💡Okay, so I need to show that if the greatest common divisor of two numbers a and b is 1, then the product of Euler's totient function of a and Euler's totient function of b equals Euler's totient function of the product ab. Hmm, that sounds a bit abstract, but let me try to break it down.First, I remember that Euler's totient function, φ(n), counts the number of integers up to n that are relatively prime to n. So, φ(a) is the count of numbers less than or equal to a that don't share any common factors with a except 1. Similarly, φ(b) is the count for b. And φ(ab) would be the count for the product ab.Since a and b are coprime, meaning their greatest common divisor is 1, I think that might simplify things. Maybe the totient function has some multiplicative property when the numbers are coprime. I recall that φ is multiplicative for coprime integers, but I need to verify that.Let me think about the prime factorization of a and b. If a and b are coprime, their prime factors are completely distinct. So, if a is, say, 2^k and b is 3^m, then ab would be 2^k * 3^m. The totient function for a prime power p^k is p^k - p^(k-1), which is p^(k-1)(p - 1). So, φ(a) would be 2^(k-1)(2 - 1) = 2^(k-1), and φ(b) would be 3^(m-1)(3 - 1) = 2*3^(m-1). Multiplying φ(a) and φ(b) would give 2^(k-1) * 2 * 3^(m-1) = 2^k * 3^(m-1). On the other hand, φ(ab) for ab = 2^k * 3^m would be (2^k - 2^(k-1))(3^m - 3^(m-1)) = 2^(k-1) * 2 * 3^(m-1) = 2^k * 3^(m-1). Hey, that's the same as φ(a)φ(b)! So, at least for this simple case with two distinct primes, it works.But I need to make sure this holds in general, not just for two primes. Let me consider a more general case where a and b have multiple prime factors, but still coprime. Suppose a = p1^k1 * p2^k2 * ... * pn^kn and b = q1^m1 * q2^m2 * ... * qm^mm, where all the pi and qj are distinct primes. Then ab would be p1^k1 * p2^k2 * ... * pn^kn * q1^m1 * q2^m2 * ... * qm^mm.Now, φ(a) would be a multiplied by (1 - 1/p1)(1 - 1/p2)...(1 - 1/pn). Similarly, φ(b) would be b multiplied by (1 - 1/q1)(1 - 1/q2)...(1 - 1/qm). So, φ(a)φ(b) would be ab multiplied by all those (1 - 1/p) and (1 - 1/q) terms.On the other hand, φ(ab) is ab multiplied by (1 - 1/p1)(1 - 1/p2)...(1 - 1/pn)(1 - 1/q1)...(1 - 1/qm). So, φ(ab) is exactly the same as φ(a)φ(b) because the product of (1 - 1/p) terms for a and the product for b just combine together when you take φ(ab).Wait, but is this always true? What if a or b is 1? If a is 1, then φ(a) is 1, and φ(ab) would be φ(b). Similarly, φ(a)φ(b) would be 1 * φ(b) = φ(b), which matches φ(ab). So, that case works too.What about if a and b are both prime? Let's say a = 5 and b = 7. Then φ(5) = 4, φ(7) = 6, and φ(5*7) = φ(35) = 24. Indeed, 4 * 6 = 24, so it works here as well.Another test case: a = 4 and b = 9. φ(4) = 2, φ(9) = 6, and φ(36) = φ(4*9) = φ(4)*φ(9) = 2*6 = 12. Wait, φ(36) is actually 12, so that works too.But wait, what if a and b are not coprime? For example, a = 4 and b = 6. φ(4) = 2, φ(6) = 2, φ(24) = 8. But φ(4)*φ(6) = 4, which is not equal to φ(24) = 8. So, the condition that a and b are coprime is essential here.So, going back, since a and b are coprime, their prime factors don't overlap, which means when we take φ(ab), it's just the product of φ(a) and φ(b). If they weren't coprime, there would be overlapping prime factors, and the totient function would account for that overlap, leading to a different result.I think I've got the gist of it. The key is that when a and b are coprime, their totient functions multiply because their prime factors are distinct, and thus the multiplicative property holds. If they weren't coprime, the overlapping factors would cause the totient function of the product to be different from the product of the totients.So, to summarize, since a and b are coprime, φ(ab) = φ(a)φ(b) because the totient function is multiplicative over coprime integers. This is due to the fact that the prime factors of a and b don't interfere with each other, allowing the totient function to simply multiply the individual totients.**Final Answer**boxed{varphi(a) varphi(b) = varphi(ab)}

💡First, I need to find the midpoint of the line segment MN. The midpoint formula is ((frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2})). Plugging in the coordinates of points M(2, 4) and N(6, 2), the midpoint is ((frac{2 + 6}{2}, frac{4 + 2}{2}) = (4, 3)).Next, I'll determine the slope of the line segment MN. The slope formula is (frac{y_2 - y_1}{x_2 - x_1}). Using the coordinates of M and N, the slope is (frac{2 - 4}{6 - 2} = frac{-2}{4} = -frac{1}{2}).The slope of the perpendicular bisector is the negative reciprocal of the slope of MN. Therefore, the slope of the perpendicular bisector is (2).Now, using the point-slope form of a line, (y - y_1 = m(x - x_1)), and substituting the midpoint (4, 3) and the slope (2), the equation becomes:[y - 3 = 2(x - 4)]Simplifying this, I get:[y - 3 = 2x - 8][y = 2x - 5]To express this in standard form, I rearrange the equation:[2x - y - 5 = 0]Finally, I compare this with the given options and find that it matches option B.

💡Alright, so I have this problem with two parts, (a) and (b). Let me start by understanding part (a).**Problem (a):** Determine all pairs ((x, y)) of real numbers with (0 < x < 1) and (0 < y < 1) for which (x + 3y) and (3x + y) are both integers. An example is given as ((x, y) = left(frac{3}{8}, frac{7}{8}right)), because (x + 3y = 3) and (3x + y = 2), both integers.Okay, so I need to find all such pairs where both expressions are integers. Let me denote (x + 3y = a) and (3x + y = b), where (a) and (b) are integers. Then, I can set up a system of equations:1. (x + 3y = a)2. (3x + y = b)I can solve this system for (x) and (y). Let me try to express (x) and (y) in terms of (a) and (b).First, from equation (1): (x = a - 3y).Substitute this into equation (2):(3(a - 3y) + y = b)Simplify:(3a - 9y + y = b)(3a - 8y = b)Now, solve for (y):(-8y = b - 3a)(y = frac{3a - b}{8})Now, substitute (y) back into equation (1) to find (x):(x + 3left(frac{3a - b}{8}right) = a)Simplify:(x + frac{9a - 3b}{8} = a)Multiply both sides by 8 to eliminate the denominator:(8x + 9a - 3b = 8a)Simplify:(8x = 8a - 9a + 3b)(8x = -a + 3b)(x = frac{3b - a}{8})So, we have expressions for both (x) and (y) in terms of (a) and (b):(x = frac{3b - a}{8})(y = frac{3a - b}{8})Now, since (0 < x < 1) and (0 < y < 1), we can set up inequalities:For (x):(0 < frac{3b - a}{8} < 1)Multiply all parts by 8:(0 < 3b - a < 8)Similarly, for (y):(0 < frac{3a - b}{8} < 1)Multiply all parts by 8:(0 < 3a - b < 8)So, we have the system of inequalities:1. (0 < 3b - a < 8)2. (0 < 3a - b < 8)Our goal is to find all integer pairs ((a, b)) that satisfy these inequalities, and then from those, get the corresponding ((x, y)) pairs.Let me analyze these inequalities.From inequality 1:(0 < 3b - a < 8)Which can be rewritten as:(a < 3b < a + 8)Similarly, from inequality 2:(0 < 3a - b < 8)Which can be rewritten as:(b < 3a < b + 8)So, we have:1. (a < 3b < a + 8)2. (b < 3a < b + 8)These inequalities must hold simultaneously.Let me try to find possible integer values for (a) and (b).Let me consider possible values for (a) and (b). Since (x) and (y) are between 0 and 1, and (x = frac{3b - a}{8}) and (y = frac{3a - b}{8}), both numerators must be positive and less than 8.So, (3b - a) and (3a - b) must be integers between 1 and 7 (since they are positive and less than 8).Therefore, (3b - a) can be 1, 2, 3, 4, 5, 6, or 7.Similarly, (3a - b) can be 1, 2, 3, 4, 5, 6, or 7.Let me denote (k = 3b - a) and (l = 3a - b), where (k, l in {1, 2, 3, 4, 5, 6, 7}).So, we have:1. (k = 3b - a)2. (l = 3a - b)We can solve for (a) and (b) in terms of (k) and (l).From equation 1: (a = 3b - k)Substitute into equation 2:(l = 3(3b - k) - b)Simplify:(l = 9b - 3k - b)(l = 8b - 3k)So, (8b = l + 3k)Thus, (b = frac{l + 3k}{8})Similarly, from (a = 3b - k):(a = 3left(frac{l + 3k}{8}right) - k)Simplify:(a = frac{3l + 9k}{8} - k)(a = frac{3l + 9k - 8k}{8})(a = frac{3l + k}{8})So, (a) and (b) must be integers, so (frac{3l + k}{8}) and (frac{l + 3k}{8}) must be integers.Therefore, (3l + k) must be divisible by 8, and (l + 3k) must be divisible by 8.So, for each (k, l in {1, 2, 3, 4, 5, 6, 7}), we need to check if (3l + k equiv 0 mod 8) and (l + 3k equiv 0 mod 8).Let me create a table for (k) and (l) from 1 to 7 and check these conditions.But this might take a while. Maybe there's a smarter way.Alternatively, since (k) and (l) are between 1 and 7, let's consider the possible values.Let me consider (k) from 1 to 7 and for each (k), find (l) such that both (3l + k) and (l + 3k) are divisible by 8.Let me start with (k = 1):For (k = 1):We need (3l + 1 equiv 0 mod 8) and (l + 3 equiv 0 mod 8).So,1. (3l equiv -1 mod 8 implies 3l equiv 7 mod 8)2. (l equiv -3 mod 8 implies l equiv 5 mod 8)So, (l) must be 5 (since (l) is between 1 and 7).Check if (3l + 1 = 15 + 1 = 16), which is divisible by 8. Yes.And (l + 3k = 5 + 3 = 8), which is divisible by 8. Yes.So, (k = 1), (l = 5) is a solution.Now, (a = frac{3l + k}{8} = frac{15 + 1}{8} = 2)(b = frac{l + 3k}{8} = frac{5 + 3}{8} = 1)So, (a = 2), (b = 1)Then, (x = frac{3b - a}{8} = frac{3 - 2}{8} = frac{1}{8})(y = frac{3a - b}{8} = frac{6 - 1}{8} = frac{5}{8})So, one solution is ((x, y) = left(frac{1}{8}, frac{5}{8}right))Wait, but in the example given, it's (left(frac{3}{8}, frac{7}{8}right)). Let me check if that's another solution.Wait, maybe I need to check other (k) and (l) values.Let me proceed.(k = 2):1. (3l + 2 equiv 0 mod 8 implies 3l equiv 6 mod 8 implies l equiv 2 mod 8) (since 3*2=6)2. (l + 6 equiv 0 mod 8 implies l equiv 2 mod 8)So, (l = 2)Check:(3l + 2 = 6 + 2 = 8), divisible by 8.(l + 3k = 2 + 6 = 8), divisible by 8.So, (k = 2), (l = 2)Then, (a = frac{3*2 + 2}{8} = frac{8}{8} = 1)(b = frac{2 + 6}{8} = 1)So, (x = frac{3*1 - 1}{8} = frac{2}{8} = frac{1}{4})(y = frac{3*1 - 1}{8} = frac{2}{8} = frac{1}{4})So, another solution is (left(frac{1}{4}, frac{1}{4}right))Wait, but the example given is (left(frac{3}{8}, frac{7}{8}right)). Let me see if that comes up.Proceeding.(k = 3):1. (3l + 3 equiv 0 mod 8 implies 3l equiv 5 mod 8)2. (l + 9 equiv 0 mod 8 implies l equiv 7 mod 8)So, (l = 7)Check:(3l + 3 = 21 + 3 = 24), which is divisible by 8? 24/8=3, yes.(l + 3k = 7 + 9 = 16), which is divisible by 8. Yes.So, (k = 3), (l = 7)Then, (a = frac{3*7 + 3}{8} = frac{21 + 3}{8} = frac{24}{8} = 3)(b = frac{7 + 9}{8} = frac{16}{8} = 2)So, (x = frac{3*2 - 3}{8} = frac{6 - 3}{8} = frac{3}{8})(y = frac{3*3 - 2}{8} = frac{9 - 2}{8} = frac{7}{8})Ah, there we go! That's the example given: (left(frac{3}{8}, frac{7}{8}right))Good, so this method works.Continuing.(k = 4):1. (3l + 4 equiv 0 mod 8 implies 3l equiv 4 mod 8)2. (l + 12 equiv 0 mod 8 implies l equiv 4 mod 8)So, (l = 4)Check:(3l + 4 = 12 + 4 = 16), divisible by 8.(l + 3k = 4 + 12 = 16), divisible by 8.So, (k = 4), (l = 4)Then, (a = frac{3*4 + 4}{8} = frac{12 + 4}{8} = frac{16}{8} = 2)(b = frac{4 + 12}{8} = frac{16}{8} = 2)So, (x = frac{3*2 - 2}{8} = frac{6 - 2}{8} = frac{4}{8} = frac{1}{2})(y = frac{3*2 - 2}{8} = frac{6 - 2}{8} = frac{4}{8} = frac{1}{2})So, another solution is (left(frac{1}{2}, frac{1}{2}right))Proceeding.(k = 5):1. (3l + 5 equiv 0 mod 8 implies 3l equiv 3 mod 8 implies l equiv 1 mod 8)2. (l + 15 equiv 0 mod 8 implies l equiv 1 mod 8)So, (l = 1)Check:(3l + 5 = 3 + 5 = 8), divisible by 8.(l + 3k = 1 + 15 = 16), divisible by 8.So, (k = 5), (l = 1)Then, (a = frac{3*1 + 5}{8} = frac{3 + 5}{8} = 1)(b = frac{1 + 15}{8} = frac{16}{8} = 2)So, (x = frac{3*2 - 1}{8} = frac{6 - 1}{8} = frac{5}{8})(y = frac{3*1 - 2}{8} = frac{3 - 2}{8} = frac{1}{8})So, another solution is (left(frac{5}{8}, frac{1}{8}right))Proceeding.(k = 6):1. (3l + 6 equiv 0 mod 8 implies 3l equiv 2 mod 8)2. (l + 18 equiv 0 mod 8 implies l equiv 6 mod 8)So, (l = 6)Check:(3l + 6 = 18 + 6 = 24), divisible by 8.(l + 3k = 6 + 18 = 24), divisible by 8.So, (k = 6), (l = 6)Then, (a = frac{3*6 + 6}{8} = frac{18 + 6}{8} = frac{24}{8} = 3)(b = frac{6 + 18}{8} = frac{24}{8} = 3)So, (x = frac{3*3 - 3}{8} = frac{9 - 3}{8} = frac{6}{8} = frac{3}{4})(y = frac{3*3 - 3}{8} = frac{9 - 3}{8} = frac{6}{8} = frac{3}{4})So, another solution is (left(frac{3}{4}, frac{3}{4}right))Proceeding.(k = 7):1. (3l + 7 equiv 0 mod 8 implies 3l equiv 1 mod 8)2. (l + 21 equiv 0 mod 8 implies l equiv 3 mod 8)So, (l = 3)Check:(3l + 7 = 9 + 7 = 16), divisible by 8.(l + 3k = 3 + 21 = 24), divisible by 8.So, (k = 7), (l = 3)Then, (a = frac{3*3 + 7}{8} = frac{9 + 7}{8} = frac{16}{8} = 2)(b = frac{3 + 21}{8} = frac{24}{8} = 3)So, (x = frac{3*3 - 2}{8} = frac{9 - 2}{8} = frac{7}{8})(y = frac{3*2 - 3}{8} = frac{6 - 3}{8} = frac{3}{8})So, another solution is (left(frac{7}{8}, frac{3}{8}right))Alright, so compiling all the solutions we've found:1. (left(frac{1}{8}, frac{5}{8}right))2. (left(frac{1}{4}, frac{1}{4}right))3. (left(frac{3}{8}, frac{7}{8}right))4. (left(frac{1}{2}, frac{1}{2}right))5. (left(frac{5}{8}, frac{1}{8}right))6. (left(frac{3}{4}, frac{3}{4}right))7. (left(frac{7}{8}, frac{3}{8}right))Wait, that's 7 solutions. Let me check if I missed any.Wait, for each (k) from 1 to 7, I found exactly one (l) that satisfies the conditions, leading to 7 solutions. So, that seems correct.But let me double-check if these are all unique and within the interval (0,1).Yes, all (x) and (y) values are between 0 and 1, and all are unique pairs.So, the conclusion for part (a) is that there are 7 such pairs.**Problem (b):** Determine the integer (m > 2) for which there are exactly 119 pairs ((x, y)) with (0 < x < 1) and (0 < y < 1) such that (x + my) and (mx + y) are integers.The remark says that if (u neq v), the pairs ((u, v)) and ((v, u)) are different.So, similar to part (a), but now with a general (m), and we need to find (m) such that the number of such pairs is 119.Let me try to generalize the approach from part (a).Let me denote:1. (x + my = a)2. (mx + y = b)Where (a, b in mathbb{Z}).Solve for (x) and (y).From equation (1): (x = a - my)Substitute into equation (2):(m(a - my) + y = b)Simplify:(ma - m^2 y + y = b)(ma - y(m^2 - 1) = b)Solve for (y):(-y(m^2 - 1) = b - ma)(y = frac{ma - b}{m^2 - 1})Similarly, substitute (y) back into equation (1):(x + mleft(frac{ma - b}{m^2 - 1}right) = a)Simplify:(x + frac{m^2 a - mb}{m^2 - 1} = a)Multiply both sides by (m^2 - 1):(x(m^2 - 1) + m^2 a - mb = a(m^2 - 1))Simplify:(x(m^2 - 1) = a(m^2 - 1) - m^2 a + mb)(x(m^2 - 1) = a m^2 - a - m^2 a + mb)(x(m^2 - 1) = -a + mb)Thus,(x = frac{mb - a}{m^2 - 1})So, we have:(x = frac{mb - a}{m^2 - 1})(y = frac{ma - b}{m^2 - 1})Now, since (0 < x < 1) and (0 < y < 1), we have:For (x):(0 < frac{mb - a}{m^2 - 1} < 1)Multiply all parts by (m^2 - 1) (which is positive since (m > 2)):(0 < mb - a < m^2 - 1)Similarly, for (y):(0 < frac{ma - b}{m^2 - 1} < 1)Multiply all parts by (m^2 - 1):(0 < ma - b < m^2 - 1)So, we have the system of inequalities:1. (0 < mb - a < m^2 - 1)2. (0 < ma - b < m^2 - 1)Let me denote:(k = mb - a)(l = ma - b)Where (k, l) are positive integers less than (m^2 - 1).So, (k, l in {1, 2, ..., m^2 - 2})From the definitions:(k = mb - a)(l = ma - b)We can solve for (a) and (b) in terms of (k) and (l).From (k = mb - a), we get (a = mb - k)Substitute into (l = ma - b):(l = m(mb - k) - b)Simplify:(l = m^2 b - mk - b)(l = (m^2 - 1)b - mk)Rearrange:((m^2 - 1)b = l + mk)Thus,(b = frac{l + mk}{m^2 - 1})Similarly, from (a = mb - k):(a = mleft(frac{l + mk}{m^2 - 1}right) - k)Simplify:(a = frac{ml + m^2 k}{m^2 - 1} - k)(a = frac{ml + m^2 k - k(m^2 - 1)}{m^2 - 1})(a = frac{ml + m^2 k - m^2 k + k}{m^2 - 1})(a = frac{ml + k}{m^2 - 1})So, (a) and (b) must be integers, so (frac{ml + k}{m^2 - 1}) and (frac{l + mk}{m^2 - 1}) must be integers.Therefore, (ml + k) must be divisible by (m^2 - 1), and (l + mk) must be divisible by (m^2 - 1).So, for each (k, l in {1, 2, ..., m^2 - 2}), we need:1. (ml + k equiv 0 mod (m^2 - 1))2. (l + mk equiv 0 mod (m^2 - 1))This seems similar to part (a), but now with a general (m). Let me see if I can find a pattern or formula.Let me consider the two congruences:1. (ml + k equiv 0 mod (m^2 - 1))2. (l + mk equiv 0 mod (m^2 - 1))Let me denote (n = m^2 - 1), so (n = (m - 1)(m + 1)).Then, the congruences become:1. (ml + k equiv 0 mod n)2. (l + mk equiv 0 mod n)Let me try to solve these congruences.From equation 1: (ml equiv -k mod n)From equation 2: (l equiv -mk mod n)Substitute equation 2 into equation 1:(m(-mk) equiv -k mod n)(-m^2 k equiv -k mod n)Multiply both sides by -1:(m^2 k equiv k mod n)Factor out (k):(k(m^2 - 1) equiv 0 mod n)But (n = m^2 - 1), so (k n equiv 0 mod n), which is always true.Therefore, the system is consistent, and the solutions are determined by equation 2: (l equiv -mk mod n)But since (l) and (k) are positive integers less than (n), we can write:(l = n - mk)But (l) must be positive, so (n - mk > 0 implies mk < n = m^2 - 1 implies k < m - frac{1}{m}). Since (k) is an integer, (k leq m - 1).Similarly, since (l = n - mk), and (l) must be less than (n), which it is because (mk geq m*1 = m), so (l = n - mk leq n - m).But (l) must also be positive, so (n - mk > 0 implies k < frac{n}{m} = frac{m^2 - 1}{m} = m - frac{1}{m}), which again gives (k leq m - 1).Therefore, for each (k) from 1 to (m - 1), there is a corresponding (l = n - mk), which will be an integer between (n - m(m - 1)) and (n - m*1).But (n = m^2 - 1), so:(l = m^2 - 1 - mk)Which must be positive and less than (n).Since (k leq m - 1), (l geq m^2 - 1 - m(m - 1) = m^2 - 1 - m^2 + m = m - 1)And (l leq m^2 - 1 - m*1 = m^2 - m - 1)So, (l) ranges from (m - 1) to (m^2 - m - 1), which is valid since (m - 1 geq 1) for (m geq 2).Therefore, for each (k) from 1 to (m - 1), there is exactly one (l) such that both congruences are satisfied.Thus, the number of solutions is equal to the number of valid (k), which is (m - 1).But wait, in part (a), when (m = 3), we had 7 solutions, but according to this, (m - 1 = 2), which contradicts. So, my reasoning must be wrong.Wait, in part (a), (m = 3), and we had 7 solutions. But according to this, it should be (m - 1 = 2). Clearly, that's not matching. So, I must have made a mistake.Wait, let me re-examine.In part (a), (m = 3), and we had 7 solutions. But according to my generalization, it should be (m - 1 = 2). That's a discrepancy.Wait, perhaps my mistake is in assuming that for each (k), there's exactly one (l). But in part (a), for each (k), there was exactly one (l), but the total number of solutions was 7, which is (m^2 - 2) when (m = 3): (3^2 - 2 = 7). So, maybe the number of solutions is (m^2 - 2).Wait, let's see:In part (a), (m = 3), and we had 7 solutions, which is (3^2 - 2 = 7).If I consider (m = 4), then the number of solutions would be (4^2 - 2 = 14).Wait, let me test this.Suppose (m = 4). Then, the number of solutions should be (16 - 2 = 14).But let me try to compute it.Alternatively, perhaps the number of solutions is (m(m - 1)). For (m = 3), that would be 6, but we had 7 solutions. So, that doesn't fit.Wait, perhaps the number of solutions is (m^2 - 2). For (m = 3), that's 7, which matches. For (m = 4), that's 14.But let me see if that's the case.Wait, in part (a), we had 7 solutions, which is (3^2 - 2 = 7). So, perhaps the general formula is (m^2 - 2).But let me think again.In part (a), for each (k) from 1 to 7, we had a unique solution. But actually, (k) was from 1 to 7, which is (m^2 - 2 = 9 - 2 = 7). Wait, no, (m = 3), (m^2 - 2 = 7). So, that matches.Wait, but in the general case, for (m), the number of solutions is (m^2 - 2). So, for part (b), we have (m^2 - 2 = 119). Therefore, (m^2 = 121), so (m = 11).Wait, that seems too straightforward. Let me verify.If the number of solutions is (m^2 - 2), then setting (m^2 - 2 = 119) gives (m^2 = 121), so (m = 11).But let me check if this formula is correct.Wait, in part (a), (m = 3), and we had 7 solutions, which is (3^2 - 2 = 7). So, that matches.If (m = 4), then the number of solutions would be (16 - 2 = 14).Let me try to compute it for (m = 4) to see if it's 14.But this might take time. Alternatively, let me think about the general case.In part (a), we saw that for each (k) from 1 to (m^2 - 2), there's a unique solution. But actually, in part (a), (m = 3), and (k) went from 1 to 7, which is (m^2 - 2 = 7). So, perhaps the number of solutions is indeed (m^2 - 2).But wait, in part (a), we had 7 solutions, but (m^2 - 2 = 7) when (m = 3). So, that seems correct.Therefore, for part (b), the number of solutions is (m^2 - 2 = 119). Solving for (m):(m^2 = 121)(m = 11) (since (m > 2))Therefore, the integer (m) is 11.But let me double-check this reasoning.In part (a), we had (m = 3), and the number of solutions was 7, which is (3^2 - 2 = 7).If (m = 4), the number of solutions should be (16 - 2 = 14).Let me try to compute it for (m = 4) to see if it's 14.But this might take time. Alternatively, let me think about the general case.From the earlier analysis, for each (k) from 1 to (m - 1), there's a corresponding (l = m^2 - 1 - mk). But in part (a), (m = 3), and (k) went up to 7, which is (m^2 - 2 = 7). So, perhaps the number of solutions is (m^2 - 2).Wait, but in the general case, we have (k) and (l) such that (k, l in {1, 2, ..., m^2 - 2}), and for each (k), there's a unique (l). Therefore, the number of solutions is (m^2 - 2).But in part (a), (m = 3), and (m^2 - 2 = 7), which matches the 7 solutions we found.Therefore, for part (b), setting (m^2 - 2 = 119), we get (m^2 = 121), so (m = 11).Yes, that seems correct.

💡Alright, so I've got this problem here about a two-digit number. It says that if you divide the number by the product of its digits, the quotient is 3 and the remainder is 8. Then, if you reverse the digits and do the same division, the quotient is 2 and the remainder is 5. I need to find this number. Hmm, okay, let's break this down step by step.First, let's define the two-digit number. Let's say the tens digit is ( x ) and the units digit is ( y ). So, the number itself can be represented as ( 10x + y ). The product of its digits would then be ( x times y ), or ( xy ) for short.According to the problem, when we divide the number by the product of its digits, we get a quotient of 3 and a remainder of 8. In mathematical terms, that means:[10x + y = 3(xy) + 8]Okay, so that's our first equation. Now, when we reverse the digits, the number becomes ( 10y + x ). The problem states that dividing this reversed number by the product of the digits gives a quotient of 2 and a remainder of 5. So, that translates to:[10y + x = 2(xy) + 5]Alright, so now we have two equations:1. ( 10x + y = 3xy + 8 )2. ( 10y + x = 2xy + 5 )Our goal is to solve for ( x ) and ( y ). Since both ( x ) and ( y ) are digits, they must be integers between 0 and 9, and ( x ) can't be 0 because it's a two-digit number.Let me start by simplifying the first equation:[10x + y = 3xy + 8]Let's rearrange it to group like terms:[10x - 3xy + y - 8 = 0]Hmm, maybe factor out ( y ) from some terms:[y(1 - 3x) + 10x - 8 = 0]So, solving for ( y ):[y(1 - 3x) = 8 - 10x][y = frac{8 - 10x}{1 - 3x}]Okay, so ( y ) is expressed in terms of ( x ). Let's keep that in mind.Now, let's look at the second equation:[10y + x = 2xy + 5]Again, let's rearrange it:[10y - 2xy + x - 5 = 0]Factor out ( y ) from the first two terms:[y(10 - 2x) + x - 5 = 0]So, solving for ( y ):[y(10 - 2x) = 5 - x][y = frac{5 - x}{10 - 2x}]Now, we have two expressions for ( y ):1. ( y = frac{8 - 10x}{1 - 3x} )2. ( y = frac{5 - x}{10 - 2x} )Since both expressions equal ( y ), we can set them equal to each other:[frac{8 - 10x}{1 - 3x} = frac{5 - x}{10 - 2x}]To solve for ( x ), we'll cross-multiply:[(8 - 10x)(10 - 2x) = (5 - x)(1 - 3x)]Let's expand both sides.First, the left side:[(8 - 10x)(10 - 2x) = 8 times 10 + 8 times (-2x) - 10x times 10 - 10x times (-2x)][= 80 - 16x - 100x + 20x^2]Combine like terms:[= 80 - 116x + 20x^2]Now, the right side:[(5 - x)(1 - 3x) = 5 times 1 + 5 times (-3x) - x times 1 - x times (-3x)][= 5 - 15x - x + 3x^2]Combine like terms:[= 5 - 16x + 3x^2]So, setting both sides equal:[80 - 116x + 20x^2 = 5 - 16x + 3x^2]Let's bring all terms to one side:[80 - 116x + 20x^2 - 5 + 16x - 3x^2 = 0]Simplify:[75 - 100x + 17x^2 = 0]Let's write it in standard quadratic form:[17x^2 - 100x + 75 = 0]Now, we need to solve this quadratic equation for ( x ). Let's use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where ( a = 17 ), ( b = -100 ), and ( c = 75 ).Calculating the discriminant:[b^2 - 4ac = (-100)^2 - 4 times 17 times 75 = 10000 - 5100 = 4900][sqrt{4900} = 70]So, the solutions are:[x = frac{100 pm 70}{34}]Calculating both possibilities:1. ( x = frac{100 + 70}{34} = frac{170}{34} = 5 )2. ( x = frac{100 - 70}{34} = frac{30}{34} approx 0.882 )Since ( x ) must be an integer digit, ( x = 5 ) is the valid solution.Now, let's find ( y ) using one of the earlier expressions. Let's use:[y = frac{5 - x}{10 - 2x}]Substituting ( x = 5 ):[y = frac{5 - 5}{10 - 10} = frac{0}{0}]Wait, that's undefined. Hmm, maybe I should use the other expression for ( y ):[y = frac{8 - 10x}{1 - 3x}]Substituting ( x = 5 ):[y = frac{8 - 50}{1 - 15} = frac{-42}{-14} = 3]Okay, so ( y = 3 ). That makes sense.Let's verify these values in both original equations to ensure they work.First equation:[10x + y = 3xy + 8]Substituting ( x = 5 ) and ( y = 3 ):[50 + 3 = 3 times 5 times 3 + 8][53 = 45 + 8][53 = 53]Good.Second equation:[10y + x = 2xy + 5]Substituting ( x = 5 ) and ( y = 3 ):[30 + 5 = 2 times 5 times 3 + 5][35 = 30 + 5][35 = 35]Perfect.So, the two-digit number is ( 10x + y = 10 times 5 + 3 = 53 ).Just to double-check, let's perform the divisions as described in the problem.First, dividing the original number by the product of its digits:[53 div (5 times 3) = 53 div 15]15 times 3 is 45, and 53 minus 45 is 8. So, quotient 3, remainder 8. That matches.Now, reversing the digits gives 35. Dividing that by the product of the digits:[35 div (5 times 3) = 35 div 15]15 times 2 is 30, and 35 minus 30 is 5. So, quotient 2, remainder 5. That also matches.Everything checks out. So, the number we're looking for is 53.

💡Alright, so I came across this interesting problem: "Determine whether there exists a prime ( q ) so that for any prime ( p ), the number ( sqrt[3]{p^2 + q} ) is never an integer." At first glance, it seems a bit tricky, but I think I can break it down step by step.Okay, let's start by understanding what the problem is asking. We need to find a prime number ( q ) such that no matter which prime number ( p ) we choose, the expression ( sqrt[3]{p^2 + q} ) never results in an integer. In other words, ( p^2 + q ) should never be a perfect cube for any prime ( p ).Hmm, so the goal is to ensure that ( p^2 + q ) isn't a cube. That means for every prime ( p ), ( p^2 + q ) shouldn't equal ( k^3 ) for some integer ( k ). So, if we can find such a prime ( q ), that would solve the problem.I remember that in number theory, modular arithmetic can be useful for proving that certain expressions can't be perfect powers. Maybe I can use that here. Let me think about how to apply it.First, let's consider the expression ( p^2 + q ). If this is a perfect cube, say ( k^3 ), then ( p^2 = k^3 - q ). So, ( p^2 ) would be equal to ( k^3 - q ). Since ( p ) is a prime, ( p^2 ) is a square of a prime, which is a specific type of number.I think a good approach might be to analyze this equation modulo some small number, like 7 or 9, where the residues of squares and cubes are limited. If I can show that ( p^2 + q ) cannot be a cube modulo that number, then it can't be a cube in general.Let me try modulo 7 first because 7 is a small prime, and the number of residues is manageable. Let's list the possible residues of squares modulo 7 and the possible residues of cubes modulo 7.For squares modulo 7:- ( 0^2 equiv 0 mod 7 )- ( 1^2 equiv 1 mod 7 )- ( 2^2 equiv 4 mod 7 )- ( 3^2 equiv 9 equiv 2 mod 7 )- ( 4^2 equiv 16 equiv 2 mod 7 )- ( 5^2 equiv 25 equiv 4 mod 7 )- ( 6^2 equiv 36 equiv 1 mod 7 )So, the quadratic residues modulo 7 are {0, 1, 2, 4}.For cubes modulo 7:- ( 0^3 equiv 0 mod 7 )- ( 1^3 equiv 1 mod 7 )- ( 2^3 equiv 8 equiv 1 mod 7 )- ( 3^3 equiv 27 equiv 6 mod 7 )- ( 4^3 equiv 64 equiv 1 mod 7 )- ( 5^3 equiv 125 equiv 6 mod 7 )- ( 6^3 equiv 216 equiv 6 mod 7 )So, the cubic residues modulo 7 are {0, 1, 6}.Alright, so if ( p^2 + q ) is a cube, then ( p^2 + q equiv 0, 1, ) or ( 6 mod 7 ). Let's denote ( q ) as a prime number, so ( q ) can be any prime, but we need to choose one such that ( p^2 + q ) never falls into the cubic residues modulo 7 for any prime ( p ).Wait, but ( q ) is fixed, and ( p ) varies over all primes. So, we need to choose ( q ) such that for any prime ( p ), ( p^2 + q ) is not congruent to 0, 1, or 6 modulo 7.Alternatively, for each prime ( p ), ( p^2 ) can be 0, 1, 2, or 4 modulo 7, as we saw earlier. So, ( p^2 + q ) modulo 7 would be ( (p^2 mod 7) + (q mod 7) ).If we can choose ( q ) such that ( (p^2 + q) mod 7 ) is never 0, 1, or 6, regardless of ( p^2 mod 7 ), then ( p^2 + q ) can't be a cube.Let me see. Let's denote ( q mod 7 ) as ( r ), where ( r ) is between 0 and 6. We need to choose ( r ) such that for all possible ( p^2 mod 7 ) (which are 0,1,2,4), ( (p^2 + r) mod 7 ) is not in {0,1,6}.So, let's consider each possible ( r ):1. If ( r = 0 ): - ( 0 + 0 = 0 ) → bad (0 is a cubic residue) - ( 1 + 0 = 1 ) → bad - ( 2 + 0 = 2 ) → good - ( 4 + 0 = 4 ) → good So, ( r = 0 ) is bad because it results in 0 and 1.2. If ( r = 1 ): - ( 0 + 1 = 1 ) → bad - ( 1 + 1 = 2 ) → good - ( 2 + 1 = 3 ) → good - ( 4 + 1 = 5 ) → good So, ( r = 1 ) is bad because it results in 1.3. If ( r = 2 ): - ( 0 + 2 = 2 ) → good - ( 1 + 2 = 3 ) → good - ( 2 + 2 = 4 ) → good - ( 4 + 2 = 6 ) → bad So, ( r = 2 ) is bad because it results in 6.4. If ( r = 3 ): - ( 0 + 3 = 3 ) → good - ( 1 + 3 = 4 ) → good - ( 2 + 3 = 5 ) → good - ( 4 + 3 = 7 equiv 0 ) → bad So, ( r = 3 ) is bad because it results in 0.5. If ( r = 4 ): - ( 0 + 4 = 4 ) → good - ( 1 + 4 = 5 ) → good - ( 2 + 4 = 6 ) → bad - ( 4 + 4 = 8 equiv 1 ) → bad So, ( r = 4 ) is bad because it results in 6 and 1.6. If ( r = 5 ): - ( 0 + 5 = 5 ) → good - ( 1 + 5 = 6 ) → bad - ( 2 + 5 = 7 equiv 0 ) → bad - ( 4 + 5 = 9 equiv 2 ) → good So, ( r = 5 ) is bad because it results in 6 and 0.7. If ( r = 6 ): - ( 0 + 6 = 6 ) → bad - ( 1 + 6 = 7 equiv 0 ) → bad - ( 2 + 6 = 8 equiv 1 ) → bad - ( 4 + 6 = 10 equiv 3 ) → good So, ( r = 6 ) is bad because it results in 6, 0, and 1.Wait a minute, so for all possible residues ( r ) modulo 7, there is at least one ( p^2 mod 7 ) that makes ( p^2 + r ) a cubic residue modulo 7. That suggests that no matter what ( q ) we choose, there will always be some prime ( p ) such that ( p^2 + q ) is a cube. But that contradicts the problem statement, which asks if such a ( q ) exists.Hmm, maybe I made a mistake in my reasoning. Let me double-check.I considered all residues ( r ) modulo 7 and found that for each ( r ), there exists some ( p^2 mod 7 ) such that ( p^2 + r ) is a cubic residue modulo 7. That would imply that for any ( q ), there exists some prime ( p ) such that ( p^2 + q ) is a cube. But the problem is asking if there exists a ( q ) such that for all primes ( p ), ( p^2 + q ) is not a cube. So, according to my analysis, such a ( q ) doesn't exist because for any ( q ), there is some ( p ) making ( p^2 + q ) a cube.But wait, maybe I'm missing something. Perhaps the problem is not asking for all primes ( p ), but for any prime ( p ), meaning that for each ( p ), ( p^2 + q ) is not a cube. But my analysis shows that for any ( q ), there exists at least one ( p ) such that ( p^2 + q ) is a cube. So, that would mean that no such ( q ) exists.But the problem is phrased as "for any prime ( p )", which could mean that for each ( p ), ( p^2 + q ) is not a cube. So, if for every ( p ), ( p^2 + q ) is not a cube, then such a ( q ) exists. But according to my earlier analysis, for any ( q ), there is some ( p ) such that ( p^2 + q ) is a cube. Therefore, such a ( q ) does not exist.Wait, but the problem is asking "whether there exists a prime ( q ) so that for any prime ( p ) the number ( sqrt[3]{p^2 + q} ) is never an integer." So, it's asking if such a ( q ) exists. Based on my analysis, it seems that for any ( q ), there is some ( p ) such that ( p^2 + q ) is a cube, meaning that ( sqrt[3]{p^2 + q} ) is an integer. Therefore, such a ( q ) does not exist.But I'm not entirely sure. Maybe I need to check specific examples. Let's try some small primes for ( q ) and see if we can find a ( p ) such that ( p^2 + q ) is a cube.Let's take ( q = 2 ):- ( p = 2 ): ( 2^2 + 2 = 6 ), not a cube.- ( p = 3 ): ( 9 + 2 = 11 ), not a cube.- ( p = 5 ): ( 25 + 2 = 27 ), which is ( 3^3 ). So, ( sqrt[3]{27} = 3 ), which is an integer. Therefore, ( q = 2 ) doesn't work.Next, ( q = 3 ):- ( p = 2 ): ( 4 + 3 = 7 ), not a cube.- ( p = 3 ): ( 9 + 3 = 12 ), not a cube.- ( p = 5 ): ( 25 + 3 = 28 ), not a cube.- ( p = 7 ): ( 49 + 3 = 52 ), not a cube.- ( p = 11 ): ( 121 + 3 = 124 ), not a cube.- ( p = 13 ): ( 169 + 3 = 172 ), not a cube.- ( p = 17 ): ( 289 + 3 = 292 ), not a cube.- ( p = 19 ): ( 361 + 3 = 364 ), not a cube.- ( p = 23 ): ( 529 + 3 = 532 ), not a cube.- ( p = 29 ): ( 841 + 3 = 844 ), not a cube.- ( p = 31 ): ( 961 + 3 = 964 ), not a cube.- ( p = 37 ): ( 1369 + 3 = 1372 ), not a cube.- ( p = 41 ): ( 1681 + 3 = 1684 ), not a cube.- ( p = 43 ): ( 1849 + 3 = 1852 ), not a cube.- ( p = 47 ): ( 2209 + 3 = 2212 ), not a cube.Hmm, up to ( p = 47 ), I haven't found any ( p ) such that ( p^2 + 3 ) is a cube. Maybe ( q = 3 ) works? But I need to check further.Wait, let's try ( p = 7 ): ( 49 + 3 = 52 ), not a cube.( p = 11 ): ( 121 + 3 = 124 ), not a cube.( p = 13 ): ( 169 + 3 = 172 ), not a cube.( p = 17 ): ( 289 + 3 = 292 ), not a cube.( p = 19 ): ( 361 + 3 = 364 ), not a cube.( p = 23 ): ( 529 + 3 = 532 ), not a cube.( p = 29 ): ( 841 + 3 = 844 ), not a cube.( p = 31 ): ( 961 + 3 = 964 ), not a cube.( p = 37 ): ( 1369 + 3 = 1372 ), not a cube.( p = 41 ): ( 1681 + 3 = 1684 ), not a cube.( p = 43 ): ( 1849 + 3 = 1852 ), not a cube.( p = 47 ): ( 2209 + 3 = 2212 ), not a cube.Still nothing. Maybe ( q = 3 ) is a candidate. But I need to be sure. Let's try ( p = 5 ): ( 25 + 3 = 28 ), not a cube.( p = 7 ): 52, not a cube.( p = 11 ): 124, not a cube.( p = 13 ): 172, not a cube.( p = 17 ): 292, not a cube.( p = 19 ): 364, not a cube.( p = 23 ): 532, not a cube.( p = 29 ): 844, not a cube.( p = 31 ): 964, not a cube.( p = 37 ): 1372, not a cube.( p = 41 ): 1684, not a cube.( p = 43 ): 1852, not a cube.( p = 47 ): 2212, not a cube.Hmm, maybe ( q = 3 ) is a possible candidate. But I need to check if there exists any prime ( p ) such that ( p^2 + 3 ) is a cube. Let's see, ( k^3 - 3 ) must be a square of a prime.Let's try small ( k ):- ( k = 2 ): ( 8 - 3 = 5 ), not a square.- ( k = 3 ): ( 27 - 3 = 24 ), not a square.- ( k = 4 ): ( 64 - 3 = 61 ), not a square.- ( k = 5 ): ( 125 - 3 = 122 ), not a square.- ( k = 6 ): ( 216 - 3 = 213 ), not a square.- ( k = 7 ): ( 343 - 3 = 340 ), not a square.- ( k = 8 ): ( 512 - 3 = 509 ), not a square.- ( k = 9 ): ( 729 - 3 = 726 ), not a square.- ( k = 10 ): ( 1000 - 3 = 997 ), not a square.- ( k = 11 ): ( 1331 - 3 = 1328 ), not a square.- ( k = 12 ): ( 1728 - 3 = 1725 ), not a square.- ( k = 13 ): ( 2197 - 3 = 2194 ), not a square.- ( k = 14 ): ( 2744 - 3 = 2741 ), not a square.- ( k = 15 ): ( 3375 - 3 = 3372 ), not a square.- ( k = 16 ): ( 4096 - 3 = 4093 ), not a square.- ( k = 17 ): ( 4913 - 3 = 4910 ), not a square.- ( k = 18 ): ( 5832 - 3 = 5829 ), not a square.- ( k = 19 ): ( 6859 - 3 = 6856 ), not a square.- ( k = 20 ): ( 8000 - 3 = 7997 ), not a square.It seems like up to ( k = 20 ), there's no solution. Maybe ( q = 3 ) is a valid candidate. But I need to be more rigorous.Alternatively, perhaps there's a specific ( q ) that avoids this. I recall that in some problems, choosing ( q ) such that ( q equiv 1 mod 7 ) can help, but I'm not sure.Wait, earlier I tried modulo 7 and found that for any ( q mod 7 ), there exists some ( p^2 mod 7 ) such that ( p^2 + q mod 7 ) is a cubic residue. That suggests that for any ( q ), there exists some ( p ) such that ( p^2 + q ) is a cube. Therefore, such a ( q ) does not exist.But in my earlier example with ( q = 3 ), I couldn't find any ( p ) up to 47 that satisfies ( p^2 + 3 ) being a cube. Maybe I need to check larger primes.Let me try ( p = 53 ): ( 53^2 = 2809 ), ( 2809 + 3 = 2812 ), not a cube.( p = 59 ): ( 3481 + 3 = 3484 ), not a cube.( p = 61 ): ( 3721 + 3 = 3724 ), not a cube.( p = 67 ): ( 4489 + 3 = 4492 ), not a cube.( p = 71 ): ( 5041 + 3 = 5044 ), not a cube.( p = 73 ): ( 5329 + 3 = 5332 ), not a cube.( p = 79 ): ( 6241 + 3 = 6244 ), not a cube.( p = 83 ): ( 6889 + 3 = 6892 ), not a cube.( p = 89 ): ( 7921 + 3 = 7924 ), not a cube.( p = 97 ): ( 9409 + 3 = 9412 ), not a cube.Still nothing. Maybe ( q = 3 ) is actually a valid answer. But according to my earlier modulo analysis, for any ( q ), there should exist some ( p ) such that ( p^2 + q ) is a cube. So, perhaps my modulo analysis was incomplete or incorrect.Wait, let's reconsider the modulo analysis. I considered ( q mod 7 ) and found that for any ( r = q mod 7 ), there exists some ( p^2 mod 7 ) such that ( p^2 + r ) is a cubic residue modulo 7. Therefore, ( p^2 + q equiv k^3 mod 7 ) for some ( k ). But this doesn't necessarily mean that ( p^2 + q ) is a cube in integers, just that it is congruent to a cube modulo 7.However, if ( p^2 + q ) is a cube, then it must satisfy ( p^2 + q equiv k^3 mod 7 ). So, if we can choose ( q ) such that ( p^2 + q ) is never congruent to a cubic residue modulo 7, then ( p^2 + q ) can't be a cube. But my earlier analysis showed that for any ( q mod 7 ), there exists some ( p^2 mod 7 ) such that ( p^2 + q mod 7 ) is a cubic residue. Therefore, for any ( q ), there exists some ( p ) such that ( p^2 + q ) is congruent to a cubic residue modulo 7, meaning that ( p^2 + q ) could potentially be a cube.But this doesn't necessarily mean that ( p^2 + q ) is a cube; it just means it could be. So, maybe there exists a ( q ) such that even though ( p^2 + q equiv k^3 mod 7 ), ( p^2 + q ) isn't actually a cube. Therefore, my earlier conclusion that such a ( q ) doesn't exist might be incorrect.Perhaps I need to look for a ( q ) such that ( p^2 + q ) is never a cube, regardless of the congruence. Maybe choosing ( q ) such that ( q equiv 1 mod 7 ) or something like that.Wait, let's try ( q = 29 ). Why 29? Because 29 is a prime, and I recall that 29 is congruent to 1 modulo 7 (since 29 - 1 = 28, which is divisible by 7). Let's see if ( q = 29 ) works.So, ( q = 29 ). Let's check if ( p^2 + 29 ) is ever a cube for any prime ( p ).First, let's try small primes:- ( p = 2 ): ( 4 + 29 = 33 ), not a cube.- ( p = 3 ): ( 9 + 29 = 38 ), not a cube.- ( p = 5 ): ( 25 + 29 = 54 ), not a cube.- ( p = 7 ): ( 49 + 29 = 78 ), not a cube.- ( p = 11 ): ( 121 + 29 = 150 ), not a cube.- ( p = 13 ): ( 169 + 29 = 198 ), not a cube.- ( p = 17 ): ( 289 + 29 = 318 ), not a cube.- ( p = 19 ): ( 361 + 29 = 390 ), not a cube.- ( p = 23 ): ( 529 + 29 = 558 ), not a cube.- ( p = 29 ): ( 841 + 29 = 870 ), not a cube.- ( p = 31 ): ( 961 + 29 = 990 ), not a cube.- ( p = 37 ): ( 1369 + 29 = 1398 ), not a cube.- ( p = 41 ): ( 1681 + 29 = 1710 ), not a cube.- ( p = 43 ): ( 1849 + 29 = 1878 ), not a cube.- ( p = 47 ): ( 2209 + 29 = 2238 ), not a cube.Hmm, still nothing. Let's try larger primes:- ( p = 53 ): ( 2809 + 29 = 2838 ), not a cube.- ( p = 59 ): ( 3481 + 29 = 3510 ), not a cube.- ( p = 61 ): ( 3721 + 29 = 3750 ), not a cube.- ( p = 67 ): ( 4489 + 29 = 4518 ), not a cube.- ( p = 71 ): ( 5041 + 29 = 5070 ), not a cube.- ( p = 73 ): ( 5329 + 29 = 5358 ), not a cube.- ( p = 79 ): ( 6241 + 29 = 6270 ), not a cube.- ( p = 83 ): ( 6889 + 29 = 6918 ), not a cube.- ( p = 89 ): ( 7921 + 29 = 7950 ), not a cube.- ( p = 97 ): ( 9409 + 29 = 9438 ), not a cube.Still no luck. Maybe ( q = 29 ) is a valid candidate. But I need to be sure. Let's try to see if there's a mathematical reason why ( q = 29 ) might work.Earlier, I considered modulo 7. Let's see what ( 29 mod 7 ) is. ( 29 ÷ 7 = 4 ) with a remainder of 1, so ( 29 equiv 1 mod 7 ).So, ( p^2 + 29 equiv p^2 + 1 mod 7 ). From earlier, the quadratic residues modulo 7 are {0,1,2,4}, so ( p^2 mod 7 ) can be 0,1,2,4. Therefore, ( p^2 + 1 mod 7 ) can be:- If ( p^2 equiv 0 mod 7 ): ( 0 + 1 = 1 mod 7 )- If ( p^2 equiv 1 mod 7 ): ( 1 + 1 = 2 mod 7 )- If ( p^2 equiv 2 mod 7 ): ( 2 + 1 = 3 mod 7 )- If ( p^2 equiv 4 mod 7 ): ( 4 + 1 = 5 mod 7 )So, ( p^2 + 29 mod 7 ) can be 1,2,3,5. The cubic residues modulo 7 are {0,1,6}. Therefore, ( p^2 + 29 mod 7 ) is either 1,2,3,5. Among these, only 1 is a cubic residue. So, if ( p^2 + 29 equiv 1 mod 7 ), it could potentially be a cube.But wait, ( p^2 + 29 equiv 1 mod 7 ) implies ( p^2 equiv 0 mod 7 ), which means ( p equiv 0 mod 7 ). But ( p ) is a prime, so the only possibility is ( p = 7 ). Let's check ( p = 7 ): ( 7^2 + 29 = 49 + 29 = 78 ), which is not a cube.Therefore, even though ( p^2 + 29 equiv 1 mod 7 ) when ( p = 7 ), the actual value 78 is not a cube. So, in this case, ( p^2 + 29 ) is not a cube.Similarly, for other primes ( p ), ( p^2 + 29 ) modulo 7 is 2,3, or 5, none of which are cubic residues. Therefore, ( p^2 + 29 ) cannot be a cube for any prime ( p ).Wait, that seems promising. So, by choosing ( q = 29 ), which is congruent to 1 modulo 7, we ensure that ( p^2 + 29 mod 7 ) is never a cubic residue unless ( p = 7 ), but in that case, the actual value isn't a cube. Therefore, ( q = 29 ) works.But earlier, I thought that for any ( q ), there exists some ( p ) such that ( p^2 + q ) is a cube. But in this case, ( q = 29 ) seems to avoid that. So, maybe my earlier conclusion was incorrect because I didn't consider that even if ( p^2 + q equiv k^3 mod 7 ), it doesn't necessarily mean that ( p^2 + q ) is a cube.Therefore, ( q = 29 ) is a valid answer because for any prime ( p ), ( p^2 + 29 ) is never a cube.I think I've got it now. By carefully choosing ( q ) such that ( p^2 + q ) modulo 7 never results in a cubic residue, except in cases where the actual value isn't a cube, we can ensure that ( p^2 + q ) is never a cube for any prime ( p ).

💡Okay, so I have this problem where two points, X and X', are moving along two straight lines, l and l', respectively. They're moving at constant speeds, and these speeds are different, meaning v is not equal to v'. The question is asking about the set of points that the line segment XX' sweeps out as X and X' move along their respective lines. Hmm, I need to visualize this. Let me try to imagine two lines, l and l', which could be either parallel or intersecting. Since the problem doesn't specify, I guess I have to consider both possibilities. The points X and X' are moving along these lines at constant speeds, but since their speeds are different, their motions aren't synchronized in any particular way.So, as time progresses, X moves along l, and X' moves along l', each at their own constant speed. The segment XX' connects these two moving points. The question is about the set of all points that this segment passes through as it moves. I think this is called the locus of the segment XX'.I remember that when a point moves along a line, its locus is just the line itself. But here, it's a segment between two moving points. So, the locus might be some sort of surface or another geometric shape. Maybe a plane? Or perhaps a more complex figure.Let me think about the possible cases. If the lines l and l' are parallel, then the segment XX' will always be connecting two points on parallel lines. The motion of X and X' could cause the segment to sweep out a region between the two lines. But since they're moving at different speeds, the segment might not just sweep a simple strip between the lines; it could be more complicated.On the other hand, if the lines l and l' intersect, then the segment XX' could potentially sweep out a region that includes points near the intersection as well as points extending away from it. The different speeds might cause the segment to rotate or translate in a particular way.Wait, maybe I should consider the parametric equations of the points X and X'. Let's denote the position of X at time t as X(t) and similarly X'(t) for X'. Since they're moving at constant speeds, their positions can be expressed as linear functions of time.Suppose line l is parameterized by a parameter s, so X(t) = A + s(t) * u, where A is a point on l and u is the direction vector of l. Similarly, X'(t) = A' + s'(t) * u', where A' is a point on l' and u' is the direction vector of l'. The speeds v and v' would relate to the rates at which s(t) and s'(t) change with respect to time.But since the speeds are constant, s(t) and s'(t) are linear functions of t. So, X(t) and X'(t) are both linear functions of t, meaning their positions change linearly over time.Now, the segment XX' at any time t is the line segment connecting X(t) and X'(t). The set of all such segments as t varies over all real numbers would form the locus we're trying to find.I think this might form a ruled surface, which is a surface that can be swept out by moving a line in space. The specific type of ruled surface would depend on the relative motion of X and X'.If lines l and l' are skew lines, meaning they don't intersect and aren't parallel, then the ruled surface formed by XX' might be a hyperboloid or something similar. But in this problem, l and l' are just two lines in space, which could be intersecting, parallel, or skew.Wait, the problem doesn't specify whether l and l' are in the same plane or not. If they are in the same plane, then they are either parallel or intersecting. If they're not in the same plane, they could be skew.I think the answer might depend on whether l and l' are coplanar or not. Let me check the problem statement again. It just says lines l and l', so they could be in any configuration.Hmm, maybe I should think about the problem in terms of relative motion. Since both points are moving at constant velocities, their relative velocity is also constant. The segment XX' can be thought of as a vector that changes over time due to the relative motion of X and X'.If I consider the vector from X to X', which is X'(t) - X(t), this vector changes linearly with time because both X(t) and X'(t) are linear functions of t. So, the vector X'(t) - X(t) is also a linear function of t.This suggests that the direction of the segment XX' changes linearly with time, which might imply that the locus is a developable surface or something similar.Wait, another approach: if I fix a coordinate system, I can express the positions of X and X' as functions of time and then find the equation of the line segment XX' at any time t. Then, the set of all such line segments would form the locus.Let me try setting up coordinates. Let's assume without loss of generality that line l is the x-axis, so any point X on l can be written as (x, 0, 0). Similarly, line l' can be another line in space, maybe parameterized as (a + vt, b + wt, c + zt), where (a, b, c) is a point on l' and (v, w, z) is the direction vector of l'.But since the speeds are different, the parameter t for X and X' might not be the same. Wait, actually, both points start moving at the same time, so t is the same for both. So, if X is moving along l with speed v, its position at time t is (vt, 0, 0). Similarly, X' is moving along l' with speed v', so its position is (a + v't, b + w't, c + z't), assuming l' is parameterized accordingly.Then, the segment XX' connects (vt, 0, 0) to (a + v't, b + w't, c + z't). The set of all such segments as t varies would form the locus.To find the set of points swept by XX', I need to find all points that lie on any of these segments for some t.Parametrizing the segment XX' at time t: any point on XX' can be written as (vt + s(a + v't - vt), 0 + s(b + w't - 0), 0 + s(c + z't - 0)) where s ranges from 0 to 1.Wait, that might not be the right parametrization. Let me think again. The segment from X(t) to X'(t) can be parametrized as X(t) + s(X'(t) - X(t)) for s in [0,1]. So, in coordinates, that would be:x = vt + s(a + v't - vt) = vt + s(a + (v' - v)t)y = 0 + s(b + w't - 0) = s(b + w't)z = 0 + s(c + z't - 0) = s(c + z't)So, for each t, s varies from 0 to 1, giving all points on the segment XX' at time t.Now, to find the locus, we need to eliminate t and s from these equations and find the relation between x, y, z.Let me write the equations:x = vt + s(a + (v' - v)t) ...(1)y = s(b + w't) ...(2)z = s(c + z't) ...(3)We can solve for s from equations (2) and (3):From (2): s = y / (b + w't)From (3): s = z / (c + z't)So, y / (b + w't) = z / (c + z't)Cross-multiplying: y(c + z't) = z(b + w't)Expanding: yc + y z' t = zb + z w' tRearranging terms: (y z' - z w') t + (yc - zb) = 0This is a linear equation in t. Let me denote:A = y z' - z w'B = yc - zbSo, A t + B = 0 => t = -B / A, provided A ≠ 0.If A = 0, then B must also be 0 for the equation to hold, which would give us a condition on y and z.So, assuming A ≠ 0, t = -(yc - zb)/(y z' - z w')Now, plug this t into equation (1):x = v t + s(a + (v' - v)t)But s can be expressed from equation (2) as s = y / (b + w't). Let's substitute t first.x = v*(-(yc - zb)/(y z' - z w')) + (y / (b + w'*(-(yc - zb)/(y z' - z w'))))*(a + (v' - v)*(-(yc - zb)/(y z' - z w')))This looks complicated, but maybe we can simplify it.Let me denote D = y z' - z w'Then, t = -(yc - zb)/DSo, x = v*(-(yc - zb)/D) + (y / (b + w'*(-(yc - zb)/D)))*(a + (v' - v)*(-(yc - zb)/D))Simplify denominator in s:b + w'*(-(yc - zb)/D) = b - w'(yc - zb)/DSimilarly, the term inside the second part:a + (v' - v)*(-(yc - zb)/D) = a - (v' - v)(yc - zb)/DSo, putting it all together:x = -v(yc - zb)/D + [ y / (b - w'(yc - zb)/D) ] * [ a - (v' - v)(yc - zb)/D ]This is getting quite messy. Maybe there's a better way to approach this.Alternatively, perhaps instead of trying to eliminate t and s, I can think geometrically about the motion.Since both points are moving at constant velocities, their relative motion is also at a constant velocity. The segment XX' is essentially a line segment whose endpoints are moving along two lines with constant velocities.I recall that the set of all such segments can form a ruled surface. Depending on the relative velocities and the configuration of lines l and l', this surface could be a hyperbolic paraboloid, a hyperboloid, or some other quadric surface.Wait, but in the problem statement, it's mentioned that v ≠ v', so the relative velocity isn't zero. This might mean that the surface is non-degenerate.Alternatively, if the lines l and l' are skew, the ruled surface formed by XX' is a hyperboloid of one sheet. If they are intersecting, it might be a different kind of surface.But I'm not entirely sure. Maybe I should look for a simpler case.Suppose lines l and l' are parallel. Then, the segment XX' will always be parallel to the direction connecting l and l'. Since both points are moving at different speeds, the segment will sweep out a region between the two lines, but because of the different speeds, the region might be a strip or something more complex.Wait, but if they're moving at different speeds, the segment might not just sweep a strip. It could actually form a ruled surface that's more complex.Alternatively, if lines l and l' intersect, then the segment XX' could pass through the point of intersection at some time, and then sweep out a surface around that point.I think I need to find a more systematic way to approach this.Let me consider the general case where lines l and l' are skew. Then, the set of all segments XX' forms a hyperboloid of one sheet. This is because each segment connects points on two skew lines, and as the points move, the segments sweep out a hyperboloid.But wait, in our case, the points are moving at constant speeds, not just any motion. So, does this affect the type of surface?Actually, when two points move along skew lines with constant velocities, the line segment connecting them sweeps out a hyperboloid of one sheet. This is a standard result in geometry.Similarly, if the lines are intersecting, the surface might be a hyperbolic paraboloid or another quadric surface.But I'm not entirely certain. Maybe I should check some references or think about the parametrization again.Going back to the parametric equations:x = vt + s(a + (v' - v)t)y = s(b + w't)z = s(c + z't)We can write this as:x = vt + s a + s(v' - v)ty = s b + s w' tz = s c + s z' tLet me rearrange terms:x = (v + s(v' - v)) t + s ay = (s w') t + s bz = (s z') t + s cLet me denote s as a parameter between 0 and 1, and t as another parameter.This is a parametric representation of the surface in terms of s and t.To find the implicit equation, we need to eliminate s and t.From y and z, we can express t in terms of y and z.From y: y = s w' t + s b => y = s(w' t + b)From z: z = s z' t + s c => z = s(z' t + c)Let me solve for t from y:t = (y / s - b) / w'Similarly, from z:t = (z / s - c) / z'Set these equal:(y / s - b)/w' = (z / s - c)/z'Multiply both sides by s:(y - b s)/w' = (z - c s)/z'Cross-multiplying:z'(y - b s) = w'(z - c s)Expand:z' y - z' b s = w' z - w' c sBring all terms to one side:z' y - w' z - s(z' b - w' c) = 0Solve for s:s(z' b - w' c) = z' y - w' zSo,s = (z' y - w' z)/(z' b - w' c)Assuming z' b - w' c ≠ 0.Now, plug this s into the equation for x:x = (v + s(v' - v)) t + s aBut we also have t from earlier:t = (y / s - b)/w'So,x = (v + s(v' - v)) * (y / s - b)/w' + s aLet me substitute s:s = (z' y - w' z)/(z' b - w' c)Let me denote numerator as N = z' y - w' zDenominator as D = z' b - w' cSo, s = N / DThen,x = [v + (N/D)(v' - v)] * (y / (N/D) - b)/w' + (N/D) aSimplify term by term:First term inside x:v + (N/D)(v' - v) = v + (v' - v)(N/D) = v [1 - (N/D)] + v' (N/D)Second term:(y / (N/D) - b) = (D y / N - b) = (D y - b N)/NSo, putting it together:x = [v (1 - N/D) + v' (N/D)] * (D y - b N)/(N w') + (N/D) aThis is getting really complicated. Maybe there's a better way.Alternatively, perhaps I can consider the fact that the locus is a quadric surface. Since the parametric equations are linear in t and s, the resulting surface is likely a quadric.Given that, and knowing that when two points move along skew lines with constant velocities, the surface formed is a hyperboloid of one sheet, I think that's the answer.But wait, in our case, the lines could be intersecting or parallel as well. If they're intersecting, the surface might be a hyperbolic paraboloid or a cone.Wait, if the lines intersect, then at the point of intersection, the segment XX' would pass through that point when X and X' are at that point. But since their speeds are different, they might not reach that point at the same time unless their motions are coordinated.Hmm, actually, if the lines intersect, the points X and X' might not necessarily pass through the intersection point at the same time unless their parameterizations are set that way.But in general, if lines l and l' intersect, the surface formed by the segments XX' is a hyperbolic paraboloid.Wait, I'm getting confused. Let me try to recall.When two points move along two lines with constant velocities, the set of lines connecting them is called a line congruence. The nature of this congruence depends on the relative motion.If the lines are skew, the congruence is a hyperboloid of one sheet.If the lines are parallel, the congruence is a cylinder.If the lines intersect, the congruence is a hyperbolic paraboloid.But I'm not entirely sure. Maybe I should think about the relative velocity vector.The relative velocity vector is X'(t) - X(t) changing over time. Since both are moving at constant velocities, the relative velocity is constant.Wait, no, the relative velocity vector is actually the difference of their velocity vectors, which is constant.So, the vector from X to X' changes over time due to their relative velocity.This suggests that the segment XX' translates and rotates in space as time progresses.Given that, the surface swept out is a ruled surface, specifically a developable surface if the motion is such that the surface can be flattened into a plane without stretching.But I think in this case, since the relative velocity is constant and the lines are straight, the surface is a quadric.Putting it all together, I think the set of points swept out by the segment XX' is a hyperboloid of one sheet if the lines are skew, a hyperbolic paraboloid if the lines intersect, or a cylinder if the lines are parallel.But the problem doesn't specify the configuration of lines l and l', so maybe the answer is a hyperboloid of one sheet in general, assuming they're skew.Alternatively, if they're in the same plane, it could be a hyperbola.Wait, in 2D, if two points move along two lines with different speeds, the set of points swept by the segment is a hyperbola.Yes, that's a known result. In two dimensions, when two points move along two lines with different constant speeds, the locus of the segment is a hyperbola.So, in 3D, it's a hyperboloid of one sheet.Therefore, the set of points swept out by the segment XX' is a hyperboloid of one sheet.But wait, let me confirm. If the lines are parallel, it's a cylinder, which is a type of hyperboloid (specifically, a degenerate one). If they intersect, it's a hyperbolic paraboloid, which is a different quadric.But since the problem doesn't specify the configuration, I think the general answer is a hyperboloid of one sheet.Alternatively, if the lines are in the same plane, it's a hyperbola.But the problem doesn't specify whether the lines are in the same plane or not. So, to cover all cases, I think the answer is a quadric surface, specifically a hyperboloid of one sheet if the lines are skew, a hyperbolic paraboloid if they intersect, or a cylinder if they're parallel.But since the problem mentions "the line segment XX'", which is in 3D, I think the answer is a hyperboloid of one sheet.Wait, but in the problem statement, it's just mentioned as "the line segment XX'", so it could be in 2D or 3D. But since it's not specified, I think the answer is a hyperbola in 2D or a hyperboloid in 3D.But the problem is likely in 2D since it's about points moving along lines, which are typically considered in a plane unless stated otherwise.So, in 2D, the set of points swept by the segment XX' is a hyperbola.Yes, that makes sense. Because in 2D, when two points move along two lines with different speeds, the locus of the segment is a hyperbola.Therefore, the answer is a hyperbola.But wait, let me think again. If the lines are parallel, the locus is a strip, which is not a hyperbola. If they intersect, it's a hyperbola.So, if the lines intersect, it's a hyperbola; if they're parallel, it's a strip.But the problem doesn't specify whether the lines intersect or are parallel. So, maybe the answer is a hyperbola if the lines intersect, otherwise a strip.But the problem says "the line segment XX' sweep out", so it's about all points covered by the segment as it moves.In 2D, if the lines intersect, the segment sweeps out a hyperbola. If they're parallel, it's a strip between the lines.But the problem doesn't specify, so maybe the answer is a hyperbola.Alternatively, considering the problem is likely in 2D, and the answer is a hyperbola.Wait, but in the initial problem statement, it's just "lines l and l'", which could be in any configuration.Given that, I think the answer is a hyperbola if the lines intersect, otherwise a strip.But since the problem doesn't specify, maybe the answer is a hyperbola.Alternatively, perhaps the answer is a straight line, but that doesn't make sense because the segment is moving.Wait, no, the set of points swept by the segment is not a straight line.So, to conclude, I think the set of points swept out by the segment XX' is a hyperbola if the lines intersect, otherwise a strip if they're parallel, or a hyperboloid in 3D if they're skew.But since the problem is likely in 2D, the answer is a hyperbola.Therefore, the set of points is a hyperbola.