💡To determine the number of multiples of 7 between 200 and 500, follow these steps:1. **Find the least multiple of 7 that is greater than or equal to 200**: - Divide 200 by 7: ( frac{200}{7} approx 28.57 ) - The smallest integer greater than 28.57 is 29. - Multiply 29 by 7: ( 29 times 7 = 203 ) - So, the smallest multiple of 7 greater than or equal to 200 is 203.2. **Find the greatest multiple of 7 that is less than or equal to 500**: - Divide 500 by 7: ( frac{500}{7} approx 71.43 ) - The largest integer less than 71.43 is 71. - Multiply 71 by 7: ( 71 times 7 = 497 ) - So, the largest multiple of 7 less than or equal to 500 is 497.3. **Determine how many multiples of 7 are between 203 and 497**: - The sequence of multiples of 7 between 203 and 497 is an arithmetic sequence where: - The first term ((a)) is 203. - The common difference ((d)) is 7. - The last term ((a_n)) is 497. - Use the formula for the (n)-th term of an arithmetic sequence: [ a_n = a + (n-1)d ] Plugging in the known values: [ 497 = 203 + (n-1) times 7 ] - Solve for (n): [ 497 - 203 = (n-1) times 7 294 = (n-1) times 7 n-1 = frac{294}{7} n-1 = 42 n = 43 ]4. **Conclusion**: - The number of multiples of 7 between 200 and 500 is 43.[boxed{43}]

💡Okay, so I have this problem: Let f(x) be continuously differentiable on (0, ∞), and suppose that the limit of f'(x) as x approaches infinity is zero. I need to prove that the limit of f(x)/x as x approaches infinity is also zero. Hmm, interesting. Alright, let me try to break this down step by step.First, let me recall what it means for a function to be continuously differentiable on (0, ∞). It means that f(x) has a derivative f'(x) that is continuous on that interval. So, f(x) is smooth enough, and its derivative doesn't have any sudden jumps or breaks. That's good to know because it ensures that f(x) behaves nicely, which might be useful later.Now, the key piece of information here is that the limit of f'(x) as x approaches infinity is zero. So, as x gets really large, the slope of f(x) is getting flatter and flatter, approaching zero. Intuitively, if the slope is approaching zero, the function f(x) isn't increasing or decreasing too rapidly as x becomes large. It's kind of leveling off.I need to show that f(x)/x approaches zero as x approaches infinity. So, even though f(x) might be growing or decaying, when I divide it by x, which is growing without bound, the whole expression should go to zero. That makes sense intuitively because if f(x) is growing slower than x, then f(x)/x would go to zero. But I need to make this precise.Let me think about how to relate f(x) and f'(x). Since f'(x) is the derivative of f(x), maybe I can use some relationship between a function and its derivative. One tool that comes to mind is integration. If I can express f(x) in terms of its derivative, perhaps I can analyze its growth.Let me recall the Fundamental Theorem of Calculus. If f is continuously differentiable, then f(x) can be expressed as the integral of its derivative plus some constant. So, f(x) = f(a) + ∫ from a to x of f'(t) dt for some a in (0, ∞). Maybe I can use this to express f(x) and then analyze its behavior as x grows.But I need to find the limit of f(x)/x. Let me write that out:lim(x→∞) f(x)/x = lim(x→∞) [f(a) + ∫ from a to x of f'(t) dt] / xHmm, so I can split this into two terms:lim(x→∞) [f(a)/x + (∫ from a to x of f'(t) dt)/x]Now, f(a) is just a constant, so f(a)/x will go to zero as x approaches infinity. That's straightforward. So, the first term is taken care of.Now, the second term is (∫ from a to x of f'(t) dt)/x. I need to evaluate the limit of this as x approaches infinity. Let me denote the integral as I(x) = ∫ from a to x of f'(t) dt. So, I need to find lim(x→∞) I(x)/x.Given that f'(x) approaches zero as x approaches infinity, I wonder how this affects the integral I(x). If f'(x) is approaching zero, then for large t, f'(t) is small. So, maybe the integral from a to x of f'(t) dt doesn't grow too fast as x increases.But I need to be more precise. Let me consider that for large t, say t > M for some M, |f'(t)| < ε for some small ε. Since lim(x→∞) f'(x) = 0, for any ε > 0, there exists an M such that for all t > M, |f'(t)| < ε.So, I can split the integral I(x) into two parts: from a to M and from M to x.I(x) = ∫ from a to M of f'(t) dt + ∫ from M to x of f'(t) dtLet me denote the first integral as C, which is a constant because it's from a fixed a to M. So, I(x) = C + ∫ from M to x of f'(t) dt.Therefore, I(x)/x = C/x + (∫ from M to x of f'(t) dt)/xAgain, C/x goes to zero as x approaches infinity. So, the limit of I(x)/x is the same as the limit of (∫ from M to x of f'(t) dt)/x.Now, let's focus on (∫ from M to x of f'(t) dt)/x. Since |f'(t)| < ε for t > M, the integral ∫ from M to x of f'(t) dt is bounded in absolute value by ∫ from M to x of |f'(t)| dt < ∫ from M to x of ε dt = ε(x - M).Therefore, |∫ from M to x of f'(t) dt| < ε(x - M)So, |(∫ from M to x of f'(t) dt)/x| < ε(x - M)/x = ε(1 - M/x)As x approaches infinity, M/x approaches zero, so ε(1 - M/x) approaches ε.Therefore, |(∫ from M to x of f'(t) dt)/x| < ε for sufficiently large x.Since ε is arbitrary, this implies that the limit of (∫ from M to x of f'(t) dt)/x is zero.Putting it all together, lim(x→∞) I(x)/x = 0, and hence lim(x→∞) f(x)/x = 0.Wait, that seems a bit hand-wavy. Let me see if I can make this more rigorous.Alternatively, maybe I can use the Mean Value Theorem. For each x > a, there exists a point c in (a, x) such that f(x) - f(a) = f'(c)(x - a). So, f(x) = f(a) + f'(c)(x - a).Then, f(x)/x = [f(a) + f'(c)(x - a)] / x = f(a)/x + f'(c)(x - a)/xAs x approaches infinity, f(a)/x approaches zero. Now, what about f'(c)(x - a)/x?Note that c is between a and x, so as x approaches infinity, c also approaches infinity. Therefore, f'(c) approaches zero because lim(x→∞) f'(x) = 0.So, f'(c) approaches zero, and (x - a)/x approaches 1. Therefore, the whole term f'(c)(x - a)/x approaches zero.Hence, lim(x→∞) f(x)/x = 0.Hmm, that seems more straightforward. So, using the Mean Value Theorem, I can express f(x) in terms of its derivative at some intermediate point c, and then analyze the limit.But wait, is this always valid? The Mean Value Theorem requires that f is continuous on [a, x] and differentiable on (a, x), which is given since f is continuously differentiable on (0, ∞). So, yes, it's applicable.Therefore, this approach seems solid. So, to summarize:1. Use the Mean Value Theorem to write f(x) = f(a) + f'(c)(x - a) for some c in (a, x).2. Divide both sides by x: f(x)/x = f(a)/x + f'(c)(x - a)/x.3. Take the limit as x approaches infinity. The first term f(a)/x goes to zero.4. The second term: f'(c) approaches zero because c approaches infinity, and (x - a)/x approaches 1.5. Therefore, the entire expression f(x)/x approaches zero.That seems to cover it. But let me double-check if there are any potential issues with this reasoning.One thing to consider is whether c approaches infinity as x approaches infinity. Since c is between a and x, and x is going to infinity, c must also go to infinity. So, f'(c) approaches zero as x approaches infinity.Another point is whether f'(c) is bounded. Since f'(x) approaches zero, for sufficiently large x, f'(x) is bounded by some small ε. So, f'(c) is also bounded by ε for large x.Therefore, the term f'(c)(x - a)/x is bounded by ε*(x - a)/x, which is less than ε*(x)/x = ε. Since ε can be made arbitrarily small, the limit must be zero.Alright, I think this reasoning holds. So, using the Mean Value Theorem provides a straightforward way to relate f(x) and its derivative f'(x), and then leveraging the given condition that f'(x) approaches zero as x approaches infinity allows us to conclude that f(x)/x also approaches zero.Alternatively, I could have used integration as I initially thought, but the Mean Value Theorem approach seems more direct and avoids dealing with the integral bounds explicitly.Just to recap:- Start with f(x) expressed via the Mean Value Theorem.- Analyze each term in the expression for f(x)/x.- Show that each term individually approaches zero as x approaches infinity.- Conclude that the entire limit is zero.Yes, that makes sense. I don't see any gaps in this reasoning. So, I think this is a valid proof.

💡First, I need to determine the total number of blankets purchased. The man bought 5 blankets at Rs. 100 each, 5 blankets at Rs. 150 each, and 2 blankets at an unknown rate. This gives a total of 5 + 5 + 2 = 12 blankets.Next, I'll calculate the total cost of the blankets with known prices. The cost of 5 blankets at Rs. 100 each is 5 * 100 = Rs. 500. The cost of 5 blankets at Rs. 150 each is 5 * 150 = Rs. 750.Let the unknown rate of the two blankets be x. Therefore, the total cost of the two blankets is 2x.The average price of all the blankets is given as Rs. 150. To find the total cost, I'll multiply the average price by the total number of blankets: 12 * 150 = Rs. 1800.Now, I'll set up the equation for the total cost: 500 + 750 + 2x = 1800.Combining the known costs, I get 1250 + 2x = 1800.Subtracting 1250 from both sides gives 2x = 550.Finally, dividing both sides by 2, I find that x = 275.Therefore, the unknown rate of the two blankets is Rs. 275 each.

💡Alright, so I have this problem here that says: "If six times the reciprocal of the circumference of a circle equals the diameter of the circle, then what is the area of the circle?" And there are five options to choose from: (A) 1/2, (B) 1, (C) 3/2, (D) 2, and (E) π.Okay, let's break this down step by step. First, I need to recall some basic formulas related to circles. The circumference of a circle is given by C = 2πr, where r is the radius. The diameter, on the other hand, is simply twice the radius, so d = 2r. And the area of a circle is A = πr². Got that down.Now, the problem says that six times the reciprocal of the circumference equals the diameter. Let me translate that into an equation. The reciprocal of the circumference would be 1/C, right? So six times that would be 6*(1/C). And that's equal to the diameter, which is d. So putting that together, I get:6*(1/C) = dNow, substituting the formulas for circumference and diameter, I can write:6*(1/(2πr)) = 2rOkay, so now I have an equation in terms of r. Let's simplify this. First, let's compute 6*(1/(2πr)). That would be 6 divided by (2πr), which simplifies to 3/(πr). So now, the equation is:3/(πr) = 2rHmm, so now I have 3/(πr) equals 2r. I need to solve for r. Let's see. Maybe I can multiply both sides by πr to get rid of the denominator. Let's try that:3 = 2r * πrSimplifying the right side, 2r * πr is 2πr². So now the equation is:3 = 2πr²Alright, so now I can solve for r² by dividing both sides by 2π:r² = 3/(2π)Okay, so r squared is 3 over 2π. But wait, the problem is asking for the area of the circle, which is πr². So if I have r², I can just multiply it by π to get the area.So, area A = πr² = π*(3/(2π)) = 3/2Wait, that simplifies nicely. The π cancels out, and I'm left with 3/2. So the area of the circle is 3/2.Let me double-check my steps to make sure I didn't make any mistakes. Starting from the given condition, I translated it into an equation, substituted the known formulas for circumference and diameter, simplified the equation, solved for r², and then used that to find the area. Each step seems logical and I don't see any calculation errors. So, I think 3/2 is the correct answer.Looking back at the options, (C) is 3/2, so that should be the right choice.**Final Answer**boxed{C}

💡First, I need to determine the most cost-effective way to purchase 20 oranges by comparing the given bulk purchase options.1. **Evaluate the cost per orange for each bulk option:** - **4 oranges for 15 cents:** [ text{Cost per orange} = frac{15 text{ cents}}{4} = 3.75 text{ cents per orange} ] - **6 oranges for 25 cents:** [ text{Cost per orange} = frac{25 text{ cents}}{6} approx 4.17 text{ cents per orange} ] - **10 oranges for 40 cents:** [ text{Cost per orange} = frac{40 text{ cents}}{10} = 4 text{ cents per orange} ]2. **Identify the cheapest option:** - The **4 oranges for 15 cents** option offers the lowest cost per orange at **3.75 cents per orange**.3. **Calculate the total cost for 20 oranges using the cheapest option:** - Number of sets of 4 oranges needed: [ frac{20 text{ oranges}}{4 text{ oranges per set}} = 5 text{ sets} ] - Total cost without discount: [ 5 text{ sets} times 15 text{ cents per set} = 75 text{ cents} ]4. **Apply the 5-cent discount for purchasing at least 20 oranges:** - Total cost after discount: [ 75 text{ cents} - 5 text{ cents} = 70 text{ cents} ]5. **Conclusion:** - The total cost to purchase 20 oranges, utilizing the most economical bulk option and including the discount, is **70 cents**.

💡Okay, so I have this problem where I need to find all possible values of ( a + b + c ) for a quadratic polynomial ( f(x) = ax^2 + bx + c ) with integer coefficients. The conditions are that ( 0 < a, b, c leq p ) where ( p ) is a prime, and ( f(x) ) is divisible by ( p ) for every positive integer ( x ). Hmm, interesting.Let me start by understanding what it means for ( f(x) ) to be divisible by ( p ) for every positive integer ( x ). That means ( f(x) equiv 0 pmod{p} ) for all ( x ). So, the polynomial ( f(x) ) is the zero polynomial modulo ( p ). But wait, ( f(x) ) isn't necessarily the zero polynomial over the integers, just that it's always divisible by ( p ) when evaluated at any integer ( x ).Since ( f(x) equiv 0 pmod{p} ) for all ( x ), this should hold for any ( x ) in the field ( mathbb{Z}/pmathbb{Z} ). In a field, a non-zero polynomial of degree ( n ) can have at most ( n ) roots. But here, the polynomial ( f(x) ) has infinitely many roots modulo ( p ), which is only possible if all the coefficients of ( f(x) ) are zero modulo ( p ). That is, ( a equiv 0 pmod{p} ), ( b equiv 0 pmod{p} ), and ( c equiv 0 pmod{p} ).But wait, the problem states that ( 0 < a, b, c leq p ). So, if ( a equiv 0 pmod{p} ), the only possibility is ( a = p ). Similarly, ( b = p ) and ( c = p ). Therefore, ( a + b + c = p + p + p = 3p ).But let me verify this step by step to make sure I'm not missing anything.First, let's consider ( x = 1 ). Then ( f(1) = a + b + c ). Since ( f(1) ) is divisible by ( p ), ( a + b + c equiv 0 pmod{p} ).Next, consider ( x = 2 ). Then ( f(2) = 4a + 2b + c ). Similarly, ( 4a + 2b + c equiv 0 pmod{p} ).Subtracting ( f(1) ) from ( f(2) ), we get ( 3a + b equiv 0 pmod{p} ).We already have ( a + b + c equiv 0 pmod{p} ). Let me write down these two congruences:1. ( a + b + c equiv 0 pmod{p} )2. ( 3a + b equiv 0 pmod{p} )From the second congruence, ( 3a + b equiv 0 pmod{p} ), we can express ( b ) as ( b equiv -3a pmod{p} ).Substituting this into the first congruence:( a + (-3a) + c equiv 0 pmod{p} )Simplifying, ( -2a + c equiv 0 pmod{p} ), which gives ( c equiv 2a pmod{p} ).So, ( c = 2a + kp ) for some integer ( k ). But since ( 0 < c leq p ), the only possible value is ( c = 2a ) if ( 2a leq p ), or ( c = 2a - p ) if ( 2a > p ).But we also know that ( a ) must satisfy ( 0 < a leq p ). Let's consider possible values of ( a ).If ( a = p ), then ( c = 2p ), but ( c leq p ), so this is not possible. Wait, that contradicts. Hmm, maybe I made a mistake here.Wait, no. If ( a = p ), then ( c equiv 2p pmod{p} ), which is ( c equiv 0 pmod{p} ). But ( c ) must be between 1 and ( p ), so ( c = p ).Similarly, from ( b equiv -3a pmod{p} ), if ( a = p ), then ( b equiv -3p pmod{p} ), which is ( b equiv 0 pmod{p} ). So ( b = p ).So, if ( a = p ), then ( b = p ) and ( c = p ). Therefore, ( a + b + c = 3p ).But wait, earlier I thought that all coefficients must be zero modulo ( p ), which would imply ( a = p ), ( b = p ), ( c = p ). So, that seems consistent.But let me check for ( a ) not equal to ( p ). Suppose ( a ) is less than ( p ). Then, ( c = 2a ) or ( c = 2a - p ).If ( c = 2a ), since ( c leq p ), ( 2a leq p ). So ( a leq p/2 ).Similarly, if ( c = 2a - p ), then ( 2a - p geq 1 ), so ( a geq (p + 1)/2 ).So, depending on the value of ( a ), ( c ) can be either ( 2a ) or ( 2a - p ).But let's see if this leads to a contradiction or not.Suppose ( a ) is less than ( p ). Then, from ( b equiv -3a pmod{p} ), ( b = p - 3a ) if ( 3a < p ), or ( b = 2p - 3a ) if ( 3a geq p ). But ( b ) must be between 1 and ( p ).Wait, let's take an example. Let me pick a small prime, say ( p = 5 ).If ( p = 5 ), then ( a ) can be 1, 2, 3, 4, or 5.Case 1: ( a = 5 ). Then ( b = 5 ), ( c = 5 ). So ( a + b + c = 15 ).Case 2: ( a = 4 ). Then ( c equiv 2*4 = 8 equiv 3 pmod{5} ). So ( c = 3 ). Then ( b equiv -3*4 = -12 equiv -12 + 15 = 3 pmod{5} ). So ( b = 3 ). So ( a + b + c = 4 + 3 + 3 = 10 ).But wait, does ( f(x) = 4x^2 + 3x + 3 ) satisfy ( f(x) equiv 0 pmod{5} ) for all ( x )?Let me check ( x = 1 ): ( 4 + 3 + 3 = 10 equiv 0 pmod{5} ).( x = 2 ): ( 16 + 6 + 3 = 25 equiv 0 pmod{5} ).( x = 3 ): ( 36 + 9 + 3 = 48 equiv 3 pmod{5} ). Wait, that's not zero. So, this doesn't work. Hmm, so my earlier reasoning might be flawed.Wait, so ( a = 4 ), ( b = 3 ), ( c = 3 ) doesn't satisfy ( f(3) equiv 0 pmod{5} ). So, that's a problem.So, perhaps my assumption that ( a ) can be less than ( p ) is incorrect. Maybe only ( a = p ) works.Wait, let's try ( a = 3 ). Then ( c = 2*3 = 6 equiv 1 pmod{5} ). So ( c = 1 ). Then ( b equiv -3*3 = -9 equiv 1 pmod{5} ). So ( b = 1 ). So ( f(x) = 3x^2 + x + 1 ).Check ( x = 1 ): 3 + 1 + 1 = 5 ≡ 0 mod 5.( x = 2 ): 12 + 2 + 1 = 15 ≡ 0 mod 5.( x = 3 ): 27 + 3 + 1 = 31 ≡ 1 mod 5. Not zero. So again, it fails.Hmm, so maybe only when ( a = p ), ( b = p ), ( c = p ), does the polynomial satisfy ( f(x) equiv 0 pmod{p} ) for all ( x ).Wait, let me test ( a = p ), ( b = p ), ( c = p ). Then ( f(x) = px^2 + px + p ). For any ( x ), ( f(x) = p(x^2 + x + 1) ), which is clearly divisible by ( p ). So that works.But when ( a ) is less than ( p ), even though the congruences for ( x = 1 ) and ( x = 2 ) are satisfied, higher ( x ) values might not satisfy the condition.Wait, so maybe the only solution is ( a = p ), ( b = p ), ( c = p ). Therefore, ( a + b + c = 3p ).But let me think again. The initial reasoning was that since ( f(x) equiv 0 pmod{p} ) for all ( x ), the coefficients must all be zero modulo ( p ). So, ( a equiv 0 pmod{p} ), ( b equiv 0 pmod{p} ), ( c equiv 0 pmod{p} ). Given ( 0 < a, b, c leq p ), the only possibility is ( a = p ), ( b = p ), ( c = p ).Therefore, ( a + b + c = 3p ).I think that's the only solution. My earlier attempt with ( a = 4 ) and ( p = 5 ) didn't work because the polynomial didn't satisfy the condition for all ( x ). So, it seems that only when all coefficients are ( p ), the polynomial is divisible by ( p ) for all ( x ).So, the final answer should be ( 3p ).**Final Answer**The only possible value of ( a + b + c ) is boxed{3p}.