🧠:<think>Okay, so I have this problem where I need to find the possible different denominators when expressing the repeating decimal (0.overline{abc}) as a fraction in its lowest terms. Here, (a), (b), and (c) are digits, and (a) is not zero. Hmm, let me think through this step by step.First, I remember that any repeating decimal can be expressed as a fraction. Specifically, for a three-digit repeating decimal like (0.overline{abc}), the fraction is (frac{abc}{999}). So, that's a good starting point. Let me write that down:[0.overline{abc} = frac{abc}{999}]Here, (abc) represents the three-digit number formed by digits (a), (b), and (c). Since (a) is not zero, (abc) ranges from 100 to 999.Now, the next step is to simplify this fraction to its lowest terms. To do that, I need to find the greatest common divisor (gcd) of the numerator ((abc)) and the denominator (999). The simplified fraction will then be:[frac{abc}{999} = frac{abc / gcd(abc, 999)}{999 / gcd(abc, 999)}]So, the denominator in the simplified fraction will be (999 / gcd(abc, 999)). Therefore, the possible denominators are the divisors of 999. Let me factorize 999 to find its divisors. 999 divided by 3 is 333, then 333 divided by 3 is 111, and 111 divided by 3 is 37. So, 999 factors into (3^3 times 37). To find all the divisors, I can consider all combinations of these prime factors:- (3^0 times 37^0 = 1)- (3^1 = 3)- (3^2 = 9)- (3^3 = 27)- (37^1 = 37)- (3^1 times 37 = 111)- (3^2 times 37 = 333)- (3^3 times 37 = 999)So, the divisors of 999 are: 1, 3, 9, 27, 37, 111, 333, and 999.But wait, can all these divisors be denominators in the simplified fraction? Let me think about that. The denominator will be (999 / gcd(abc, 999)). So, for each divisor (d) of 999, there must exist some (abc) such that (gcd(abc, 999) = 999 / d). Let me check each divisor:1. 1: For the denominator to be 1, the gcd must be 999. That would mean (abc = 999). But (abc) is a three-digit number, so (abc = 999) is possible. However, in the context of the problem, (a) is not zero, but 999 is allowed since (a = 9). So, 1 is a possible denominator.2. 3: For the denominator to be 3, the gcd must be 333. That means (abc) must be a multiple of 333. Since 333 is a three-digit number, (abc) can be 333, 666, or 999. All these are valid three-digit numbers with (a neq 0). So, 3 is a possible denominator.3. 9: For the denominator to be 9, the gcd must be 111. So, (abc) must be a multiple of 111. The multiples of 111 within three digits are 111, 222, 333, 444, 555, 666, 777, 888, 999. All these are valid, so 9 is a possible denominator.4. 27: For the denominator to be 27, the gcd must be 37. So, (abc) must be a multiple of 37. Since 37 is a prime number, the multiples of 37 within three digits are 037, 074, 111, 148, 185, 222, 259, 296, 333, 370, 407, 444, 481, 518, 555, 592, 629, 666, 693, 730, 767, 804, 841, 878, 915, 952, 989. However, (abc) must be a three-digit number with (a neq 0), so we exclude 037 and 074. The rest are valid, so 27 is a possible denominator.5. 37: For the denominator to be 37, the gcd must be 27. So, (abc) must be a multiple of 27. The multiples of 27 within three digits are 027, 054, 081, 108, 135, 162, 189, 216, 243, 270, 297, 324, 351, 378, 405, 432, 459, 486, 513, 540, 567, 594, 621, 648, 675, 702, 729, 756, 783, 810, 837, 864, 891, 918, 945, 972, 999. Again, (abc) must be a three-digit number with (a neq 0), so we exclude 027, 054, 081. The rest are valid, so 37 is a possible denominator.6. 111: For the denominator to be 111, the gcd must be 9. So, (abc) must be a multiple of 9. Since all three-digit numbers with (a neq 0) can potentially be multiples of 9, as long as their digits sum to a multiple of 9. For example, 108, 117, 126, etc., are all valid. So, 111 is a possible denominator.7. 333: For the denominator to be 333, the gcd must be 3. So, (abc) must be a multiple of 3. Again, since any three-digit number where the sum of digits is a multiple of 3 is valid, and there are plenty of such numbers, 333 is a possible denominator.8. 999: For the denominator to be 999, the gcd must be 1. That means (abc) and 999 must be coprime. Since 999 factors into (3^3 times 37), (abc) must not be divisible by 3 or 37. There are many three-digit numbers that are not divisible by 3 or 37, so 999 is a possible denominator.Wait a minute, earlier I thought 1 was a possible denominator because (abc = 999) would make the fraction (999/999 = 1/1). But in the context of the problem, the fraction is expressed in lowest terms, so 1 is technically a possible denominator. However, the problem might be considering only denominators greater than 1, as denominators of 1 are trivial. Let me check the original problem statement again.It says, "expressed as a fraction in lowest terms. What are the possible different denominators?" It doesn't specify excluding 1, so technically, 1 is a possible denominator. But in the initial thought process, I considered that 1 can't be a denominator because (abc) can't be 999. Wait, no, (abc) can be 999, which would make the fraction 1/1. So, 1 is indeed a possible denominator.But then, in the initial answer, it was stated that 1 can't be a denominator because (abc) will never be 999. That seems contradictory. Let me clarify.If (abc = 999), then the fraction is (999/999 = 1/1), so the denominator is 1. Therefore, 1 is a possible denominator. However, in the initial answer, it was said that 1 can't be a denominator because (abc) will never be 999. That's incorrect because (abc) can indeed be 999. So, 1 is a valid denominator.But wait, in the problem statement, it says "the repeating decimal (0.overline{abc})". If (abc = 999), then (0.overline{999}) is equal to 1.000..., which is an integer, not a decimal. So, in that case, the fraction would be 1/1, which is an integer, not a proper fraction. Therefore, perhaps the problem is considering only proper fractions, where the numerator is less than the denominator. If that's the case, then 1 would not be a possible denominator because the fraction would reduce to 1/1, which is not a proper fraction.But the problem doesn't specify that it's a proper fraction, just that it's expressed in lowest terms. So, technically, 1 is a possible denominator. However, in the context of repeating decimals, (0.overline{abc}) is a repeating decimal less than 1, so when expressed as a fraction, it must be a proper fraction (numerator less than denominator). Therefore, the denominator cannot be 1 because that would imply the fraction is 1, which is not a proper fraction. So, 1 is excluded.Therefore, the possible denominators are 3, 9, 27, 37, 111, 333, and 999. That's 7 denominators.Wait, but earlier I thought 1 was a possible denominator, but now considering the context of the problem, it's excluded. So, the answer should be 7 denominators.Let me summarize:- The fraction is (frac{abc}{999}).- Simplify by dividing both numerator and denominator by (gcd(abc, 999)).- The denominator becomes (999 / gcd(abc, 999)).- The possible values of (gcd(abc, 999)) are the divisors of 999: 1, 3, 9, 27, 37, 111, 333, 999.- Therefore, the possible denominators are (999 / d) where (d) is a divisor of 999.- This gives denominators: 999, 333, 111, 37, 27, 9, 3, 1.- Excluding 1 because the fraction must be less than 1 (proper fraction), the denominators are 3, 9, 27, 37, 111, 333, 999.So, there are 7 possible different denominators.</think>

🧠:<think>Alright, so I have this trigonometry problem here: If (2tanalpha = 3tan frac{pi}{8}), then I need to find (tanleft(alpha - frac{pi}{8}right)). Hmm, okay. Let me try to figure this out step by step.First, I remember that (frac{pi}{8}) is 22.5 degrees. I don't know the exact value of (tan frac{pi}{8}) off the top of my head, but maybe I can derive it using some trigonometric identities. I recall that there's a double-angle formula for tangent, which is:[tan(2theta) = frac{2tantheta}{1 - tan^2theta}]Since (frac{pi}{4}) is twice (frac{pi}{8}), maybe I can use this identity to find (tan frac{pi}{8}). Let me set (theta = frac{pi}{8}), so (2theta = frac{pi}{4}). I know that (tan frac{pi}{4} = 1), so plugging into the formula:[1 = frac{2tan frac{pi}{8}}{1 - tan^2 frac{pi}{8}}]Let me denote (t = tan frac{pi}{8}) to simplify the equation:[1 = frac{2t}{1 - t^2}]Multiplying both sides by (1 - t^2) gives:[1 - t^2 = 2t]Rearranging terms:[t^2 + 2t - 1 = 0]This is a quadratic equation in (t). Using the quadratic formula (t = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = 2), and (c = -1):[t = frac{-2 pm sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = frac{-2 pm sqrt{4 + 4}}{2} = frac{-2 pm sqrt{8}}{2}]Simplify (sqrt{8}) as (2sqrt{2}):[t = frac{-2 pm 2sqrt{2}}{2} = -1 pm sqrt{2}]Since (tan frac{pi}{8}) is positive (as (frac{pi}{8}) is in the first quadrant), we take the positive solution:[tan frac{pi}{8} = -1 + sqrt{2}]Wait, that seems a bit odd because (sqrt{2} approx 1.414), so (-1 + 1.414 approx 0.414), which is positive. Okay, that makes sense. So, (tan frac{pi}{8} = sqrt{2} - 1).Now, going back to the original equation:[2tanalpha = 3tan frac{pi}{8}]Substituting the value of (tan frac{pi}{8}):[2tanalpha = 3(sqrt{2} - 1)]Divide both sides by 2:[tanalpha = frac{3}{2}(sqrt{2} - 1)]So, (tanalpha = frac{3sqrt{2} - 3}{2}).Now, I need to find (tanleft(alpha - frac{pi}{8}right)). I remember the tangent subtraction formula:[tan(A - B) = frac{tan A - tan B}{1 + tan A tan B}]Let me set (A = alpha) and (B = frac{pi}{8}). Then:[tanleft(alpha - frac{pi}{8}right) = frac{tanalpha - tan frac{pi}{8}}{1 + tanalpha tan frac{pi}{8}}]I already know (tanalpha = frac{3sqrt{2} - 3}{2}) and (tan frac{pi}{8} = sqrt{2} - 1). Let's plug these values in.First, compute the numerator:[tanalpha - tan frac{pi}{8} = left(frac{3sqrt{2} - 3}{2}right) - (sqrt{2} - 1)]Let me get a common denominator for the terms:[= frac{3sqrt{2} - 3}{2} - frac{2sqrt{2} - 2}{2} = frac{3sqrt{2} - 3 - 2sqrt{2} + 2}{2}]Simplify the numerator:[= frac{(3sqrt{2} - 2sqrt{2}) + (-3 + 2)}{2} = frac{sqrt{2} - 1}{2}]Okay, so the numerator simplifies to (frac{sqrt{2} - 1}{2}).Now, the denominator:[1 + tanalpha tan frac{pi}{8} = 1 + left(frac{3sqrt{2} - 3}{2}right)(sqrt{2} - 1)]Let me compute the product (left(frac{3sqrt{2} - 3}{2}right)(sqrt{2} - 1)):First, factor out the 3 in the numerator:[= frac{3(sqrt{2} - 1)}{2} times (sqrt{2} - 1)]Multiply the terms:[= frac{3(sqrt{2} - 1)^2}{2}]Expand ((sqrt{2} - 1)^2):[(sqrt{2} - 1)^2 = (sqrt{2})^2 - 2sqrt{2} times 1 + (1)^2 = 2 - 2sqrt{2} + 1 = 3 - 2sqrt{2}]So, the product becomes:[frac{3(3 - 2sqrt{2})}{2} = frac{9 - 6sqrt{2}}{2}]Now, add 1 to this:[1 + frac{9 - 6sqrt{2}}{2} = frac{2}{2} + frac{9 - 6sqrt{2}}{2} = frac{11 - 6sqrt{2}}{2}]So, the denominator is (frac{11 - 6sqrt{2}}{2}).Putting it all together, we have:[tanleft(alpha - frac{pi}{8}right) = frac{frac{sqrt{2} - 1}{2}}{frac{11 - 6sqrt{2}}{2}} = frac{sqrt{2} - 1}{11 - 6sqrt{2}}]Now, to simplify this expression, I should rationalize the denominator. Multiply numerator and denominator by the conjugate of the denominator, which is (11 + 6sqrt{2}):[= frac{(sqrt{2} - 1)(11 + 6sqrt{2})}{(11 - 6sqrt{2})(11 + 6sqrt{2})}]First, compute the denominator:[(11 - 6sqrt{2})(11 + 6sqrt{2}) = 11^2 - (6sqrt{2})^2 = 121 - 72 = 49]So, the denominator is 49.Now, compute the numerator:[(sqrt{2} - 1)(11 + 6sqrt{2}) = sqrt{2} times 11 + sqrt{2} times 6sqrt{2} - 1 times 11 - 1 times 6sqrt{2}]Simplify each term:- (sqrt{2} times 11 = 11sqrt{2})- (sqrt{2} times 6sqrt{2} = 6 times 2 = 12)- (-1 times 11 = -11)- (-1 times 6sqrt{2} = -6sqrt{2})Combine these:[11sqrt{2} + 12 - 11 - 6sqrt{2} = (11sqrt{2} - 6sqrt{2}) + (12 - 11) = 5sqrt{2} + 1]So, the numerator is (5sqrt{2} + 1).Putting it all together:[tanleft(alpha - frac{pi}{8}right) = frac{5sqrt{2} + 1}{49}]Therefore, the value of (tanleft(alpha - frac{pi}{8}right)) is (frac{5sqrt{2} + 1}{49}).Final AnswerThe value of (tanleft(alpha - frac{pi}{8}right)) is boxed{dfrac{5sqrt{2} + 1}{49}}.</think>

🧠:<think>Okay, so I have this summation here, right? It's written as the sum from i equals 1 to n of x_i times y_i. Hmm, I think that means I'm supposed to add up all these products of x and y terms, starting from the first pair up to the nth pair. So, like, x1 times y1 plus x2 times y2 and so on, all the way to xn times yn. That makes sense.Now, I need to look at the options given and see which one matches this summation. Let's go through them one by one.Option A says (x1 + x2 + ... + xn) times y1. Wait, that doesn't sound right. If I add up all the x terms first and then multiply by y1, that's not the same as multiplying each x_i by its corresponding y_i and then adding them all up. For example, if I have x1=1, x2=2, y1=3, y2=4, then the summation would be 1*3 + 2*4 = 3 + 8 = 11. But option A would be (1+2)*3 = 3*3 = 9, which is different. So A isn't equivalent.Option B is similar but with the y terms. It says (y1 + y2 + ... + yn) times x1. Using the same numbers, that would be (3+4)*1 = 7*1 = 7, which is also different from the summation result of 11. So B isn't equivalent either.Option C is x1y1 + x2y2 + ... It looks like it's starting to do the right thing, but it ends with an ellipsis without specifying up to xnyn. That could be a problem because if it doesn't include all the terms up to n, it's not the same as the summation. For example, if n is 2, and C only goes up to x2y2, that's fine, but if n is larger, say 3, and C stops at x2y2, then it's missing x3y3. So C might not be complete.Option D is x1y1 + x2y2 + ... + xnyn. This one explicitly includes all the terms up to n, which matches the summation perfectly. Using my earlier example, it would correctly calculate 1*3 + 2*4 = 11. So D seems to be the right choice.Wait, but let me double-check. If I have more terms, say n=3, with x3=5 and y3=6, then the summation would be 1*3 + 2*4 + 5*6 = 3 + 8 + 30 = 41. Option D would be 1*3 + 2*4 + 5*6 = 41, which matches. Option C, if it only goes up to x2y2, would give 3 + 8 = 11, which is incorrect. So yeah, D is definitely the correct one.I think I've got it. The summation is just the sum of each x_i multiplied by its corresponding y_i, and option D is the only one that does that completely.</think>

🧠:<think>Okay, so I have this function f(x) = x⁴ + e^{|x|}, and I need to find all real numbers t that satisfy the inequality 2f(ln t) - f(ln(1/t)) ≤ f(2). Hmm, let me try to unpack this step by step.First, I should understand the function f(x). It's x to the fourth power plus e raised to the absolute value of x. So, f(x) = x⁴ + e^{|x|}. That means for any x, whether positive or negative, the function is going to be positive because both x⁴ and e^{|x|} are always positive. Also, since it's using the absolute value of x, the function is symmetric about the y-axis. So, f(-x) = f(x). That makes it an even function. That might be useful later.Now, the inequality is 2f(ln t) - f(ln(1/t)) ≤ f(2). Let me see if I can simplify this expression. I notice that ln(1/t) is the same as -ln t because ln(1/t) = ln(t^{-1}) = -ln t. So, f(ln(1/t)) is f(-ln t). But since f is even, f(-ln t) = f(ln t). So, f(ln(1/t)) = f(ln t). That's helpful.So, substituting that back into the inequality, I have 2f(ln t) - f(ln t) ≤ f(2). Simplifying the left side, that's just f(ln t) ≤ f(2). So, the inequality reduces to f(ln t) ≤ f(2).Now, since f is an even function, f(ln t) = f(|ln t|). So, f(|ln t|) ≤ f(2). Now, I need to figure out when f(|ln t|) is less than or equal to f(2). Since f is a function of |ln t|, which is always non-negative, I can consider the behavior of f on the non-negative real numbers.Looking at f(x) = x⁴ + e^{|x|}, for x ≥ 0, this simplifies to f(x) = x⁴ + e^{x}. Now, I need to see if f is increasing or decreasing for x ≥ 0. Let's take the derivative to check.The derivative f’(x) for x ≥ 0 is 4x³ + e^{x}. Both terms, 4x³ and e^{x}, are always positive for x ≥ 0. So, f’(x) > 0 for all x ≥ 0. That means f is strictly increasing on [0, ∞). Since f is even, it's decreasing on (-∞, 0] and increasing on [0, ∞).So, since f is increasing for x ≥ 0, f(|ln t|) ≤ f(2) implies that |ln t| ≤ 2. Because if the function is increasing, then the argument inside the function must be less than or equal to 2 in absolute value.So, |ln t| ≤ 2. That means -2 ≤ ln t ≤ 2. Now, to solve for t, I can exponentiate all parts of the inequality because the exponential function is strictly increasing.So, e^{-2} ≤ t ≤ e^{2}. Therefore, t must be between e^{-2} and e^{2}.Wait, let me double-check that. If |ln t| ≤ 2, then ln t is between -2 and 2. So, t is between e^{-2} and e^{2}. Yes, that makes sense.But hold on, t has to be positive because ln t is only defined for t > 0. So, t must be in (0, ∞), but our solution is e^{-2} ≤ t ≤ e^{2}, which is within (0, ∞). So, that's fine.Let me recap:1. Given f(x) = x⁴ + e^{|x|}, which is even and increasing for x ≥ 0.2. The inequality simplifies to f(ln t) ≤ f(2).3. Since f is increasing for x ≥ 0, this implies |ln t| ≤ 2.4. Solving |ln t| ≤ 2 gives e^{-2} ≤ t ≤ e^{2}.So, the set of real numbers t that satisfy the inequality is [e^{-2}, e^{2}].I think that's it. Let me just make sure I didn't make any mistakes in my reasoning.First, I correctly identified that f is even and increasing for x ≥ 0. Then, I correctly substituted ln(1/t) as -ln t and used the evenness of f to replace f(-ln t) with f(ln t). That led me to simplify the inequality correctly to f(ln t) ≤ f(2). Then, since f is increasing, I concluded that |ln t| ≤ 2, which is correct because f is increasing, so the argument must be less than or equal to 2 in absolute value. Finally, solving for t gave me e^{-2} ≤ t ≤ e^{2}, which is the correct interval.I don't see any errors in my steps, so I think the solution is correct.</think>