rationals. So all roots are either real or complex. But since it's a cubic, there must be at least one real root. Let me try to approximate it.Let me compute P(1) = 1 -3 -1 = -3,P(2) = 8 -6 -1 = 1,So there's a root between 1 and 2.Similarly, P(-1) = -1 +3 -1 = 1,P(-2) = -8 +6 -1 = -3,So another root between -2 and -1. Since it's a cubic, there are three real roots? Or maybe one real and two complex?Wait, but if there are two turning points. Let's compute the derivative P’ = 3X² -3. Setting to zero, 3X² -3 = 0 => X² =1 => X=1 or X=-1. So critical points at X=1 and X=-1. Let's compute P(1) = -3 and P(-1)=1. So the graph decreases to X=1, reaches -3, then increases to X=-1, reaches 1, then decreases again. Wait, that might not be the case. Wait, the derivative is positive when |X|>1, and negative when |X|<1. So the function is increasing when X >1 or X < -1, decreasing between -1 and 1.So let's see: as X approaches -∞, P(X) approaches -∞, at X=-1, P(-1)=1, then decreases from X=-1 to X=1, reaching P(1)=-3, then increases again towards infinity as X approaches ∞. Therefore, there are three real roots: one less than -1, one between -1 and 1, and one greater than 1? Wait, but between -1 and 1, the function goes from P(-1)=1 to P(1)=-3, so crosses the x-axis once in that interval. Then, as X approaches -∞, P(X) approaches -∞ and at X=-1 it's 1, so crosses once to the left of -1. Similarly, as X approaches ∞, P(X) approaches ∞ and at X=1 it's -3, so crosses once to the right of 1. Hence, three real roots. Therefore, all roots are real. So, even though the polynomial is irreducible over the rationals, it has three real roots.But regardless of whether the roots are real or complex, Vieta's formulas still hold. So my previous calculation using Vieta's should still be valid.Wait, but the answer seems 9. Let me check again. Let me compute sum 1/α² + 1/β² + 1/γ².From Vieta:α + β + γ = 0,αβ + αγ + βγ = -3,αβγ = 1.Let me compute S = 1/α² + 1/β² + 1/γ².Express S in terms of the known sums. Alternatively, note that 1/α² is (1/α)^2. Let me consider variables y_i = 1/α_i, where α_i are the roots. Then, the sum we need is y₁² + y₂² + y₃².We can relate this to the polynomial whose roots are y_i. Since y_i = 1/α_i, the polynomial with roots y_i is obtained by reversing the coefficients of the original polynomial. Let me see. If P(X) = X³ - 3X -1, then the polynomial with roots 1/α, 1/β, 1/γ is X³P(1/X) = X³(1/X³ - 3/X -1) = 1 - 3X² - X³ = -X³ -3X² +1. To make it monic, divide by -1: X³ +3X² -1. Wait, but the coefficients would correspond to Vieta's formulas for the y_i.Let me confirm. Let Q(X) be the polynomial with roots 1/α, 1/β, 1/γ. Then Q(X) = (X - 1/α)(X - 1/β)(X - 1/γ) = X³ - (1/α + 1/β +1/γ)X² + (1/(αβ) + 1/(αγ) +1/(βγ))X -1/(αβγ).From Vieta, since 1/α +1/β +1/γ = (αβ + αγ + βγ)/(αβγ) = (-3)/1 = -3,and 1/(αβ) +1/(αγ) +1/(βγ) = (α + β + γ)/(αβγ) = 0/1 = 0,and 1/(αβγ) = 1/1 =1,So Q(X) = X³ - (-3)X² +0X -1 = X³ +3X² -1.Wait, but earlier when I did X³P(1/X), I got -X³ -3X² +1. So that would be equivalent to -Q(X). Because Q(X) is X³ +3X² -1. So X³P(1/X) = -Q(X). Therefore, Q(X) = -X³P(1/X).But regardless, the coefficients of Q(X) give the sums for the reciprocals. So for Q(X) = X³ +3X² -1, the sum of roots is -3 (coefficient of X²), which matches 1/α +1/β +1/γ = -3. The sum of products two at a time is 0 (coefficient of X), which is 1/(αβ) +1/(αγ) +1/(βγ) =0, and the product of the roots is 1 (constant term, with sign (-1)^3 * constant term / leading coefficient = -(-1)/1 =1). So that checks out.Now, if I want to compute the sum of the squares of the roots of Q(X), which are y₁² + y₂² + y₃² where y_i =1/α_i. The sum of squares of the roots of Q(X) can be found using the formula:sum y_i² = (sum y_i)^2 - 2 sum y_i y_j.From Vieta's formulas on Q(X):sum y_i = -3,sum y_i y_j = 0,Therefore, sum y_i² = (-3)^2 - 2*0 =9 -0=9.Thus, the sum of the inverses of the squares of the roots of P is 9. Therefore, the answer is 9.But let me check once again by another method to be safe. Let me consider that S = 1/α² +1/β² +1/γ². I can also express this as (α²β² + α²γ² + β²γ²)/(αβγ)^2. Since the denominator is (αβγ)^2 =1^2=1, so S = α²β² +α²γ² +β²γ².So now I need to compute α²β² +α²γ² +β²γ². How can I express this in terms of the known sums? Let me recall that (αβ + αγ + βγ)^2 = α²β² + α²γ² + β²γ² + 2αβγ(α + β + γ).Wait, let's compute (αβ + αγ + βγ)^2:= α²β² + α²γ² + β²γ² + 2α²βγ + 2αβ²γ + 2αβγ²= α²β² + α²γ² + β²γ² + 2αβγ(α + β + γ)Yes, so (αβ + αγ + βγ)^2 = sum α²β² + 2αβγ(sum α).Therefore, sum α²β² = (αβ + αγ + βγ)^2 - 2αβγ(α + β + γ).We know αβ + αγ + βγ = -3,αβγ =1,and α + β + γ=0,Therefore, sum α²β² = (-3)^2 -2*1*0=9 -0=9.Therefore, S = sum α²β² =9. Since the denominator was 1, this gives S=9. Therefore, the sum of inverses squared is indeed 9.So both methods confirm that the answer is 9. Therefore, I think that's the correct result. Let me just recap to ensure I haven't missed anything.Original polynomial: X³ -3X -1. We need sum 1/α² +1/β² +1/γ². Using Vieta's formulas:1. Express sum of inverses squared as (sum of inverses)^2 - 2(sum of product of inverses two at a time).2. Sum of inverses is (αβ + αγ + βγ)/αβγ = -3/1 = -3.3. Sum of product of inverses two at a time is (α + β + γ)/αβγ =0/1=0.4. Therefore, sum of inverses squared = (-3)^2 -2*0=9.Alternatively, notice that sum of inverses squared is equal to sum α²β² (since (1/α² +1/β² +1/γ²)= (α²β² + α²γ² + β²γ²)/(αβγ)^2 = sum α²β² as denominator is 1). Then, compute sum α²β² using (αβ + αγ + βγ)^2 -2αβγ(α + β + γ)=9 -0=9.Either way, same result. Therefore, the answer is 9. I think this is solid.Final Answerboxed{9}</think>To calculate the sum of the inverses of the squares of the roots of the polynomial ( P = X^3 - 3X - 1 ), we use Vieta's formulas and algebraic identities.Given the polynomial ( P = X^3 - 3X - 1 ), the roots are ( alpha, beta, gamma ). By Vieta's formulas, we have:[alpha + beta + gamma = 0,][alphabeta + betagamma + gammaalpha = -3,][alphabetagamma = 1.]We need to find ( frac{1}{alpha^2} + frac{1}{beta^2} + frac{1}{gamma^2} ). This can be expressed using the identity:[frac{1}{alpha^2} + frac{1}{beta^2} + frac{1}{gamma^2} = left( frac{1}{alpha} + frac{1}{beta} + frac{1}{gamma} right)^2 - 2 left( frac{1}{alphabeta} + frac{1}{betagamma} + frac{1}{gammaalpha} right).]First, we calculate ( frac{1}{alpha} + frac{1}{beta} + frac{1}{gamma} ):[frac{1}{alpha} + frac{1}{beta} + frac{1}{gamma} = frac{alphabeta + betagamma + gammaalpha}{alphabetagamma} = frac{-3}{1} = -3.]Next, we calculate ( frac{1}{alphabeta} + frac{1}{betagamma} + frac{1}{gammaalpha} ):[frac{1}{alphabeta} + frac{1}{betagamma} + frac{1}{gammaalpha} = frac{alpha + beta + gamma}{alphabetagamma} = frac{0}{1} = 0.]Substituting these into the identity, we get:[left( -3 right)^2 - 2 cdot 0 = 9 - 0 = 9.]Thus, the sum of the inverses of the squares of the roots is:[boxed{9}]]

🔑:Let's denote r_{1}, r_{2}, r_{3} as these roots. Then by Viète's formulas:begin{aligned}frac{1}{r_{1}^{2}}+frac{1}{r_{2}^{2}}+frac{1}{r_{3}^{2}} & =left(frac{1}{r_{1}}+frac{1}{r_{2}}+frac{1}{r_{3}}right)^{2}-2 cdotleft(frac{1}{r_{2} r_{3}} frac{1}{r_{1} r_{2}} frac{1}{r_{1} r_{3}}right) & =left(frac{r_{1} r_{2}+r_{1} r_{3}+r_{2} r_{3}}{r_{1} r_{2} r_{3}}right)^{2}-2 cdotleft(frac{r_{1}+r_{2}+r_{3}}{r_{1} r_{2} r_{3}}right) & =(-3)^{2}-2 cdot 0=9end{aligned}However, it is not possible to express the roots simply in terms of the coefficients in the general case (only for degrees 1,2,3,4, and it is impossible for degrees greater than 5 (Galois groups)).The Lagrange interpolation polynomial answers the question: can we find a polynomial whose graph passes through certain points?Take n distinct real numbers x_{1}, ldots, x_{n} and n real numbers y_{1}, ldots, y_{n}.The initial idea is to start by solving an easier problem: find a polynomial L_{k} of degree n-1 that vanishes at n-1 distinct real numbers x_{1}, ldots, x_{k-1}, x_{k+1}, ldots, x_{n} and takes a certain value y_{k} at x_{k}: L_{k}=y_{k} cdot prod_{i neq k} frac{X-x_{i}}{x_{k}-x_{i}} works. It suffices to sum these polynomials to obtain the desired result.The interpolation polynomial of degree n-1 is unique. It follows that the polynomial solutions are exactly theQ(X) cdot prod_{k=1}^{n}left(X-x_{k}right)+sum_{k=1}^{n} L_{k}(X)with Q(X) any polynomial.All these results also hold for rationals and complexes.

🧠:[<think>Alright, let's tackle this problem step by step. So, we have a sequence {a_n} where the sum of the first n terms is S_n. The first term a₁ is given as 1/2, and there's a recurrence relation: 2S_n = S_{n-1} + 1 for n ≥ 2. First, part 1 asks for the general formula of the sequence {a_n}. Hmm. Let me recall that S_n is the sum of the first n terms, so S_n = a₁ + a₂ + ... + a_n. Similarly, S_{n-1} would be the sum of the first n-1 terms. The given relation is 2S_n = S_{n-1} + 1. Let me write that down:2S_n = S_{n-1} + 1 for n ≥ 2.I need to find a formula for a_n. Since a_n = S_n - S_{n-1}, right? Because the sum up to n minus the sum up to n-1 gives the nth term. So maybe if I can express S_n in terms of S_{n-1} and then find a recurrence relation for S_n, I can find a formula for a_n.Let's start by solving the recurrence relation for S_n. Let's rearrange the given equation:2S_n = S_{n-1} + 1Divide both sides by 2:S_n = (1/2)S_{n-1} + 1/2So this is a linear recurrence relation for S_n. It looks like a non-homogeneous linear recurrence. The general solution for such equations is the sum of the homogeneous solution and a particular solution.First, solve the homogeneous equation:S_n - (1/2)S_{n-1} = 0The characteristic equation would be r - 1/2 = 0, so r = 1/2. Therefore, the homogeneous solution is S_n^h = C*(1/2)^n, where C is a constant.Now, find a particular solution. Since the non-homogeneous term is a constant (1/2), let's assume a constant particular solution S_n^p = K.Substitute into the recurrence:K = (1/2)K + 1/2Solving for K:K - (1/2)K = 1/2 => (1/2)K = 1/2 => K = 1Therefore, the general solution is S_n = S_n^h + S_n^p = C*(1/2)^n + 1Now, apply the initial condition to find C. Wait, what is the initial condition? For n=1, S_1 = a₁ = 1/2. Let's check when n=1. Wait, the recurrence is given for n ≥ 2. So n starts from 2. Therefore, we need to use S_1 to find the constant C.So when n=1, S_1 = 1/2. Let's plug n=1 into the general solution:S_1 = C*(1/2)^1 + 1 = C*(1/2) + 1 = 1/2Therefore,(1/2)C + 1 = 1/2 => (1/2)C = -1/2 => C = -1So the general formula for S_n is:S_n = - (1/2)^n + 1Simplify that: 1 - (1/2)^nSo S_n = 1 - (1/2)^nNow, to find a_n, which is S_n - S_{n-1} for n ≥ 2, and a₁ = 1/2.Let's compute a_n for n ≥ 2:a_n = S_n - S_{n-1} = [1 - (1/2)^n] - [1 - (1/2)^{n-1}] = - (1/2)^n + (1/2)^{n-1} = (1/2)^{n-1} - (1/2)^n = (1/2)^n (2 - 1) = (1/2)^nWait, that can't be right. Wait, let's compute:(1/2)^{n-1} - (1/2)^n = (1/2)^{n-1} (1 - 1/2) = (1/2)^{n-1} * 1/2 = (1/2)^nYes, that's correct. So for n ≥ 2, a_n = (1/2)^n. But let's check for n=2. If n=2, S_2 should be a₁ + a₂ = 1/2 + a₂. From the recurrence relation, when n=2, 2S_2 = S_1 + 1 = 1/2 + 1 = 3/2, so S_2 = 3/4. Then a₂ = S_2 - S₁ = 3/4 - 1/2 = 1/4, which is (1/2)^2 = 1/4. Correct. Similarly for n=3, a₃ = (1/2)^3 = 1/8. Then S_3 = 1 - (1/2)^3 = 7/8, which is 1/2 + 1/4 + 1/8 = 7/8. Correct. So indeed, for n ≥ 2, a_n = (1/2)^n. But for n=1, a₁ = 1/2, which is also (1/2)^1. So actually, the general formula is a_n = (1/2)^n for all n ≥ 1. Wait, that's interesting. So even though we derived that for n ≥ 2, a_n = (1/2)^n, and for n=1, a₁ = 1/2 which is also (1/2)^1, so the sequence is a geometric sequence with first term 1/2 and common ratio 1/2. Therefore, the general formula is a_n = (1/2)^n. Is that possible?Wait, let me check S_n. If a_n = (1/2)^n, then S_n would be the sum from k=1 to n of (1/2)^k. That is a geometric series with sum (1/2)(1 - (1/2)^n)/(1 - 1/2) ) = (1/2)(1 - (1/2)^n)/(1/2) ) = 1 - (1/2)^n, which matches the S_n we found earlier. So yes, the entire sequence is a geometric sequence with first term 1/2 and ratio 1/2. Therefore, the general formula is indeed a_n = (1/2)^n for all n ≥ 1.So part 1 answer is a_n = (1/2)^n.Moving on to part 2. Let b_n = log_{1/2} a_n. Then we need to find the sum T_n of the first n terms of the sequence {1/(b_n b_{n+1})}.First, let's find b_n. Since a_n = (1/2)^n, then log base 1/2 of a_n is log_{1/2} (1/2)^n. Remember that log_b (b^k) = k. So here, log_{1/2} (1/2)^n = n. Because (1/2)^n is the argument, and the base is 1/2, so the logarithm is just n. Therefore, b_n = n.Therefore, the sequence {1/(b_n b_{n+1})} becomes 1/(n(n+1)).So T_n is the sum from k=1 to n of 1/(k(k+1)).This is a telescoping series. The general term 1/(k(k+1)) can be written as 1/k - 1/(k+1). Let's verify:1/k - 1/(k+1) = (k+1 - k)/(k(k+1)) = 1/(k(k+1)). Correct.Therefore, the sum T_n = Σ_{k=1}^n [1/k - 1/(k+1)] = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1)). All the intermediate terms cancel out, leaving T_n = 1 - 1/(n+1) = n/(n+1).Therefore, the sum T_n is n/(n+1).Let me double-check with n=1: 1/(1*2) = 1/2, and T_1 = 1 - 1/2 = 1/2. Correct.For n=2: 1/(1*2) + 1/(2*3) = 1/2 + 1/6 = 2/3, and T_2 = 1 - 1/3 = 2/3. Correct.So yes, the sum is n/(n+1).Final Answer1. The general formula for the sequence is boxed{a_n = left(dfrac{1}{2}right)^n}.2. The sum of the first (n) terms (T_n) is boxed{T_n = dfrac{n}{n + 1}}.</think>Given a sequence ({a_n}) with the sum of its first (n) terms denoted as (S_n), where (a_1 = frac{1}{2}) and (2S_n = S_{n-1} + 1) for (n geq 2):1. To find the general formula for the sequence ({a_n}): - The recurrence relation for (S_n) is (2S_n = S_{n-1} + 1), which simplifies to (S_n = frac{1}{2}S_{n-1} + frac{1}{2}). - Solving this linear recurrence relation, we find the homogeneous solution (S_n^h = Cleft(frac{1}{2}right)^n) and the particular solution (S_n^p = 1). Thus, the general solution is (S_n = Cleft(frac{1}{2}right)^n + 1). - Using the initial condition (S_1 = frac{1}{2}), we determine (C = -1), leading to (S_n = 1 - left(frac{1}{2}right)^n). - The (n)-th term (a_n = S_n - S_{n-1}) results in (a_n = left(frac{1}{2}right)^n) for all (n geq 1).2. For (b_n = log_{frac{1}{2}} a_n) and the sum (T_n) of the sequence (left{frac{1}{b_n b_{n+1}}right}): - Since (a_n = left(frac{1}{2}right)^n), we have (b_n = log_{frac{1}{2}} left(frac{1}{2}right)^n = n). - The sequence (frac{1}{b_n b_{n+1}}) becomes (frac{1}{n(n+1)}), which is a telescoping series. - The sum (T_n = sum_{k=1}^n frac{1}{k(k+1)} = sum_{k=1}^n left(frac{1}{k} - frac{1}{k+1}right)) simplifies to (1 - frac{1}{n+1} = frac{n}{n+1}). Final Answer1. The general formula for the sequence is (boxed{a_n = left(dfrac{1}{2}right)^n}).2. The sum of the first (n) terms (T_n) is (boxed{T_n = dfrac{n}{n + 1}}).,<think>Okay, let's tackle this problem step by step. So, we have a sequence {a_n} where the sum of the first n terms is S_n. We know that a₁ = 1/2, and for n ≥ 2, 2S_n = S_{n-1} + 1. The questions are to find the general formula for a_n and then, using that, find the sum T_n for another sequence related to b_n, which is defined as the logarithm base 1/2 of a_n.Starting with part 1. The first thing I notice is that the problem gives a recursive relation involving S_n and S_{n-1}. Since S_n is the sum of the first n terms, we can express S_n in terms of S_{n-1} plus a_n. That is, S_n = S_{n-1} + a_n. But the problem states that 2S_n = S_{n-1} + 1. Let me write that down:Given: 2S_n = S_{n-1} + 1 for n ≥ 2.But we also know that S_n = S_{n-1} + a_n. So maybe we can substitute S_n from the second equation into the first one. Let's try that.From S_n = S_{n-1} + a_n, multiply both sides by 2:2S_n = 2S_{n-1} + 2a_n.But according to the problem, 2S_n = S_{n-1} + 1. So:2S_{n-1} + 2a_n = S_{n-1} + 1.Subtract S_{n-1} from both sides:S_{n-1} + 2a_n = 1.Therefore, S_{n-1} = 1 - 2a_n.Hmm. But S_{n-1} is also equal to S_{n-2} + a_{n-1} for n ≥ 2. Wait, but when n=2, S_{1} = a₁ = 1/2. Let me check for n=2 first. Maybe plugging in n=2 can help us see a pattern.For n=2:2S_2 = S_1 + 1. Since S₁ = a₁ = 1/2, then 2S₂ = 1/2 + 1 = 3/2 ⇒ S₂ = 3/4.Then a₂ = S₂ - S₁ = 3/4 - 1/2 = 1/4.Similarly, for n=3:2S₃ = S₂ + 1 = 3/4 + 1 = 7/4 ⇒ S₃ = 7/8.Thus, a₃ = S₃ - S₂ = 7/8 - 3/4 = 7/8 - 6/8 = 1/8.n=4:2S₄ = S₃ + 1 = 7/8 + 1 = 15/8 ⇒ S₄ = 15/16.a₄ = S₄ - S₃ = 15/16 - 7/8 = 15/16 - 14/16 = 1/16.Hmm, so the terms are a₁ = 1/2, a₂ = 1/4, a₃ = 1/8, a₄ = 1/16... Looks like a pattern here. Each term is half of the previous one. So is this a geometric sequence with ratio 1/2?Wait, but let's check if this holds for n=5:2S₅ = S₄ + 1 = 15/16 + 1 = 31/16 ⇒ S₅ = 31/32.a₅ = S₅ - S₄ = 31/32 - 15/16 = 31/32 - 30/32 = 1/32.Yes, so aₙ = (1/2)^n. Wait, but a₁ is (1/2)^1 = 1/2, which matches. a₂ = (1/2)^2 = 1/4, which matches. So seems like the general term is aₙ = (1/2)^n.But wait, let's confirm this using the recursive relation. Suppose that aₙ = (1/2)^n. Then Sₙ = sum_{k=1}^n (1/2)^k. The sum of a geometric series: Sₙ = (1/2)(1 - (1/2)^n)/(1 - 1/2)) = (1 - (1/2)^n). Let me check that.Yes, sum_{k=1}^n (1/2)^k = 1 - (1/2)^n. Then 2Sₙ = 2(1 - (1/2)^n) = 2 - 2*(1/2)^n = 2 - (1/2)^{n-1}.But according to the given relation, 2Sₙ = S_{n-1} + 1. Let's compute S_{n-1} + 1. If S_{n-1} = 1 - (1/2)^{n-1}, then S_{n-1} + 1 = 1 - (1/2)^{n-1} + 1 = 2 - (1/2)^{n-1}, which is exactly 2Sₙ. So yes, it satisfies the equation. Therefore, the general formula is aₙ = (1/2)^n.So part 1 is solved: aₙ = 1/2^n.Now part 2. Let bₙ = log_{1/2} aₙ. So since aₙ = (1/2)^n, then log_{1/2} (1/2)^n = n. Because log base 1/2 of (1/2)^n is n. Because (1/2)^n = (1/2)^n, so exponent is n. Therefore, bₙ = n.Then the sequence we need to find the sum T_n for is 1/(bₙ b_{n+1}).Since bₙ = n, then 1/(bₙ b_{n+1}) = 1/(n(n+1)).So T_n is the sum from k=1 to n of 1/(k(k+1)).This is a telescoping series. Recall that 1/(k(k+1)) = 1/k - 1/(k+1). Therefore, the sum telescopes:T_n = sum_{k=1}^n (1/k - 1/(k+1)) = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1)) = 1 - 1/(n+1) = n/(n+1).Therefore, the sum T_n is n/(n+1).Let me verify with n=1: 1/(1*2) = 1/2. T₁ = 1/2. According to the formula, 1/(1+1) = 1/2. Correct.n=2: 1/(1*2) + 1/(2*3) = 1/2 + 1/6 = 2/3. Formula: 2/(2+1) = 2/3. Correct.n=3: 1/2 + 1/6 + 1/12 = (6 + 2 + 1)/12 = 9/12 = 3/4. Formula: 3/4. Correct.So yes, the formula holds.Therefore, part 2's answer is T_n = n/(n+1).Final Answer1. The general formula for the sequence is boxed{a_n = dfrac{1}{2^n}}.2. The sum of the first ( n ) terms is boxed{T_n = dfrac{n}{n + 1}}.</think>Given a sequence ({a_n}) with the sum of its first (n) terms denoted as (S_n), where (a_1 = frac{1}{2}), and (2S_n = S_{n-1} + 1) for (n geq 2):1. To find the general formula for the sequence ({a_n}): - We start with the given recurrence relation (2S_n = S_{n-1} + 1). - Using (S_n = S_{n-1} + a_n), we substitute into the recurrence relation to get (2(S_{n-1} + a_n) = S_{n-1} + 1). - Simplifying this, we find (S_{n-1} + 2a_n = 1), leading to (S_{n-1} = 1 - 2a_n). - By calculating the first few terms and recognizing the pattern, we observe that (a_n) forms a geometric sequence with the first term (frac{1}{2}) and common ratio (frac{1}{2}). - Confirming this pattern using the sum of a geometric series, we find (a_n = frac{1}{2^n}).2. For (b_n = log_{frac{1}{2}} a_n) and finding the sum (T_n) of the first (n) terms of the sequence (left{frac{1}{b_n b_{n+1}}right}): - Since (a_n = frac{1}{2^n}), we have (b_n = log_{frac{1}{2}} left(frac{1}{2^n}right) = n). - The sequence becomes (frac{1}{b_n b_{n+1}} = frac{1}{n(n+1)}), which is a telescoping series. - The sum (T_n) is calculated as (sum_{k=1}^n left(frac{1}{k} - frac{1}{k+1}right)), which simplifies to (1 - frac{1}{n+1} = frac{n}{n+1}). Final Answer1. The general formula for the sequence is (boxed{a_n = dfrac{1}{2^n}}).2. The sum of the first ( n ) terms is (boxed{T_n = dfrac{n}{n + 1}}).]

🔑:1. When n=2, using 2S_n = S_{n-1} + 1 and a_1 = frac{1}{2}, we obtain 2S_2 = S_1 + 1, which simplifies to 2a_1 + 2a_2 = a_1 + 1, yielding a_2 = frac{1}{4}.From 2S_n = S_{n-1} + 1, we can derive:2S_{n+1} = S_n + 1.Subtracting the two equations, we get 2a_{n+1} = a_n, which gives us a_{n+1} = frac{1}{2}a_n (n geq 2).Since frac{a_2}{a_1} = frac{1}{2} holds true, sequence {a_n} is a geometric sequence with the first term frac{1}{2} and the common ratio frac{1}{2}. Thus, we have a_n = frac{1}{2^n} (n in mathbb{N}^*).2. Using the result from (1) and b_n = log_{frac{1}{2}}a_n (n in mathbb{N}^*), we can deduce that b_n = log_{frac{1}{2}}(frac{1}{2})^n = n. Consequently, frac{1}{b_nb_{n+1}} = frac{1}{n(n+1)} = frac{1}{n} - frac{1}{n+1}.The sum of the first n terms for sequence {frac{1}{b_nb_{n+1}}} is:T_n = frac{1}{b_1b_2} + frac{1}{b_2b_3} + dots + frac{1}{b_nb_{n+1}}=left[(1 - frac{1}{2}) + (frac{1}{2} - frac{1}{3}) + dots + (frac{1}{n} - frac{1}{n+1})right]= 1 - frac{1}{n+1} = boxed{frac{n}{n+1}}.

🧠:[<think>Okay, so I need to figure out how to transform the graph of y = sin 2x into the graph of y = cos 2x - 1. The options involve shifting it left or right and then up or down. Let me start by recalling some trigonometric identities and transformation rules.First, I know that the basic sine and cosine functions are related by a phase shift. Specifically, cos(x) is equal to sin(x + π/2). So, if I have a cosine function, it can be written as a sine function shifted to the left by π/2 units. But in this problem, both functions have an argument of 2x instead of just x. Let me see how that affects things.The given function is y = cos 2x - 1. Let's compare this to the original function y = sin 2x. If I can express cos 2x in terms of a shifted sine function, then I can figure out the necessary horizontal shift. Then, subtracting 1 would just be a vertical shift downward by 1 unit.Starting with cos 2x. Since cos θ = sin(θ + π/2), replacing θ with 2x gives cos 2x = sin(2x + π/2). Let me check that:sin(2x + π/2) = sin(2x)cos(π/2) + cos(2x)sin(π/2) = sin(2x)*0 + cos(2x)*1 = cos(2x). Yes, that works. So cos 2x is equal to sin(2x + π/2). Therefore, the function y = cos 2x is equivalent to y = sin(2x + π/2). But how does that translate to a shift? When dealing with horizontal shifts, if you have sin(Bx + C), the phase shift is -C/B. In this case, comparing to sin(2x), we have sin(2x + π/2) = sin[2(x + π/4)]. Let me verify that:sin[2(x + π/4)] = sin(2x + π/2). Yes, that's correct. So the phase shift here is π/4 units to the left. Wait, but the phase shift formula is -C/B. For sin(Bx + C) = sin(B(x + C/B)), so the shift is -C/B. So in this case, C is +π/2. So the phase shift would be -π/2 divided by 2, which is -π/4. Hmm, negative phase shift would mean shifting to the right by π/4.Wait, hold on. Let me clarify this. Let's take the general form: sin(Bx + C). To write this as sin(B(x - h)), then h is the phase shift. Let's set B(x - h) = Bx - Bh. So comparing to Bx + C, we have -Bh = C, so h = -C/B. So the phase shift is h = -C/B. In our case, we have sin(2x + π/2) = sin(2(x - h)). Therefore, solving for h: 2x + π/2 = 2(x - h) => 2x + π/2 = 2x - 2h => π/2 = -2h => h = -π/4. Therefore, the phase shift is h = -π/4. A negative phase shift means shifting to the left by π/4? Wait, no. Wait, actually, phase shift h is calculated as h = -C/B. If h is positive, that means shifting to the right by h units. If h is negative, shifting to the left by |h| units. But here, h = -C/B. Let me take C = π/2. So h = - (π/2)/2 = -π/4. So h is negative, which would indicate a shift to the left by π/4 units. But when we have sin(2x + π/2) = sin(2(x - h)), so if h is negative, then x - h = x + |h|, so it's a shift to the left by |h|. Let me verify with numbers. Suppose x = 0. Then sin(2*0 + π/2) = sin(π/2) = 1. If we shift sin(2x) left by π/4, then we have sin(2(x + π/4)) = sin(2x + π/2), which is the same as the original expression. So yes, shifting left by π/4 units gives sin(2x + π/2). Therefore, cos 2x is obtained by shifting sin 2x left by π/4. But the answer options don't have left by π/4. Wait, wait. Wait, the answer options involve moving right or left by π/4 or π/2, then up or down. Let me check again.Wait, but the question is about transforming sin 2x into cos 2x - 1. So first, we need to get from sin 2x to cos 2x, which we just saw can be done by shifting left by π/4. Then, subtract 1, which is a vertical shift down by 1. So combining these two transformations: shift left by π/4 and then down by 1. Looking at the options, option B is moving left by π/4 and then down by 1. Option A is right by π/4 and up 1. Option B is left π/4 and down 1. So is option B correct?But wait, hold on. Let me double-check. Because sometimes there's ambiguity with phase shifts. Let me think again. If we have sin(2x + π/2) = sin[2(x + π/4)]. So replacing x with x + π/4 shifts the graph to the left by π/4. Therefore, starting with sin(2x), shifting left by π/4 gives sin(2(x + π/4)) = sin(2x + π/2) = cos 2x. Then subtract 1, which is shifting down 1. So yes, left π/4 and down 1. Therefore, option B. But let me check the answer options again:A: right by π/4, up 1B: left by π/4, down 1C: right by π/2, up 1D: left by π/2, down 1So according to this, B is left π/4, down 1, which matches our conclusion. But let me check if there's another way to represent the phase shift. For example, sometimes when you have a coefficient inside the argument, the direction of the shift can be counterintuitive.Alternatively, maybe using another identity. We know that cos(2x) can also be written as sin(2x + π/2), which is equivalent to a phase shift. Alternatively, is there a way to represent it as a shift to the right?Wait, let me recall that cos(θ) = sin(θ + π/2). So cos(2x) = sin(2x + π/2). Alternatively, if I factor out the 2, it's sin[2(x + π/4)], which is a shift to the left by π/4. If I want to write cos(2x) as sin(2(x - h)), then:sin(2(x - h)) = sin(2x - 2h) = cos(2x). So we need sin(2x - 2h) = cos(2x). This would require that 2x - 2h = 2x + π/2 - 2πn (since sine has a period of 2π, so we can add multiples of 2π). Therefore, -2h = π/2 - 2πn. Solving for h: h = (2πn - π/2)/2. The principal solution (n=0) gives h = -π/4, which is again a shift to the left by π/4. If n=1, h = (2π - π/2)/2 = (3π/2)/2 = 3π/4, which would be a shift to the right by 3π/4, but that's not necessary here. So the minimal shift is left by π/4.Therefore, shifting left by π/4 and then down by 1. So option B. But let me check once more because sometimes when dealing with horizontal shifts and scaling, it's easy to get confused.Alternatively, maybe there's a different approach. Let's consider the graphs of sin(2x) and cos(2x). The period of both functions is π, since the period is 2π divided by the coefficient of x, which is 2. So their periods are π.The graph of sin(2x) starts at (0,0), goes up to (π/4, 1), down to (π/2, 0), etc. The graph of cos(2x) starts at (0,1), goes down to (π/2, -1), back to (π, 1), etc. So to go from sin(2x) to cos(2x), you need to shift the graph so that the point at (0,0) moves to where cos(2x) starts, which is (0,1). But how does this happen?If we shift sin(2x) left by π/4, then the original point at (0,0) would move to (-π/4, 0). But the cos(2x) graph at x=0 is 1. Wait, maybe this is getting too detailed. Alternatively, think about critical points.Take the maximum of sin(2x) which occurs at x = π/4. The maximum of cos(2x) occurs at x = 0. So to make the maximum of sin(2x) at π/4 move to x=0, we need to shift left by π/4. Therefore, shifting the graph of sin(2x) left by π/4 would align its maximum with the maximum of cos(2x). Then, after that shift, the function becomes cos(2x), and then subtracting 1 shifts it down by 1. That seems to make sense.Therefore, the correct transformation is shifting left by π/4 and then down by 1, which is option B.But let me check the answer options again. The options are:A: right π/4, up 1B: left π/4, down 1C: right π/2, up 1D: left π/2, down 1So yes, B is left π/4 and down 1, which matches our conclusion. Therefore, the answer should be B.Wait, but let me verify by plugging in a point. Let's take x = 0.Original function: y = sin(2*0) = 0.After shifting left by π/4: x becomes x + π/4. So new function is sin(2(x + π/4)) = sin(2x + π/2) = cos(2x). At x = 0, this is cos(0) = 1. Then subtract 1, so y = 1 - 1 = 0. So at x = 0, the transformed function is 0. The original target function is y = cos(2*0) -1 = 1 -1 = 0. So that matches.Another point: x = π/4.Original function: sin(2*(π/4)) = sin(π/2) = 1.Shift left by π/4: x = π/4 - π/4 = 0. Then y = sin(2*0 + π/2) = sin(π/2) = 1. Then subtract 1: y = 0. The target function at x = π/4: cos(2*(π/4)) -1 = cos(π/2) -1 = 0 -1 = -1. Wait, but transformed function gives 0 here. Wait, that's a problem.Wait, no. Wait, let me clarify. If we shift left by π/4, then the new function is sin(2(x + π/4)) -1. So at x = π/4, it's sin(2*(π/4 + π/4)) -1 = sin(2*(π/2)) -1 = sin(π) -1 = 0 -1 = -1. Which matches the target function at x = π/4: cos(2*(π/4)) -1 = cos(π/2) -1 = 0 -1 = -1. So that works.Wait, but earlier, when I considered shifting left by π/4, the point x = 0 in the original becomes x = -π/4 in the shifted graph. But in the transformed function, after shifting left by π/4 and then down 1, the point at x = -π/4 would be sin(2*(-π/4 + π/4)) -1 = sin(0) -1 = -1. Which corresponds to the target function at x = -π/4: cos(2*(-π/4)) -1 = cos(-π/2) -1 = 0 -1 = -1. So that matches.Alternatively, if we take the original function at x = π/8: sin(2*(π/8)) = sin(π/4) = √2/2 ≈ 0.707.After shifting left by π/4, x becomes π/8 + π/4 = 3π/8. The transformed function before vertical shift is sin(2*(3π/8)) = sin(3π/4) = √2/2. Then subtract 1: √2/2 -1 ≈ -0.292. The target function at x = 3π/8: cos(2*(3π/8)) -1 = cos(3π/4) -1 = -√2/2 -1 ≈ -1.707. Wait, that doesn't match. Wait, what's wrong here?Wait, no. Wait, after shifting left by π/4, the transformed function is sin(2(x + π/4)) -1. So at x = 3π/8, it's sin(2*(3π/8 + π/4)) -1. Let's compute 3π/8 + π/4 = 3π/8 + 2π/8 = 5π/8. Then 2*(5π/8) = 5π/4. sin(5π/4) = -√2/2. So subtract 1: -√2/2 -1 ≈ -1.707. The target function at x = 3π/8 is cos(2*(3π/8)) -1 = cos(3π/4) -1 = -√2/2 -1 ≈ -1.707. So that does match. So my earlier confusion was because I miscalculated the point. So it does check out.Therefore, shifting left by π/4 and then down by 1 unit does transform sin(2x) into cos(2x) -1. Therefore, the correct answer is B.But let me check the other options just in case. Option A: moving right π/4 and up 1. If we shift right by π/4, the function becomes sin(2(x - π/4)) = sin(2x - π/2). Then shift up 1: sin(2x - π/2) +1. Let's compare that to the target function cos(2x) -1. Is sin(2x - π/2) +1 equal to cos(2x) -1? Let's see: sin(2x - π/2) = -cos(2x), so sin(2x - π/2) +1 = -cos(2x) +1. Which is not the same as cos(2x) -1. So A is incorrect.Option C: moving right π/2 and up 1. Shifting right π/2: sin(2(x - π/2)) = sin(2x - π). Then up 1: sin(2x - π) +1. Sin(2x - π) = -sin(2x), so this becomes -sin(2x) +1. Not equal to cos(2x) -1. Not correct.Option D: moving left π/2 and down 1. Shifting left π/2: sin(2(x + π/2)) = sin(2x + π) = -sin(2x). Then down 1: -sin(2x) -1. Which is not equal to cos(2x) -1. So D is incorrect.Therefore, only option B works. So the answer is B.Final Answerboxed{B}</think>To transform the graph of ( y = sin 2x ) into the graph of ( y = cos 2x - 1 ), we need to use trigonometric identities and understand the necessary shifts.1. Phase Shift: The cosine function can be expressed as a shifted sine function. Specifically, ( cos 2x = sin(2x + frac{pi}{2}) ). This can be rewritten as ( sin(2(x + frac{pi}{4})) ), which indicates a horizontal shift. - The term ( 2(x + frac{pi}{4}) ) shows that the graph of ( sin 2x ) is shifted left by ( frac{pi}{4} ) units.2. Vertical Shift: After obtaining ( cos 2x ) from the phase shift, we need to subtract 1. This results in a vertical shift downward by 1 unit.Combining these transformations:- A horizontal shift left by ( frac{pi}{4} ) units.- A vertical shift downward by 1 unit.Thus, the correct answer is (boxed{B}).,<think>Okay, so I need to figure out how to transform the graph of y = sin 2x into the graph of y = cos 2x - 1. The question is asking about the necessary shifts: whether we move it left or right by some units and then up or down. The options are A, B, C, D with different combinations. Let me think step by step.First, I remember that sine and cosine functions are related through a phase shift. Specifically, cos(θ) is equal to sin(θ + π/2). So, cos(2x) can be written as sin(2x + π/2). Wait, but if I factor out the 2 inside the sine function, that would be sin[2(x + π/4)]. Because when you have sin(B(x + C)), the phase shift is -C. So, sin(2x + π/2) is equivalent to sin[2(x + π/4)], which means shifting the graph of sin(2x) to the left by π/4 units. But wait, is that right?Hold on. Let me recall the phase shift formula. For a function y = sin(Bx + C), the phase shift is -C/B. So in this case, if we have sin(2x + π/2), then the phase shift would be -π/2 divided by 2, which is -π/4. A negative phase shift means moving to the left, so it's a shift left by π/4. Hmm, but then cos(2x) is sin(2x + π/2), which is a left shift of π/4. Therefore, to go from sin(2x) to cos(2x), we shift left by π/4.But the problem here is wanting to get from sin(2x) to cos(2x) - 1. So first, we need to get to cos(2x), which is a left shift of π/4, and then subtract 1, which is a vertical shift down by 1 unit. So combining those two transformations: left by π/4 and then down by 1. Let me check the options.Looking at the options:A: right by π/4, then up 1B: left by π/4, then down 1C: right by π/2, then up 1D: left by π/2, then down 1So based on my previous reasoning, it's B. But wait, let me verify again because sometimes phase shifts can be confusing.Alternative approach: Maybe instead of converting sin(2x) to cos(2x) first, think of the vertical shift. The target function is cos(2x) - 1. If I can express this in terms of sin(2x) shifted appropriately and then shifted vertically.As we know, cos(2x) is sin(2x + π/2). So cos(2x) - 1 is sin(2x + π/2) - 1. So starting from sin(2x), shift left by π/2/2 = π/4 (because the phase shift is calculated as -C/B where the function is sin(Bx + C)), so here, C is π/2, B is 2, so phase shift is -π/2 / 2 = -π/4. So left by π/4, then down by 1. So that's option B.Wait, but hold on. Let's check this again. Suppose we have y = sin(2x). Let's perform a phase shift. To get cos(2x), we need to shift sin(2x) by some amount. Since cos(θ) = sin(θ + π/2), so replacing θ with 2x, we get cos(2x) = sin(2x + π/2). So to write this as a function of x, we have sin(2(x + π/4)) because 2(x + π/4) = 2x + π/2. So shifting x to x + π/4, which is a shift to the left by π/4. Therefore, the graph of cos(2x) is obtained by shifting sin(2x) left by π/4. Then, subtracting 1 would shift it down by 1. So the total transformation is left π/4 and down 1, which is option B. So that seems correct.But just to be absolutely sure, let's test specific points. Take a key point on the sin(2x) graph. For example, at x = 0, sin(0) = 0. After shifting left by π/4, the new x-coordinate would be x = -π/4, but the point (0, 0) would move to (-π/4, 0). Then shifting down by 1, it becomes (-π/4, -1). Now, let's check what's the corresponding point on cos(2x) - 1. At x = -π/4, cos(2*(-π/4)) -1 = cos(-π/2) -1 = 0 -1 = -1. So that point is indeed on the graph. Another point: original sin(2x) has a maximum at x = π/4, sin(2*(π/4)) = sin(π/2) = 1. Shifting left by π/4 would move that maximum to x = π/4 - π/4 = 0. Then shifting down 1, the point becomes (0, 0). Let's check cos(2*0) -1 = 1 -1 = 0. That's correct. Similarly, the minimum of sin(2x) is at x = 3π/4, sin(3π/2) = -1. Shifting left by π/4 gives x = 3π/4 - π/4 = π/2. Then down 1: (π/2, -2). Check cos(2*(π/2)) -1 = cos(π) -1 = -1 -1 = -2. Correct. So that seems to check out.Alternatively, suppose we thought about shifting right by π/4. Then cos(2x) would be sin(2(x - π/4)) = sin(2x - π/2). But cos(2x) is equal to sin(2x + π/2), not sin(2x - π/2). So shifting right would not give the correct function. So shifting left is the right direction.Another way: Let's write both functions in terms of sine.We have y = cos(2x) - 1 = sin(2x + π/2) - 1. So starting from sin(2x), to get sin(2x + π/2), we replace x with x + π/4, which is a shift to the left by π/4. Then subtract 1, which is shifting down by 1. So again, left π/4, down 1. So option B.Alternatively, if we tried shifting right, we would have sin(2(x - h)) + k. Let's suppose we shift right by h and down by k. Then the equation would be sin(2(x - h)) - k. We need this to equal cos(2x) - 1. So:sin(2(x - h)) - k = sin(2x - 2h) - k = cos(2x) - 1So, sin(2x - 2h) must equal cos(2x) and -k must equal -1, so k =1. But sin(2x - 2h) = cos(2x). Let's solve for h. sin(θ - 2h) = cosθ. We know that cosθ = sin(θ + π/2). So sin(θ - 2h) = sin(θ + π/2). Therefore, the arguments must differ by a multiple of 2π, or be supplementary angles differing by a multiple of 2π.So, θ - 2h = θ + π/2 + 2πnSubtract θ: -2h = π/2 + 2πnTherefore, h = -π/4 - πnTaking n = 0, h = -π/4. But h is the shift. Since we have sin(2(x - h)) = sin(2x - 2h). If h is -π/4, then the shift is h = -π/4, which would mean shifting right by π/4. Wait, because in the expression x - h, if h is negative, then it becomes x - (-π/4) = x + π/4, which is a shift left by π/4. So actually, h is the shift to the right. So if h = -π/4, that would imply shifting right by -π/4, which is equivalent to shifting left by π/4. So that's consistent with the previous result. Therefore, shifting left by π/4 and then shifting down by 1 (since k =1, but in our equation we have -k = -1, so k=1. Wait, the original equation is sin(2(x - h)) - k. So if we set that equal to cos(2x) -1, then sin(2(x - h)) must equal cos(2x), and -k must equal -1, so k=1. So we shift right by h and down by 1. But h turned out to be -π/4, which is a left shift. So combining, moving left by π/4 and then down by 1. Therefore, option B.Alternatively, if someone thought that h is positive, they might have thought that h = π/4, but that would result in sin(2x - π/2). Which is not equal to cos(2x). Let's check sin(2x - π/2) = sin[2x - π/2] = -cos(2x). Because sin(θ - π/2) = -cosθ. Therefore, sin(2x - π/2) = -cos(2x). Therefore, if we had shifted right by π/4 (h=π/4), we would have sin(2x - π/2), which is -cos(2x). Then subtract 1 would be -cos(2x) -1. Which is not equal to cos(2x) -1. So that's not correct. Therefore, shifting right by π/4 and then down by 1 gives -cos(2x) -1, which is not the desired function. So that's why shifting left by π/4 is necessary.Therefore, the correct answer must be option B: moving left by π/4 and then down by 1. Let me just check the options again.Options:A: right π/4, up 1B: left π/4, down 1C: right π/2, up 1D: left π/2, down 1So yes, B is the correct answer.But to ensure no mistakes, let's also consider another approach using graphing.Imagine the graph of y = sin(2x). It has a period of π, with peaks at π/4, 3π/4, etc. The graph of cos(2x) is similar but shifted. Since cos(0) = 1, the peak at x=0 for cos(2x). The equivalent point in sin(2x) is at x = π/4 (since sin(2*(π/4)) = sin(π/2) = 1). So to get the peak from x=π/4 to x=0, you need to shift the graph left by π/4. So that's the horizontal shift. Then, subtracting 1 would lower the entire graph by 1 unit. Therefore, shifting left π/4 and down 1, which is option B.Alternatively, if we thought of shifting right by π/4, the peak at x=π/4 would move to x=π/2, which is not where cos(2x) has its peak. So that's incorrect.Therefore, the answer is B.Final Answerboxed{B}</think>To determine how to shift the graph of ( y = sin 2x ) to obtain the graph of ( y = cos 2x - 1 ), we start by recognizing the relationship between sine and cosine functions. The cosine function can be expressed as a phase-shifted sine function:[cos(2x) = sinleft(2x + frac{pi}{2}right)]This can be rewritten by factoring out the 2 inside the sine function:[sinleft(2left(x + frac{pi}{4}right)right)]This indicates a horizontal shift of the sine graph to the left by ( frac{pi}{4} ) units. Therefore, the graph of ( cos(2x) ) is obtained by shifting the graph of ( sin(2x) ) left by ( frac{pi}{4} ) units. Next, we need to account for the vertical shift in the function ( cos(2x) - 1 ). The term (-1) indicates a vertical shift downward by 1 unit. Combining these transformations, we first shift the graph of ( sin(2x) ) left by ( frac{pi}{4} ) units to get ( cos(2x) ), and then shift it down by 1 unit to obtain ( cos(2x) - 1 ).Thus, the correct answer is moving the graph left by ( frac{pi}{4} ) units and then down by 1 unit, which corresponds to option B.[boxed{B}]]

🔑:To obtain the graph of y=cos 2x-1=sin left(2x+ dfrac{pi}{2}right)-1=sin 2left(x+ dfrac{pi}{4}right)-1, one only needs to shift the graph of the function y=sin 2x left by dfrac{pi}{4} units, then down by 1 unit, therefore, the correct answer is boxed{B}. By using the formula for trigonometric transformations, the transformation rule for the graph of y=Asin (omega x+varphi) is concluded. This question mainly tests the formula for trigonometric transformations, the transformation rule for the graph of y=Asin (omega x+varphi), and is considered a basic question.

🧠:[<think>Okay, so I have this problem here: Two ordered positive integers have a sum of 915 and their greatest common divisor is 61. I need to find how many possible pairs there are, and the options are A:4, B:6, C:8, D:14. Let me try to work through this step by step.First, the problem mentions two numbers with a sum of 915 and their GCD is 61. Since the GCD is 61, that means both numbers must be multiples of 61. Let me denote the two numbers as 61a and 61b, where a and b are positive integers. Since they are ordered pairs, (61a, 61b) and (61b, 61a) would count as two different pairs if a ≠ b.Now, their sum is 915. So, 61a + 61b = 915. Let me factor out 61: 61(a + b) = 915. Then, dividing both sides by 61 gives a + b = 915 / 61. Let me calculate that. Hmm, 61 times 15 is 915, right? 61*10=610, 61*5=305, so 610+305=915. Yes, so 915 divided by 61 is 15. Therefore, a + b = 15.But since a and b are positive integers, and they must be coprime, right? Because the GCD of 61a and 61b is 61, so the GCD of a and b must be 1. Otherwise, if a and b had a common divisor greater than 1, say d, then the GCD of 61a and 61b would be 61*d, which contradicts the given GCD of 61. So, a and b need to be coprime, i.e., GCD(a, b) = 1.So the problem reduces to finding the number of ordered pairs (a, b) of positive integers such that a + b = 15 and GCD(a, b) = 1. Then, the number of such pairs would be the answer we're looking for. But since the original numbers are ordered, each pair (a, b) and (b, a) would count as two different pairs unless a = b, but in this case, a + b = 15, which is odd, so a can't equal b. Therefore, all pairs are distinct when ordered.Wait, but let me check that. If a and b are positive integers adding up to 15, and coprime, then how many ordered pairs are there? Each unordered pair {a, b} where a < b would correspond to two ordered pairs: (a, b) and (b, a). So the number of ordered pairs would be twice the number of unordered pairs where a < b and GCD(a, b) = 1.Alternatively, since the problem says "ordered positive integers," so (a, b) and (b, a) are considered different unless a = b. But since a + b = 15, and 15 is odd, a and b can't be equal because that would require a + a = 15, which would mean a = 7.5, not an integer. Therefore, all pairs are distinct when ordered, so the total number is twice the number of unordered pairs where a < b and GCD(a, b) = 1.So first, let's find the number of unordered pairs (a, b) with a < b, a + b = 15, and GCD(a, b) = 1. Then multiply by 2 to get the ordered pairs. But actually, maybe there's a formula for the number of coprime pairs (a, b) such that a + b = n. Let me think.Alternatively, since a and b are positive integers with a + b = 15 and GCD(a, b) = 1, then for each a from 1 to 14 (since b = 15 - a), we can check if GCD(a, 15 - a) = 1. But GCD(a, 15 - a) = GCD(a, 15). Because GCD(a, 15 - a) = GCD(a, 15 - a + a) = GCD(a, 15). Therefore, the condition is that GCD(a, 15) = 1. Because GCD(a, 15 - a) = GCD(a, 15).Therefore, the number of a in [1, 14] such that GCD(a, 15) = 1. But since a and 15 - a must both be positive integers less than 15, but a can be from 1 to 14. So, the number of a in 1 to 14 where GCD(a,15) =1. The number of integers less than 15 and coprime to 15 is given by Euler's totient function φ(15). φ(15) = φ(3*5) = φ(3)*φ(5) = (3-1)*(5-1) = 2*4 = 8. But φ(n) counts numbers from 1 to n-1 coprime to n. So φ(15)=8. Therefore, there are 8 numbers between 1 and 14 that are coprime to 15. But since a ranges from 1 to 14, each coprime a will give a coprime pair (a, 15 - a). However, note that in this case, a and 15 - a are distinct, so each coprime pair is counted twice in the range 1 to 14. Except when a = 15 - a, which would imply a = 7.5, which isn't an integer, so all pairs are distinct. Therefore, the number of unordered pairs is φ(15)/2 = 8/2 = 4. Wait, but φ(15) counts the numbers from 1 to 14 coprime to 15, each coprime a less than 15/2 (since a < b) would give unique unordered pairs. Wait, maybe another way.Wait, φ(15) = 8, which counts the numbers a where 1 ≤ a ≤14 and GCD(a,15)=1. Each such a corresponds to a unique pair (a,15 -a). But since (a,15 -a) is the same as (15 -a, a), unless a =15 -a, which isn't possible here. So, the number of unordered pairs would be φ(15)/2 = 4. Therefore, the number of ordered pairs would be 8. But wait, this is conflicting with my initial thought. Let me check.Let's list all a from 1 to 14 where GCD(a,15) =1.15 is 3*5, so numbers not divisible by 3 or 5.Numbers from 1 to 14 coprime to 15:1: GCD(1,15)=12: GCD(2,15)=14: GCD(4,15)=17: GCD(7,15)=18: GCD(8,15)=111: GCD(11,15)=113: GCD(13,15)=114: GCD(14,15)=1So that's 8 numbers. So each of these a's gives a pair (a,15 -a). But since (a,15 -a) and (15 -a, a) are different ordered pairs, but same unordered pair. So the number of ordered pairs is 8, since each of the 8 a's gives a distinct ordered pair. Wait, no. If we have a from 1 to 14, coprime to 15, then for each a, the pair (a,15 -a) is an ordered pair, but if a <15 -a, then a <7.5, so a from 1 to7. But wait, our list includes a from 1 to14, so for example, a=1 and a=14 both are coprime to 15, but (1,14) and (14,1) are different ordered pairs. Similarly, a=2 and a=13: (2,13) and (13,2). So all 8 numbers would correspond to 8 ordered pairs. Wait, but 8 numbers, each a and 15 -a. But if a is in the list, then 15 -a is also in the list, except when a=15 -a, which is not possible here. So, for each coprime a less than 7.5, there's a corresponding a' =15 -a greater than 7.5. So the total number is actually 8 ordered pairs, because each coprime a gives a distinct ordered pair (a,15 -a), and since a ranges from 1 to14, coprime to 15, the ordered pairs are 8. Wait, but φ(15)=8, which counts the number of a's, but each a corresponds to a unique ordered pair (a,15 -a). However, if a and 15 -a are both in the list, then (a,15 -a) and (15 -a, a) are two different ordered pairs, but both are counted in the 8 a's. Wait, but when you list all the coprime a's from 1 to14, you get 8 numbers. For example, take a=1: coprime, so pair (1,14). Similarly, a=2: (2,13). a=4: (4,11). a=7: (7,8). Then a=8: (8,7). a=11: (11,4). a=13: (13,2). a=14: (14,1). So each of these 8 a's gives a unique ordered pair, but they come in reciprocal pairs. So (1,14) and (14,1) are two different ordered pairs, both included in the 8. So the total number of ordered pairs is 8, right?But earlier I thought maybe φ(n) counts the numbers, and since each pair is counted twice, the number of unordered pairs would be φ(n)/2. But in this case, since n=15, φ(15)=8, so unordered pairs would be 4, and ordered pairs would be 8. But the question is asking for ordered pairs, so the answer would be 8. But wait, the options are A:4, B:6, C:8, D:14. So C:8 is an option. But let me make sure.Wait, but the original problem states "two ordered positive integers", so the order matters. So (61a, 61b) and (61b, 61a) are considered different pairs unless a = b, but since a + b =15, and a and b are positive integers, a and b can't be equal. So each pair is ordered, so the total number of ordered pairs is equal to the number of coprime pairs (a,b) with a + b =15. Since for each a coprime to 15, there's a corresponding b =15 -a, which is also coprime to15, as GCD(a,15)=1 implies GCD(15 -a,15)=GCD(a,15)=1. Therefore, the number of ordered pairs is φ(15) =8. So answer is C:8.But wait, the original problem is phrased as "two ordered positive integers". So if they are ordered, then yes, each permutation counts. So the answer is 8. Therefore, option C. But let me verify once again by enumerating the pairs.List of a where GCD(a,15)=1: 1,2,4,7,8,11,13,14. That's 8 numbers. So the ordered pairs (a,15 -a) would be:(1,14), (2,13), (4,11), (7,8), (8,7), (11,4), (13,2), (14,1). So that's 8 ordered pairs. Therefore, when multiplied by 61, they become:(61*1, 61*14) = (61, 854)(61*2, 61*13) = (122, 793)(61*4, 61*11) = (244, 671)(61*7, 61*8) = (427, 488)Then the reverse pairs:(854, 61), (793, 122), (671, 244), (488, 427). But since the problem says "ordered pairs", all of these are distinct. Therefore, there are 8 ordered pairs. Therefore, the answer should be C:8.But wait, the problem says "ordered positive integers have a sum of 915". So ordered pairs are considered different if the order is different. Therefore, yes, the total is 8. So answer is C.But wait, the answer options given include C:8, which matches. Let me check again my steps.1. Express numbers as 61a and 61b, so GCD(61a,61b)=61*GCD(a,b). Since GCD is given as 61, so GCD(a,b)=1.2. Their sum is 61(a + b) =915 ⇒ a + b=15.3. Find the number of ordered pairs (a,b) with a + b=15 and GCD(a,b)=1.4. Since ordered, each permutation counts.5. The number of such pairs is equal to φ(15)*2 / something? Wait, no. Wait, how does φ(15) relate here?Wait, φ(n) is the number of integers from 1 to n-1 coprime to n. But here, we have a + b=15, and we are looking for coprime pairs (a,b). But since GCD(a,b)=1 and a + b=15, then GCD(a,15)=1. As shown earlier, since GCD(a,b)=GCD(a,15 -a)=GCD(a,15). So if GCD(a,15)=1, then GCD(a,b)=1.Therefore, the number of integers a from 1 to14 where GCD(a,15)=1 is φ(15)=8. Each such a gives a unique ordered pair (a,15 -a). Therefore, the number of ordered pairs is 8. So answer is C:8.Yes, this seems correct. Therefore, the answer should be C:8.Final Answerboxed{C}</think>Given two ordered positive integers with a sum of 915 and their greatest common divisor (GCD) is 61, we need to find the number of possible pairs of these numbers.1. Express the numbers in terms of their GCD: Let the two numbers be (61a) and (61b) where (a) and (b) are coprime positive integers (since their GCD is 61).2. Sum of the numbers: The sum of the numbers is (61a + 61b = 915). Factoring out 61, we get: [ 61(a + b) = 915 implies a + b = frac{915}{61} = 15 ]3. Finding coprime pairs: We need to find the number of ordered pairs ((a, b)) such that (a + b = 15) and (gcd(a, b) = 1). Since (a) and (b) are coprime, (gcd(a, 15) = 1) because (gcd(a, b) = gcd(a, 15 - a) = gcd(a, 15)).4. Using Euler's totient function: The number of integers (a) from 1 to 14 that are coprime to 15 is given by Euler's totient function (phi(15)). Since (15 = 3 times 5), [ phi(15) = phi(3) times phi(5) = (3 - 1)(5 - 1) = 2 times 4 = 8 ]5. Counting ordered pairs: Each integer (a) coprime to 15 gives a pair ((a, 15 - a)). Since the pairs ((a, 15 - a)) and ((15 - a, a)) are considered distinct ordered pairs, the total number of ordered pairs is 8.Thus, the number of possible ordered pairs of the two numbers is (boxed{C}).,<think>Okay, let's see. The problem is about two positive integers that add up to 915, and their greatest common divisor (GCD) is 61. We need to find how many possible pairs there are, and the options are A:4, B:6, C:8, D:14. Hmm, okay. Alright, let's break this down step by step.First, I remember that if two numbers have a GCD of 61, that means both numbers are multiples of 61. So maybe I can simplify the problem by dividing both numbers by 61. That way, their GCD should become 1, right? Because we're factoring out the common divisor.So let me denote the two numbers as 61a and 61b, where a and b are positive integers. Then, according to the problem, their sum is 915. So:61a + 61b = 915We can factor out 61:61(a + b) = 915Then divide both sides by 61:a + b = 915 / 61Let me compute that. 61 times 15 is 915 because 60*15=900 and 1*15=15, so 900+15=915. So 915 divided by 61 is 15. Therefore:a + b = 15Now, since the GCD of the original numbers is 61, the GCD of a and b must be 1. Because if a and b had a common divisor greater than 1, then 61 times that divisor would be a greater common divisor of the original numbers, which contradicts the given GCD of 61. So GCD(a, b) = 1.Therefore, the problem reduces to finding the number of pairs (a, b) such that:1. a and b are positive integers2. a + b = 153. GCD(a, b) = 1And since the pairs (a, b) and (b, a) are considered the same pair for the original problem (since the two numbers are just ordered pairs, but the question is about ordered pairs?), wait, wait. Wait, the problem says "two ordered positive integers". So (61a, 61b) and (61b, 61a) are considered different pairs if a ≠ b. But since a and b are positive integers and ordered, we need to consider both orders unless a = b. However, in our case, since a + b =15, a and b can't be equal because 15 is odd. So each pair (a, b) and (b, a) are distinct ordered pairs.But wait, hold on. The original question is in Chinese, and the translation says "two ordered positive integers". So in Chinese math problems, sometimes "ordered" just means that the order matters, so (x, y) and (y, x) are considered different if x ≠ y. So we need to account for that.But before jumping into that, let me first figure out how many unordered pairs there are, and then maybe double it (except when a = b, but in this case, since a + b =15, which is odd, a can't equal b). So first, find the number of unordered pairs (a, b) where a + b =15 and GCD(a, b) =1, then multiply by 2 to get ordered pairs.Alternatively, maybe it's better to think in terms of ordered pairs directly. Let me check.Given that a and b are positive integers, and a + b =15. The number of ordered pairs (a, b) is 14, because a can range from 1 to 14, and b is 15 -a. But we need to count only those pairs where GCD(a, b) =1. So how many such pairs are there?But since GCD(a, b) = GCD(a, 15 -a) = GCD(a, 15). Because GCD(a, 15 -a) = GCD(a, 15 -a +a) = GCD(a, 15). So GCD(a, 15) must be 1. Therefore, the number of integers a between 1 and 14 inclusive such that GCD(a,15) =1.This is Euler's totient function φ(15). φ(15) = φ(3*5) = φ(3)*φ(5) = (3-1)*(5-1) = 2*4 =8. So φ(15)=8. Therefore, there are 8 numbers a between 1 and 14 such that GCD(a,15)=1. Therefore, there are 8 ordered pairs (a, b) where a + b =15 and GCD(a, b)=1. Therefore, the number of ordered pairs (61a, 61b) would be 8, so the answer is 8, which is option C.Wait, but hold on. Let me verify this again, because the answer options include 14, which is the total number of ordered pairs without considering the GCD. So if we just take a from 1 to14, there are 14 ordered pairs. But with GCD(a, b)=1, it's 8. Therefore, the number of ordered pairs (a, b) with GCD 1 is 8, so the original numbers would be (61a, 61b), which would also be 8 ordered pairs.But the answer choices are A:4, B:6, C:8, D:14. So 8 is option C. Therefore, the answer should be C:8.But let me check again. Because sometimes when problems say "ordered pairs", they might be considering unordered pairs. Wait, but the problem explicitly says "ordered positive integers", so order matters. So (61a,61b) and (61b,61a) are different unless a =b. But since a +b =15, and a and b are positive integers, they can't be equal. Therefore, each pair (a,b) and (b,a) are distinct ordered pairs. However, when we compute φ(15), which is 8, does that count ordered pairs or unordered pairs?Wait, φ(n) counts the numbers less than n and coprime to n. But in our case, we're considering a from 1 to14, and a must be coprime to 15. Each such a gives a unique ordered pair (a,15 -a). So since each a is unique and corresponds to one ordered pair, the total number is φ(15) =8. So these are ordered pairs where a is the first element and b is the second. But since the problem says "ordered pairs", meaning that (a,b) and (b,a) are different, even though they might have the same GCD. Wait, but in our case, if (a,b) is coprime, then (b,a) is also coprime. So the total number of ordered pairs is indeed 8, since each a gives a distinct ordered pair. But let's check with actual numbers.Take for example, a=1: then b=14, GCD(1,14)=1. So this is a valid pair. Similarly, a=2, b=13: GCD(2,13)=1. a=4, b=11: GCD(4,11)=1. a=7, b=8: GCD(7,8)=1. Similarly, a=8, b=7: but wait, if we reverse them, that's a different ordered pair. Wait, no. Wait, when we take a from 1 to14, each a gives a unique ordered pair. For example, when a=1, we get (1,14); when a=14, we get (14,1). Both have GCD 1. So in this case, the number of ordered pairs is 8, because there are 8 values of a where GCD(a,15)=1. Each such a gives an ordered pair (a,15 -a). Since a ranges from 1 to14, each coprime a gives a distinct ordered pair. Therefore, since φ(15)=8, there are 8 ordered pairs. Therefore, the answer is 8, which is option C.But wait, let me verify by listing all pairs:First, list all a from 1 to14 where GCD(a,15)=1.15 factors are 3 and 5. So numbers not divisible by 3 or 5.Numbers from 1 to14:1: coprime with 152: coprime3: divisible by 34: coprime5: divisible by 56: divisible by 37: coprime8: coprime9: divisible by 310: divisible by 511: coprime12: divisible by 313: coprime14: coprimeSo the coprime a's are:1,2,4,7,8,11,13,14. That's 8 numbers. So each of these a's gives an ordered pair (a,15 -a):(1,14), (2,13), (4,11), (7,8), (8,7), (11,4), (13,2), (14,1). So that's 8 ordered pairs. Indeed, 8 pairs. So the answer is 8, which is option C.Therefore, the correct answer is C:8.Final Answerboxed{C}</think>Given two ordered positive integers with a sum of 915 and their greatest common divisor (GCD) is 61, we need to find the number of possible pairs.1. Denote the two numbers as (61a) and (61b) where (a) and (b) are positive integers.2. The sum of these numbers is (915), so we have: [ 61a + 61b = 915 ]3. Factoring out 61, we get: [ 61(a + b) = 915 implies a + b = frac{915}{61} = 15 ]4. The GCD of (a) and (b) must be 1 because the GCD of the original numbers is 61. Therefore, we need pairs ((a, b)) such that (a + b = 15) and (gcd(a, b) = 1).5. The number of such pairs is given by Euler's totient function (phi(15)), which counts the numbers less than 15 and coprime to 15. The prime factors of 15 are 3 and 5, so: [ phi(15) = phi(3 cdot 5) = phi(3) cdot phi(5) = (3 - 1)(5 - 1) = 2 cdot 4 = 8 ]6. The coprime pairs ((a, b)) are ((1, 14), (2, 13), (4, 11), (7, 8), (8, 7), (11, 4), (13, 2), (14, 1)), which are 8 ordered pairs.Thus, the number of possible pairs of the two numbers is (boxed{C}).]

🔑:Let the two numbers be a and b. Then, we have:a + b = 915, quad gcd(a, b) = 61,Let a = 61x and b = 61y,Given 1 leq x leq 14, 1 leq y leq 14, gcd(x, y) = 1, and x + y = 15, we find the pairs:(x, y) = (1, 14), (14, 1), (2, 13), (13, 2), (4, 11), (11, 4), (7, 8), (8, 7),which totals to 8 pairs.Therefore, the correct choice is boxed{C}.

🧠:[<think>Alright, so I have this problem about construction workers and their specialties. Let me try to parse it out step by step. First, the problem says that each worker has at least one construction specialty. There are three specialties mentioned: mason, painter, and plasterer. The numbers given are 10 people have the mason specialty, 9 have painter, and 8 have plasterer. Then there are some overlapping conditions: at least four people have both mason and plasterer, at least five have both mason and painter, and at least three have both painter and plasterer. The question is asking for the maximum number of workers that can be on the site under these conditions.Hmm. So this is a problem about sets and their intersections. It reminds me of the principle of inclusion-exclusion. But since we need the maximum number of workers, we probably have to minimize the overlaps beyond the given constraints. Wait, but the problem states "at least" for the overlaps, so those are minimums. So to maximize the total number of workers, we need to consider the minimum possible overlaps beyond those required, right? Because overlapping more would reduce the total number, while overlapping just the required amount would keep the total higher.Let me recall the inclusion-exclusion principle for three sets. For three sets A, B, C, the total number of elements is |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.In this case, the total number of workers would be the union of masons, painters, and plasterers. So if we denote M as the set of masons, P as painters, and L as plasterers, then:Total Workers = |M ∪ P ∪ L| = |M| + |P| + |L| - |M ∩ P| - |M ∩ L| - |P ∩ L| + |M ∩ P ∩ L|.Given that |M| = 10, |P| = 9, |L| = 8. The overlaps |M ∩ L| ≥ 4, |M ∩ P| ≥ 5, |P ∩ L| ≥ 3. We need to find the maximum possible |M ∪ P ∪ L|. To maximize this, we need to minimize the overlaps (since they are subtracted) and maximize the triple overlap (since it is added). But wait, but the overlaps have minimums. So we can't set |M ∩ L| to 4, |M ∩ P| to 5, |P ∩ L| to 3. But then the triple overlap |M ∩ P ∩ L| also affects the individual pairwise overlaps. For example, if some workers have all three specialties, they are counted in all three pairwise overlaps.Therefore, to minimize the pairwise overlaps beyond the required, we need to maximize the number of workers who have all three specialties, because each person in all three specialties contributes to all three pairwise overlaps. So by increasing the triple overlap, we can satisfy the minimum required pairwise overlaps with fewer "unique" workers in each pairwise overlap. So let me formalize that. Let’s denote x = |M ∩ P ∩ L|. Then, the pairwise overlaps can be expressed as:|M ∩ P| = x + a, where a ≥ 0 (the number of workers who are both M and P but not L)|M ∩ L| = x + b, where b ≥ 0|P ∩ L| = x + c, where c ≥ 0Given that |M ∩ P| ≥ 5, so x + a ≥ 5Similarly, |M ∩ L| ≥ 4 ⇒ x + b ≥ 4|P ∩ L| ≥ 3 ⇒ x + c ≥ 3Our goal is to maximize the total workers:Total = |M| + |P| + |L| - |M ∩ P| - |M ∩ L| - |P ∩ L| + |M ∩ P ∩ L|= 10 + 9 + 8 - (x + a) - (x + b) - (x + c) + x= 27 - 2x - a - b - cTo maximize Total, we need to minimize 2x + a + b + c. However, we have constraints:x + a ≥ 5 ⇒ a ≥ 5 - xx + b ≥ 4 ⇒ b ≥ 4 - xx + c ≥ 3 ⇒ c ≥ 3 - xSince a, b, c can't be negative, the lower bounds for a, b, c are max(0, 5 - x), max(0, 4 - x), max(0, 3 - x) respectively.Therefore, the minimal possible values for a, b, c given x are:a = max(0, 5 - x)b = max(0, 4 - x)c = max(0, 3 - x)Therefore, substituting into Total:Total = 27 - 2x - [max(0, 5 - x) + max(0, 4 - x) + max(0, 3 - x)]We need to find x that minimizes 2x + [max(0, 5 - x) + max(0, 4 - x) + max(0, 3 - x)]. So, we need to analyze this expression for different values of x.Let me consider possible values of x. Since x is the number of workers with all three specialties, it can't be more than the smallest of the pairwise overlaps. Wait, actually, the constraints are |M ∩ P| ≥ 5, |M ∩ L| ≥ 4, |P ∩ L| ≥ 3. But x must satisfy x ≤ |M ∩ P|, x ≤ |M ∩ L|, x ≤ |P ∩ L|. So x is at most the minimum of these pairwise overlaps. But since those pairwise overlaps have lower bounds, but their actual values can be higher. However, since we are trying to maximize x (to minimize a, b, c), but x can't exceed the actual pairwise overlaps. Wait, but since the pairwise overlaps are x + a, x + b, x + c, and a, b, c are non-negative, x can be as large as possible, but if x is too large, then the required a, b, c can become negative, which is not allowed. Hence, x is limited by the lower bounds on the pairwise overlaps.Wait, let's see. If x increases, then the required a, b, c can decrease. For example, if x is 5, then a ≥ 0 (since 5 - 5 = 0), b ≥ -1 (but since b can't be negative, b ≥ 0), and c ≥ -2 (also c ≥ 0). So in that case, x can be up to 5? Wait, but the problem is that the actual overlaps can't be less than the given minimums. So if x is 5, then |M ∩ P| = 5 + a. But |M ∩ P| is given to be at least 5. So if x is 5, then a can be 0, so |M ∩ P| = 5. Similarly, |M ∩ L| would be 5 + b. Since |M ∩ L| needs to be at least 4, 5 + b ≥ 4 ⇒ b ≥ -1, but since b ≥ 0, that's okay. So x can be up to 5? Wait, but if x is 5, then |M ∩ L| must be at least 4. But if x is 5, then |M ∩ L| = 5 + b. So 5 + b ≥ 4 ⇒ b ≥ -1, but b can't be negative. Therefore, b can be 0. So |M ∩ L| = 5, which is okay because the minimum required is 4. Similarly, |P ∩ L| = 5 + c ≥ 3. So 5 + c ≥ 3 ⇒ c ≥ -2. But c can't be negative, so c = 0. So |P ∩ L| = 5, which is okay because the minimum is 3. So in that case, with x = 5, all the pairwise overlaps meet the minimums. But x can't be higher than 5 because if x were 6, then |M ∩ L| would be 6 + b, but |M ∩ L| is only required to be at least 4, which would be okay, but |P ∩ L| would be 6 + c. However, |P ∩ L| is required to be at least 3. If x = 6, then c ≥ 3 - 6 = -3, but c ≥ 0, so c = 0. Therefore, |P ∩ L| = 6, which is fine. Wait, but |P ∩ L| only needs to be at least 3. So why can't x be higher? Wait, but x is the number of people in all three specialties, so it's limited by the sizes of each individual set. For example, |M| = 10, |P| = 9, |L| = 8. So the intersection of all three can't exceed the size of the smallest set, which is 8. But actually, the intersection can't exceed the size of any individual set. Wait, no, the intersection of all three can't exceed the size of any of the individual sets. Since |M| = 10, |P| = 9, |L| = 8, so the maximum possible x is 8. But we might have constraints from the pairwise overlaps. Wait, perhaps not. Let's see.Wait, actually, the maximum possible value of x (the triple overlap) is the minimum of the sizes of each set, but since each set is of size 10, 9, 8. So x ≤ 8. But also, the pairwise overlaps can affect this. For example, if |M ∩ P ∩ L| is x, then |M ∩ P| must be at least x (since x is part of it). Similarly for the others. But the problem states that |M ∩ P| is at least 5, |M ∩ L| at least 4, |P ∩ L| at least 3. So x can be up to the minimum of |M ∩ P|, |M ∩ L|, |P ∩ L|. But since those overlaps have minimums, but no maximums. So theoretically, x could be as high as the minimum of the sizes of the individual sets. But maybe in this case, even higher? Wait, perhaps not. Let's think.But in reality, the maximum x can't exceed the size of any individual set, so x ≤ 8. But even that might not be possible because of overlaps. But given that we need to maximize x to minimize a, b, c, let's see what's the maximum x we can have such that a, b, c can be zero. Because if x is 5, then a can be 0, b can be 0 (since 5 ≥ 4), c can be 0 (since 5 ≥ 3). Wait, if x is 5, then:|M ∩ P| = 5 + a, and since |M ∩ P| must be at least 5, so a ≥ 0. If we set a = 0, then |M ∩ P| = 5. Similarly, |M ∩ L| = 5 + b. Since the minimum required is 4, b can be 0. |P ∩ L| = 5 + c. Minimum required is 3, so c can be 0. Therefore, if x = 5, then a = b = c = 0. That is, all the pairwise overlaps are exactly the minimum required, and the triple overlap is 5. But is that possible? Let's check if this is feasible.Wait, but if the triple overlap is 5, then in the set M (masons) which has 10 people, 5 are also in P and L. Then, the remaining masons are 10 - 5 (in all three) - a (in M and P only) - b (in M and L only) - d (in M only). Wait, but in this case, if a, b, c are zero, then the only overlaps are the triple overlap. Wait, but if a, b, c are zero, then |M ∩ P| = x = 5, |M ∩ L| = x = 5, |P ∩ L| = x = 5. So how does that affect the counts in each set?Let me try to visualize the Venn diagram. If x = 5, then:- The number of workers who are only M is |M| - |M ∩ P| - |M ∩ L| + |M ∩ P ∩ L|. Wait, no. Wait, the formula for the number only in M is |M| - |M ∩ P| - |M ∩ L| + |M ∩ P ∩ L|. Wait, is that right? Let me recall that in inclusion-exclusion, the number only in M is |M| - |M ∩ P| - |M ∩ L| + |M ∩ P ∩ L|. Wait, actually, no. Wait, the standard formula for elements only in M is |M| - |M ∩ P| - |M ∩ L| + |M ∩ P ∩ L|. Wait, is that correct?Wait, actually, the number of elements only in M is |M| - (number in M and P but not L) - (number in M and L but not P) - (number in all three). So:Only M = |M| - (a) - (b) - xSimilarly, since |M ∩ P| = a + x, and |M ∩ L| = b + x.But in our case, if a = 0 and b = 0, then:Only M = |M| - 0 - 0 - x = 10 - 5 = 5Similarly, Only P = |P| - (a) - (c) - x = 9 - 0 - 0 -5 = 4Only L = |L| - (b) - (c) - x = 8 -0 -0 -5 = 3Then, the overlaps:M and P only: a = 0M and L only: b = 0P and L only: c = 0All three: x =5So total workers would be Only M + Only P + Only L + a + b + c + x = 5 + 4 + 3 + 0 +0 +0 +5=17But let's check if that satisfies all the original counts:- Total masons: Only M + a + b + x =5 +0 +0 +5=10 ✔️- Total painters: Only P + a + c + x=4 +0 +0 +5=9 ✔️- Total plasterers: Only L + b + c + x=3 +0 +0 +5=8 ✔️Yes, that works. So in this case, if we set x=5 and a=b=c=0, we get total workers 17.But the question is asking for the maximum number of workers. Is 17 the maximum? Wait, but maybe if x is higher, we can get more workers? Wait, but if x is higher, then a, b, c can be lower, but since they are already zero, we can't go lower. Wait, but when x increases, the expression for Total is 27 -2x - (sum of a,b,c). If x increases beyond 5, but a, b, c can't be negative, so their minimal values would be max(0, 5 -x), etc.Wait, let's test x=5:Total = 27 - 2*5 - (0 +0 +0)=27 -10=17If x=6:Then a= max(0,5-6)=0b= max(0,4-6)=0c= max(0,3-6)=0So Total=27 -2*6 -0=27-12=15Which is less than 17. So increasing x beyond 5 reduces the total. Therefore, 17 is higher.What if x=4:Then:a= max(0,5-4)=1b= max(0,4-4)=0c= max(0,3-4)=0Total=27 -2*4 -1 -0 -0=27-8-1=18Ah, so 18. That's higher. So perhaps x=4 gives a higher total.Wait, so when x=4:a=1, b=0, c=0So then:Total workers=27 -8 -1=18Let me verify if that's feasible.So, x=4 (workers with all three specialties)Then:|M ∩ P|=x + a=4 +1=5 (which meets the minimum of 5)|M ∩ L|=x + b=4 +0=4 (meets minimum of 4)|P ∩ L|=x + c=4 +0=4 (which is above the minimum of 3)So that's acceptable.Now, let's compute the number of workers in each category.Only M: |M| - |M ∩ P| - |M ∩ L| + |M ∩ P ∩ L| =10 -5 -4 +4=5Wait, wait, that formula again. Wait, maybe I should break it down:Only M: |M| - (M∩P only) - (M∩L only) - (all three)But (M∩P only) is a=1(M∩L only) is b=0(all three)=x=4Therefore:Only M =10 -1 -0 -4=5Similarly:Only P= |P| - (M∩P only) - (P∩L only) - (all three)=9 -1 -0 -4=4Only L= |L| - (M∩L only) - (P∩L only) - (all three)=8 -0 -0 -4=4Then:Workers in M∩P only:1Workers in M∩L only:0Workers in P∩L only:0Workers in all three:4Total workers=5+4+4 +1+0+0 +4= 5+4+4+1+4=18Checking the totals:Masons:5 +1 +0 +4=10 ✔️Painters:4 +1 +0 +4=9 ✔️Plasterers:4 +0 +0 +4=8 ✔️Perfect. So that works, and the total is 18, which is higher than when x=5.So x=4 gives a higher total. Let's check x=3.If x=3:a= max(0,5-3)=2b= max(0,4-3)=1c= max(0,3-3)=0Total=27 -2*3 -2 -1 -0=27 -6 -3=18Same as x=4. Wait, so Total=18.But let's check if this is feasible.x=3, a=2, b=1, c=0.So:|M ∩ P|=3 +2=5|M ∩ L|=3 +1=4|P ∩ L|=3 +0=3So all meet the minimums.Calculating the workers:Only M=10 -2 -1 -3=4Only P=9 -2 -0 -3=4Only L=8 -1 -0 -3=4Workers in M∩P only:2Workers in M∩L only:1Workers in P∩L only:0Workers in all three:3Total=4+4+4 +2+1+0 +3=4+4+4+2+1+3=18Check totals:Masons:4 +2 +1 +3=10 ✔️Painters:4 +2 +0 +3=9 ✔️Plasterers:4 +1 +0 +3=8 ✔️Yes, works. So total workers 18 here as well.So x=3 and x=4 both give 18. Interesting. Let me check x=2.x=2:a= max(0,5-2)=3b= max(0,4-2)=2c= max(0,3-2)=1Total=27 -2*2 -3 -2 -1=27 -4 -6=17Lower than 18. So x=2 gives 17.Similarly, x=1:a=4, b=3, c=2Total=27 -2*1 -4 -3 -2=27 -2 -9=16x=0:a=5, b=4, c=3Total=27 -0 -5 -4 -3=15So the maximum occurs at x=3 and x=4 with Total=18.But can we go higher? Let's check x=5. Wait, already checked x=5 gives 17, which is less.Wait, so both x=3 and x=4 give 18. Is that the maximum?Wait, let's check x=6. Wait, x=6:a= max(0,5-6)=0b= max(0,4-6)=0c= max(0,3-6)=0But then |M ∩ P|=6 +0=6 (which is okay, since minimum is 5)|M ∩ L|=6 +0=6 (okay, min 4)|P ∩ L|=6 +0=6 (okay, min 3)Total workers=27 -2*6 -0 -0 -0=27 -12=15But 15 is lower than 18. So x=6 is worse.So the maximum seems to be 18. But let's check x=4 and x=3. Wait, both give 18, but let's see if there's a way to get higher.Wait, maybe if x is higher than 4 but with some a, b, c not minimal. Wait, but according to our formula, the total is 27 -2x - (a + b + c). To maximize total, we need to minimize 2x + a + b + c. But given the constraints a ≥5 -x, etc., the minimal possible value of a + b + c is (5 -x) + (4 -x) + (3 -x) =12 -3x, but only if x ≤3, since otherwise some terms would be zero.Wait, let me clarify. For x ≤3:Then a ≥5 -x, b ≥4 -x, c ≥3 -x, so the minimal a + b + c= (5 -x)+(4 -x)+(3 -x)=12 -3xFor 3 <x ≤4:Then a ≥5 -x, b ≥4 -x, c ≥0 (since 3 -x becomes negative). So minimal a + b + c= (5 -x)+(4 -x)+0=9 -2xFor 4 <x ≤5:a ≥5 -x, b ≥0, c ≥0, so minimal a + b + c= (5 -x) +0 +0=5 -xFor x >5:a ≥0, b ≥0, c ≥0, so minimal a + b +c=0Therefore, the expression for minimal a + b + c is:- If x ≤3: 12 -3x- If 3 <x ≤4: 9 -2x- If 4 <x ≤5:5 -x- If x >5:0Therefore, the expression for Total=27 -2x - (a + b + c). Substituting minimal a + b +c:For x ≤3: Total=27 -2x - (12 -3x)=27 -12 +x=15 +xFor 3 <x ≤4: Total=27 -2x - (9 -2x)=27 -9=18For 4 <x ≤5: Total=27 -2x - (5 -x)=27 -5 -x=22 -xFor x >5: Total=27 -2x -0=27 -2xSo now, analyzing these intervals:- For x ≤3: Total=15 +x, which increases as x increases, reaching 18 at x=3- For 3 <x ≤4: Total=18, constant- For 4 <x ≤5: Total=22 -x, which decreases as x increases, from 22 -4=18 to 22 -5=17- For x >5: Total=27 -2x, decreasing as x increasesTherefore, the maximum Total is 18, achieved when 3 ≤x ≤4. Since x must be an integer (number of workers), x=3 or x=4.So the maximum number of workers is 18.But let's confirm with another approach. Let's use the inclusion-exclusion formula with the minimal overlaps and see.But the problem is that the overlaps can't be less than the given numbers. So:Total workers = M + P + L - (MP + ML + PL) + MPLWhere MP ≥5, ML ≥4, PL ≥3.To maximize the total workers, we need to minimize MP, ML, PL, and maximize MPL.So minimal MP=5, ML=4, PL=3. Then, Total workers=10+9+8 -5 -4 -3 + MPL=27 -12 + MPL=15 + MPL.But the maximum possible MPL is constrained by the overlaps. Because MPL can't exceed any of the pairwise overlaps. Since MP=5, ML=4, PL=3, then MPL can be at most the minimum of these, which is 3. Therefore, Total workers=15 +3=18.Yes, that's another way to see it. If we set the pairwise overlaps to their minimal values, then the triple overlap can be at most the minimum of the pairwise overlaps, which is 3. Then, the total workers=15 +3=18.But wait, but if we set the pairwise overlaps to their minimal values, can the triple overlap be 3? Let's check.If MP=5, ML=4, PL=3, then the maximum possible MPL is the minimum of MP, ML, PL, so min(5,4,3)=3. So yes, MPL=3 is possible. Let's verify:Using inclusion-exclusion:Total workers=10+9+8 -5 -4 -3 +3=27 -12 +3=18.But need to check if such a distribution is possible. Let's see.If MPL=3, then:Only MP (M and P but not L)=MP - MPL=5 -3=2Only ML (M and L but not P)=ML - MPL=4 -3=1Only PL (P and L but not M)=PL - MPL=3 -3=0Then:Only M=10 - (Only MP) - (Only ML) - MPL=10 -2 -1 -3=4Only P=9 - (Only MP) - (Only PL) - MPL=9 -2 -0 -3=4Only L=8 - (Only ML) - (Only PL) - MPL=8 -1 -0 -3=4Total workers=4+4+4 +2+1+0 +3=18, which matches.So that works, and gives 18 workers. Alternatively, if we set MPL=4, but that would require increasing some overlaps. Wait, but if MPL=4, then MP must be at least 4, but the minimal MP is 5, so MP=5. Then, MPL=4 would mean Only MP=5 -4=1. Similarly, ML=4, so Only ML=4 -4=0. PL must be at least 3, but if MPL=4, then PL must be at least 4 (since MPL is part of PL). So PL=4. Then:Total workers=10+9+8 -5 -4 -4 +4=27 -13 +4=18. Same total.Indeed, in this case:Only M=10 -1 -0 -4=5Only P=9 -1 -0 -4=4Only L=8 -0 -0 -4=4Workers in Only MP=1, Only ML=0, Only PL=0, MPL=4Total=5+4+4 +1+0+0 +4=18.So both scenarios where MPL=3 or MPL=4 give total workers 18. Therefore, the maximum possible number of workers is 18.Therefore, the answer should be boxed{18}.Final Answerboxed{18}</think>Given the problem of determining the maximum number of workers on a construction site with specific specialties and overlap conditions, we use the principle of inclusion-exclusion for three sets. The specialties are mason (10 workers), painter (9 workers), and plasterer (8 workers). The overlaps must be at least 4 for mason and plasterer, 5 for mason and painter, and 3 for painter and plasterer.To maximize the total number of workers, we need to minimize the overlaps beyond the given constraints and consider the triple overlap (workers with all three specialties). The inclusion-exclusion principle for three sets is:[|M cup P cup L| = |M| + |P| + |L| - |M cap P| - |M cap L| - |P cap L| + |M cap P cap L|]We denote the number of workers with all three specialties as ( x ). The pairwise overlaps can be expressed as ( |M cap P| = x + a ), ( |M cap L| = x + b ), and ( |P cap L| = x + c ), where ( a, b, c geq 0 ). The constraints are ( x + a geq 5 ), ( x + b geq 4 ), and ( x + c geq 3 ).To find the maximum total number of workers, we need to minimize the expression ( 2x + a + b + c ). We analyze different values of ( x ):1. For ( x leq 3 ): - ( a geq 5 - x ), ( b geq 4 - x ), ( c geq 3 - x ) - Total workers: ( 27 - 2x - (12 - 3x) = 15 + x )2. For ( 3 < x leq 4 ): - ( a geq 5 - x ), ( b geq 4 - x ), ( c geq 0 ) - Total workers: ( 27 - 2x - (9 - 2x) = 18 )3. For ( 4 < x leq 5 ): - ( a geq 5 - x ), ( b geq 0 ), ( c geq 0 ) - Total workers: ( 27 - 2x - (5 - x) = 22 - x )4. For ( x > 5 ): - Total workers: ( 27 - 2x )The maximum total workers is achieved when ( 3 leq x leq 4 ), giving a total of 18. Verifying with specific values of ( x = 3 ) and ( x = 4 ), we find both scenarios satisfy the constraints and result in 18 workers.Thus, the maximum number of workers that can be on the construction site is (boxed{18}).,<think>Okay, let's try to figure out this problem step by step. So, we have a construction site with workers who each have at least one specialty: mason, painter, or plasterer. The numbers given are 10 people have mason specialty, 9 painter, and 8 plasterer. Then there are some overlaps mentioned: at least 4 people have both mason and plasterer, at least 5 have both mason and painter, and at least 3 have both painter and plasterer. The question is asking for the maximum number of workers possible under these conditions.Hmm, okay. So, this is a problem about sets and their intersections. It sounds like we need to use the principle of inclusion-exclusion here. But since we are asked for the maximum number of workers, we need to consider the overlaps as much as possible to minimize the total count. Wait, actually, inclusion-exclusion usually gives the total number by adding individual sets and subtracting overlaps. But if we want the maximum number of workers, we need to minimize the overlaps as much as possible, right? Because if more people have multiple specialties, that would reduce the total number of unique workers. So, to maximize the number of workers, we should have as few overlaps as possible. However, the problem states that there are at least some overlaps: at least 4 with both mason and plasterer, at least 5 with mason and painter, and at least 3 with painter and plasterer. So, those overlaps can't be lower than those numbers. So, perhaps the minimal overlaps are given, but we might have to account for possible overlaps in all three specialties. Hmmm.Let me recall the inclusion-exclusion principle for three sets. The formula is:Total = |M ∪ P ∪ Pl| = |M| + |P| + |Pl| - |M ∩ P| - |M ∩ Pl| - |P ∩ Pl| + |M ∩ P ∩ Pl|Where M is masons, P is painters, Pl is plasterers.So, if we want to maximize the total number of workers, given that the individual set sizes are fixed (|M|=10, |P|=9, |Pl|=8), but the intersections have minimums. So, to maximize Total, we need to minimize the overlaps, but since overlaps have minimums, we need to set the intersections to their minimums. However, we also have to consider the triple overlap, |M ∩ P ∩ Pl|, because if we have people who have all three specialties, that affects all three pairwise overlaps. So, if someone is in all three, they are counted in all three pairwise intersections. Therefore, if we have people with all three specialties, the pairwise overlaps can be reduced by those people. But since the pairwise overlaps have minimums, the number of people with all three specialties cannot be so large that it causes the pairwise overlaps to drop below their required minimums.So, perhaps the strategy is:1. Start with the minimum required pairwise overlaps.2. Account for the triple overlaps, which can reduce the required pairwise overlaps.Wait, let me think. Let’s denote:a = |M ∩ P| (minimum 5)b = |M ∩ Pl| (minimum 4)c = |P ∩ Pl| (minimum 3)d = |M ∩ P ∩ Pl|So, in the inclusion-exclusion formula, the total number of workers is:Total = |M| + |P| + |Pl| - (a + b + c) + dBut since we need to maximize Total, and we have a lower bound on a, b, c, then to maximize Total, we need to minimize (a + b + c) and maximize d. However, since a, b, c can't be lower than their given minimums, but d can be as large as possible without forcing a, b, c below their minimums.Wait, but how does d relate to a, b, c? The maximum possible d is the minimum of the overlaps. For example, the number of people who have all three specialties cannot exceed the number of people in any of the pairwise overlaps. So, d cannot exceed a, b, or c. But since a, b, c have minimums, d is constrained by them.Let me think. Let's suppose that we set a = 5, b = 4, c = 3, which are the minimum required overlaps. Then, the maximum possible d would be the minimum of a, b, c, which is 3. But if we set d = 3, then the actual pairwise overlaps would be a' = a - d = 5 - 3 = 2, which is less than the required minimum of 5. Wait, no. Wait, the pairwise overlaps include those who have all three. So, if the pairwise overlap a = |M ∩ P| is the number of people who are both M and P, regardless of whether they are also Pl. So, if we have d people in all three, then the pairwise overlaps are a = (M ∩ P only) + d. Similarly, b = (M ∩ Pl only) + d, and c = (P ∩ Pl only) + d.Therefore, to have the minimal a, b, c, we need to have the overlaps being exactly the minimum, so:a = (M ∩ P only) + d >= 5Similarly,b = (M ∩ Pl only) + d >= 4c = (P ∩ Pl only) + d >= 3Therefore, to minimize a, b, c, we set (M ∩ P only) = 5 - d, (M ∩ Pl only) = 4 - d, (P ∩ Pl only) = 3 - d. But all these "only" intersections must be non-negative. So, 5 - d >= 0 => d <= 54 - d >=0 => d <=43 - d >=0 => d <=3Therefore, the maximum possible d is 3, since otherwise (P ∩ Pl only) would become negative.Therefore, if we set d = 3, then:(M ∩ P only) = 5 - 3 = 2(M ∩ Pl only) = 4 - 3 = 1(P ∩ Pl only) = 3 - 3 = 0So, these are non-negative.Therefore, the total number of workers would be:Total = |M| + |P| + |Pl| - (a + b + c) + dBut a = |M ∩ P| = (M ∩ P only) + d = 2 + 3 = 5Similarly, b = 1 + 3 = 4c = 0 + 3 = 3Therefore,Total = 10 + 9 + 8 - (5 + 4 + 3) + 3 = 27 - 12 + 3 = 18Wait, but let me check that.Alternatively, maybe constructing the Venn diagram step by step.Let me try to compute each region:1. Only M: M only2. Only P: P only3. Only Pl: Pl only4. M ∩ P only5. M ∩ Pl only6. P ∩ Pl only7. M ∩ P ∩ PlTotal workers = 1 + 2 + 3 + 4 + 5 + 6 + 7We know that:|M| = 10 = (M only) + (M ∩ P only) + (M ∩ Pl only) + (M ∩ P ∩ Pl)Similarly,|P| = 9 = (P only) + (M ∩ P only) + (P ∩ Pl only) + (M ∩ P ∩ Pl)|Pl| = 8 = (Pl only) + (M ∩ Pl only) + (P ∩ Pl only) + (M ∩ P ∩ Pl)Also, given the minimum overlaps:|M ∩ P| = (M ∩ P only) + (M ∩ P ∩ Pl) >=5|M ∩ Pl| = (M ∩ Pl only) + (M ∩ P ∩ Pl) >=4|P ∩ Pl| = (P ∩ Pl only) + (M ∩ P ∩ Pl) >=3To maximize the total number of workers, we need to minimize the overlaps, so set |M ∩ P| =5, |M ∩ Pl|=4, |P ∩ Pl|=3, and maximize the triple overlap d = |M ∩ P ∩ Pl|. However, as above, the maximum possible d is 3, since |P ∩ Pl| >=3, so d <=3. Let's set d=3. Then:From |M ∩ P| =5 = (M ∩ P only) + d => (M ∩ P only) =5 -3=2From |M ∩ Pl|=4= (M ∩ Pl only) +d => (M ∩ Pl only)=4-3=1From |P ∩ Pl|=3= (P ∩ Pl only)+d => (P ∩ Pl only)=3-3=0Now, let's compute the only parts:For M: M only = |M| - (M ∩ P only) - (M ∩ Pl only) - d =10 -2 -1 -3=4For P: P only = |P| - (M ∩ P only) - (P ∩ Pl only) -d=9 -2 -0 -3=4For Pl: Pl only= |Pl| - (M ∩ Pl only) - (P ∩ Pl only) -d=8 -1 -0 -3=4Then total workers =4 (M only) +4 (P only) +4 (Pl only) +2 (M∩P only)+1 (M∩Pl only)+0 (P∩Pl only)+3 (all three) =4+4+4+2+1+0+3=18So, total workers=18.Is that the maximum? Let's check if increasing d beyond 3 is possible. If d=4, then:From |P ∩ Pl|=3= (P ∩ Pl only) +4 => (P ∩ Pl only)= -1, which is impossible. Therefore, d cannot exceed 3.Alternatively, if we set d=2, then:(M ∩ P only)=5-2=3(M ∩ Pl only)=4-2=2(P ∩ Pl only)=3-2=1Then M only=10 -3 -2 -2=3P only=9 -3 -1 -2=3Pl only=8 -2 -1 -2=3Total workers=3+3+3+3+2+1+2=17, which is less than 18. So, that's worse.Wait, actually, let me compute that again:If d=2:M ∩ P only=5-2=3M ∩ Pl only=4-2=2P ∩ Pl only=3-2=1Then:M only=10 -3 -2 -2=3P only=9 -3 -1 -2=3Pl only=8 -2 -1 -2=3Total workers=3+3+3 +3+2+1 +2=17.Yes, that's correct. So, total is 17. Which is lower. So, increasing d up to 3 gives us a higher total. So, 18 is better.Similarly, if d=1:M ∩ P only=5-1=4M ∩ Pl only=4-1=3P ∩ Pl only=3-1=2Then:M only=10 -4 -3 -1=2P only=9 -4 -2 -1=2Pl only=8 -3 -2 -1=2Total workers=2+2+2+4+3+2+1=16Even less.And if d=0:Then, M ∩ P only=5, M ∩ Pl only=4, P ∩ Pl only=3So,M only=10 -5 -4 -0=1Wait, no, wait:Wait, if d=0, then:|M ∩ P|=5 = M ∩ P only|M ∩ Pl|=4 = M ∩ Pl only|P ∩ Pl|=3 = P ∩ Pl onlyThen,M only=10 -5 -4 -0=1P only=9 -5 -3 -0=1Pl only=8 -4 -3 -0=1Total workers=1+1+1+5+4+3+0=15So, even lower. So, indeed, the maximum is achieved when d=3, giving total workers=18.Therefore, the answer should be 18. But let me check again if there's another way to get a higher total.Wait, another thought: maybe the triple overlap isn't necessary. If we don't have any triple overlaps, then the pairwise overlaps have to be at least the given minimums. So, if we set d=0, but then the pairwise overlaps need to be at least 5,4,3. But then, as above, total workers would be 15. Which is lower. So, definitely, having triple overlap allows us to satisfy the minimum pairwise overlaps with less "dedicated" overlap, thereby saving on the total count.Alternatively, is there a way to have some overlaps higher than the minimum to allow a higher total? Wait, but if overlaps are higher than the minimum, that would mean more people are shared between specialties, thus reducing the total number of workers. So, to maximize workers, we need the minimal possible overlaps. So, the minimal overlaps are given as constraints, so we need to set overlaps to exactly those minimums. But as the overlaps include the triple overlap, so we can have some people in all three overlaps, which allows the pairwise overlaps to be met with fewer "unique" overlaps.Therefore, in order to minimize the sum (a + b + c), which would maximize the total, we need to maximize d, since Total = |M| + |P| + |Pl| - (a + b + c) + d = 27 - (a + b - c) + d. Wait, no:Wait, the formula is Total = |M| + |P| + |Pl| - |M ∩ P| - |M ∩ Pl| - |P ∩ Pl| + |M ∩ P ∩ Pl|. So, Total = 27 - (a + b + c) + d. Therefore, to maximize Total, we need to minimize (a + b + c) and maximize d. Since a, b, c have lower bounds, we can set them to their minimums, and set d as large as possible given those minimums. So, as we have:a >=5, b >=4, c >=3But a = (M ∩ P only) + dSimilarly, (M ∩ P only) >=0 => a >=dSimilarly, b >=d, c >=dTherefore, d <= min(a, b, c). So, given that a >=5, b >=4, c >=3, then d <=3. Therefore, maximum d is 3, which gives a=5, b=4, c=3, and Total=27 - (5+4+3) +3=27-12+3=18.So, 18 seems correct.But let me think again if there's another angle. Suppose we use another formula or approach.Alternatively, the maximum number of workers is achieved when the overlaps are as small as possible, but not smaller than the given minima. So, the minimal possible overlaps are 5,4,3. However, these overlaps might have a common part (the triple overlap). So, the maximum number of workers would be |M| + |P| + |Pl| - |M ∩ P| - |M ∩ Pl| - |P ∩ Pl| + |M ∩ P ∩ Pl|. Since |M ∩ P| >=5, |M ∩ Pl|>=4, |P ∩ Pl|>=3, and |M ∩ P ∩ Pl| <= min(|M ∩ P|, |M ∩ Pl|, |P ∩ Pl|).Therefore, if we set |M ∩ P|=5, |M ∩ Pl|=4, |P ∩ Pl|=3, then the maximum possible |M ∩ P ∩ Pl| is 3. Therefore, using the formula:Total =10 +9 +8 -5 -4 -3 +3=27 -12 +3=18.Therefore, 18 is indeed the maximum number.So, I think that's the answer. Let me check once more with an example.Suppose we have:- 4 workers who are only Masons.- 4 workers who are only Painters.- 4 workers who are only Plasterers.Then:- 2 workers who are both Mason and Painter (but not Plasterers).- 1 worker who is both Mason and Plasterer (but not Painter).- 0 workers who are both Painter and Plasterer (but not Masons).- 3 workers who are all three.Total workers: 4+4+4+2+1+0+3=18.Check the counts:Masons: 4 +2 +1 +3=10 ✅Painters:4 +2 +0 +3=9 ✅Plasterers:4 +1 +0 +3=8 ✅Overlaps:Mason ∩ Painter:2 +3=5 ✅Mason ∩ Plasterer:1 +3=4 ✅Painter ∩ Plasterer:0 +3=3 ✅All overlaps meet the minimum requirements. So, this example works. Therefore, the maximum number is indeed 18.Final Answerboxed{18}</think>To determine the maximum number of workers on the construction site, we use the principle of inclusion-exclusion for three sets. Given the specialties and their overlaps, we need to minimize the overlaps while satisfying the given minimums. The formula for the total number of workers is:[|M cup P cup Pl| = |M| + |P| + |Pl| - |M cap P| - |M cap Pl| - |P cap Pl| + |M cap P cap Pl|]Given:- (|M| = 10)- (|P| = 9)- (|Pl| = 8)- Minimum overlaps: (|M cap P| geq 5), (|M cap Pl| geq 4), (|P cap Pl| geq 3)To maximize the total number of workers, we set the overlaps to their minimum values and maximize the triple overlap (|M cap P cap Pl|). The maximum possible triple overlap is 3, as it cannot exceed the smallest minimum pairwise overlap (3).Calculating the individual components:- (|M cap P| = 5)- (|M cap Pl| = 4)- (|P cap Pl| = 3)- (|M cap P cap Pl| = 3)Substituting these values into the inclusion-exclusion formula:[|M cup P cup Pl| = 10 + 9 + 8 - 5 - 4 - 3 + 3 = 27 - 12 + 3 = 18]Verification:- Only Masons: (10 - 2 - 1 - 3 = 4)- Only Painters: (9 - 2 - 0 - 3 = 4)- Only Plasterers: (8 - 1 - 0 - 3 = 4)- Mason and Painter only: (5 - 3 = 2)- Mason and Plasterer only: (4 - 3 = 1)- Painter and Plasterer only: (3 - 3 = 0)- All three specialties: (3)Summing these gives (4 + 4 + 4 + 2 + 1 + 0 + 3 = 18).Thus, the maximum number of workers is (boxed{18}).]

🔑:# Answer 18.Solution: According to the principle of inclusion-exclusion, the total number of workers is K+M+S - KM-KS-MS + KMS = 10+9+8-4-5-3 + KMS = 15 + KMS. Note that the number of workers proficient in all three specialties cannot exceed 3.