💡Okay, so I have this problem where I need to find the sum of the reciprocals of all positive integers ( n ) such that ( 3 cdot varphi(n) = n ). Euler's totient function ( varphi(n) ) counts the number of integers up to ( n ) that are relatively prime to ( n ). First, I remember that for any integer ( n ) with prime factorization ( n = p_1^{e_1} p_2^{e_2} cdots p_k^{e_k} ), the totient function is given by:[varphi(n) = n left(1 - frac{1}{p_1}right)left(1 - frac{1}{p_2}right) cdots left(1 - frac{1}{p_k}right)]So, substituting this into the equation ( 3 cdot varphi(n) = n ), we get:[3 cdot n left(1 - frac{1}{p_1}right)left(1 - frac{1}{p_2}right) cdots left(1 - frac{1}{p_k}right) = n]Dividing both sides by ( n ), we simplify to:[3 left(1 - frac{1}{p_1}right)left(1 - frac{1}{p_2}right) cdots left(1 - frac{1}{p_k}right) = 1]So, the product of these terms must equal ( frac{1}{3} ). I need to figure out which primes ( p_i ) satisfy this condition. Let me consider small primes first. If I take ( p_1 = 2 ), then ( 1 - frac{1}{2} = frac{1}{2} ). If I take another prime, say ( p_2 = 3 ), then ( 1 - frac{1}{3} = frac{2}{3} ). Multiplying these together:[frac{1}{2} cdot frac{2}{3} = frac{1}{3}]Perfect, that gives me exactly ( frac{1}{3} ). So, it seems that the primes involved must be 2 and 3. Therefore, any ( n ) that satisfies ( 3 cdot varphi(n) = n ) must be of the form ( 2^a cdot 3^b ) where ( a ) and ( b ) are positive integers (since if either exponent were zero, we wouldn't have both primes contributing to the product). Now, I need to find the sum of the reciprocals of all such ( n ). That is:[sum_{n in S} frac{1}{n} = sum_{a=1}^{infty} sum_{b=1}^{infty} frac{1}{2^a cdot 3^b}]This is a double sum, which can be separated into two geometric series:[left( sum_{a=1}^{infty} frac{1}{2^a} right) cdot left( sum_{b=1}^{infty} frac{1}{3^b} right)]I know that the sum of a geometric series ( sum_{k=0}^{infty} ar^k ) is ( frac{a}{1 - r} ) when ( |r| < 1 ). However, our sums start from ( a=1 ) and ( b=1 ), so we need to adjust accordingly.For the series ( sum_{a=1}^{infty} frac{1}{2^a} ), the first term is ( frac{1}{2} ) and the common ratio is ( frac{1}{2} ). So, the sum is:[frac{frac{1}{2}}{1 - frac{1}{2}} = frac{frac{1}{2}}{frac{1}{2}} = 1]Similarly, for the series ( sum_{b=1}^{infty} frac{1}{3^b} ), the first term is ( frac{1}{3} ) and the common ratio is ( frac{1}{3} ). The sum is:[frac{frac{1}{3}}{1 - frac{1}{3}} = frac{frac{1}{3}}{frac{2}{3}} = frac{1}{2}]Multiplying these two results together gives:[1 cdot frac{1}{2} = frac{1}{2}]So, the sum of the reciprocals of all such ( n ) is ( frac{1}{2} ).I should double-check if there are any other primes that could satisfy the condition ( 3 cdot varphi(n) = n ). Suppose there was another prime ( p ) involved. Then, the product ( left(1 - frac{1}{2}right)left(1 - frac{1}{3}right)left(1 - frac{1}{p}right) ) would have to equal ( frac{1}{3} ). But:[frac{1}{2} cdot frac{2}{3} cdot left(1 - frac{1}{p}right) = frac{1}{3} cdot left(1 - frac{1}{p}right)]This would need to equal ( frac{1}{3} ), which implies ( left(1 - frac{1}{p}right) = 1 ), meaning ( p ) would have to be infinity, which isn't possible. Therefore, no other primes can be part of the factorization of ( n ).Thus, my conclusion is that the sum is indeed ( frac{1}{2} ).**Final Answer**boxed{dfrac{1}{2}}

💡Okay, so I have this function f(x) = sin⁴x - sinx cosx + cos⁴x, and I need to find its range. Hmm, I remember that the range of a function is the set of all possible output values it can take. So, I need to figure out the minimum and maximum values that f(x) can have as x varies over all real numbers.First, let me write down the function again to make sure I have it correctly:f(x) = sin⁴x - sinx cosx + cos⁴xHmm, that's a combination of sin⁴x, cos⁴x, and a term with sinx cosx. I think I can simplify this expression using some trigonometric identities. I remember that sin²x + cos²x = 1, which is the Pythagorean identity. Maybe I can use that to rewrite sin⁴x and cos⁴x.Let me recall that sin⁴x can be written as (sin²x)², and similarly, cos⁴x is (cos²x)². So, maybe I can express sin⁴x + cos⁴x in terms of sin²x and cos²x. Let me try that.I know that (sin²x + cos²x)² = sin⁴x + 2 sin²x cos²x + cos⁴x. Since sin²x + cos²x = 1, then (1)² = 1 = sin⁴x + 2 sin²x cos²x + cos⁴x. Therefore, sin⁴x + cos⁴x = 1 - 2 sin²x cos²x.So, substituting back into f(x), I get:f(x) = (1 - 2 sin²x cos²x) - sinx cosxSo, f(x) = 1 - 2 sin²x cos²x - sinx cosxHmm, that's a bit simpler. Now, I have f(x) expressed in terms of sinx cosx and sin²x cos²x. Maybe I can express everything in terms of a single trigonometric function. I remember that sin(2x) = 2 sinx cosx, so perhaps I can let t = sin(2x). That might help.Let me set t = sin(2x). Then, sinx cosx = t/2. Also, sin²x cos²x = (sinx cosx)² = (t/2)² = t²/4.So, substituting back into f(x):f(x) = 1 - 2*(t²/4) - t/2Simplify that:f(x) = 1 - (t²/2) - (t/2)So, f(x) = 1 - (t² + t)/2Hmm, that's a quadratic in terms of t. Let me write it as:f(x) = - (t² + t)/2 + 1I can rewrite this as:f(x) = - (1/2)t² - (1/2)t + 1Now, since t = sin(2x), and the sine function has a range of [-1, 1], t must be in [-1, 1]. So, t ∈ [-1, 1]. Therefore, f(x) is a quadratic function in terms of t, where t is between -1 and 1.To find the range of f(x), I need to find the maximum and minimum values of this quadratic function over the interval t ∈ [-1, 1].Quadratic functions have their extrema either at the vertex or at the endpoints of the interval. So, I should find the vertex of this parabola and evaluate the function at t = -1, t = 1, and at the vertex if it lies within the interval.First, let me write the quadratic function in standard form:f(t) = - (1/2)t² - (1/2)t + 1This is a quadratic function of the form f(t) = at² + bt + c, where a = -1/2, b = -1/2, and c = 1.The vertex of a parabola given by f(t) = at² + bt + c is at t = -b/(2a). Let me compute that:t_vertex = -b/(2a) = -(-1/2)/(2*(-1/2)) = (1/2)/(-1) = -1/2So, the vertex is at t = -1/2. Since -1/2 is within the interval [-1, 1], the maximum or minimum occurs at this point.Since the coefficient of t² is negative (a = -1/2), the parabola opens downward, meaning the vertex is a maximum point.Therefore, the maximum value of f(t) occurs at t = -1/2, and the minimum values occur at the endpoints t = -1 and t = 1.Let me compute f(t) at t = -1, t = 1, and t = -1/2.First, at t = -1:f(-1) = - (1/2)(-1)² - (1/2)(-1) + 1 = - (1/2)(1) - (-1/2) + 1 = -1/2 + 1/2 + 1 = 1Wait, that seems off. Let me recalculate:f(-1) = - (1/2)(-1)^2 - (1/2)(-1) + 1= - (1/2)(1) - (-1/2) + 1= -1/2 + 1/2 + 1= (-1/2 + 1/2) + 1= 0 + 1= 1Okay, so f(-1) = 1.Now, at t = 1:f(1) = - (1/2)(1)^2 - (1/2)(1) + 1= - (1/2)(1) - (1/2) + 1= -1/2 - 1/2 + 1= (-1/2 - 1/2) + 1= -1 + 1= 0So, f(1) = 0.Now, at t = -1/2:f(-1/2) = - (1/2)(-1/2)^2 - (1/2)(-1/2) + 1First, compute (-1/2)^2 = 1/4So,= - (1/2)(1/4) - (1/2)(-1/2) + 1= - (1/8) - (-1/4) + 1= -1/8 + 1/4 + 1Convert to eighths:= -1/8 + 2/8 + 8/8= (-1 + 2 + 8)/8= 9/8So, f(-1/2) = 9/8.Therefore, the function f(t) reaches its maximum value of 9/8 at t = -1/2 and its minimum value of 0 at t = 1. At t = -1, the function is 1.So, putting this all together, the range of f(x) is from 0 to 9/8. Therefore, the range of f(x) is [0, 9/8].Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, f(-1):f(-1) = - (1/2)(1) - (1/2)(-1) + 1 = -1/2 + 1/2 + 1 = 1. Correct.f(1):f(1) = - (1/2)(1) - (1/2)(1) + 1 = -1/2 -1/2 +1 = -1 +1 = 0. Correct.f(-1/2):f(-1/2) = - (1/2)(1/4) - (1/2)(-1/2) +1 = -1/8 +1/4 +1 = (-1/8 + 2/8) +8/8 = (1/8) +8/8=9/8. Correct.So, yes, the maximum is 9/8 and the minimum is 0. Therefore, the range is [0, 9/8].I think that makes sense. Let me see if there's another way to approach this problem to confirm.Alternatively, I could express f(x) in terms of multiple angles or use calculus to find the maximum and minimum. But since I've already transformed it into a quadratic in t and found the range, and the steps seem correct, I think this is solid.Just to recap:1. Expressed sin⁴x + cos⁴x in terms of sin²x cos²x.2. Substituted t = sin(2x) to simplify the expression.3. Rewrote f(x) as a quadratic in t.4. Found the vertex and evaluated at endpoints.5. Concluded the range is [0, 9/8].Yes, that seems thorough. I don't see any errors in my reasoning. So, I'm confident that the range of f(x) is from 0 to 9/8.

💡To determine the maximum number of students who can each receive the same number of pens and pencils, we need to find the greatest common divisor (GCD) of the total number of pens and pencils.**Step 1:** Identify the total number of pens and pencils.- Pens: 3540- Pencils: 2860**Step 2:** Apply the Euclidean algorithm to find the GCD of 3540 and 2860.1. Divide 3540 by 2860: - 3540 ÷ 2860 = 1 with a remainder of 680.2. Replace 3540 with 2860 and 2860 with 680: - 2860 ÷ 680 = 4 with a remainder of 120.3. Replace 2860 with 680 and 680 with 120: - 680 ÷ 120 = 5 with a remainder of 80.4. Replace 680 with 120 and 120 with 80: - 120 ÷ 80 = 1 with a remainder of 40.5. Replace 120 with 80 and 80 with 40: - 80 ÷ 40 = 2 with a remainder of 0.**Step 3:** The last non-zero remainder is 40, which is the GCD of 3540 and 2860.**Conclusion:** The maximum number of students is 40.

💡Alright, so I've got this problem here about an arithmetic sequence. Let me try to break it down step by step. First, the problem says that we have an arithmetic sequence {a_n} with a common difference d, which isn't zero, and the first term a_1 is equal to 2d. Okay, so that's our starting point.Now, the problem mentions that a_k is the geometric mean of a_1 and a_{2k+1}. Hmm, geometric mean. I remember that the geometric mean of two numbers, say x and y, is the square root of their product. So, if a_k is the geometric mean of a_1 and a_{2k+1}, that should mean that a_k squared equals a_1 times a_{2k+1}. Let me write that down:a_k² = a_1 * a_{2k+1}Alright, now I need to express a_k and a_{2k+1} in terms of d and k. Since it's an arithmetic sequence, the general term is given by a_n = a_1 + (n - 1)d. So, let's compute a_k and a_{2k+1}.First, a_k = a_1 + (k - 1)d. But we know that a_1 is 2d, so substituting that in:a_k = 2d + (k - 1)dLet me simplify that. 2d plus (k - 1)d is the same as (2 + k - 1)d, which simplifies to (k + 1)d. So, a_k = (k + 1)d.Next, let's find a_{2k+1}. Using the same formula:a_{2k+1} = a_1 + (2k + 1 - 1)dSimplify that. The -1 and +1 cancel out, so we have:a_{2k+1} = a_1 + 2k dAgain, since a_1 is 2d, substitute that in:a_{2k+1} = 2d + 2k dFactor out the d:a_{2k+1} = (2 + 2k)d = 2(k + 1)dSo, a_{2k+1} is 2(k + 1)d.Now, going back to the geometric mean equation:a_k² = a_1 * a_{2k+1}Substitute the expressions we found:[(k + 1)d]² = (2d) * [2(k + 1)d]Let me compute both sides. On the left side, squaring (k + 1)d gives:(k + 1)² d²On the right side, multiplying 2d and 2(k + 1)d:2d * 2(k + 1)d = 4(k + 1)d²So, now we have:(k + 1)² d² = 4(k + 1)d²Hmm, both sides have a d² term, and since d isn't zero, we can divide both sides by d² to simplify:(k + 1)² = 4(k + 1)That looks better. Now, let's expand the left side:(k + 1)² = k² + 2k + 1So, the equation becomes:k² + 2k + 1 = 4(k + 1)Let me bring all terms to one side to solve for k. Subtract 4(k + 1) from both sides:k² + 2k + 1 - 4k - 4 = 0Simplify the terms:k² - 2k - 3 = 0Alright, so we have a quadratic equation: k² - 2k - 3 = 0. Let me try to factor this. I need two numbers that multiply to -3 and add up to -2. Hmm, 1 and -3? Yes, because 1 * (-3) = -3 and 1 + (-3) = -2.So, factoring:(k + 1)(k - 3) = 0Setting each factor equal to zero:k + 1 = 0 => k = -1k - 3 = 0 => k = 3But k represents a term in the sequence, so it has to be a positive integer. Therefore, k = 3 is the valid solution.Let me double-check to make sure I didn't make any mistakes. Starting from the beginning:a_1 = 2da_k = (k + 1)da_{2k+1} = 2(k + 1)dThen, geometric mean condition:[(k + 1)d]^2 = 2d * 2(k + 1)dWhich simplifies to:(k + 1)^2 d^2 = 4(k + 1)d^2Divide both sides by d^2:(k + 1)^2 = 4(k + 1)Subtract 4(k + 1):(k + 1)^2 - 4(k + 1) = 0Factor:(k + 1)(k + 1 - 4) = 0Which is:(k + 1)(k - 3) = 0So, k = -1 or k = 3. Since k must be positive, k = 3. That seems correct.I think I've got it right. So, the answer should be 3, which is option B.**Final Answer**boxed{B}

💡Okay, so I have these two inequalities to solve. Let me take them one at a time and think through each step carefully. I want to make sure I understand what I'm doing and not make any mistakes.Starting with the first inequality: -x² + x + 6 ≤ 0. Hmm, this is a quadratic inequality. I remember that to solve quadratic inequalities, it's helpful to rewrite them in standard form, which is ax² + bx + c. So, let me rearrange this inequality to make it easier to work with.If I move all the terms to the other side, I get: -x² + x + 6 ≤ 0. Wait, that's the same as it was. Maybe I should factor out the negative sign to make it easier to factor. Let me try that. If I factor out a -1, I get: -(x² - x - 6) ≤ 0. Okay, now I can multiply both sides by -1, but I have to remember that multiplying both sides of an inequality by a negative number reverses the inequality sign. So, that would give me: x² - x - 6 ≥ 0.Alright, now I have x² - x - 6 ≥ 0. Let me try to factor this quadratic. I need two numbers that multiply to -6 and add up to -1. Let's see: 2 and -3. Yes, because 2 * (-3) = -6 and 2 + (-3) = -1. So, the factored form is (x + 2)(x - 3) ≥ 0.Now, to solve this inequality, I need to find the intervals where the product of (x + 2) and (x - 3) is greater than or equal to zero. First, I'll find the critical points where each factor is zero. Setting each factor equal to zero:x + 2 = 0 ⇒ x = -2x - 3 = 0 ⇒ x = 3These critical points divide the number line into three intervals:1. x < -22. -2 < x < 33. x > 3I'll test each interval to see if the product is positive or negative in that interval.For the first interval, x < -2: Let's pick x = -3.Plugging into (x + 2)(x - 3): (-3 + 2)(-3 - 3) = (-1)(-6) = 6, which is positive. So, the product is positive in this interval.For the second interval, -2 < x < 3: Let's pick x = 0.Plugging into (x + 2)(x - 3): (0 + 2)(0 - 3) = (2)(-3) = -6, which is negative. So, the product is negative in this interval.For the third interval, x > 3: Let's pick x = 4.Plugging into (x + 2)(x - 3): (4 + 2)(4 - 3) = (6)(1) = 6, which is positive. So, the product is positive in this interval.Since the inequality is ≥ 0, we're looking for where the product is positive or zero. That happens in the intervals x ≤ -2 and x ≥ 3. So, the solution to the first inequality is x ∈ (-∞, -2] ∪ [3, ∞).Okay, moving on to the second inequality: x² - 2x - 5 < 2x. Again, this is a quadratic inequality, so I'll start by bringing all terms to one side to get it into standard form.Subtracting 2x from both sides: x² - 4x - 5 < 0. Now, I need to factor this quadratic. Let me see if it factors nicely. I'm looking for two numbers that multiply to -5 and add up to -4. Hmm, -5 and +1. Yes, because (-5) * 1 = -5 and (-5) + 1 = -4. So, the factored form is (x - 5)(x + 1) < 0.Now, to solve this inequality, I'll find the critical points where each factor is zero:x - 5 = 0 ⇒ x = 5x + 1 = 0 ⇒ x = -1These critical points divide the number line into three intervals:1. x < -12. -1 < x < 53. x > 5I'll test each interval to determine the sign of the product in that interval.For the first interval, x < -1: Let's pick x = -2.Plugging into (x - 5)(x + 1): (-2 - 5)(-2 + 1) = (-7)(-1) = 7, which is positive. So, the product is positive in this interval.For the second interval, -1 < x < 5: Let's pick x = 0.Plugging into (x - 5)(x + 1): (0 - 5)(0 + 1) = (-5)(1) = -5, which is negative. So, the product is negative in this interval.For the third interval, x > 5: Let's pick x = 6.Plugging into (x - 5)(x + 1): (6 - 5)(6 + 1) = (1)(7) = 7, which is positive. So, the product is positive in this interval.Since the inequality is < 0, we're looking for where the product is negative. That happens in the interval -1 < x < 5. So, the solution to the second inequality is x ∈ (-1, 5).Wait, let me double-check my work to make sure I didn't make any mistakes.For the first inequality, I had -x² + x + 6 ≤ 0, which I converted to x² - x - 6 ≥ 0. Factored it to (x + 2)(x - 3) ≥ 0, found critical points at x = -2 and x = 3, tested intervals, and concluded x ≤ -2 or x ≥ 3. That seems correct.For the second inequality, x² - 2x - 5 < 2x, I moved 2x to the left to get x² - 4x - 5 < 0, factored to (x - 5)(x + 1) < 0, found critical points at x = -1 and x = 5, tested intervals, and concluded -1 < x < 5. That also seems correct.I think I did everything right. Maybe I should quickly sketch the graphs of these quadratics to visualize the solutions.For the first inequality, the quadratic x² - x - 6 is a parabola opening upwards with roots at x = -2 and x = 3. Since it's a parabola opening upwards, it will be above the x-axis (i.e., y ≥ 0) outside the interval between the roots. So, x ≤ -2 or x ≥ 3. That matches my solution.For the second inequality, the quadratic x² - 4x - 5 is also a parabola opening upwards with roots at x = -1 and x = 5. Since it's opening upwards, it will be below the x-axis (i.e., y < 0) between the roots. So, -1 < x < 5. That also matches my solution.Alright, I feel confident about these answers now.

💡First, I need to determine the total lifetime of Jack, denoted as ( L ).Jack spent ( frac{1}{6}L ) of his life in adolescence.After that, his facial hair started growing after an additional ( frac{1}{12}L ) of his life.He married Diana after another ( frac{1}{7}L ) of his life.Their son was born 5 years after their marriage.The son lived half as long as Jack, so the son's lifespan is ( frac{1}{2}L ).Jack died 4 years after his son's death.To find ( L ), I'll set up an equation that accounts for all these periods:[frac{1}{6}L + frac{1}{12}L + frac{1}{7}L + 5 + frac{1}{2}L + 4 = L]Next, I'll find a common denominator for the fractions, which is 84:[frac{14}{84}L + frac{7}{84}L + frac{12}{84}L + 5 + frac{42}{84}L + 4 = L]Combining the fractions:[frac{75}{84}L + 9 = L]Subtracting ( frac{75}{84}L ) from both sides:[9 = frac{9}{84}L]Solving for ( L ):[L = 84]Therefore, Jack lived for 84 years.