💡First, I need to find the conjugate of the complex number ( z = -1 - i ). The conjugate of a complex number ( z = a + bi ) is ( overline{z} = a - bi ). Therefore, ( overline{z} = -1 + i ).Next, I need to compute ( (1 - z) cdot overline{z} ). Substituting the value of ( z ), we have:[1 - z = 1 - (-1 - i) = 1 + 1 + i = 2 + i]Now, multiply ( (2 + i) ) by ( overline{z} = -1 + i ):[(2 + i) cdot (-1 + i) = 2 cdot (-1) + 2 cdot i + i cdot (-1) + i cdot i = -2 + 2i - i + i^2]Since ( i^2 = -1 ), this simplifies to:[-2 + 2i - i - 1 = -3 + i]Finally, to find the magnitude of ( -3 + i ), we use the formula for the magnitude of a complex number ( a + bi ):[|a + bi| = sqrt{a^2 + b^2}]Substituting ( a = -3 ) and ( b = 1 ):[|-3 + i| = sqrt{(-3)^2 + 1^2} = sqrt{9 + 1} = sqrt{10}]

💡Okay, so I have this equation: 1/2022 = 1/x + 1/y. I need to find how many integer solutions exist for x and y. Hmm, let me think about how to approach this.First, I remember that when dealing with equations involving reciprocals, it's often helpful to combine the fractions. So, maybe I can rewrite the equation to have a common denominator. Let's see:1/2022 = (y + x) / (xy)So, cross-multiplying both sides to eliminate the denominators, I get:xy = 2022(y + x)Hmm, that seems manageable. Let me rearrange this equation to bring all terms to one side:xy - 2022x - 2022y = 0Now, I notice that this looks similar to something I've seen before, where you can factor it by adding a term to both sides. Maybe I can add 2022 squared to both sides to complete the rectangle or something. Let me try that:xy - 2022x - 2022y + 2022^2 = 2022^2Now, the left side looks like it can be factored. Let me check:(x - 2022)(y - 2022) = 2022^2Yes, that works! So, now the equation is transformed into a product of two terms equal to 2022 squared. This means that (x - 2022) and (y - 2022) are factors of 2022 squared.So, to find integer solutions, I need to find all pairs of integers (a, b) such that a * b = 2022^2. Each such pair will correspond to a solution (x, y) where x = a + 2022 and y = b + 2022.Therefore, the number of integer solutions is equal to the number of factor pairs of 2022 squared.Now, I need to find how many factors 2022 squared has. First, let's factorize 2022.2022 is an even number, so divisible by 2: 2022 / 2 = 1011.1011: Let's check divisibility by 3. 1 + 0 + 1 + 1 = 3, which is divisible by 3. So, 1011 / 3 = 337.Now, 337 is a prime number because it doesn't divide by any primes up to its square root (which is around 18.35). Checking primes like 2, 3, 5, 7, 11, 13, 17: none divide 337.So, the prime factorization of 2022 is 2 * 3 * 337.Therefore, 2022 squared is (2 * 3 * 337)^2 = 2^2 * 3^2 * 337^2.The number of factors of a number is found by adding 1 to each of the exponents in its prime factorization and then multiplying those together.So, for 2022 squared, the exponents are 2, 2, and 2 for primes 2, 3, and 337 respectively.Thus, the number of factors is (2 + 1) * (2 + 1) * (2 + 1) = 3 * 3 * 3 = 27.Each factor corresponds to a pair (a, b) such that a * b = 2022^2. However, since a and b are ordered pairs (meaning (a, b) is different from (b, a) unless a = b), the number of ordered pairs is 27 * 2 = 54.But wait, I need to be careful here. If 2022 squared is a perfect square, then there is a middle factor where a = b. In this case, 2022 squared is indeed a perfect square, so there is one pair where a = b = 2022.Therefore, the total number of ordered pairs is 27 * 2 - 1 = 53. Wait, no, actually, since each factor less than the square root pairs with a factor greater than the square root, except for the square root itself which pairs with itself.So, the number of unordered pairs is (number of factors + 1)/2 if the number is a perfect square. But since we are considering ordered pairs, it's actually 2 * (number of factors / 2) if the number is a perfect square.Wait, maybe I'm overcomplicating it. Let me think again.Each factor a of 2022 squared gives a unique factor b = 2022 squared / a. So, for each of the 27 factors, there is a corresponding pair (a, b). However, since (a, b) and (b, a) are considered different ordered pairs unless a = b, the total number of ordered pairs is 27 * 2 = 54.But wait, in our case, since 2022 squared is a perfect square, one of these pairs is (2022, 2022). So, in the 54 ordered pairs, this is only counted once. Therefore, the total number of ordered pairs is indeed 54.However, we need to check if all these pairs lead to valid integer solutions for x and y. Specifically, we need to ensure that x and y are integers and not equal to zero because division by zero is undefined.Looking back at our substitution, x = a + 2022 and y = b + 2022. Since a and b are factors of 2022 squared, which is positive, a and b can be positive or negative. Therefore, x and y can be positive or negative integers, but we need to ensure that x and y are not zero.Wait, but if a = -2022, then x = a + 2022 = 0, which is invalid. Similarly, if b = -2022, y = 0, which is invalid. So, we need to exclude the cases where a = -2022 or b = -2022.But in our factorization, we considered all factors, both positive and negative. So, how many of these 54 ordered pairs result in x or y being zero?Let's see: For x = 0, we need a = -2022. Similarly, for y = 0, we need b = -2022. So, how many such pairs are there?In the factor pairs, (a, b) = (-2022, -2022) would give x = 0 and y = 0, but actually, 2022 squared is positive, so (-2022, -2022) is a valid factor pair because (-2022)*(-2022) = 2022 squared.Wait, but in our earlier step, we have (x - 2022)(y - 2022) = 2022 squared, which is positive. Therefore, both (x - 2022) and (y - 2022) must be both positive or both negative.So, if (x - 2022) is positive, then (y - 2022) is positive, meaning x > 2022 and y > 2022.If (x - 2022) is negative, then (y - 2022) is negative, meaning x < 2022 and y < 2022.But in both cases, x and y cannot be zero because if x were zero, then (x - 2022) would be -2022, which is allowed, but y would be (2022^2)/(-2022) + 2022 = (-2022) + 2022 = 0, which is invalid. Similarly, if y were zero, x would be zero.Therefore, the pair where a = -2022 and b = -2022 leads to x = 0 and y = 0, which is invalid. So, we need to exclude this particular pair.How many such invalid pairs are there? Only one: (a, b) = (-2022, -2022). Therefore, out of the 54 ordered pairs, one leads to x = 0 and y = 0, which is invalid.Therefore, the total number of valid integer solutions is 54 - 1 = 53.Wait, but let me double-check. Are there any other pairs where x or y could be zero?Suppose a = -2022, then x = a + 2022 = 0, and b = 2022^2 / a = 2022^2 / (-2022) = -2022, so y = b + 2022 = 0. So, only this pair leads to x = 0 and y = 0.Similarly, if a = 2022, then x = 2022 + 2022 = 4044, and b = 2022^2 / 2022 = 2022, so y = 2022 + 2022 = 4044. This is a valid solution because both x and y are non-zero.Therefore, the only invalid solution is when a = -2022 and b = -2022, leading to x = 0 and y = 0. So, we subtract one from the total.Hence, the total number of integer solutions is 54 - 1 = 53.Wait, but let me think again. Are there any other cases where x or y could be zero? For example, if a = something else, could x or y be zero?Suppose a = something else, say a = 1, then x = 1 + 2022 = 2023, which is fine. Similarly, a = -1, then x = -1 + 2022 = 2021, which is also fine. So, only when a = -2022 do we get x = 0, which is invalid.Therefore, only one pair leads to invalid solutions, so 54 - 1 = 53.But wait, another thought: when we consider all factor pairs, including both positive and negative, do we have to consider that some pairs might lead to x or y being zero? But as we saw, only the pair where a = -2022 and b = -2022 leads to x = 0 and y = 0. All other pairs, even if a or b is negative, will result in x and y being non-zero because:If a is negative but not equal to -2022, then x = a + 2022 is not zero. Similarly, b = 2022^2 / a, which would also not be -2022 unless a = -2022.Therefore, only one pair leads to x and y being zero, so we subtract one.Hence, the total number of integer solutions is 53.Wait, but let me confirm the number of factor pairs. 2022 squared has 27 factors, so 27 positive factors and 27 negative factors, making a total of 54 factors. Each factor a corresponds to a unique factor b = 2022^2 / a. Therefore, the number of ordered pairs is 54.But since (a, b) and (b, a) are considered different unless a = b, and since 2022 squared is a perfect square, there is one pair where a = b = 2022. So, in the 54 ordered pairs, 26 pairs are where a ≠ b, each contributing two ordered pairs, and one pair where a = b.But regardless, the total is 54 ordered pairs. Excluding the one pair where a = -2022 and b = -2022, which leads to x = 0 and y = 0, we have 53 valid solutions.Therefore, the number of integer solutions is 53.Wait, but I'm a bit confused because sometimes when dealing with equations, we might have to consider positive and negative solutions separately. Let me think about it again.Given that (x - 2022)(y - 2022) = 2022^2, which is positive, so both (x - 2022) and (y - 2022) must be both positive or both negative.Case 1: Both positive.In this case, x > 2022 and y > 2022. The number of such solutions is equal to the number of positive factor pairs of 2022 squared, which is 27. But since we are considering ordered pairs, it's 27 * 2 = 54, but wait, no. Actually, for each positive factor a, there is a corresponding positive factor b = 2022^2 / a. So, the number of positive ordered pairs is equal to the number of positive factors, which is 27.Wait, no, actually, for each factor a, there is a unique b, so the number of ordered pairs is equal to the number of factors, which is 27. But since we are considering ordered pairs, it's actually 27 ordered pairs where a and b are positive.Similarly, for negative factors, since (x - 2022) and (y - 2022) are both negative, we have (x - 2022) = -a and (y - 2022) = -b, where a and b are positive factors of 2022 squared. Therefore, the number of negative ordered pairs is also 27.So, in total, we have 27 positive ordered pairs and 27 negative ordered pairs, making 54 ordered pairs.But in the negative ordered pairs, one of them is when a = 2022, so (x - 2022) = -2022, leading to x = 0, which is invalid. Similarly, y would also be 0. So, only one of the 27 negative ordered pairs is invalid.Therefore, the total number of valid solutions is 54 - 1 = 53.Yes, that makes sense. So, the number of integer solutions is 53.Wait, but let me think about it differently. Suppose I consider all possible factor pairs, both positive and negative, and count the number of solutions where x and y are integers, excluding x = 0 and y = 0.Given that (x - 2022)(y - 2022) = 2022^2, and 2022^2 is positive, so both (x - 2022) and (y - 2022) must be both positive or both negative.Case 1: Both positive.Here, x > 2022 and y > 2022. The number of such solutions is equal to the number of positive factor pairs of 2022^2, which is 27. Each factor a of 2022^2 gives a unique solution (x, y) = (a + 2022, (2022^2)/a + 2022). Since we are considering ordered pairs, each factor a corresponds to a unique ordered pair, so there are 27 solutions in this case.Case 2: Both negative.Here, x < 2022 and y < 2022. Let (x - 2022) = -a and (y - 2022) = -b, where a and b are positive factors of 2022^2. So, x = 2022 - a and y = 2022 - b. The number of such solutions is also 27, as each positive factor a gives a unique solution.However, in this case, we need to ensure that x and y are not zero. So, if a = 2022, then x = 2022 - 2022 = 0, which is invalid. Similarly, if b = 2022, y = 0, which is invalid. But since a and b are factors of 2022^2, and 2022 is a factor, there is exactly one such pair where a = 2022 and b = 2022, leading to x = 0 and y = 0.Therefore, in the negative case, we have 27 solutions, but one of them is invalid, leaving us with 26 valid solutions.Therefore, the total number of valid solutions is 27 (positive case) + 26 (negative case) = 53.Yes, that confirms the earlier result. So, the number of integer solutions is 53.I think I've covered all the cases and checked for any possible overlaps or invalid solutions. Therefore, I'm confident that the answer is 53.

💡First, I recognize that if two vectors are parallel, one must be a scalar multiple of the other. This means that the components of one vector are proportional to the components of the other vector.Given the vectors (overrightarrow{a} = (2, 3)) and (overrightarrow{b} = (4, y + 1)), I can set up a proportion based on their corresponding components. Specifically, the ratio of the first components should be equal to the ratio of the second components.So, I write the proportion as:[frac{2}{4} = frac{3}{y + 1}]Next, I simplify the fraction on the left side:[frac{2}{4} = frac{1}{2}]Now, the proportion becomes:[frac{1}{2} = frac{3}{y + 1}]To solve for (y), I cross-multiply:[1 times (y + 1) = 2 times 3][y + 1 = 6]Finally, I subtract 1 from both sides to isolate (y):[y = 6 - 1][y = 5]Therefore, the value of (y) that makes the vectors (overrightarrow{a}) and (overrightarrow{b}) parallel is 5.

💡Okay, so I have this problem here where I need to find the value of f(8) given some conditions about the function f(x). Let me try to break it down step by step.First, the problem says that f(x) is a monotonous function on the interval (0, +∞). Monotonous means it's either always increasing or always decreasing. That's good to know because it tells me that f(x) doesn't have any peaks or valleys; it's consistently going up or down. This property might help later on.Next, the problem states that for any real number x in (0, +∞), the equation f[f(x) - log₂x - 1] = 2 holds true. Hmm, that's a bit complex. Let me try to parse it.So, inside the function f, we have another expression: f(x) - log₂x - 1. Let's call this expression something, maybe t. So, t = f(x) - log₂x - 1. Then, the equation becomes f(t) = 2. So, no matter what x we pick, when we plug t into f, we always get 2.Since f is monotonous, it's either strictly increasing or strictly decreasing. If f is strictly increasing, then if t1 < t2, f(t1) < f(t2). Similarly, if it's decreasing, then t1 < t2 implies f(t1) > f(t2). But in our case, f(t) is always equal to 2, regardless of x. That suggests that t must be a constant because if t varied, f(t) would vary as well, but it's given that f(t) is always 2.So, if t is a constant, let's denote it as c. That means f(x) - log₂x - 1 = c for all x in (0, +∞). Therefore, f(x) = log₂x + c + 1.Now, since t = c, and f(t) = 2, we can substitute t into f. So, f(c) = 2. But f(c) is also equal to log₂c + c + 1, based on our earlier expression for f(x). So, we have:log₂c + c + 1 = 2Let me write that equation down:log₂c + c + 1 = 2Simplify it:log₂c + c = 1Hmm, this is an equation in terms of c. Let me try to solve for c.Let me denote log₂c as y. Then, c = 2^y. So, substituting back into the equation:y + 2^y = 1So, we have:y + 2^y = 1This is a transcendental equation, meaning it can't be solved with simple algebraic manipulations. I might need to solve it numerically or see if there's an integer solution.Let me test y = 0:0 + 2^0 = 0 + 1 = 1Hey, that works! So, y = 0 is a solution.Is there another solution? Let's check y = 1:1 + 2^1 = 1 + 2 = 3 ≠ 1Too big.What about y = -1:-1 + 2^{-1} = -1 + 0.5 = -0.5 ≠ 1Too small.So, y = 0 is the only solution. Therefore, c = 2^y = 2^0 = 1.So, c = 1.Therefore, f(x) = log₂x + c + 1 = log₂x + 1 + 1 = log₂x + 2.So, f(x) = log₂x + 2.Now, the problem asks for f(8). Let's compute that.f(8) = log₂8 + 2I know that log₂8 is 3 because 2^3 = 8.So, f(8) = 3 + 2 = 5.Therefore, the answer is 5.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I assumed f is monotonous, so f(t) = 2 implies t is constant because f is injective (since it's monotonic). That seems correct.2. I set t = f(x) - log₂x - 1 = c, so f(x) = log₂x + c + 1. Then, since f(c) = 2, I substituted and got log₂c + c + 1 = 2, leading to log₂c + c = 1.3. I solved for c by setting y = log₂c, so c = 2^y, leading to y + 2^y = 1. Testing y=0 gives 0 + 1 = 1, which works. So c=1.4. Thus, f(x) = log₂x + 2. Then, f(8) = log₂8 + 2 = 3 + 2 = 5.Everything seems to check out. So, I think the answer is 5.**Final Answer**boxed{5}

💡Alright, so I have this equation: (2^x + 5^y - 31^z = n!). I need to prove that there are only finitely many non-negative integer solutions for (x, y, z, n). Hmm, okay. Let me try to break this down.First, I know that (n!) grows very rapidly as (n) increases. On the other hand, (2^x), (5^y), and (31^z) are exponential functions as well, but with different bases. So maybe for large (n), (n!) will outpace the sum (2^x + 5^y - 31^z), making the equation impossible? That might be a direction to explore.Let me consider different cases based on the value of (n). Maybe starting with small (n) and then moving to larger (n).**Case 1: Small values of (n)**Let's start with (n = 0). Then (n! = 0! = 1). So the equation becomes:[2^x + 5^y - 31^z = 1]I need to find non-negative integers (x, y, z) satisfying this.Looking at modulo 2: (2^x) is 0 mod 2, (5^y) is 1 mod 2, and (31^z) is 1 mod 2. So:[0 + 1 - 1 equiv 0 pmod{2}]Which is consistent with the right-hand side being 1. Hmm, not much help.Maybe modulo 3? Let's see:(2^x) mod 3 cycles every 2: 2, 1, 2, 1,...(5^y) mod 3 is 2^y, which cycles every 2: 2, 1, 2, 1,...(31^z) mod 3 is 1^z = 1.So:[2^x + 5^y - 31^z equiv 2^x + 2^y - 1 pmod{3}]This should be congruent to 1 mod 3 (since RHS is 1). So:[2^x + 2^y - 1 equiv 1 pmod{3}]Which simplifies to:[2^x + 2^y equiv 2 pmod{3}]So either both (2^x) and (2^y) are 1 mod 3, or one is 2 and the other is 0, but since 2^k mod 3 is never 0, both must be 1 mod 3. That means (x) and (y) must both be even.So (x = 2a), (y = 2b). Then the equation becomes:[2^{2a} + 5^{2b} - 31^z = 1]Which is:[4^a + 25^b - 31^z = 1]Hmm, this seems tricky. Maybe trying small values.If (a = 0), then (4^0 = 1), so:[1 + 25^b - 31^z = 1]Which simplifies to:[25^b = 31^z]Since 25 and 31 are coprime, the only solution is (b = z = 0). So (x = 0), (y = 0), (z = 0). That gives a solution: (0, 0, 0, 0).If (a = 1), then (4^1 = 4):[4 + 25^b - 31^z = 1]So:[25^b - 31^z = -3]Looking for non-negative integers (b, z). Let's try (b = 0): (1 - 31^z = -3) → (31^z = 4). No solution. (b = 1): (25 - 31^z = -3) → (31^z = 28). No solution. (b = 2): (625 - 31^z = -3) → (31^z = 628). 31^2 = 961, which is too big. So no solution here.Similarly, (a = 2): (16 + 25^b - 31^z = 1) → (25^b - 31^z = -15). Trying (b = 1): (25 - 31^z = -15) → (31^z = 40). No solution. (b = 2): (625 - 31^z = -15) → (31^z = 640). 31^2 = 961, too big. So no solution.It seems like for (n = 0), the only solution is (0, 0, 0, 0).**Case 2: (n = 1)**Then (n! = 1). So same equation as (n = 0):[2^x + 5^y - 31^z = 1]We already saw that the only solution is (0, 0, 0, 1).**Case 3: (n = 2)**Then (n! = 2). So:[2^x + 5^y - 31^z = 2]Again, let's try modulo 2:(2^x) is 0 mod 2, (5^y) is 1 mod 2, (31^z) is 1 mod 2. So:[0 + 1 - 1 equiv 0 pmod{2}]Which is consistent with RHS being 2, which is 0 mod 2.Modulo 3:(2^x) cycles 2,1; (5^y) cycles 2,1; (31^z) is 1 mod 3.So:[2^x + 2^y - 1 equiv 2 pmod{3}]Which simplifies to:[2^x + 2^y equiv 0 pmod{3}]So either both (2^x) and (2^y) are 0 mod 3, which they never are, or one is 1 and the other is 2. But since (2^k) mod 3 cycles between 2 and 1, this means one exponent is even and the other is odd.So either (x) even and (y) odd, or vice versa.Let me try small values.If (x = 0): (1 + 5^y - 31^z = 2) → (5^y - 31^z = 1). Let's see:If (y = 0): (1 - 31^z = 1) → (31^z = 0). No solution.If (y = 1): (5 - 31^z = 1) → (31^z = 4). No solution.If (y = 2): (25 - 31^z = 1) → (31^z = 24). No solution.If (y = 3): (125 - 31^z = 1) → (31^z = 124). 31^2 = 961, too big. So no solution.If (x = 1): (2 + 5^y - 31^z = 2) → (5^y - 31^z = 0). So (5^y = 31^z). Since 5 and 31 are primes, only solution is (y = z = 0). So (x = 1), (y = 0), (z = 0). That's a solution: (1, 0, 0, 2).If (x = 2): (4 + 5^y - 31^z = 2) → (5^y - 31^z = -2). Let's see:(y = 0): (1 - 31^z = -2) → (31^z = 3). No solution.(y = 1): (5 - 31^z = -2) → (31^z = 7). No solution.(y = 2): (25 - 31^z = -2) → (31^z = 27). No solution.(y = 3): (125 - 31^z = -2) → (31^z = 127). 31^2 = 961, too big. So no solution.If (x = 3): (8 + 5^y - 31^z = 2) → (5^y - 31^z = -6).(y = 0): (1 - 31^z = -6) → (31^z = 7). No solution.(y = 1): (5 - 31^z = -6) → (31^z = 11). No solution.(y = 2): (25 - 31^z = -6) → (31^z = 31). So (z = 1). Therefore, (x = 3), (y = 2), (z = 1). That's another solution: (3, 2, 1, 2).If (x = 4): (16 + 5^y - 31^z = 2) → (5^y - 31^z = -14).(y = 0): (1 - 31^z = -14) → (31^z = 15). No solution.(y = 1): (5 - 31^z = -14) → (31^z = 19). No solution.(y = 2): (25 - 31^z = -14) → (31^z = 39). No solution.(y = 3): (125 - 31^z = -14) → (31^z = 139). 31^2 = 961, too big. So no solution.If (x = 5): (32 + 5^y - 31^z = 2) → (5^y - 31^z = -30).(y = 0): (1 - 31^z = -30) → (31^z = 31) → (z = 1). So (x = 5), (y = 0), (z = 1). That's a solution: (5, 0, 1, 2).(y = 1): (5 - 31^z = -30) → (31^z = 35). No solution.(y = 2): (25 - 31^z = -30) → (31^z = 55). No solution.(y = 3): (125 - 31^z = -30) → (31^z = 155). 31^2 = 961, too big. So no solution.It seems like for (n = 2), we have solutions: (1, 0, 0, 2), (3, 2, 1, 2), and (5, 0, 1, 2).**Case 4: (n = 3)**Then (n! = 6). So:[2^x + 5^y - 31^z = 6]Modulo 2:(2^x) is 0 mod 2, (5^y) is 1 mod 2, (31^z) is 1 mod 2. So:[0 + 1 - 1 equiv 0 pmod{2}]Which is consistent with RHS being 6, which is 0 mod 2.Modulo 3:(2^x) cycles 2,1; (5^y) cycles 2,1; (31^z) is 1 mod 3.So:[2^x + 2^y - 1 equiv 0 pmod{3}]Which simplifies to:[2^x + 2^y equiv 1 pmod{3}]So either one of (2^x) or (2^y) is 1 and the other is 0, but since they can't be 0 mod 3, both must be 1 mod 3. So both (x) and (y) must be even.So (x = 2a), (y = 2b). Then the equation becomes:[4^a + 25^b - 31^z = 6]Let me try small values.If (a = 0): (1 + 25^b - 31^z = 6) → (25^b - 31^z = 5).(b = 1): (25 - 31^z = 5) → (31^z = 20). No solution.(b = 2): (625 - 31^z = 5) → (31^z = 620). 31^2 = 961, too big. So no solution.If (a = 1): (4 + 25^b - 31^z = 6) → (25^b - 31^z = 2).(b = 1): (25 - 31^z = 2) → (31^z = 23). No solution.(b = 2): (625 - 31^z = 2) → (31^z = 623). 31^2 = 961, too big. So no solution.If (a = 2): (16 + 25^b - 31^z = 6) → (25^b - 31^z = -10).(b = 1): (25 - 31^z = -10) → (31^z = 35). No solution.(b = 2): (625 - 31^z = -10) → (31^z = 635). 31^2 = 961, too big. So no solution.If (a = 3): (64 + 25^b - 31^z = 6) → (25^b - 31^z = -58).(b = 1): (25 - 31^z = -58) → (31^z = 83). No solution.(b = 2): (625 - 31^z = -58) → (31^z = 683). 31^2 = 961, too big. So no solution.Hmm, maybe trying different approach. Let's consider modulo 5.Original equation:[2^x + 5^y - 31^z = 6]Modulo 5:(2^x) mod 5 cycles every 4: 2,4,3,1,...(5^y) is 0 mod 5.(31^z) mod 5 is 1^z = 1.So:[2^x + 0 - 1 equiv 6 pmod{5}]Which simplifies to:[2^x - 1 equiv 1 pmod{5}]So:[2^x equiv 2 pmod{5}]Which means (x equiv 1 pmod{4}). So (x = 4k + 1).So let's set (x = 1): Then equation becomes:[2 + 5^y - 31^z = 6) → (5^y - 31^z = 4).Let me see:(y = 1): (5 - 31^z = 4) → (31^z = 1) → (z = 0). So solution: (1, 1, 0, 3).(y = 2): (25 - 31^z = 4) → (31^z = 21). No solution.(y = 3): (125 - 31^z = 4) → (31^z = 121). 31^2 = 961, too big. So no solution.If (x = 5): (32 + 5^y - 31^z = 6) → (5^y - 31^z = -26).(y = 1): (5 - 31^z = -26) → (31^z = 31) → (z = 1). So solution: (5, 1, 1, 3).(y = 2): (25 - 31^z = -26) → (31^z = 51). No solution.(y = 3): (125 - 31^z = -26) → (31^z = 151). 31^2 = 961, too big. So no solution.If (x = 9): (512 + 5^y - 31^z = 6) → (5^y - 31^z = -506). This seems too big, but let's check:(y = 3): (125 - 31^z = -506) → (31^z = 631). 31^2 = 961, too big.(y = 4): (625 - 31^z = -506) → (31^z = 1131). 31^2 = 961, 31^3 = 29791, way too big. So no solution.So for (n = 3), solutions are: (1, 1, 0, 3) and (5, 1, 1, 3).**Case 5: (n geq 4)**Now, (n! geq 24). Let's see if the equation can hold.First, modulo 3:[2^x + 5^y - 31^z equiv 2^x + 2^y - 1 pmod{3}]This should be congruent to (n! pmod{3}). For (n geq 3), (n! equiv 0 pmod{3}). So:[2^x + 2^y - 1 equiv 0 pmod{3}]Which simplifies to:[2^x + 2^y equiv 1 pmod{3}]So either one of (2^x) or (2^y) is 1 and the other is 0, but since they can't be 0 mod 3, both must be 1 mod 3. So both (x) and (y) must be even.So (x = 2a), (y = 2b). Then the equation becomes:[4^a + 25^b - 31^z = n!]Now, modulo 5:[4^a + 25^b - 31^z equiv 4^a + 0 - 1 pmod{5}]Which simplifies to:[4^a - 1 equiv n! pmod{5}]For (n geq 5), (n! equiv 0 pmod{5}). So:[4^a - 1 equiv 0 pmod{5}]Which means:[4^a equiv 1 pmod{5}]Since 4 mod 5 is 4, and 4^2 = 16 ≡ 1 mod 5. So (a) must be even. Let (a = 2c). Then (x = 4c).So now, the equation is:[16^c + 25^b - 31^z = n!]This is getting complicated. Maybe consider modulo 7.[16^c + 25^b - 31^z equiv (2^c) + (4^b) - (3^z) pmod{7}]Because 16 ≡ 2 mod 7, 25 ≡ 4 mod 7, 31 ≡ 3 mod 7.So:[2^c + 4^b - 3^z equiv n! pmod{7}]For (n geq 7), (n! equiv 0 pmod{7}). So:[2^c + 4^b - 3^z equiv 0 pmod{7}]This is getting too involved. Maybe instead, consider the growth rates.For large (n), (n!) grows much faster than (2^x + 5^y - 31^z). So beyond a certain (n), the equation can't hold. But I need to make this precise.Alternatively, use bounds. Suppose (n geq 7). Then (n! geq 5040). Let's see if (2^x + 5^y - 31^z) can reach 5040.But (2^x) and (5^y) are positive, and (31^z) is positive. So (2^x + 5^y) must be greater than (31^z + 5040). But (31^z) grows exponentially as well. For (z = 4), (31^4 = 923521), which is way bigger than 5040. So maybe for (z geq 4), (31^z) is too big.Wait, but (2^x + 5^y) needs to be bigger than (31^z + n!). If (z) is small, say (z = 0), then (31^0 = 1), so equation becomes (2^x + 5^y - 1 = n!). For (n geq 7), (n! geq 5040), so (2^x + 5^y = n! + 1). But (2^x + 5^y) is much smaller than (n!) for large (n). For example, (2^x + 5^y leq 2^{x} + 5^{y}), but (n!) is factorial, which grows faster than any exponential function.Similarly, for (z = 1), (31^1 = 31), so (2^x + 5^y = n! + 31). Again, for large (n), (n! + 31) is way bigger than (2^x + 5^y).For (z = 2), (31^2 = 961), so (2^x + 5^y = n! + 961). Still, (n!) dominates.For (z = 3), (31^3 = 29791), so (2^x + 5^y = n! + 29791). Again, (n!) is way bigger.So for (z geq 0), (n!) is too big for (2^x + 5^y) to catch up when (n) is large. Therefore, there must be only finitely many solutions.To make this rigorous, I can argue that for (n) sufficiently large, (n!) exceeds any possible value of (2^x + 5^y - 31^z). Since (n!) grows faster than exponential functions, beyond a certain (n), the equation can't hold. Therefore, only finitely many (n) can satisfy the equation, and for each such (n), there are only finitely many (x, y, z) because exponentials grow quickly and can't be too large without making the left side too big.So putting it all together, we've checked small (n) and found specific solutions, and for large (n), the equation can't hold because (n!) is too big. Therefore, there are only finitely many solutions.

💡First, identify the proportions given: 3, 5, and 7/2. To simplify the ratio, multiply each term by 2 to eliminate the fraction, resulting in 6, 10, and 7. The sum of these terms is 23. Next, determine the value of each part by dividing the total amount, 150, by the sum of the ratio terms: 150 ÷ 23 ≈ 6.5217. Finally, calculate the smallest part by multiplying this value by the smallest ratio term: 6.5217 × 6 ≈ 39.13. Since none of the provided answer choices match this result, the correct answer is E) None of these answers.