answer：To find the other root of the quadratic equation ( x^2 - 6x + m = 0 ) when one root is 2, I can use the relationship between the roots and the coefficients of the equation.First, I recall that for a quadratic equation ( x^2 + bx + c = 0 ), the sum of the roots is ( -b ) and the product of the roots is ( c ).In this equation, the sum of the roots is ( 6 ) because ( -b = -(-6) = 6 ). Let the other root be ( r ). Since one root is ( 2 ), the sum of the roots is ( 2 + r = 6 ). Solving for ( r ), I subtract 2 from both sides to get ( r = 4 ).Therefore, the other root is 4.

answer：First, I need to solve the equation ((a + 25)^2 = 1000). Taking the square root of both sides gives (a + 25 = pm sqrt{1000}).Next, I'll simplify (sqrt{1000}) to (10sqrt{10}), so the equation becomes (a + 25 = pm 10sqrt{10}).Solving for (a), I subtract 25 from both sides, resulting in (a = -25 pm 10sqrt{10}).Now, I need to find the value of ((a + 15)(a + 35)). Expanding this expression gives (a^2 + 50a + 525).I know that ((a + 25)^2 = a^2 + 50a + 625), and since this equals 1000, I can substitute to find (a^2 + 50a = 375).Substituting back into the expanded expression, ((a + 15)(a + 35) = 375 + 525 = 900).

answer：First, determine how many sit-ups Nathan can do. Since Nathan can do twice as many sit-ups as Ken, and Ken can do 20 sit-ups, Nathan can do 20 multiplied by 2, which equals 40 sit-ups.Next, calculate the combined number of sit-ups that Ken and Nathan can do together. Adding Ken's 20 sit-ups to Nathan's 40 sit-ups gives a total of 60 sit-ups.Then, find out how many sit-ups Bob can do. Bob can do half of the combined sit-ups of Ken and Nathan, so half of 60 is 30 sit-ups.Finally, determine how many more sit-ups Bob can do compared to Ken by subtracting Ken's sit-ups from Bob's sit-ups. Subtracting 20 from 30 results in 10 more sit-ups that Bob can do than Ken.

answer：First, I recognize that 60% of the students are girls, which means that 40% are boys because the total percentage must add up to 100%.Given that the number of boys is 300, I can set up the equation 0.4 * Total = 300 to find the total number of students.Solving for Total, I divide 300 by 0.4, which gives me 750 students in total.Finally, to find the number of girls, I calculate 60% of the total number of students: 0.6 * 750 = 450.Therefore, there are 450 girls at the school.

answer：First, I need to calculate the volume of the sidewalk in cubic yards. The dimensions provided are in feet and inches, so I'll start by converting all measurements to yards.The width is 3 feet. Since 1 yard equals 3 feet, the width in yards is 1 yard.The length is 60 feet. Converting this to yards, I divide by 3, which gives 20 yards.The thickness is 3 inches. There are 36 inches in a yard, so the thickness in yards is 3 divided by 36, which simplifies to 1/12 yards.Now, I'll calculate the volume by multiplying the width, length, and thickness:1 yard (width) × 20 yards (length) × (1/12) yard (thickness) = 20/12 cubic yards, which simplifies to 5/3 cubic yards.Since concrete must be ordered in whole cubic yards, I'll round up 5/3 to the nearest whole number, which is 2 cubic yards.Therefore, the contractor needs to order 2 cubic yards of concrete.

answer：Okay, so I have this problem here where I need to show that there's a positive constant C such that for all positive real numbers a and b with a + b being an integer, the sum of the fractional parts of a³ and b³ plus C divided by (a + b)⁶ is less than or equal to 2. Hmm, that sounds a bit tricky, but let me try to break it down.First, let's recall that the fractional part of a number x, denoted {x}, is x minus the floor of x. So, {x} is always between 0 (inclusive) and 1 (exclusive). That means {a³} and {b³} are each less than 1, so their sum is less than 2. But the problem is asking us to add something else, C/(a + b)^6, and still have the total be less than or equal to 2. So, essentially, we need to make sure that the sum of these fractional parts doesn't get too close to 2, and if it does, the term C/(a + b)^6 will compensate for it.Since a and b are positive reals with a + b being an integer, let's denote n = a + b, where n is a positive integer. So, n is at least 1, right? Because a and b are positive, so their sum is at least greater than 0, but since it's an integer, the smallest n can be is 1.So, we can rewrite the problem in terms of n. Let me define a = n - b, so that a + b = n. Then, the fractional parts {a³} and {b³} can be written in terms of n and b. Hmm, but maybe it's better to express both a and b in terms of n and another variable. Let's say, let’s set a = n - t, where t is a positive real number less than n. Then, b = t. So, a = n - t and b = t, with t ∈ (0, n).So, substituting, we have {a³} = {(n - t)³} and {b³} = {t³}. So, the expression becomes:{(n - t)³} + {t³} + C/n⁶ ≤ 2.Our goal is to find a constant C such that this inequality holds for all positive integers n and for all t ∈ (0, n).Since the fractional parts are always less than 1, their sum is less than 2, but we need to ensure that even when their sum is close to 2, the term C/n⁶ is enough to bring the total down to 2. So, perhaps the maximum of {(n - t)³} + {t³} occurs at some point where t is such that both (n - t)³ and t³ are just below an integer, making their fractional parts close to 1 each. So, the sum would be close to 2, and then we need to subtract something to make sure it doesn't exceed 2.Wait, but actually, the fractional parts can't both be exactly 1, because that would mean that (n - t)³ and t³ are integers, which is only possible if t is such that both t and n - t are cube roots of integers. But since n is an integer, t is a real number, so it's not necessarily the case. So, maybe the maximum of the sum of fractional parts is less than 2, but we need to bound it such that when we add C/n⁶, it's still less than or equal to 2.Alternatively, perhaps the maximum of the sum of fractional parts is 2 minus something, and we need to find C such that 2 minus something plus C/n⁶ is still less than or equal to 2. Hmm, that might not make sense. Maybe I'm overcomplicating.Let me think about the behavior of {a³} + {b³}. Since a + b = n, which is an integer, maybe we can express a³ + b³ in terms of n and some other terms. Let's compute a³ + b³.a³ + b³ = (a + b)³ - 3ab(a + b) = n³ - 3abn.So, a³ + b³ = n³ - 3abn. Therefore, {a³} + {b³} = {n³ - 3abn}.Wait, but n is an integer, so n³ is an integer, and 3abn is a real number. So, {n³ - 3abn} is equal to {-3abn} because the fractional part of an integer plus a real number is just the fractional part of the real number. But wait, actually, {x + k} = {x} for any integer k. So, {n³ - 3abn} = {-3abn} = { -3abn }.But fractional parts are always non-negative, so { -3abn } is equal to 1 - {3abn} if 3abn is not an integer, right? Because the fractional part of a negative number is 1 minus the fractional part of its absolute value. Wait, is that correct?Let me recall: for any real number x, {x} = x - floor(x). So, if x is negative, say x = -y where y > 0, then floor(x) is the greatest integer less than or equal to x, which would be -ceil(y). So, {x} = x - floor(x) = -y - (-ceil(y)) = ceil(y) - y. But ceil(y) - y is equal to 1 - {y} if y is not an integer. So, yes, { -y } = 1 - { y } when y is not an integer. If y is an integer, then { -y } = 0.So, in our case, { -3abn } is equal to 1 - {3abn} if 3abn is not an integer, and 0 otherwise. So, {a³} + {b³} = {n³ - 3abn} = { -3abn } = 1 - {3abn} if 3abn is not an integer, and 0 otherwise.Wait, but {a³} + {b³} is equal to {n³ - 3abn}. But n³ is integer, so {n³ - 3abn} = { -3abn } = 1 - {3abn} if 3abn is not integer, else 0.But {a³} + {b³} is equal to {n³ - 3abn}. So, depending on whether 3abn is integer or not, {a³} + {b³} is either 1 - {3abn} or 0. Hmm, that seems a bit odd because {a³} and {b³} are both non-negative, so their sum should be non-negative. If 3abn is integer, then {a³} + {b³} = 0, which would imply that both {a³} and {b³} are 0, meaning a³ and b³ are integers. But a + b is integer, so if a³ and b³ are integers, then a and b must be integers as well? Wait, not necessarily. For example, a could be the cube root of an integer, but a + b is integer.Wait, but if a + b is integer and a³ is integer, does that imply that a is rational? Maybe not necessarily. For example, a could be irrational, but a³ could be integer. But in that case, b = n - a would be irrational as well, but b³ would also have to be integer? Hmm, that might not be possible unless a and b are both integers.Wait, actually, if a + b is integer and a³ is integer, then b³ = (n - a)³ = n³ - 3n²a + 3na² - a³. Since n and a³ are integers, then 3n²a - 3na² must be integer as well. So, 3n²a - 3na² = 3na(n - a) = 3na b. So, 3nab must be integer. So, if a³ is integer, then 3nab must be integer as well.But if a³ is integer, and 3nab is integer, does that imply that a is rational? Maybe. Let me think. Suppose a³ is integer, so a is either integer or irrational. If a is integer, then b is integer as well, since a + b is integer. If a is irrational, then a³ is integer, which is a special kind of irrational number, like the cube root of 2, but then 3nab would have to be integer. If a is irrational, then b = n - a is also irrational, so 3nab would be 3n times an irrational number, which is irrational, so it can't be integer. Therefore, the only possibility is that a and b are integers. So, in that case, {a³} + {b³} = 0 + 0 = 0. So, {a³} + {b³} is either 0 or 1 - {3abn}.Wait, but that seems conflicting with the initial thought that {a³} and {b³} are each less than 1. So, if {a³} + {b³} is either 0 or 1 - {3abn}, which is between 0 and 1, then how can their sum be close to 2? That doesn't make sense. Maybe my earlier reasoning is flawed.Wait, let's double-check. I said that {a³} + {b³} = {n³ - 3abn} = { -3abn } = 1 - {3abn} if 3abn is not integer, else 0. But {a³} + {b³} is equal to {n³ - 3abn}, but n³ is integer, so {n³ - 3abn} = { -3abn }.But { -3abn } is equal to 1 - {3abn} if 3abn is not integer, else 0. So, that would mean that {a³} + {b³} is either 0 or 1 - {3abn}, which is between 0 and 1. But that contradicts the fact that {a³} and {b³} can each be close to 1, so their sum can be close to 2. So, my mistake must be in the step where I equate {a³} + {b³} to {n³ - 3abn}.Wait, no, actually, {a³} + {b³} is not necessarily equal to {a³ + b³}. Because {x} + {y} is not equal to {x + y} in general. For example, if x = 0.5 and y = 0.5, then {x} + {y} = 1, but {x + y} = {1} = 0. So, in our case, {a³} + {b³} is not equal to {a³ + b³}, which is {n³ - 3abn}. So, my earlier reasoning was incorrect.So, scratch that. I need to think differently.Since {a³} + {b³} can be as large as just under 2, we need to find a C such that when we add C/n⁶, the total is at most 2. So, the maximum of {a³} + {b³} is less than 2, but how close can it get to 2?Perhaps the maximum of {a³} + {b³} is 2 - ε for some ε > 0, and then we can choose C such that 2 - ε + C/n⁶ ≤ 2, which would require that C/n⁶ ≤ ε. But since n is a positive integer, the smallest n can be is 1, so C must be less than or equal to ε. But ε depends on n, so maybe this approach isn't directly applicable.Wait, maybe we can find a lower bound on {a³} + {b³} and then relate it to C/n⁶. Alternatively, perhaps we can analyze when {a³} + {b³} is close to 2 and then bound the term C/n⁶ accordingly.Let me consider the case where {a³} and {b³} are both close to 1. That would mean that a³ is just below an integer, say k, and b³ is just below an integer, say m. So, a³ = k - δ and b³ = m - ε, where δ and ε are small positive numbers. Then, {a³} = 1 - δ and {b³} = 1 - ε, so their sum is 2 - (δ + ε). So, we need to ensure that 2 - (δ + ε) + C/n⁶ ≤ 2, which implies that C/n⁶ ≤ δ + ε.So, if we can bound δ + ε from below, then C can be chosen accordingly. But how can we relate δ and ε to n?Since a + b = n, and a³ + b³ = n³ - 3abn, as we had earlier. So, a³ + b³ = n³ - 3abn. But a³ = k - δ and b³ = m - ε, so a³ + b³ = (k + m) - (δ + ε). Therefore, n³ - 3abn = (k + m) - (δ + ε). So, 3abn = n³ - (k + m) + (δ + ε).But 3abn must be a real number, so n³ - (k + m) + (δ + ε) must be equal to 3abn. Since δ and ε are small, n³ - (k + m) is approximately 3abn.But n is an integer, so n³ is an integer, and k and m are integers as well, so n³ - (k + m) is an integer. Let's denote this integer as l, so l = n³ - (k + m). Then, 3abn = l + (δ + ε). Since δ and ε are small, l must be close to 3abn.But l is an integer, so 3abn is approximately an integer. So, 3abn is close to an integer, say l. Therefore, 3abn = l + η, where η is small. Then, δ + ε = η.So, in this case, δ + ε = η, which is the difference between 3abn and the integer l. So, η = 3abn - l, which is less than 1 in absolute value because l is the nearest integer to 3abn.But we need to relate η to something else. Maybe we can bound η in terms of a and b.Wait, since a + b = n, we can express ab in terms of a or b. Let's say, ab = a(n - a). So, ab = na - a². So, 3abn = 3n(na - a²) = 3n²a - 3na².So, 3abn is a quadratic in a. Maybe we can find a lower bound on |3abn - l|, which is η, using some form of the theory of Diophantine approximation.Alternatively, perhaps we can use the fact that if 3abn is close to an integer, then a must be close to some rational number with denominator dividing 3n² or something like that.Wait, maybe I can use the concept of fractional parts and their properties. Let me recall that for any real number x, the fractional part {x} is the distance from x to the nearest integer below x. So, {x} = x - floor(x). Similarly, the distance to the nearest integer is min({x}, 1 - {x}).But in our case, we have 3abn ≈ l, so {3abn} ≈ 0 or 1. So, the distance from 3abn to the nearest integer is small, say less than η. So, |3abn - l| < η.But how can we bound η from below? Maybe using the theory of linear forms in logarithms or something like that, but that might be too advanced.Alternatively, perhaps we can use the fact that a and b are positive reals with a + b = n, so a and b are in (0, n). So, ab is maximized when a = b = n/2, giving ab = n²/4. So, 3abn is maximized at 3*(n²/4)*n = 3n³/4. So, 3abn ranges from 0 to 3n³/4.But 3abn is a real number, so its fractional part can be anywhere between 0 and 1. But we are interested in the case where the fractional part is very small, meaning 3abn is very close to an integer.So, in such cases, we can say that 3abn is within η of an integer l, where η is small. So, 3abn = l + η, with η < 1.But we need to relate η to something else. Maybe we can use the fact that a and b are real numbers, so ab can be expressed in terms of a or b, and then we can write 3abn as a function of a, and then find the minimal distance from this function to the nearest integer.Alternatively, perhaps we can use the concept of the minimal gap between 3abn and integers. Since a and b are real numbers, 3abn can take any value in (0, 3n³/4). So, the minimal distance from 3abn to an integer can be as small as we like, but in reality, due to the distribution of real numbers, the minimal distance is bounded below by some function of n.Wait, but I think that for any real number x, the fractional part {x} can be arbitrarily small, meaning that η can be as small as we want. So, in theory, {a³} + {b³} can approach 2 as η approaches 0. So, in that case, the term C/n⁶ must compensate for the fact that {a³} + {b³} is approaching 2.But how can we ensure that C/n⁶ is large enough to make the total less than or equal to 2? Since when {a³} + {b³} is close to 2, C/n⁶ must be at least the difference between 2 and {a³} + {b³}. So, C/n⁶ ≥ 2 - ({a³} + {b³}).But 2 - ({a³} + {b³}) = 2 - (1 - δ + 1 - ε) = δ + ε. So, C/n⁶ ≥ δ + ε.But δ and ε are related to η, which is the distance from 3abn to the nearest integer. So, η = δ + ε, as we saw earlier.So, we have C/n⁶ ≥ η.But η is the distance from 3abn to the nearest integer, so η ≤ 1/2, because the maximum distance from a real number to the nearest integer is 1/2.But we need a lower bound on η. Wait, actually, we need an upper bound on η in terms of n. Because if η can be as small as we like, then C/n⁶ must be at least as large as η, but since η can be made arbitrarily small, C must be zero, which contradicts the requirement that C is positive.Wait, that suggests that my approach is flawed. Maybe I need to think differently.Perhaps instead of looking at the worst case where {a³} + {b³} is close to 2, I can find an upper bound for {a³} + {b³} in terms of n, and then choose C such that the upper bound plus C/n⁶ is less than or equal to 2.But how can I find such an upper bound?Let me consider the function f(a) = {a³} + {(n - a)³}. I need to find the maximum of f(a) over a ∈ (0, n). Then, if I can show that f(a) ≤ 2 - C/n⁶, then adding C/n⁶ would give me f(a) + C/n⁶ ≤ 2.So, perhaps I can find the maximum of f(a) and then express it as 2 - something, and then set C/n⁶ to be at least that something.Alternatively, maybe I can use the mean value theorem or some calculus to find the maximum of f(a).Let me try taking the derivative of f(a). Since f(a) = {a³} + {(n - a)³}, which is equal to a³ - floor(a³) + (n - a)³ - floor((n - a)³). So, f(a) = a³ + (n - a)³ - [floor(a³) + floor((n - a)³)].So, the derivative f’(a) is 3a² - 3(n - a)². Setting this equal to zero to find critical points:3a² - 3(n - a)² = 0a² = (n - a)²Taking square roots, |a| = |n - a|. Since a and n - a are positive, this simplifies to a = n - a, so 2a = n, so a = n/2.So, the function f(a) has a critical point at a = n/2. Let's check the second derivative to see if it's a maximum or a minimum.The second derivative f''(a) is 6a + 6(n - a) = 6n, which is positive, so the critical point at a = n/2 is a minimum. So, the maximum of f(a) occurs at the endpoints of the interval, i.e., as a approaches 0 or n.Wait, but when a approaches 0, {a³} approaches 0, and {b³} approaches {n³}, which is 0 because n is integer. So, f(a) approaches 0 + 0 = 0. Similarly, when a approaches n, {a³} approaches {n³} = 0, and {b³} approaches 0 as b approaches 0. So, f(a) approaches 0 in both cases.Wait, that doesn't make sense because earlier I thought that {a³} + {b³} can approach 2. Maybe I made a mistake in the derivative.Wait, no, actually, when a approaches 0, a³ approaches 0, so {a³} approaches 0, and b approaches n, so b³ approaches n³, which is integer, so {b³} approaches 0. Similarly, when a approaches n, {a³} approaches {n³} = 0, and {b³} approaches 0 as b approaches 0. So, f(a) approaches 0 at both ends.But earlier, I thought that {a³} + {b³} can be close to 2. So, maybe the maximum occurs somewhere in the middle, not at the endpoints.Wait, but according to the derivative, the only critical point is at a = n/2, which is a minimum. So, maybe the function f(a) has its maximum somewhere else. Wait, but if the derivative is 3a² - 3(n - a)², which is 3(a² - (n - a)²) = 3(a - (n - a))(a + (n - a)) = 3(2a - n)(n). So, f’(a) = 3n(2a - n). So, when a < n/2, f’(a) is negative, and when a > n/2, f’(a) is positive. So, the function is decreasing on (0, n/2) and increasing on (n/2, n). So, the minimum is at a = n/2, and the maximum is at the endpoints, but as we saw, the endpoints give f(a) approaching 0. So, that suggests that f(a) is always less than or equal to some value less than 2.Wait, but that contradicts the initial thought that {a³} + {b³} can be close to 2. So, perhaps my function f(a) is not the right one to consider.Wait, no, f(a) is {a³} + {b³}, which is the sum of two fractional parts. Each fractional part is less than 1, so their sum is less than 2. But when can their sum be close to 2? It would require that both {a³} and {b³} are close to 1, meaning that a³ and b³ are just below integers. So, a³ = k - δ and b³ = m - ε, with δ, ε small.But given that a + b = n, which is integer, is it possible for both a³ and b³ to be just below integers? Let's see.Suppose a³ is just below k, so a ≈ k^{1/3}, and b³ is just below m, so b ≈ m^{1/3}. Then, a + b ≈ k^{1/3} + m^{1/3} = n, which is integer.So, for some integers k and m, we have k^{1/3} + m^{1/3} ≈ n. So, if we can find integers k and m such that k^{1/3} + m^{1/3} is very close to an integer n, then a and b can be chosen such that a³ ≈ k and b³ ≈ m, making {a³} + {b³} ≈ 2.But how close can k^{1/3} + m^{1/3} be to an integer n? That depends on the Diophantine approximation properties of cube roots.I think that for any integer n, we can find integers k and m such that k^{1/3} + m^{1/3} is arbitrarily close to n, but I'm not sure. If that's the case, then {a³} + {b³} can be made arbitrarily close to 2, which would mean that the term C/n⁶ must be able to compensate for that, i.e., C/n⁶ must be at least 2 - ({a³} + {b³}), which can be made as small as we like. But since C is a constant, independent of n, this seems problematic because for large n, C/n⁶ becomes very small, so it can't compensate for the sum approaching 2.Wait, but the problem states that a and b are positive reals with a + b being an integer. So, n is a positive integer, but it can be any integer. So, for each n, we need to find C such that for all a, b with a + b = n, the inequality holds. So, perhaps C can be chosen such that for each n, C/n⁶ is at least the maximum of 2 - ({a³} + {b³}).But if for each n, the maximum of {a³} + {b³} is less than 2 - C/n⁶, then adding C/n⁶ would make the total less than or equal to 2.But how can we find such a C?Alternatively, maybe we can use the fact that when {a³} + {b³} is close to 2, then 3abn is close to an integer, as we saw earlier. So, perhaps we can bound the distance from 3abn to the nearest integer in terms of n, and then relate that to C/n⁶.Let me recall that for any real number x, the distance from x to the nearest integer is at least 1/(k + 1) for some k related to the continued fraction expansion of x. But I'm not sure if that helps here.Alternatively, perhaps we can use the theory of simultaneous Diophantine approximations. Since a and b are related by a + b = n, maybe we can find a lower bound on |3abn - l| for integers l.Wait, let's consider that 3abn is close to an integer l. So, |3abn - l| < η, where η is small. Then, we can write ab = (l + η)/(3n). Since a + b = n, we can write a and b as roots of the quadratic equation x² - nx + (l + η)/(3n) = 0. The discriminant of this equation is n² - 4*(l + η)/(3n). For real roots, the discriminant must be non-negative, so n² - 4*(l + η)/(3n) ≥ 0. So, 4*(l + η)/(3n) ≤ n², which is always true since l + η ≤ 3abn + η, but I'm not sure.Alternatively, perhaps we can use the AM-GM inequality. Since a + b = n, the maximum of ab is n²/4, achieved when a = b = n/2. So, ab ≤ n²/4. Therefore, 3abn ≤ 3n³/4. So, l must be less than or equal to 3n³/4. So, l is an integer less than or equal to 3n³/4.But how does that help?Wait, maybe we can use the fact that if 3abn is close to an integer l, then ab is close to l/(3n). Since ab ≤ n²/4, l/(3n) ≤ n²/4, so l ≤ 3n³/4, which is consistent with what we had earlier.But I'm not sure how to proceed from here.Wait, perhaps I can use the concept of the minimal distance from 3abn to the nearest integer. Let me denote this minimal distance as δ. So, δ = min_{l ∈ ℤ} |3abn - l|. Then, we have {a³} + {b³} = 2 - δ, as we saw earlier. So, we need to ensure that 2 - δ + C/n⁶ ≤ 2, which simplifies to C/n⁶ ≥ δ.So, we need to find a lower bound for δ in terms of n, such that C/n⁶ ≥ δ. Then, C must be at least δ * n⁶. So, if we can find that δ ≥ K / n⁶ for some constant K, then choosing C = K would satisfy the inequality.Therefore, our task reduces to showing that δ ≥ K / n⁶ for some constant K, which would imply that {a³} + {b³} ≤ 2 - K / n⁶, and hence {a³} + {b³} + C / n⁶ ≤ 2 with C = K.So, how can we find such a K?This seems related to the theory of Diophantine approximation, specifically to how well real numbers can be approximated by rationals. In our case, 3abn is a real number, and we want to bound its distance to the nearest integer from below.I recall that for any real number α, there are infinitely many rationals p/q such that |α - p/q| < 1/q². This is Dirichlet's approximation theorem. However, in our case, we have a specific form: 3abn is close to an integer l. So, |3abn - l| < η. We need a lower bound on η.But I think that for algebraic numbers, there is a result called Liouville's theorem which states that algebraic numbers cannot be too well approximated by rationals. Specifically, if α is an algebraic number of degree d, then there exists a constant C such that |α - p/q| > C / q^d for all rationals p/q.In our case, 3abn is a real number, but a and b are related by a + b = n. So, a and b are roots of the quadratic equation x² - nx + ab = 0. So, if ab is rational, then a and b are either both rational or conjugate irrationals. But 3abn is a real number, and if ab is rational, then 3abn is rational. If ab is irrational, then 3abn is irrational.But I'm not sure if 3abn is algebraic. If a and b are algebraic, then 3abn is algebraic, but if a and b are transcendental, then 3abn could be transcendental.Wait, but in our case, a and b are arbitrary positive reals with a + b = n. So, they could be transcendental or algebraic. So, perhaps we can't directly apply Liouville's theorem.Alternatively, maybe we can use the fact that if 3abn is close to an integer, then ab is close to l/(3n). So, ab ≈ l/(3n). But since a + b = n, we can write a and b as the roots of the quadratic equation x² - nx + l/(3n) ≈ 0. The discriminant of this equation is n² - 4*(l/(3n)) = n² - 4l/(3n). For real roots, this discriminant must be non-negative, so n² ≥ 4l/(3n), which implies l ≤ 3n³/4, which is consistent with earlier findings.But how does this help us bound δ?Alternatively, perhaps we can use the theory of continued fractions. For any real number x, there exists a convergent p/q such that |x - p/q| < 1/q². But again, I'm not sure how to apply this here.Wait, maybe I can consider that if 3abn is very close to an integer l, then ab is very close to l/(3n). So, ab ≈ l/(3n). But since a + b = n, we can write a ≈ (n ± sqrt(n² - 4ab))/2 ≈ (n ± sqrt(n² - 4l/(3n)))/2.But if ab is very close to l/(3n), then sqrt(n² - 4l/(3n)) is approximately sqrt(n² - 4l/(3n)) ≈ n - 2l/(3n²). So, a ≈ (n ± (n - 2l/(3n²)))/2. So, a ≈ n/2 ± (n - 2l/(3n²))/2 ≈ n/2 ± (n/2 - l/(3n³)).So, a ≈ n/2 + n/2 - l/(3n³) = n - l/(3n³) or a ≈ n/2 - n/2 + l/(3n³) = l/(3n³). So, a is either approximately n - l/(3n³) or approximately l/(3n³). So, in either case, a is very close to n or very close to 0, which contradicts the earlier thought that {a³} and {b³} can both be close to 1 when a and b are both away from 0 and n.Wait, maybe I made a mistake in the approximation. Let me try again.If ab ≈ l/(3n), then a ≈ l/(3n²) or b ≈ l/(3n²), since a + b = n. So, if a ≈ l/(3n²), then b ≈ n - l/(3n²). So, a is very small, and b is close to n. Then, a³ is very small, so {a³} ≈ a³, and b³ ≈ (n - l/(3n²))³ ≈ n³ - 3n²*(l/(3n²)) + 3n*(l/(3n²))² - (l/(3n²))³ ≈ n³ - l + l²/(3n²) - l³/(27n⁶). So, {b³} ≈ {n³ - l + l²/(3n²) - l³/(27n⁶)}. Since n³ - l is integer, {b³} ≈ { - l + l²/(3n²) - l³/(27n⁶) } ≈ { l²/(3n²) - l³/(27n⁶) }.But l is an integer, so l²/(3n²) - l³/(27n⁶) is a real number. Its fractional part is just itself if it's less than 1, otherwise, it wraps around. But since l ≤ 3n³/4, l²/(3n²) ≤ (9n⁶/16)/(3n²) = 3n⁴/16, which is much larger than 1 for n ≥ 2. So, {b³} ≈ { l²/(3n²) - l³/(27n⁶) } is just the fractional part of that expression.But this seems complicated. Maybe instead of trying to compute {b³}, I can note that if a is very small, then {a³} ≈ a³, and {b³} ≈ {n³ - 3abn}. Since a is small, ab ≈ a*n, so 3abn ≈ 3a n². So, {b³} ≈ {n³ - 3a n²}.But n³ is integer, so {n³ - 3a n²} = { -3a n² } = 1 - {3a n²} if 3a n² is not integer, else 0.So, {b³} ≈ 1 - {3a n²}.Therefore, {a³} + {b³} ≈ a³ + 1 - {3a n²}.But a is very small, so a³ is negligible compared to 1. So, {a³} + {b³} ≈ 1 - {3a n²}.But {3a n²} is the fractional part of 3a n². Since a is small, 3a n² is small, so {3a n²} ≈ 3a n². Therefore, {a³} + {b³} ≈ 1 - 3a n².But a is approximately l/(3n²), so 3a n² ≈ l. Therefore, {a³} + {b³} ≈ 1 - l. But l is an integer, so 1 - l is either 0 or negative, which doesn't make sense because {a³} + {b³} is always non-negative.Wait, I think I messed up the approximation. Let me try again.If a is very small, then 3abn ≈ 3a n². So, 3abn ≈ 3a n² = l + η, where η is small. So, a ≈ (l + η)/(3n²). Then, {b³} ≈ {n³ - 3abn} ≈ {n³ - l - η} = { - l - η } = 1 - { l + η } if l + η is not integer. But l is integer, so { l + η } = { η } = η. Therefore, {b³} ≈ 1 - η.Similarly, {a³} ≈ a³ ≈ ((l + η)/(3n²))³ ≈ l³/(27n⁶). So, {a³} + {b³} ≈ l³/(27n⁶) + 1 - η.But we need {a³} + {b³} ≤ 2 - C/n⁶. So, 1 - η + l³/(27n⁶) ≤ 2 - C/n⁶. Rearranging, we get C/n⁶ ≤ 1 + η - l³/(27n⁶).But η = 3abn - l ≈ 3*(l/(3n²))*n² - l = l - l = 0. Wait, that can't be right. Maybe η is the difference between 3abn and l, so η = 3abn - l. Since a ≈ l/(3n²), then 3abn ≈ 3*(l/(3n²))*n² = l. So, η ≈ 0.Therefore, {a³} + {b³} ≈ 1 - 0 + l³/(27n⁶) = 1 + l³/(27n⁶). But this is greater than 1, which contradicts the earlier thought that {a³} + {b³} is less than 2. Wait, but 1 + l³/(27n⁶) is less than 2 because l³/(27n⁶) is less than ( (3n³/4)³ )/(27n⁶) ) = (27n⁹/64)/(27n⁶) = n³/64, which for n ≥ 1 is less than 1. So, 1 + l³/(27n⁶) < 2.Wait, but if l is close to 3n³/4, then l³/(27n⁶) ≈ (27n⁹/64)/(27n⁶) = n³/64, which for n=1 is 1/64, for n=2 is 8/64=1/8, etc. So, {a³} + {b³} ≈ 1 + n³/64, which for n=1 is 1 + 1/64 ≈ 1.0156, which is still less than 2.Wait, but this seems inconsistent because earlier I thought that {a³} + {b³} can approach 2. Maybe my approximation is too rough.Alternatively, perhaps when a is close to n/2, both {a³} and {b³} can be close to 1, making their sum close to 2. Let's explore this case.Let me set a = n/2 + t and b = n/2 - t, where t is small. Then, a³ = (n/2 + t)³ = n³/8 + 3n² t/4 + 3n t²/2 + t³, and b³ = (n/2 - t)³ = n³/8 - 3n² t/4 + 3n t²/2 - t³.So, a³ + b³ = n³/4 + 3n t². Therefore, {a³} + {b³} = {n³/8 + 3n² t/4 + 3n t²/2 + t³} + {n³/8 - 3n² t/4 + 3n t²/2 - t³}.But n³/8 is integer if n is even, otherwise it's a multiple of 1/8. Wait, n is an integer, so n³ is integer, so n³/8 is integer if n is even, otherwise it's a fraction with denominator 8.Similarly, 3n² t/4 is a real number, and 3n t²/2 is a real number, and t³ is a real number.So, {a³} = {n³/8 + 3n² t/4 + 3n t²/2 + t³} and {b³} = {n³/8 - 3n² t/4 + 3n t²/2 - t³}.If n is even, say n = 2k, then n³/8 = k³, which is integer. So, {a³} = {3n² t/4 + 3n t²/2 + t³} and {b³} = { -3n² t/4 + 3n t²/2 - t³}.But {x} + {y} is not necessarily equal to {x + y}, so we can't directly add them. However, if t is very small, then 3n² t/4 is small, 3n t²/2 is very small, and t³ is negligible. So, {a³} ≈ {3n² t/4} and {b³} ≈ { -3n² t/4 }.But { -3n² t/4 } = 1 - {3n² t/4} if 3n² t/4 is not integer. So, {a³} + {b³} ≈ {3n² t/4} + 1 - {3n² t/4} = 1.Wait, so when n is even and t is small, {a³} + {b³} ≈ 1. So, in this case, the sum is about 1, which is less than 2.But if n is odd, say n = 2k + 1, then n³/8 = (2k + 1)³ /8 = (8k³ + 12k² + 6k + 1)/8 = k³ + 3k²/2 + 3k/4 + 1/8. So, n³/8 is not integer, it's k³ + 3k²/2 + 3k/4 + 1/8.So, {a³} = {n³/8 + 3n² t/4 + 3n t²/2 + t³} = {1/8 + 3n² t/4 + 3n t²/2 + t³} because the integer part is k³ + 3k²/2 + 3k/4.Similarly, {b³} = {n³/8 - 3n² t/4 + 3n t²/2 - t³} = {1/8 - 3n² t/4 + 3n t²/2 - t³}.So, {a³} + {b³} ≈ {1/8 + 3n² t/4} + {1/8 - 3n² t/4}.If 3n² t/4 is small, then {1/8 + 3n² t/4} ≈ 1/8 + 3n² t/4 and {1/8 - 3n² t/4} ≈ 1/8 - 3n² t/4. So, their sum is approximately 1/8 + 3n² t/4 + 1/8 - 3n² t/4 = 1/4.But if 3n² t/4 is close to 1/8, then {1/8 + 3n² t/4} could be close to 1/2, and {1/8 - 3n² t/4} could be close to 1/2 as well, making their sum close to 1.Wait, let's suppose that 3n² t/4 = 1/8 - δ, where δ is small. Then, {1/8 + 3n² t/4} = {1/8 + 1/8 - δ} = {1/4 - δ} ≈ 1/4 - δ, and {1/8 - 3n² t/4} = {1/8 - (1/8 - δ)} = {δ} ≈ δ. So, their sum is approximately 1/4 - δ + δ = 1/4.Alternatively, if 3n² t/4 = 1/8 + δ, then {1/8 + 3n² t/4} = {1/8 + 1/8 + δ} = {1/4 + δ} ≈ 1/4 + δ, and {1/8 - 3n² t/4} = {1/8 - 1/8 - δ} = { -δ } ≈ 1 - δ. So, their sum is approximately 1/4 + δ + 1 - δ = 5/4.Wait, that's interesting. So, if 3n² t/4 is slightly larger than 1/8, then {a³} + {b³} ≈ 5/4. If it's slightly smaller, it's ≈ 1/4. So, the maximum in this case is 5/4, which is still less than 2.Hmm, so maybe when a is near n/2, the sum {a³} + {b³} is bounded away from 2. So, perhaps the maximum occurs when a is near 0 or near n, but as we saw earlier, when a is near 0, {a³} is near 0, and {b³} is near {n³}, which is 0. So, the sum is near 0.Wait, but earlier I thought that when a is near 0, {a³} is near 0, and {b³} is near {n³}, which is 0, so the sum is near 0. But when a is near n, similar thing happens.But earlier, I considered the case where a is near 0, but b is near n, and tried to make {b³} near 1. But that didn't work because b³ ≈ n³, which is integer, so {b³} ≈ 0.Wait, maybe I need to consider a different approach. Let's think about the function f(a) = {a³} + {b³} where b = n - a. We need to find the maximum of f(a) over a ∈ (0, n).We saw that f(a) has a minimum at a = n/2, but the maximum might occur somewhere else. Maybe near the endpoints, but as a approaches 0, f(a) approaches 0, and as a approaches n, f(a) approaches 0 as well. So, perhaps the maximum occurs somewhere in the middle.Wait, but earlier when I set a = n/2 + t, I found that f(a) ≈ 1 or 5/4 depending on the value of t. So, maybe the maximum is around 5/4.But wait, let's test with n=1. If n=1, then a + b =1. Let's set a =1 - ε, b=ε, where ε is small. Then, a³ = (1 - ε)³ ≈1 - 3ε, so {a³} ≈1 - 3ε. Similarly, b³ = ε³, so {b³} ≈ ε³. So, {a³} + {b³} ≈1 - 3ε + ε³. As ε approaches 0, this approaches 1. So, for n=1, the maximum of {a³} + {b³} is approaching 1.Similarly, for n=2, let's set a=2 - ε, b=ε. Then, a³ ≈8 - 12ε, so {a³}≈1 - 12ε. b³≈ε³, so {b³}≈ε³. So, {a³} + {b³}≈1 -12ε + ε³, approaching 1 as ε approaches 0.Wait, so for n=1 and n=2, the maximum of {a³} + {b³} is approaching 1. So, in these cases, adding C/n⁶ would give us 1 + C/n⁶ ≤2, which is true for any C ≥1.But for larger n, maybe the maximum of {a³} + {b³} is larger.Wait, let's try n=3. Let me set a=3 - ε, b=ε. Then, a³≈27 - 27ε, so {a³}≈1 -27ε. b³≈ε³, so {b³}≈ε³. So, {a³} + {b³}≈1 -27ε + ε³, approaching 1 as ε approaches 0.Alternatively, let's set a=1.5 + t, b=1.5 - t, so a³ + b³= (1.5 + t)³ + (1.5 - t)³= 2*(1.5³ + 3*(1.5)*(t²))=2*(3.375 + 4.5 t²)=6.75 +9 t². So, {a³} + {b³}= {6.75 +9 t² - floor(6.75 +9 t²)}.But 6.75 is 6 + 0.75, so if 9 t² is small, then 6.75 +9 t² is between 6.75 and 7.75. So, floor(6.75 +9 t²)=6 if 9 t² <0.25, i.e., t² <1/36, t <1/6≈0.1667. So, if t <1/6, then {a³} + {b³}=0.75 +9 t².If t=0, {a³} + {b³}=0.75. If t approaches 1/6, then {a³} + {b³} approaches 0.75 +9*(1/36)=0.75 +0.25=1.So, for n=3, the maximum of {a³} + {b³} is approaching 1.Wait, so for n=1,2,3, the maximum is approaching 1. Is this a pattern? Maybe for all n, the maximum of {a³} + {b³} is approaching 1 as a approaches n or 0.But earlier, I thought that {a³} + {b³} can approach 2, but maybe that's not the case. Maybe the maximum is actually 1.Wait, let's test with n=4. Let me set a=4 - ε, b=ε. Then, a³≈64 - 48ε, so {a³}≈1 -48ε. b³≈ε³, so {b³}≈ε³. So, {a³} + {b³}≈1 -48ε + ε³, approaching 1 as ε approaches 0.Alternatively, set a=2 + t, b=2 - t. Then, a³ + b³= (2 + t)³ + (2 - t)³=2*(8 + 12 t²)=16 +24 t². So, {a³} + {b³}= {16 +24 t² - floor(16 +24 t²)}.Since 16 is integer, {16 +24 t²}= {24 t²}. So, {a³} + {b³}= {24 t²}.If t is small, {24 t²}≈24 t². So, the maximum of {a³} + {b³} is less than 1, achieved when 24 t² approaches 1, i.e., t≈sqrt(1/24)≈0.204. So, when t≈0.204, {a³} + {b³}≈1.So, again, the maximum approaches 1.Wait, so maybe for any n, the maximum of {a³} + {b³} is approaching 1 as a approaches n or 0, and when a is near n/2, the sum is around 1 or less.But earlier, I thought that when a is near n/2, the sum could be up to 5/4, but maybe that's not the case.Wait, let's take n=2 and set a=1 + t, b=1 - t. Then, a³ + b³= (1 + t)³ + (1 - t)³=2 + 6 t². So, {a³} + {b³}= {2 +6 t² - floor(2 +6 t²)}.Since 2 is integer, {2 +6 t²}= {6 t²}. So, {a³} + {b³}= {6 t²}.If t is small, {6 t²}≈6 t². So, the maximum is less than 1, achieved when 6 t² approaches 1, i.e., t≈sqrt(1/6)≈0.408. So, when t≈0.408, {a³} + {b³}≈1.So, again, the maximum is 1.Wait, so maybe for any n, the maximum of {a³} + {b³} is 1, achieved when a is near 0 or n, or when a is near n/2 with t such that 3abn is near an integer.But earlier, when I considered a near n/2, I thought that {a³} + {b³} could be up to 5/4, but that seems incorrect because when a is near n/2, the sum is actually less than 1.Wait, maybe I made a mistake in that earlier calculation. Let me re-examine it.When n is odd, say n=3, and a=1.5 + t, b=1.5 - t, then a³ + b³=6.75 +9 t². So, {a³} + {b³}= {6.75 +9 t²}= {0.75 +9 t²}.If t is small, this is approximately 0.75 +9 t². So, when t=0, it's 0.75, and as t increases, it approaches 1 when 9 t²=0.25, i.e., t=1/6≈0.1667.So, the maximum is 1.Similarly, for n=5, setting a=2.5 + t, b=2.5 - t, a³ + b³= (2.5 + t)³ + (2.5 - t)³=2*(15.625 + 15 t²)=31.25 +30 t². So, {a³} + {b³}= {31.25 +30 t²}= {0.25 +30 t²}.If t is small, this is approximately 0.25 +30 t². So, as t increases, it approaches 1 when 30 t²=0.75, i.e., t= sqrt(0.75/30)=sqrt(1/40)=≈0.158.So, again, the maximum is 1.Therefore, it seems that for any n, the maximum of {a³} + {b³} is 1, achieved when a is near 0 or n, or when a is near n/2 with t such that 3abn is near an integer.Wait, but earlier, when I considered a near n/2, I thought that {a³} + {b³} could be up to 5/4, but now I see that it's actually approaching 1. So, maybe my initial thought was wrong.Therefore, if the maximum of {a³} + {b³} is 1, then adding C/n⁶ would give us 1 + C/n⁶ ≤2, which is true for any C≥1. So, choosing C=1 would suffice.But wait, the problem states that the sum {a³} + {b³} + C/(a + b)^6 ≤2. Since a + b =n, this is {a³} + {b³} + C/n⁶ ≤2.If the maximum of {a³} + {b³} is 1, then 1 + C/n⁶ ≤2 implies that C/n⁶ ≤1, which is true for any C≤n⁶. But since C must be a constant independent of n, we can choose C=1, because for all n≥1, 1/n⁶ ≤1, so 1 +1/n⁶ ≤2.But wait, when n=1, 1 +1/1=2, so equality holds. For n>1, 1 +1/n⁶ <2. So, C=1 works.But wait, earlier I thought that {a³} + {b³} can approach 1, but in reality, when a is near n/2, the sum can approach 1, but when a is near 0 or n, the sum approaches 0. So, the maximum is 1.But wait, let me test with n=1. If n=1, a approaches 1, b approaches 0. Then, {a³} approaches {1}=0, {b³} approaches 0. So, the sum approaches 0. But earlier, when I set a=1 - ε, b=ε, I found that {a³} + {b³} approaches 1. Wait, that seems contradictory.Wait, no, when n=1, a=1 - ε, b=ε. Then, a³=(1 - ε)³≈1 -3ε, so {a³}=1 -3ε. b³=ε³, so {b³}=ε³. So, {a³} + {b³}=1 -3ε + ε³. As ε approaches 0, this approaches 1. So, for n=1, the maximum of {a³} + {b³} is approaching 1.Similarly, for n=2, when a approaches 2, {a³} approaches {8}=0, but when a=2 - ε, {a³}=1 -12ε, and {b³}=ε³, so the sum approaches 1.Wait, so for n=1,2,3,4,5, the maximum of {a³} + {b³} is approaching 1 as a approaches n or 0. So, in all cases, the maximum is 1.Therefore, {a³} + {b³} ≤1 for all a,b with a + b =n. So, adding C/n⁶, we get {a³} + {b³} + C/n⁶ ≤1 + C/n⁶ ≤2, provided that C/n⁶ ≤1, which is true for C=1 and n≥1.But wait, when n=1, 1 +1=2, so equality holds. For n>1, 1 +1/n⁶ <2. So, C=1 works.But the problem states that we need to show that there exists a positive constant C such that the inequality holds for all positive reals a and b with a + b integer. So, choosing C=1 would suffice.But wait, earlier I thought that when a is near n/2, the sum {a³} + {b³} could be up to 5/4, but now I see that it's actually approaching 1. So, maybe my initial concern was unfounded.Therefore, the maximum of {a³} + {b³} is 1, so adding C/n⁶ with C=1 ensures that the total is ≤2.But wait, let me check for n=1. When a=1 - ε, b=ε, {a³} + {b³}=1 -3ε + ε³. So, as ε approaches 0, this approaches 1. So, {a³} + {b³} +1/n⁶=1 +1=2. So, equality holds.For n=2, when a=2 - ε, {a³} + {b³}=1 -12ε + ε³, so {a³} + {b³} +1/64≈1 +1/64 <2.Similarly, for n=3, the sum approaches 1, so 1 +1/729 <2.Therefore, choosing C=1 works.But wait, the problem says "there exists a positive constant C", so C=1 is sufficient. But maybe a smaller C would also work, but the question only asks to show existence, so C=1 is fine.Therefore, the answer is C=1.But wait, let me think again. When n=1, the sum {a³} + {b³} +1/n⁶=2, which is exactly the upper bound. For n>1, it's less than 2. So, C=1 works.But the problem didn't specify that the inequality is strict, so equality is allowed. Therefore, C=1 is a valid choice.So, I think the answer is C=1.**Final Answer**boxed{1}