💡First, I need to determine how much water the Cyclone unit pumps in one hour, which is given as 500 gallons.Next, I should convert the 30 minutes into a fraction of an hour. Since there are 60 minutes in an hour, 30 minutes is half of an hour, or 0.5 hours.Finally, I'll calculate the amount of water pumped in 30 minutes by multiplying the hourly rate by the fraction of the hour:500 gallons/hour × 0.5 hours = 250 gallons.

💡Alright, so I have this problem about an isosceles trapezoid that has both an inscribed circle and a circumscribed circle. The ratio of the height of the trapezoid to the radius of the circumscribed circle is given as sqrt(2/3). I need to find the angles of the trapezoid.First, let me recall some properties of trapezoids. An isosceles trapezoid has the legs equal in length, and the base angles are equal. Also, if a trapezoid is both cyclic (has a circumscribed circle) and tangential (has an inscribed circle), it must be isosceles. That makes sense because only in an isosceles trapezoid can the sum of the lengths of the two bases equal the sum of the lengths of the two legs, which is a requirement for a trapezoid to be tangential.Given that, let me denote the trapezoid as ABCD, with AB and CD as the two bases, and AD and BC as the legs. Since it's isosceles, AD = BC, and the base angles at A and B are equal, and the base angles at C and D are equal.Let me denote the height of the trapezoid as h, and the radius of the circumscribed circle as R. The ratio h/R is given as sqrt(2/3), so h = R * sqrt(2/3).Now, since the trapezoid is cyclic, it can be inscribed in a circle. For a cyclic trapezoid, which is isosceles, the sum of the squares of the sides equals the sum of the squares of the diagonals. But I'm not sure if that's directly useful here.Alternatively, since it's cyclic, the opposite angles sum to 180 degrees. So, if one of the base angles is alpha, the adjacent angle will be 180 - alpha. But in an isosceles trapezoid, the base angles are equal, so actually, both base angles at the larger base are equal, and both base angles at the smaller base are equal. So, if the acute angle is alpha, the obtuse angle will be 180 - alpha.Wait, but in an isosceles trapezoid, the base angles are equal, so if one base has angles alpha, the other base will have angles 180 - alpha. So, the trapezoid has two angles of alpha and two angles of 180 - alpha.Now, since the trapezoid is also tangential, it has an inscribed circle. For a tangential quadrilateral, the sum of the lengths of the two opposite sides are equal. So, in this case, AB + CD = AD + BC. But since it's isosceles, AD = BC, so AB + CD = 2 * AD.Let me denote AB as the longer base and CD as the shorter base. Let me denote the lengths of AB as a, CD as b, and the legs AD and BC as c.So, from the tangential property, a + b = 2c.Also, the height h can be expressed in terms of the legs and the angles. In an isosceles trapezoid, the height h = c * sin(alpha), where alpha is the acute angle at the larger base.Additionally, since the trapezoid is cyclic, the radius R of the circumscribed circle can be related to the sides. For a cyclic quadrilateral, the formula for the radius is R = sqrt((ab + cd)(ac + bd)(ad + bc)) / (4 * area). But that seems complicated.Alternatively, for a cyclic trapezoid, which is isosceles, the radius can be related to the sides and the height. Let me think.The diameter of the circumscribed circle would be the diagonal of the trapezoid. In an isosceles trapezoid, the diagonals are equal in length. So, the length of the diagonal can be found using the Pythagorean theorem in terms of the bases and the height.Wait, the diagonal length d can be found as d = sqrt(c^2 + (a - b)^2 / 4 + h^2). Hmm, not sure.Alternatively, since the trapezoid is cyclic, the formula for the radius R is R = d / (2 * sin(theta)), where theta is the angle subtended by the diagonal at the center of the circle. But I'm not sure about that.Wait, maybe it's better to use the formula for the radius of the circumscribed circle around a cyclic quadrilateral. The formula is R = sqrt((ab + cd)(ac + bd)(ad + bc)) / (4K), where K is the area of the quadrilateral.But in our case, since it's an isosceles trapezoid, the sides are a, b, c, c. So, substituting, we get R = sqrt((ab + c^2)(ac + bc)(ad + bc)) / (4K). Hmm, this seems messy.Alternatively, maybe I can express the radius in terms of the sides and the height.Wait, another approach: in a cyclic trapezoid, the radius can be found using the formula R = (sqrt(a^2 + c^2)) / (2 * sin(alpha)), where alpha is the acute angle. But I'm not sure.Wait, maybe I can use the fact that in a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. But in an isosceles trapezoid, the diagonals are equal, so that might not help directly.Alternatively, let's consider the triangle formed by the radius, the height, and the leg of the trapezoid.Wait, maybe I can use the fact that the radius R is related to the sides and the angles.Let me denote the acute angle as alpha. Then, the obtuse angle is 180 - alpha.In an isosceles trapezoid, the legs are equal, and the height h = c * sin(alpha).Also, the difference between the bases is 2 * c * cos(alpha). So, a - b = 2c * cos(alpha).From the tangential property, a + b = 2c.So, we have two equations:1. a + b = 2c2. a - b = 2c * cos(alpha)We can solve these two equations for a and b.Adding the two equations: 2a = 2c + 2c * cos(alpha) => a = c(1 + cos(alpha))Subtracting the two equations: 2b = 2c - 2c * cos(alpha) => b = c(1 - cos(alpha))So, a = c(1 + cos(alpha)) and b = c(1 - cos(alpha)).Now, the area K of the trapezoid is (a + b)/2 * h = (2c)/2 * h = c * h.But h = c * sin(alpha), so K = c * c * sin(alpha) = c^2 * sin(alpha).Now, for a cyclic quadrilateral, the area can also be expressed as K = sqrt((s - a)(s - b)(s - c)(s - d)), where s is the semiperimeter.But in our case, since it's an isosceles trapezoid, the sides are a, b, c, c. So, s = (a + b + 2c)/2 = (2c + 2c)/2 = 2c.Wait, that can't be right. Wait, a + b = 2c, so s = (a + b + 2c)/2 = (2c + 2c)/2 = 2c.So, the semiperimeter s = 2c.Then, the area K = sqrt((s - a)(s - b)(s - c)(s - d)) = sqrt((2c - a)(2c - b)(2c - c)(2c - c)) = sqrt((2c - a)(2c - b)(c)(c)).But a = c(1 + cos(alpha)) and b = c(1 - cos(alpha)), so 2c - a = 2c - c(1 + cos(alpha)) = c(1 - cos(alpha)) and 2c - b = 2c - c(1 - cos(alpha)) = c(1 + cos(alpha)).So, K = sqrt((c(1 - cos(alpha)))(c(1 + cos(alpha)))(c)(c)) = sqrt(c^2(1 - cos^2(alpha)) * c^2) = sqrt(c^4 sin^2(alpha)) = c^2 sin(alpha).Which matches our earlier expression for K. So, that's consistent.Now, let's think about the radius R of the circumscribed circle.For a cyclic quadrilateral, the radius R can be expressed as R = sqrt((ab + cd)(ac + bd)(ad + bc)) / (4K).But in our case, sides are a, b, c, c.So, substituting, we get:R = sqrt((ab + c^2)(ac + bc)(ad + bc)) / (4K)But ad and bc are both c, so ad + bc = c + c = 2c.Similarly, ac + bc = c(a + b) = c * 2c = 2c^2.And ab + c^2 = ab + c^2.So, R = sqrt((ab + c^2)(2c^2)(2c)) / (4K) = sqrt((ab + c^2) * 4c^3) / (4K) = sqrt(4c^3(ab + c^2)) / (4K).Simplify numerator: sqrt(4c^3(ab + c^2)) = 2c^(3/2) sqrt(ab + c^2).Denominator: 4K = 4c^2 sin(alpha).So, R = (2c^(3/2) sqrt(ab + c^2)) / (4c^2 sin(alpha)) = (c^(3/2) sqrt(ab + c^2)) / (2c^2 sin(alpha)) = sqrt(ab + c^2) / (2c^(1/2) sin(alpha)).Hmm, this is getting complicated. Maybe there's a better way.Alternatively, since the trapezoid is cyclic, the radius can be related to the sides and the angles.Wait, in a cyclic trapezoid, the radius can be found using the formula R = d / (2 sin(theta)), where d is the length of the diagonal and theta is the angle subtended by the diagonal at the center.But in an isosceles trapezoid, the diagonals are equal, and the angle theta would be related to the angles of the trapezoid.Alternatively, maybe I can use the formula for the radius in terms of the sides and the area.Wait, another formula for the radius of a cyclic quadrilateral is R = (a b c d) / (4 K). But in our case, sides are a, b, c, c, so R = (a b c^2) / (4 K).But K = c^2 sin(alpha), so R = (a b c^2) / (4 c^2 sin(alpha)) = (a b) / (4 sin(alpha)).So, R = (a b) / (4 sin(alpha)).But from earlier, a = c(1 + cos(alpha)) and b = c(1 - cos(alpha)), so a b = c^2(1 - cos^2(alpha)) = c^2 sin^2(alpha).Therefore, R = (c^2 sin^2(alpha)) / (4 sin(alpha)) = (c^2 sin(alpha)) / 4.So, R = (c^2 sin(alpha)) / 4.But we also have h = c sin(alpha), so R = (c^2 sin(alpha)) / 4 = (c h) / 4.But h = R sqrt(2/3), so substituting, R = (c * R sqrt(2/3)) / 4.Simplify: R = (c R sqrt(2/3)) / 4 => 1 = (c sqrt(2/3)) / 4 => c = 4 / sqrt(2/3) = 4 * sqrt(3/2) = 4 * sqrt(3)/sqrt(2) = 4 * sqrt(6)/2 = 2 sqrt(6).Wait, that seems odd. Let me check.Wait, from R = (c h) / 4, and h = R sqrt(2/3), so R = (c * R sqrt(2/3)) / 4.Divide both sides by R: 1 = (c sqrt(2/3)) / 4 => c = 4 / sqrt(2/3) = 4 * sqrt(3/2) = 4 * sqrt(3)/sqrt(2) = (4 sqrt(6)) / 2 = 2 sqrt(6).So, c = 2 sqrt(6).But I don't know if that's useful yet.Alternatively, let's express c in terms of R.From R = (c h) / 4, and h = R sqrt(2/3), so R = (c * R sqrt(2/3)) / 4 => 1 = (c sqrt(2/3)) / 4 => c = 4 / sqrt(2/3) = 4 * sqrt(3/2) = 2 sqrt(6).So, c = 2 sqrt(6).But I need to find alpha.Wait, let's go back to the expression for R.We have R = (a b) / (4 sin(alpha)).But a = c(1 + cos(alpha)) and b = c(1 - cos(alpha)), so a b = c^2 (1 - cos^2(alpha)) = c^2 sin^2(alpha).Therefore, R = (c^2 sin^2(alpha)) / (4 sin(alpha)) = (c^2 sin(alpha)) / 4.But we also have R = (c h) / 4, and h = c sin(alpha), so R = (c * c sin(alpha)) / 4 = (c^2 sin(alpha)) / 4, which is consistent.But we also have h = R sqrt(2/3), so h = R sqrt(2/3).But h = c sin(alpha), so c sin(alpha) = R sqrt(2/3).But R = (c^2 sin(alpha)) / 4, so substituting:c sin(alpha) = (c^2 sin(alpha)) / 4 * sqrt(2/3).Simplify:c sin(alpha) = (c^2 sin(alpha) sqrt(2/3)) / 4.Divide both sides by c sin(alpha) (assuming c sin(alpha) ≠ 0, which it isn't):1 = (c sqrt(2/3)) / 4 => c = 4 / sqrt(2/3) = 4 * sqrt(3/2) = 2 sqrt(6).So, c = 2 sqrt(6).Now, let's express h in terms of R.h = R sqrt(2/3).But h = c sin(alpha) = 2 sqrt(6) sin(alpha).So, 2 sqrt(6) sin(alpha) = R sqrt(2/3).But R = (c^2 sin(alpha)) / 4 = ( (2 sqrt(6))^2 sin(alpha) ) / 4 = (24 sin(alpha)) / 4 = 6 sin(alpha).So, R = 6 sin(alpha).Substituting back into h = R sqrt(2/3):h = 6 sin(alpha) * sqrt(2/3) = 6 sin(alpha) * sqrt(2)/sqrt(3) = 6 sin(alpha) * sqrt(6)/3 = 2 sqrt(6) sin(alpha).But h is also equal to 2 sqrt(6) sin(alpha), which matches. So, that's consistent.But I need to find alpha.Wait, let's go back to the expression for R.We have R = 6 sin(alpha).But we also have h = 2 sqrt(6) sin(alpha).But h = R sqrt(2/3), so substituting R = 6 sin(alpha):h = 6 sin(alpha) * sqrt(2/3) = 6 sin(alpha) * sqrt(2)/sqrt(3) = 6 sin(alpha) * sqrt(6)/3 = 2 sqrt(6) sin(alpha).Which is consistent with h = 2 sqrt(6) sin(alpha).So, that doesn't give us new information.Wait, maybe I can use the fact that in an isosceles trapezoid, the radius R is related to the sides and the angles.Alternatively, let's consider the triangle formed by the radius, the height, and the leg.Wait, in the circumscribed circle, the center is equidistant from all vertices. So, the distance from the center to each vertex is R.Let me consider the coordinates of the trapezoid to model this.Let me place the trapezoid in a coordinate system such that the center of the circumscribed circle is at the origin.Let me denote the coordinates of the vertices as follows:- A at (-a/2, h/2)- B at (a/2, h/2)- C at (b/2, -h/2)- D at (-b/2, -h/2)But since it's an isosceles trapezoid, the legs AD and BC are equal, and the trapezoid is symmetric about the y-axis.Now, the distance from the center (0,0) to each vertex is R.So, for point A (-a/2, h/2), the distance to the center is sqrt( (a/2)^2 + (h/2)^2 ) = R.Similarly, for point C (b/2, -h/2), the distance is sqrt( (b/2)^2 + (h/2)^2 ) = R.So, we have:(a/2)^2 + (h/2)^2 = R^2and(b/2)^2 + (h/2)^2 = R^2Therefore, (a/2)^2 = (b/2)^2 => a = b.But that's a contradiction because in a trapezoid, the two bases must be of different lengths. So, this suggests that my coordinate system assumption is incorrect.Wait, maybe the center of the circumscribed circle is not at the midpoint of the height. Hmm.Alternatively, perhaps the center is not at the origin. Maybe I need to adjust the coordinate system.Wait, in a cyclic trapezoid, which is isosceles, the center of the circumscribed circle lies along the line of symmetry, which is the y-axis in my previous coordinate system.But the distance from the center to each vertex must be equal.Let me denote the center as (0, k), somewhere along the y-axis.Then, the distance from the center to point A (-a/2, h/2) is sqrt( (a/2)^2 + (h/2 - k)^2 ) = R.Similarly, the distance from the center to point C (b/2, -h/2) is sqrt( (b/2)^2 + (-h/2 - k)^2 ) = R.So, we have two equations:1. (a/2)^2 + (h/2 - k)^2 = R^22. (b/2)^2 + (-h/2 - k)^2 = R^2Subtracting equation 1 from equation 2:(b/2)^2 - (a/2)^2 + [(-h/2 - k)^2 - (h/2 - k)^2] = 0Let me expand the squares:[ (b^2)/4 - (a^2)/4 ] + [ (h^2/4 + h k + k^2) - (h^2/4 - h k + k^2) ] = 0Simplify:( (b^2 - a^2)/4 ) + (2 h k) = 0So, (b^2 - a^2)/4 + 2 h k = 0 => (b^2 - a^2) + 8 h k = 0.But from earlier, we have a = c(1 + cos(alpha)) and b = c(1 - cos(alpha)).So, a^2 = c^2(1 + 2 cos(alpha) + cos^2(alpha)) and b^2 = c^2(1 - 2 cos(alpha) + cos^2(alpha)).Therefore, b^2 - a^2 = c^2(1 - 2 cos(alpha) + cos^2(alpha) - 1 - 2 cos(alpha) - cos^2(alpha)) = c^2(-4 cos(alpha)).So, b^2 - a^2 = -4 c^2 cos(alpha).Substituting back into the equation:-4 c^2 cos(alpha) + 8 h k = 0 => 8 h k = 4 c^2 cos(alpha) => 2 h k = c^2 cos(alpha).But h = c sin(alpha), so:2 * c sin(alpha) * k = c^2 cos(alpha) => 2 k sin(alpha) = c cos(alpha) => k = (c cos(alpha)) / (2 sin(alpha)) = (c / 2) cot(alpha).So, the y-coordinate of the center is k = (c / 2) cot(alpha).Now, let's substitute back into equation 1:(a/2)^2 + (h/2 - k)^2 = R^2.We have a = c(1 + cos(alpha)), h = c sin(alpha), and k = (c / 2) cot(alpha).So, a/2 = c(1 + cos(alpha))/2.h/2 - k = (c sin(alpha))/2 - (c / 2) cot(alpha) = (c / 2)(sin(alpha) - cot(alpha)).But cot(alpha) = cos(alpha)/sin(alpha), so:h/2 - k = (c / 2)(sin(alpha) - cos(alpha)/sin(alpha)) = (c / 2)( (sin^2(alpha) - cos(alpha)) / sin(alpha) ).Wait, that seems messy. Let me compute it step by step.h/2 - k = (c sin(alpha))/2 - (c / 2)(cos(alpha)/sin(alpha)) = (c / 2)(sin(alpha) - cos(alpha)/sin(alpha)).Let me write sin(alpha) as sin(alpha)/1, so:= (c / 2)( (sin^2(alpha) - cos(alpha)) / sin(alpha) ).Wait, no, that's not correct. Let me find a common denominator:= (c / 2)( (sin^2(alpha) - cos(alpha)) / sin(alpha) ).Wait, actually, sin(alpha) - cos(alpha)/sin(alpha) = (sin^2(alpha) - cos(alpha)) / sin(alpha).But that doesn't seem right. Let me re-express:sin(alpha) - cos(alpha)/sin(alpha) = (sin^2(alpha) - cos(alpha)) / sin(alpha).Wait, actually, no. Let me compute:sin(alpha) - cos(alpha)/sin(alpha) = (sin^2(alpha) - cos(alpha)) / sin(alpha).Wait, no, that's not correct. Let me compute:sin(alpha) - cos(alpha)/sin(alpha) = (sin^2(alpha) - cos(alpha)) / sin(alpha).Wait, actually, that's correct.So, h/2 - k = (c / 2) * (sin^2(alpha) - cos(alpha)) / sin(alpha).Now, let's plug into equation 1:(a/2)^2 + (h/2 - k)^2 = R^2.So,[ c^2(1 + cos(alpha))^2 / 4 ] + [ (c^2 / 4) * (sin^2(alpha) - cos(alpha))^2 / sin^2(alpha) ] = R^2.Simplify:c^2(1 + 2 cos(alpha) + cos^2(alpha)) / 4 + c^2 (sin^2(alpha) - cos(alpha))^2 / (4 sin^2(alpha)) = R^2.Factor out c^2 / 4:c^2 / 4 [ (1 + 2 cos(alpha) + cos^2(alpha)) + (sin^2(alpha) - cos(alpha))^2 / sin^2(alpha) ] = R^2.Let me compute the term inside the brackets:Term1 = 1 + 2 cos(alpha) + cos^2(alpha)Term2 = (sin^2(alpha) - cos(alpha))^2 / sin^2(alpha)So, Term1 + Term2.Let me compute Term2:(sin^2(alpha) - cos(alpha))^2 = sin^4(alpha) - 2 sin^2(alpha) cos(alpha) + cos^2(alpha).So, Term2 = [sin^4(alpha) - 2 sin^2(alpha) cos(alpha) + cos^2(alpha)] / sin^2(alpha) = sin^2(alpha) - 2 cos(alpha) + cos^2(alpha)/sin^2(alpha).Therefore, Term1 + Term2 = [1 + 2 cos(alpha) + cos^2(alpha)] + [sin^2(alpha) - 2 cos(alpha) + cos^2(alpha)/sin^2(alpha)].Simplify:1 + 2 cos(alpha) + cos^2(alpha) + sin^2(alpha) - 2 cos(alpha) + cos^2(alpha)/sin^2(alpha).Notice that 2 cos(alpha) - 2 cos(alpha) cancels out.Also, cos^2(alpha) + sin^2(alpha) = 1.So, Term1 + Term2 = 1 + 1 + cos^2(alpha)/sin^2(alpha) = 2 + cot^2(alpha).Therefore, the equation becomes:c^2 / 4 [2 + cot^2(alpha)] = R^2.But we also have R = (c^2 sin(alpha)) / 4 from earlier.So, R^2 = (c^4 sin^2(alpha)) / 16.Therefore, we have:c^2 / 4 [2 + cot^2(alpha)] = (c^4 sin^2(alpha)) / 16.Multiply both sides by 16:4 c^2 [2 + cot^2(alpha)] = c^4 sin^2(alpha).Divide both sides by c^2 (assuming c ≠ 0):4 [2 + cot^2(alpha)] = c^2 sin^2(alpha).But from earlier, R = (c^2 sin(alpha)) / 4, so c^2 sin(alpha) = 4 R.Therefore, c^2 sin^2(alpha) = 4 R sin(alpha).So, substituting back:4 [2 + cot^2(alpha)] = 4 R sin(alpha).Divide both sides by 4:2 + cot^2(alpha) = R sin(alpha).But we also have h = R sqrt(2/3), and h = c sin(alpha).From earlier, c = 2 sqrt(6).So, h = 2 sqrt(6) sin(alpha) = R sqrt(2/3).Therefore, R = (2 sqrt(6) sin(alpha)) / sqrt(2/3) = 2 sqrt(6) sin(alpha) * sqrt(3/2) = 2 sqrt(6) * sqrt(3/2) sin(alpha).Simplify sqrt(6) * sqrt(3/2) = sqrt(6 * 3 / 2) = sqrt(9) = 3.So, R = 2 * 3 sin(alpha) = 6 sin(alpha).Therefore, R sin(alpha) = 6 sin^2(alpha).So, going back to the equation:2 + cot^2(alpha) = R sin(alpha) = 6 sin^2(alpha).But cot(alpha) = cos(alpha)/sin(alpha), so cot^2(alpha) = cos^2(alpha)/sin^2(alpha).Therefore, 2 + cos^2(alpha)/sin^2(alpha) = 6 sin^2(alpha).Multiply both sides by sin^2(alpha):2 sin^2(alpha) + cos^2(alpha) = 6 sin^4(alpha).But cos^2(alpha) = 1 - sin^2(alpha), so:2 sin^2(alpha) + 1 - sin^2(alpha) = 6 sin^4(alpha).Simplify:sin^2(alpha) + 1 = 6 sin^4(alpha).Let me denote x = sin^2(alpha). Then, the equation becomes:x + 1 = 6 x^2.Rearranged:6 x^2 - x - 1 = 0.Solve for x:Using quadratic formula: x = [1 ± sqrt(1 + 24)] / 12 = [1 ± 5] / 12.So, x = (1 + 5)/12 = 6/12 = 1/2, or x = (1 - 5)/12 = -4/12 = -1/3.Since x = sin^2(alpha) must be between 0 and 1, we discard the negative solution.So, sin^2(alpha) = 1/2 => sin(alpha) = sqrt(2)/2 or sin(alpha) = -sqrt(2)/2. But since alpha is an acute angle, sin(alpha) = sqrt(2)/2.Therefore, alpha = 45 degrees.Thus, the acute angles of the trapezoid are 45 degrees, and the obtuse angles are 180 - 45 = 135 degrees.So, the angles are 45° and 135°.

💡First, calculate the total distance to the school using Dave's walking details. Dave walks at a rate of 85 steps per minute, with each step being 70 cm long. Therefore, his speed is 85 steps/min multiplied by 70 cm/step, which equals 5950 cm/min. Since it takes him 20 minutes to get to school, the total distance is 5950 cm/min multiplied by 20 min, resulting in 119,000 cm.Next, determine Mike's walking speed. Mike averages 95 steps per minute, with each step being 65 cm long. Thus, his speed is 95 steps/min multiplied by 65 cm/step, which equals 6175 cm/min.Finally, calculate the time it takes for Mike to walk the same distance to school. Divide the total distance of 119,000 cm by Mike's speed of 6175 cm/min. This gives approximately 19.27 minutes, which can be expressed as 19 and 2/7 minutes.

💡First, Lizzy starts with 30.She loans out 15 to her friend, leaving her with 15.Her friend returns the 15 loan plus 20% interest. The interest is calculated as 15 multiplied by 20%, which equals 3.Therefore, the total amount returned is 15 plus 3, totaling 18.Finally, adding the returned amount to Lizzy's remaining money: 15 plus 18 equals 33.Thus, Lizzy will have 33 after her friend returns the loan with interest.

💡Okay, so I have this problem: Find all pairs of natural numbers (a, b) such that b^a divides a^b - 1. Hmm, that sounds a bit tricky, but let me try to break it down.First, let me understand what it means for b^a to divide a^b - 1. If b^a divides a^b - 1, then a^b ≡ 1 mod b^a. That means when I raise a to the power of b and subtract 1, the result should be a multiple of b^a. So, a^b - 1 is a multiple of b^a.Let me start by checking some small values for a and b to see if I can find any patterns or possible solutions.If a = 1, then the equation becomes b^1 divides 1^b - 1, which simplifies to b divides 0. Since any number divides 0, this means that for a = 1, any natural number b works. So, (1, b) is a solution for any b in natural numbers.If b = 1, then the equation becomes 1^a divides a^1 - 1, which simplifies to 1 divides (a - 1). Since 1 divides any integer, this means that for b = 1, any natural number a works. So, (a, 1) is a solution for any a in natural numbers.Okay, so the trivial cases where a or b is 1 are covered. Now, let's consider cases where both a and b are greater than 1.Let me try a = 2 and see what b can be.If a = 2, then the equation becomes b^2 divides 2^b - 1. Let's test some small values of b:- b = 2: 2^2 = 4. Does 4 divide 2^2 - 1 = 4 - 1 = 3? No, because 3 is not divisible by 4.- b = 3: 3^2 = 9. Does 9 divide 2^3 - 1 = 8 - 1 = 7? No.- b = 4: 4^2 = 16. Does 16 divide 2^4 - 1 = 16 - 1 = 15? No.- b = 5: 5^2 = 25. Does 25 divide 2^5 - 1 = 32 - 1 = 31? No.- b = 6: 6^2 = 36. Does 36 divide 2^6 - 1 = 64 - 1 = 63? 63 divided by 36 is 1.75, which is not an integer. So, no.Hmm, seems like for a = 2, there are no solutions where b > 1.Let me try a = 3.If a = 3, then the equation becomes b^3 divides 3^b - 1. Let's test some small values of b:- b = 2: 2^3 = 8. Does 8 divide 3^2 - 1 = 9 - 1 = 8? Yes, 8 divides 8. So, (3, 2) is a solution.- b = 3: 3^3 = 27. Does 27 divide 3^3 - 1 = 27 - 1 = 26? No.- b = 4: 4^3 = 64. Does 64 divide 3^4 - 1 = 81 - 1 = 80? 80 divided by 64 is 1.25, which is not an integer. So, no.- b = 5: 5^3 = 125. Does 125 divide 3^5 - 1 = 243 - 1 = 242? 242 divided by 125 is approximately 1.936, not an integer. So, no.So, for a = 3, only b = 2 works.Let me try a = 4.If a = 4, then the equation becomes b^4 divides 4^b - 1. Testing small b:- b = 2: 2^4 = 16. Does 16 divide 4^2 - 1 = 16 - 1 = 15? No.- b = 3: 3^4 = 81. Does 81 divide 4^3 - 1 = 64 - 1 = 63? No.- b = 4: 4^4 = 256. Does 256 divide 4^4 - 1 = 256 - 1 = 255? No.- b = 5: 5^4 = 625. Does 625 divide 4^5 - 1 = 1024 - 1 = 1023? 1023 divided by 625 is approximately 1.6368, not an integer. So, no.Hmm, no solutions for a = 4.Let me try a = 5.If a = 5, then the equation becomes b^5 divides 5^b - 1. Testing small b:- b = 2: 2^5 = 32. Does 32 divide 5^2 - 1 = 25 - 1 = 24? No.- b = 3: 3^5 = 243. Does 243 divide 5^3 - 1 = 125 - 1 = 124? No.- b = 4: 4^5 = 1024. Does 1024 divide 5^4 - 1 = 625 - 1 = 624? No.- b = 5: 5^5 = 3125. Does 3125 divide 5^5 - 1 = 3125 - 1 = 3124? No.No solutions for a = 5 either.Wait, maybe I should try a different approach instead of testing each a and b. Let's think about the properties of a and b.Given that b^a divides a^b - 1, we can write a^b ≡ 1 mod b^a. This implies that the multiplicative order of a modulo b^a divides b. But the multiplicative order of a modulo b^a must also divide φ(b^a), where φ is Euler's totient function.Since φ(b^a) = b^a - b^{a-1} = b^{a-1}(b - 1). So, the multiplicative order of a modulo b^a divides both b and φ(b^a). Therefore, the order must divide gcd(b, φ(b^a)).But φ(b^a) = b^{a-1}(b - 1). So, gcd(b, φ(b^a)) = gcd(b, b^{a-1}(b - 1)) = gcd(b, b - 1) = 1, since b and b - 1 are consecutive integers and thus coprime.Therefore, the multiplicative order of a modulo b^a must divide 1, which implies that a ≡ 1 mod b^a.But a ≡ 1 mod b^a implies that a = 1 + k b^a for some integer k ≥ 0. However, since a and b are natural numbers, and b ≥ 2 (since we've already considered b = 1), this would mean that a is at least 1 + b^a, which is much larger than b^a for a ≥ 1 and b ≥ 2.But if a is at least 1 + b^a, then a^b would be at least (1 + b^a)^b, which is way larger than b^a. However, a^b - 1 must be divisible by b^a, which is possible only if a is not too large. This seems contradictory unless a is very small.Wait, maybe I made a mistake here. Let me think again.If a ≡ 1 mod b^a, then a = 1 + k b^a for some integer k. But if k is at least 1, then a is at least 1 + b^a, which is indeed very large. However, if k = 0, then a = 1, which is the trivial case we already considered.So, this suggests that the only possible solution is when a = 1, which we already know works for any b. But earlier, we found that (3, 2) is also a solution. So, perhaps my reasoning here is missing something.Maybe I need to consider the case when b is a prime number. Let's suppose that b is prime. Then, φ(b^a) = b^a - b^{a - 1} = b^{a - 1}(b - 1). So, the multiplicative order of a modulo b^a divides both b and φ(b^a). Since b is prime, φ(b^a) = b^{a - 1}(b - 1). So, gcd(b, φ(b^a)) = gcd(b, b^{a - 1}(b - 1)) = gcd(b, b - 1) = 1, as before.So, again, the multiplicative order must divide 1, implying a ≡ 1 mod b^a. Which again suggests that a = 1 + k b^a. But this seems too restrictive.Wait, perhaps I need to consider the case when b is not prime. Let's take b = 2, which is prime, but let's see:If b = 2, then φ(2^a) = 2^a - 2^{a - 1} = 2^{a - 1}. So, the multiplicative order of a modulo 2^a must divide both 2 and 2^{a - 1}. So, the order divides gcd(2, 2^{a - 1}) = 2.Therefore, the multiplicative order of a modulo 2^a is either 1 or 2.If the order is 1, then a ≡ 1 mod 2^a. If the order is 2, then a^2 ≡ 1 mod 2^a.Let's consider a ≡ 1 mod 2^a. Then, a = 1 + k 2^a. For a ≥ 1, this would mean a is at least 1 + 2^a, which is quite large. However, for a = 1, we have the trivial solution.Alternatively, if a^2 ≡ 1 mod 2^a, then a^2 - 1 ≡ 0 mod 2^a. So, 2^a divides (a - 1)(a + 1). Since a is odd (because if a were even, a^2 would be 0 mod 4, but 1 mod 4 is 1, so a must be odd), both a - 1 and a + 1 are even. Let's see:If a is odd, then a - 1 and a + 1 are consecutive even numbers. So, one of them is divisible by 2 and the other by 4. Therefore, their product is divisible by 8. So, 2^a divides (a - 1)(a + 1) implies that 2^a ≤ (a - 1)(a + 1).Let's test for small a:- a = 1: 2^1 = 2. (1 - 1)(1 + 1) = 0. Not applicable since a = 1 is trivial.- a = 3: 2^3 = 8. (3 - 1)(3 + 1) = 2 * 4 = 8. So, 8 divides 8. This works. So, a = 3, b = 2 is a solution.- a = 5: 2^5 = 32. (5 - 1)(5 + 1) = 4 * 6 = 24. 32 does not divide 24.- a = 7: 2^7 = 128. (7 - 1)(7 + 1) = 6 * 8 = 48. 128 does not divide 48.So, only a = 3 works for b = 2.This suggests that (3, 2) is a solution. Are there any others?Let me try b = 4.If b = 4, then the equation becomes 4^a divides a^4 - 1. Let's test small a:- a = 1: 4^1 = 4. 1^4 - 1 = 0. 4 divides 0, so (1, 4) is a solution.- a = 2: 4^2 = 16. 2^4 - 1 = 16 - 1 = 15. 16 does not divide 15.- a = 3: 4^3 = 64. 3^4 - 1 = 81 - 1 = 80. 64 divides 80? 80 / 64 = 1.25. No.- a = 4: 4^4 = 256. 4^4 - 1 = 256 - 1 = 255. 256 does not divide 255.- a = 5: 4^5 = 1024. 5^4 - 1 = 625 - 1 = 624. 1024 does not divide 624.So, only a = 1 works for b = 4.What about b = 3?If b = 3, then the equation becomes 3^a divides a^3 - 1. Let's test small a:- a = 1: 3^1 = 3. 1^3 - 1 = 0. 3 divides 0, so (1, 3) is a solution.- a = 2: 3^2 = 9. 2^3 - 1 = 8 - 1 = 7. 9 does not divide 7.- a = 3: 3^3 = 27. 3^3 - 1 = 27 - 1 = 26. 27 does not divide 26.- a = 4: 3^4 = 81. 4^3 - 1 = 64 - 1 = 63. 81 does not divide 63.- a = 5: 3^5 = 243. 5^3 - 1 = 125 - 1 = 124. 243 does not divide 124.So, only a = 1 works for b = 3.Hmm, so far, the only non-trivial solution I've found is (3, 2). Let me see if there are any others.Let me try b = 6.If b = 6, then the equation becomes 6^a divides a^6 - 1. Testing small a:- a = 1: 6^1 = 6. 1^6 - 1 = 0. 6 divides 0, so (1, 6) is a solution.- a = 2: 6^2 = 36. 2^6 - 1 = 64 - 1 = 63. 36 does not divide 63.- a = 3: 6^3 = 216. 3^6 - 1 = 729 - 1 = 728. 216 divides 728? 728 / 216 ≈ 3.37. No.- a = 4: 6^4 = 1296. 4^6 - 1 = 4096 - 1 = 4095. 1296 divides 4095? 4095 / 1296 ≈ 3.16. No.- a = 5: 6^5 = 7776. 5^6 - 1 = 15625 - 1 = 15624. 7776 divides 15624? 15624 / 7776 ≈ 2.01. No.So, only a = 1 works for b = 6.What about b = 5?If b = 5, then the equation becomes 5^a divides a^5 - 1. Testing small a:- a = 1: 5^1 = 5. 1^5 - 1 = 0. 5 divides 0, so (1, 5) is a solution.- a = 2: 5^2 = 25. 2^5 - 1 = 32 - 1 = 31. 25 does not divide 31.- a = 3: 5^3 = 125. 3^5 - 1 = 243 - 1 = 242. 125 does not divide 242.- a = 4: 5^4 = 625. 4^5 - 1 = 1024 - 1 = 1023. 625 does not divide 1023.- a = 5: 5^5 = 3125. 5^5 - 1 = 3125 - 1 = 3124. 3125 does not divide 3124.So, only a = 1 works for b = 5.Hmm, it seems like the only non-trivial solution is (3, 2). Let me check if there are any other possibilities.Let me consider a = 3 and b = 2, which we already saw works. Let's see if there are any other pairs where a and b are both greater than 1.Suppose a = 4 and b = 2. Then, 2^4 = 16. Does 16 divide 4^2 - 1 = 16 - 1 = 15? No.a = 5, b = 2: 2^5 = 32. 5^2 - 1 = 25 - 1 = 24. 32 does not divide 24.a = 2, b = 3: 3^2 = 9. 2^3 - 1 = 8 - 1 = 7. 9 does not divide 7.a = 4, b = 3: 3^4 = 81. 4^3 - 1 = 64 - 1 = 63. 81 does not divide 63.a = 5, b = 3: 3^5 = 243. 5^3 - 1 = 125 - 1 = 124. 243 does not divide 124.a = 2, b = 4: 4^2 = 16. 2^4 - 1 = 16 - 1 = 15. 16 does not divide 15.a = 3, b = 4: 4^3 = 64. 3^4 - 1 = 81 - 1 = 80. 64 divides 80? 80 / 64 = 1.25. No.a = 4, b = 4: 4^4 = 256. 4^4 - 1 = 256 - 1 = 255. 256 does not divide 255.a = 5, b = 4: 4^5 = 1024. 5^4 - 1 = 625 - 1 = 624. 1024 does not divide 624.a = 2, b = 5: 5^2 = 25. 2^5 - 1 = 32 - 1 = 31. 25 does not divide 31.a = 3, b = 5: 5^3 = 125. 3^5 - 1 = 243 - 1 = 242. 125 does not divide 242.a = 4, b = 5: 5^4 = 625. 4^5 - 1 = 1024 - 1 = 1023. 625 does not divide 1023.a = 5, b = 5: 5^5 = 3125. 5^5 - 1 = 3125 - 1 = 3124. 3125 does not divide 3124.Hmm, seems like no other pairs work. Let me try a = 3 and b = 2 again to confirm.a = 3, b = 2: 2^3 = 8. 3^2 - 1 = 9 - 1 = 8. 8 divides 8, so yes, it works.Is there any other pair where a and b are both greater than 1? Let me think about a = 3 and b = 4.a = 3, b = 4: 4^3 = 64. 3^4 - 1 = 81 - 1 = 80. 64 divides 80? 80 / 64 = 1.25. No.a = 3, b = 6: 6^3 = 216. 3^6 - 1 = 729 - 1 = 728. 216 divides 728? 728 / 216 ≈ 3.37. No.a = 3, b = 5: 5^3 = 125. 3^5 - 1 = 243 - 1 = 242. 125 does not divide 242.a = 3, b = 7: 7^3 = 343. 3^7 - 1 = 2187 - 1 = 2186. 343 divides 2186? 2186 / 343 ≈ 6.37. No.a = 3, b = 8: 8^3 = 512. 3^8 - 1 = 6561 - 1 = 6560. 512 divides 6560? 6560 / 512 ≈ 12.8125. No.Hmm, no luck there either.Let me try a = 4 and b = 2 again.a = 4, b = 2: 2^4 = 16. 4^2 - 1 = 16 - 1 = 15. 16 does not divide 15.a = 4, b = 3: 3^4 = 81. 4^3 - 1 = 64 - 1 = 63. 81 does not divide 63.a = 4, b = 4: 4^4 = 256. 4^4 - 1 = 256 - 1 = 255. 256 does not divide 255.a = 4, b = 5: 5^4 = 625. 4^5 - 1 = 1024 - 1 = 1023. 625 does not divide 1023.a = 4, b = 6: 6^4 = 1296. 4^6 - 1 = 4096 - 1 = 4095. 1296 divides 4095? 4095 / 1296 ≈ 3.16. No.a = 4, b = 7: 7^4 = 2401. 4^7 - 1 = 16384 - 1 = 16383. 2401 divides 16383? 16383 / 2401 ≈ 6.82. No.Hmm, still nothing.What about a = 5 and b = 2?a = 5, b = 2: 2^5 = 32. 5^2 - 1 = 25 - 1 = 24. 32 does not divide 24.a = 5, b = 3: 3^5 = 243. 5^3 - 1 = 125 - 1 = 124. 243 does not divide 124.a = 5, b = 4: 4^5 = 1024. 5^4 - 1 = 625 - 1 = 624. 1024 does not divide 624.a = 5, b = 5: 5^5 = 3125. 5^5 - 1 = 3125 - 1 = 3124. 3125 does not divide 3124.a = 5, b = 6: 6^5 = 7776. 5^6 - 1 = 15625 - 1 = 15624. 7776 divides 15624? 15624 / 7776 ≈ 2.01. No.a = 5, b = 7: 7^5 = 16807. 5^7 - 1 = 78125 - 1 = 78124. 16807 divides 78124? 78124 / 16807 ≈ 4.64. No.Okay, I'm starting to think that (3, 2) might be the only non-trivial solution. Let me try to see if there's a theoretical reason why this is the case.Earlier, I considered the multiplicative order of a modulo b^a. If b is prime, then φ(b^a) = b^{a - 1}(b - 1). The multiplicative order of a modulo b^a must divide both b and φ(b^a). Since b is prime, gcd(b, φ(b^a)) = gcd(b, b^{a - 1}(b - 1)) = gcd(b, b - 1) = 1. Therefore, the multiplicative order must be 1, implying a ≡ 1 mod b^a. But this would mean a is very large, which contradicts our earlier solution (3, 2).Wait, but in the case of b = 2, which is prime, we found that a = 3 works. Let me see why.For b = 2, φ(2^a) = 2^{a - 1}. So, the multiplicative order of a modulo 2^a must divide both 2 and 2^{a - 1}. Therefore, the order divides gcd(2, 2^{a - 1}) = 2. So, the order is either 1 or 2.If the order is 1, then a ≡ 1 mod 2^a, which would require a to be very large, as before. However, if the order is 2, then a^2 ≡ 1 mod 2^a. This is possible for a = 3, since 3^2 = 9 ≡ 1 mod 8 (since 9 - 1 = 8). Indeed, 3^2 = 9 ≡ 1 mod 8, so the order is 2.Therefore, for b = 2, a = 3 is a solution because the multiplicative order of 3 modulo 8 is 2, which divides b = 2. This is a special case where the order is 2, allowing a smaller a to satisfy the condition.This suggests that for b = 2, a = 3 is a solution because the multiplicative order is 2, which divides b. For other primes b, the multiplicative order would have to divide 1, forcing a to be 1, which is trivial.Therefore, the only non-trivial solution is (3, 2). All other solutions involve either a = 1 or b = 1, which are trivial.To summarize:- For a = 1, any b works.- For b = 1, any a works.- The only non-trivial solution is (3, 2).So, the complete set of solutions is all pairs where a = 1 or b = 1, plus the pair (3, 2).

💡First, I need to determine the sum and difference of the numbers 555 and 445.The sum of 555 and 445 is:555 + 445 = 1000The difference between 555 and 445 is:555 - 445 = 110Next, I'll divide the given number, 220050, by the sum (1000) to find the quotient and remainder.220050 ÷ 1000 = 220 with a remainder of 50This means that the quotient is 220 and the remainder is 50.According to the problem, the quotient is a certain multiple of the difference (110). Let's denote this multiple as ( x ).So, we have:x * 110 = 220To find ( x ), I'll divide both sides of the equation by 110:x = 220 ÷ 110x = 2Therefore, the multiplier for the difference is 2.