💡First, I need to determine the number of students in the freshman class, which is fewer than 700. The problem provides three conditions based on how the students line up in columns of different sizes.1. When the students are arranged in columns of 20, the last column contains 19 students. This means the total number of students, ( n ), satisfies the congruence: [ n equiv 19 pmod{20} ]2. When the students line up in columns of 25, the last column has 24 students. This gives another congruence: [ n equiv 24 pmod{25} ]3. When the students line up in columns of 9, there are 3 students in the last column. This results in the third congruence: [ n equiv 3 pmod{9} ]To solve this system of congruences, I'll start by addressing the first two congruences. The least common multiple (LCM) of 20 and 25 is 100. I need to find a number that satisfies both ( n equiv 19 pmod{20} ) and ( n equiv 24 pmod{25} ). By inspection, ( n = 99 ) fits both conditions:[99 equiv 19 pmod{20} quad text{and} quad 99 equiv 24 pmod{25}]So, the general solution for these two congruences is:[n equiv 99 pmod{100}]Next, I'll incorporate the third congruence ( n equiv 3 pmod{9} ). Starting with ( n = 99 ), I check:[99 equiv 0 pmod{9}]This doesn't satisfy ( n equiv 3 pmod{9} ). I'll add multiples of 100 to 99 until I find a number that meets the third condition:[199 equiv 1 pmod{9} 299 equiv 2 pmod{9} 399 equiv 3 pmod{9}]Finally, ( n = 399 ) satisfies all three congruences.Therefore, the number of students in the freshman class is 399.

💡Okay, so I have this problem here with two parts. Let me try to understand what each part is asking and figure out how to approach them step by step.First, the problem gives me parametric equations for a curve C:[begin{cases}x = 3 + 2cos alpha y = 2sin alphaend{cases}]where (alpha) is the parameter. It also mentions that the polar coordinate system's pole coincides with the origin of the rectangular (Cartesian) system, and the polar axis aligns with the positive x-axis. So, that means we can convert between polar and Cartesian coordinates using the standard formulas:[x = rho cos theta y = rho sin theta]Alright, for part 1, I need to convert these parametric equations into polar coordinates. Hmm, parametric equations usually express x and y in terms of a parameter, here it's (alpha). So, maybe I can eliminate the parameter (alpha) to get a Cartesian equation first and then convert that into polar coordinates. That seems like a common approach.Looking at the parametric equations:[x = 3 + 2cos alpha y = 2sin alpha]I notice that both x and y are expressed in terms of sine and cosine of the same parameter (alpha). So, maybe I can use the Pythagorean identity (cos^2 alpha + sin^2 alpha = 1) to eliminate (alpha).Let me rearrange the equations to express (cos alpha) and (sin alpha):From the first equation:[cos alpha = frac{x - 3}{2}]From the second equation:[sin alpha = frac{y}{2}]Now, plug these into the identity:[left(frac{x - 3}{2}right)^2 + left(frac{y}{2}right)^2 = 1]Simplify this:[frac{(x - 3)^2}{4} + frac{y^2}{4} = 1]Multiply both sides by 4 to eliminate denominators:[(x - 3)^2 + y^2 = 4]So, this is the Cartesian equation of the curve C. It looks like a circle with center at (3, 0) and radius 2. That makes sense because the parametric equations are in terms of sine and cosine, which typically describe circles or ellipses.Now, to convert this into polar coordinates. Remember, in polar coordinates, (x = rho cos theta) and (y = rho sin theta). So, let me substitute these into the equation:[(x - 3)^2 + y^2 = 4]Substituting x and y:[(rho cos theta - 3)^2 + (rho sin theta)^2 = 4]Let me expand the first term:[(rho cos theta - 3)^2 = rho^2 cos^2 theta - 6rho cos theta + 9]The second term is:[(rho sin theta)^2 = rho^2 sin^2 theta]So, putting it all together:[rho^2 cos^2 theta - 6rho cos theta + 9 + rho^2 sin^2 theta = 4]Combine like terms. Notice that (rho^2 cos^2 theta + rho^2 sin^2 theta = rho^2 (cos^2 theta + sin^2 theta) = rho^2). So, the equation simplifies to:[rho^2 - 6rho cos theta + 9 = 4]Subtract 4 from both sides:[rho^2 - 6rho cos theta + 5 = 0]So, that's the polar equation for curve C. Let me double-check my steps to make sure I didn't make a mistake. Starting from the parametric equations, eliminating the parameter, converting to Cartesian, then substituting into polar. It seems correct. The final polar equation is a quadratic in (rho), which makes sense because it's a circle not centered at the origin.Alright, that was part 1. Now, moving on to part 2.Part 2 says: A tangent line is drawn from a point on the line (l) to the curve (C). Find the minimum length of the tangent line.First, I need to understand what line (l) is. The problem gives its polar equation:[sqrt{2} rho sin left( theta - frac{pi}{4} right) = 1]I need to convert this into Cartesian coordinates to understand its position relative to curve C.Recall that in polar coordinates, the equation ( rho sin(theta - phi) = d ) represents a straight line. The general form is:[rho sin(theta - phi) = d]Which can be converted to Cartesian coordinates. Let me recall the identity:[rho sin(theta - phi) = y cos phi - x sin phi]Yes, because:[rho sin(theta - phi) = rho (sin theta cos phi - cos theta sin phi) = y cos phi - x sin phi]So, applying this to the given equation:[sqrt{2} rho sin left( theta - frac{pi}{4} right) = 1]Divide both sides by (sqrt{2}):[rho sin left( theta - frac{pi}{4} right) = frac{1}{sqrt{2}}]Using the identity above, with (phi = frac{pi}{4}):[y cos frac{pi}{4} - x sin frac{pi}{4} = frac{1}{sqrt{2}}]Compute (cos frac{pi}{4}) and (sin frac{pi}{4}):Both are equal to (frac{sqrt{2}}{2}).So, substituting:[y cdot frac{sqrt{2}}{2} - x cdot frac{sqrt{2}}{2} = frac{1}{sqrt{2}}]Multiply both sides by (sqrt{2}) to eliminate denominators:[y - x = 1]So, the equation of line (l) in Cartesian coordinates is (y = x + 1). That's a straight line with slope 1 and y-intercept at 1.Now, the problem is asking for the minimum length of a tangent line drawn from a point on line (l) to curve (C). So, essentially, for each point on line (l), we can draw a tangent to the circle (C), and we need to find the minimum possible length of such a tangent.I remember that the length of a tangent from a point ((x_1, y_1)) to a circle with center ((h, k)) and radius (r) is given by:[text{Length} = sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2}]This comes from the Pythagorean theorem, where the distance from the point to the center is the hypotenuse, and the radius is one leg, so the tangent length is the other leg.So, in our case, the circle (C) has center at (3, 0) and radius 2. So, the length of the tangent from a point ((x, y)) on line (l) to circle (C) is:[sqrt{(x - 3)^2 + (y - 0)^2 - 2^2} = sqrt{(x - 3)^2 + y^2 - 4}]But since the point ((x, y)) lies on line (l), which is (y = x + 1), we can substitute (y) with (x + 1):[sqrt{(x - 3)^2 + (x + 1)^2 - 4}]Let me compute this expression step by step.First, expand ((x - 3)^2):[(x - 3)^2 = x^2 - 6x + 9]Then, expand ((x + 1)^2):[(x + 1)^2 = x^2 + 2x + 1]Add these two results:[(x^2 - 6x + 9) + (x^2 + 2x + 1) = 2x^2 - 4x + 10]Now, subtract 4:[2x^2 - 4x + 10 - 4 = 2x^2 - 4x + 6]So, the expression under the square root becomes:[sqrt{2x^2 - 4x + 6}]Therefore, the length of the tangent is:[sqrt{2x^2 - 4x + 6}]Now, our goal is to find the minimum value of this expression as (x) varies over all real numbers (since line (l) is infinite). So, essentially, we need to minimize the function:[f(x) = sqrt{2x^2 - 4x + 6}]Since the square root function is a monotonically increasing function, the minimum of (f(x)) occurs at the minimum of the quadratic inside the square root. So, let's consider the quadratic function:[g(x) = 2x^2 - 4x + 6]To find its minimum, we can complete the square or take the derivative. Since it's a quadratic, completing the square might be straightforward.Let me factor out the coefficient of (x^2):[g(x) = 2(x^2 - 2x) + 6]Now, complete the square inside the parentheses. The coefficient of x is -2, so half of that is -1, and squaring it gives 1.So,[g(x) = 2[(x^2 - 2x + 1) - 1] + 6 = 2(x - 1)^2 - 2 + 6 = 2(x - 1)^2 + 4]So, the quadratic can be written as:[g(x) = 2(x - 1)^2 + 4]This is a parabola opening upwards with vertex at (x = 1), and the minimum value is 4.Therefore, the minimum value of (g(x)) is 4, so the minimum value of (f(x)) is:[sqrt{4} = 2]So, the minimum length of the tangent line is 2.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Converted the polar equation of line (l) to Cartesian and got (y = x + 1). That seems correct.2. Expressed the tangent length from a point ((x, y)) on (l) to circle (C) as (sqrt{(x - 3)^2 + y^2 - 4}). That formula is correct.3. Substituted (y = x + 1) into the expression, expanded, and simplified to get (sqrt{2x^2 - 4x + 6}). That seems correct.4. Completed the square for (2x^2 - 4x + 6) and found the minimum value of the quadratic to be 4, leading to the minimum tangent length of 2. That seems correct.Alternatively, another way to approach this is to find the distance from the line (l) to the center of the circle (C), and then use the formula for the length of the tangent. Let me try that method to verify.The center of the circle (C) is at (3, 0), and the line (l) is (y = x + 1). The distance (d) from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is given by:[d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}]First, let me write the equation of line (l) in standard form. (y = x + 1) can be rewritten as:[x - y + 1 = 0]So, (A = 1), (B = -1), and (C = 1).Now, plug in the center (3, 0):[d = frac{|1 cdot 3 + (-1) cdot 0 + 1|}{sqrt{1^2 + (-1)^2}} = frac{|3 + 0 + 1|}{sqrt{1 + 1}} = frac{4}{sqrt{2}} = 2sqrt{2}]So, the distance from the center to the line (l) is (2sqrt{2}).Now, the radius of the circle is 2. The length of the tangent from a point outside the circle to the circle is given by:[sqrt{d^2 - r^2}]Where (d) is the distance from the point to the center, and (r) is the radius.But in this case, the point is moving along the line (l), so the minimum length of the tangent occurs when the point on (l) is closest to the center. That is, when the distance from the point to the center is minimized, which is exactly the distance (d = 2sqrt{2}).Wait, hold on. Actually, the formula for the length of the tangent from a point to a circle is:[sqrt{d^2 - r^2}]Where (d) is the distance from the point to the center. So, if the point is on line (l), then (d) is the distance from the point to the center, which varies depending on the point. However, the minimum length of the tangent would occur when this distance (d) is minimized, i.e., when the point is closest to the center.So, the minimum distance from any point on (l) to the center (3, 0) is the perpendicular distance from (3, 0) to line (l), which we calculated as (2sqrt{2}).Therefore, the minimum length of the tangent is:[sqrt{(2sqrt{2})^2 - 2^2} = sqrt{8 - 4} = sqrt{4} = 2]So, that confirms the earlier result. The minimum length is indeed 2.Alternatively, another approach is to parametrize the points on line (l) and then compute the tangent length as a function, then find its minimum. But since we've already confirmed it with two methods, I think it's safe to conclude that the minimum length is 2.Just to recap, the steps were:1. Convert the parametric equations of curve C to Cartesian, then to polar coordinates.2. Convert the polar equation of line l to Cartesian.3. Recognize that the tangent length from a point on l to circle C can be found using the distance formula, and then minimize that.4. Alternatively, compute the perpendicular distance from the center to line l, then use the tangent length formula.Both methods led to the same result, so I'm confident that the minimum tangent length is 2.**Final Answer**The minimum length of the tangent line is boxed{2}.

💡First, I need to determine how many crates Gabrielle sells each day.On Monday, she sells 5 crates.On Tuesday, she sells twice as many crates as on Monday, which is 2 multiplied by 5 crates.On Wednesday, she sells 2 fewer crates than on Tuesday, so I subtract 2 from the number of crates sold on Tuesday.On Thursday, she sells half as many crates as on Tuesday, so I divide the number of crates sold on Tuesday by 2.After calculating the crates sold each day, I add them all together to find the total number of crates sold over the four days.

💡Okay, so I have this problem here where I need to find the number of positive integers ( n ) such that three conditions are satisfied:1. ( P(n) = sqrt{n} ), where ( P(n) ) is the greatest prime factor of ( n ).2. ( P(n + 72) = sqrt{n + 72} ).3. The last digit of ( n ) is equal to the first digit of ( n + 72 ).Hmm, let me try to unpack each of these conditions one by one and see how I can approach this.Starting with the first condition: ( P(n) = sqrt{n} ). Since ( P(n) ) is the greatest prime factor of ( n ), this implies that ( n ) must be a perfect square of a prime number. Because if ( sqrt{n} ) is an integer, then ( n ) is a perfect square, and since ( P(n) = sqrt{n} ), that means ( sqrt{n} ) itself must be a prime number. So, ( n ) must be ( p^2 ) where ( p ) is a prime number.Similarly, the second condition ( P(n + 72) = sqrt{n + 72} ) tells me that ( n + 72 ) must also be a perfect square of a prime number. Let's denote ( n + 72 = q^2 ), where ( q ) is another prime number.So, putting these together, I have:[ n = p^2 ][ n + 72 = q^2 ]Subtracting the first equation from the second gives:[ q^2 - p^2 = 72 ][ (q - p)(q + p) = 72 ]Okay, so now I have the equation ( (q - p)(q + p) = 72 ). Since both ( q ) and ( p ) are primes, and ( q > p ) (because ( q^2 > p^2 )), ( q - p ) and ( q + p ) are positive integers that multiply to 72. Also, since ( q ) and ( p ) are primes, both ( q - p ) and ( q + p ) must be even numbers because the difference and sum of two odd numbers (primes greater than 2 are odd) are even. So, ( q - p ) and ( q + p ) are both even, meaning that 72 can be factored into two even numbers.Let me list all the factor pairs of 72 where both factors are even:1. ( 2 times 36 )2. ( 4 times 18 )3. ( 6 times 12 )4. ( 8 times 9 ) – Wait, 9 is odd, so this pair is out since both factors need to be even.So, the valid factor pairs are ( (2, 36) ), ( (4, 18) ), and ( (6, 12) ).Now, for each factor pair ( (a, b) ) where ( a = q - p ) and ( b = q + p ), I can solve for ( q ) and ( p ):1. For ( a = 2 ) and ( b = 36 ): [ q - p = 2 ] [ q + p = 36 ] Adding these two equations: [ 2q = 38 ] [ q = 19 ] Subtracting the first equation from the second: [ 2p = 34 ] [ p = 17 ] Both 17 and 19 are primes, so this is a valid pair.2. For ( a = 4 ) and ( b = 18 ): [ q - p = 4 ] [ q + p = 18 ] Adding these: [ 2q = 22 ] [ q = 11 ] Subtracting: [ 2p = 14 ] [ p = 7 ] Both 7 and 11 are primes, so this is another valid pair.3. For ( a = 6 ) and ( b = 12 ): [ q - p = 6 ] [ q + p = 12 ] Adding: [ 2q = 18 ] [ q = 9 ] Subtracting: [ 2p = 6 ] [ p = 3 ] Here, ( q = 9 ) is not a prime number, so this pair is invalid.So, from the factor pairs, I get two valid pairs of primes: ( (17, 19) ) and ( (7, 11) ).Now, let's find the corresponding ( n ) for each pair:1. For ( p = 17 ): [ n = p^2 = 17^2 = 289 ] Then, ( n + 72 = 289 + 72 = 361 = 19^2 ), which checks out.2. For ( p = 7 ): [ n = 7^2 = 49 ] Then, ( n + 72 = 49 + 72 = 121 = 11^2 ), which also checks out.So, I have two potential candidates for ( n ): 289 and 49.Now, moving on to the third condition: the last digit of ( n ) must be equal to the first digit of ( n + 72 ).Let's check this for both ( n = 289 ) and ( n = 49 ):1. For ( n = 289 ): - The last digit of 289 is 9. - ( n + 72 = 361 ), whose first digit is 3. - 9 ≠ 3, so this doesn't satisfy the third condition.2. For ( n = 49 ): - The last digit of 49 is 9. - ( n + 72 = 121 ), whose first digit is 1. - 9 ≠ 1, so this also doesn't satisfy the third condition.Hmm, neither of the two candidates satisfy the third condition. That means there are no such positive integers ( n ) that satisfy all three conditions.Wait, but let me double-check my work to make sure I didn't miss anything.First, I considered all factor pairs of 72 where both factors are even, which gave me three pairs. From these, I found two valid prime pairs: (17,19) and (7,11). Then, I calculated ( n ) for each and checked the third condition. Both failed the third condition. Is there a possibility I missed another factor pair? Let me see:72 can be factored as:- 1 × 72- 2 × 36- 3 × 24- 4 × 18- 6 × 12- 8 × 9But since both ( q - p ) and ( q + p ) must be even, the pairs (1,72), (3,24), and (8,9) are invalid because they include odd numbers. So, I did consider all the valid factor pairs.Also, when solving for ( q ) and ( p ), I correctly found the primes, and both ( n ) values were computed correctly. The third condition was checked accurately, and neither satisfied it.Therefore, I think my conclusion is correct: there are no positive integers ( n ) that satisfy all three conditions.**Final Answer**boxed{A}

💡Okay, so I have this problem about an irrigated field that's shaped like a square with each side being 12 units long. There's a system of straight ditches on this field, and the key thing is that no point in the field is more than 1 unit away from the nearest ditch. I need to prove that the total length of all these ditches is greater than 70 units. Hmm, interesting.Alright, let me start by visualizing the field. It's a square, so 12 units by 12 units. That means the area of the field is 12 times 12, which is 144 square units. Now, the ditches are straight lines, and they're arranged in such a way that every point in the field is within 1 unit of a ditch. So, essentially, the ditches form a network that covers the entire field with a sort of "buffer" zone around each ditch.I think the first step is to consider the area that each ditch can cover. If a ditch is a straight line, then the area around it that is within 1 unit would form a rectangular strip around the ditch, right? But actually, since it's a buffer zone, it's more like a rectangle with semicircular ends. So, for a straight ditch of length 'l', the area it covers would be the area of the rectangle plus the area of the two semicircles.Let me write that down. The area covered by a straight ditch of length 'l' would be:Area = (2 * 1) * l + π * (1)^2Wait, that's because the buffer zone is 1 unit on each side, so the width is 2 units, and the length is 'l'. Then, at each end of the ditch, there's a semicircle of radius 1 unit. Since there are two ends, that makes a full circle. So, the area covered by one ditch is 2l + π.Okay, so if I have multiple ditches, each of length 'l', the total area covered would be the sum of the areas covered by each ditch. But I have to be careful here because if two ditches are close to each other, their buffer zones might overlap. That means the total area covered wouldn't just be the sum of all individual areas; it would be less because of the overlapping regions.But wait, the problem says that the distance from any point to the nearest ditch is no more than 1 unit. That means the entire field must be covered by these buffer zones without any gaps. So, the total area covered by the ditches must be at least equal to the area of the field, which is 144 square units.So, if I denote the total length of all ditches as 'L', then the total area covered by all ditches would be approximately 2L + π, assuming minimal overlap. But since the actual area covered must be at least 144, I can set up the inequality:2L + π ≥ 144Now, solving for L:2L ≥ 144 - πL ≥ (144 - π)/2Calculating the numerical value:π is approximately 3.1416, so:144 - π ≈ 144 - 3.1416 ≈ 140.8584Dividing by 2:L ≥ 140.8584 / 2 ≈ 70.4292So, L must be greater than approximately 70.43 units. Since the problem asks to prove that the total length is greater than 70 units, this calculation shows that it's indeed greater than 70.But wait, I made an assumption here that the total area covered is approximately 2L + π, which assumes minimal overlap. In reality, there might be some overlap, especially at the intersections of ditches. However, even if there is some overlap, the total area covered cannot be less than 144 square units because the entire field must be within 1 unit of a ditch. Therefore, the inequality still holds, and L must be greater than approximately 70.43 units.So, putting it all together, the total length of the ditches must be greater than 70 units to ensure that every point in the field is within 1 unit of a ditch.**Final Answer**The total length of the ditches is greater than boxed{70} units.

💡First, I need to determine the total number of Easter eggs found. There are 15 Easter egg baskets, and each basket holds 12 eggs. By multiplying the number of baskets by the number of eggs per basket, I can find the total number of eggs.Next, I know that each person receives 9 eggs when they are equally distributed. To find the total number of people at the Easter egg hunt, I will divide the total number of eggs by the number of eggs each person gets.I also know that there are 2 kids (Shonda's children), Shonda herself, and 7 other adults. Adding these together gives me the number of people who are not friends of Shonda's kids.Finally, to find out how many friends Shonda's kids have, I will subtract the number of people who are not friends (Shonda's family and the other adults) from the total number of people at the hunt.