answer：Okay, so I have this problem where I need to find the largest positive integer ( n ) such that the sum of the floor of the base-2 logarithms from 1 to ( n ) equals 1994. That is,[lfloor log_2 1 rfloor + lfloor log_2 2 rfloor + lfloor log_2 3 rfloor + cdots + lfloor log_2 n rfloor = 1994]Hmm, let me start by understanding what the floor function does here. The floor of a logarithm, specifically ( lfloor log_2 x rfloor ), gives the greatest integer less than or equal to ( log_2 x ). So, for each integer ( x ), this floor function essentially tells me the exponent of the highest power of 2 that is less than or equal to ( x ).For example, ( lfloor log_2 1 rfloor = 0 ) because ( log_2 1 = 0 ). Similarly, ( lfloor log_2 2 rfloor = 1 ), ( lfloor log_2 3 rfloor = 1 ), ( lfloor log_2 4 rfloor = 2 ), and so on. So, it seems that for numbers between ( 2^k ) and ( 2^{k+1} - 1 ), the floor value will be ( k ).That makes sense because ( 2^k leq x < 2^{k+1} ) implies ( k leq log_2 x < k+1 ), so the floor is ( k ).Therefore, if I can figure out how many numbers fall into each interval ( [2^k, 2^{k+1} - 1] ), I can compute the sum by multiplying the number of terms in each interval by ( k ) and then adding them all up.Let me formalize this. For each integer ( k geq 0 ), the numbers ( x ) such that ( lfloor log_2 x rfloor = k ) are ( x = 2^k, 2^k + 1, ldots, 2^{k+1} - 1 ). So, there are ( 2^{k+1} - 1 - 2^k + 1 = 2^k ) numbers in each interval.Therefore, the contribution to the sum from each ( k ) is ( k times 2^k ).But wait, this is only true for ( k ) starting from 0 up to some maximum value. However, when ( n ) is not exactly ( 2^m - 1 ) for some ( m ), the last interval won't be complete. So, I need to handle that case separately.Let me denote ( S_n = sum_{i=1}^n lfloor log_2 i rfloor ). Then, if ( n = 2^r - 1 ) for some integer ( r ), the sum ( S_n ) can be written as:[S_{2^r - 1} = sum_{k=0}^{r-1} k times 2^k]I remember that there is a formula for the sum ( sum_{k=0}^{m} k times 2^k ). Let me recall it. I think it is ( (m - 1) times 2^{m+1} + 2 ). Let me verify this.Let me compute the sum ( sum_{k=0}^{m} k times 2^k ).Let ( S = sum_{k=0}^{m} k times 2^k ).Then, multiplying both sides by 2:( 2S = sum_{k=0}^{m} k times 2^{k+1} = sum_{k=1}^{m+1} (k-1) times 2^k )Subtracting the original S:( 2S - S = S = sum_{k=1}^{m+1} (k - 1) times 2^k - sum_{k=0}^{m} k times 2^k )Simplify the right-hand side:( S = sum_{k=1}^{m+1} (k - 1) times 2^k - sum_{k=0}^{m} k times 2^k )Breaking it down:( S = sum_{k=1}^{m} (k - 1) times 2^k + m times 2^{m+1} - sum_{k=0}^{m} k times 2^k )Notice that ( sum_{k=1}^{m} (k - 1) times 2^k = sum_{k=0}^{m-1} k times 2^{k+1} = 2 sum_{k=0}^{m-1} k times 2^k )So, substituting back:( S = 2 sum_{k=0}^{m-1} k times 2^k + m times 2^{m+1} - sum_{k=0}^{m} k times 2^k )Simplify further:( S = 2 sum_{k=0}^{m-1} k times 2^k + m times 2^{m+1} - sum_{k=0}^{m-1} k times 2^k - m times 2^m )Which simplifies to:( S = sum_{k=0}^{m-1} k times 2^k + m times 2^{m+1} - m times 2^m )Factor out ( m times 2^m ):( S = sum_{k=0}^{m-1} k times 2^k + m times 2^m (2 - 1) )So,( S = sum_{k=0}^{m-1} k times 2^k + m times 2^m )But notice that ( S = sum_{k=0}^{m} k times 2^k ), so:( sum_{k=0}^{m} k times 2^k = sum_{k=0}^{m-1} k times 2^k + m times 2^m )Which is consistent. So, perhaps I need a different approach to find the closed-form formula.Alternatively, I recall that the sum ( sum_{k=0}^{m} k r^k ) is given by ( frac{r - (m+1) r^{m+1} + m r^{m+2}}{(1 - r)^2} ) for ( r neq 1 ). Let me plug in ( r = 2 ):So,( sum_{k=0}^{m} k 2^k = frac{2 - (m+1) 2^{m+1} + m 2^{m+2}}{(1 - 2)^2} )Simplify denominator: ( (1 - 2)^2 = 1 )So,( sum_{k=0}^{m} k 2^k = 2 - (m+1) 2^{m+1} + m 2^{m+2} )Simplify numerator:( 2 - (m+1) 2^{m+1} + m 2^{m+2} = 2 - (m+1) 2^{m+1} + 4 m 2^{m} )Factor out ( 2^m ):( 2 + 2^m [ - (m+1) 2 + 4 m ] = 2 + 2^m [ -2(m + 1) + 4m ] = 2 + 2^m [ -2m - 2 + 4m ] = 2 + 2^m (2m - 2) )Factor further:( 2 + 2^{m+1} (m - 1) )So,( sum_{k=0}^{m} k 2^k = 2 + (m - 1) 2^{m+1} )Wait, let me verify this for a small ( m ). Let's take ( m = 1 ):Left-hand side: ( 0 times 2^0 + 1 times 2^1 = 0 + 2 = 2 )Right-hand side: ( 2 + (1 - 1) 2^{2} = 2 + 0 = 2 ). Okay, that works.For ( m = 2 ):Left-hand side: ( 0 times 1 + 1 times 2 + 2 times 4 = 0 + 2 + 8 = 10 )Right-hand side: ( 2 + (2 - 1) 2^{3} = 2 + 8 = 10 ). Good.For ( m = 3 ):Left-hand side: ( 0 + 2 + 8 + 3 times 8 = 0 + 2 + 8 + 24 = 34 )Right-hand side: ( 2 + (3 - 1) 2^{4} = 2 + 2 times 16 = 2 + 32 = 34 ). Perfect.So, the formula holds:[sum_{k=0}^{m} k 2^k = 2 + (m - 1) 2^{m+1}]Therefore, in our case, when ( n = 2^r - 1 ), the sum ( S_n ) is:[S_{2^r - 1} = sum_{k=0}^{r - 1} k 2^k = 2 + (r - 2) 2^{r}]Wait, let me check this. If ( m = r - 1 ), then:[sum_{k=0}^{r - 1} k 2^k = 2 + ( (r - 1) - 1 ) 2^{(r - 1) + 1} = 2 + (r - 2) 2^{r}]Yes, that's correct.So, ( S_{2^r - 1} = 2 + (r - 2) 2^r ).Let me compute this for some small ( r ):- For ( r = 1 ): ( S_{1} = 0 ). Plugging into the formula: ( 2 + (-1) times 2^1 = 2 - 2 = 0 ). Correct.- For ( r = 2 ): ( S_{3} = 0 + 1 + 1 = 2 ). Formula: ( 2 + (0) times 4 = 2 ). Correct.- For ( r = 3 ): ( S_{7} = 0 + 1 + 1 + 2 + 2 + 2 + 2 = 0 + 1 + 1 + 2 + 2 + 2 + 2 = 10 ). Formula: ( 2 + (1) times 8 = 10 ). Correct.- For ( r = 4 ): ( S_{15} ). Let's compute it:Numbers from 1 to 15:- 1: 0- 2-3: 1 each (2 numbers)- 4-7: 2 each (4 numbers)- 8-15: 3 each (8 numbers)Sum: 0 + 2*1 + 4*2 + 8*3 = 0 + 2 + 8 + 24 = 34Formula: ( 2 + (4 - 2) times 16 = 2 + 2 times 16 = 34 ). Correct.Great, so the formula works.Now, I need to find the largest ( n ) such that ( S_n = 1994 ).First, let me find for which ( r ) the sum ( S_{2^r - 1} ) is just below 1994.Compute ( S_{2^r - 1} = 2 + (r - 2) 2^r ).Let me compute this for increasing ( r ):- ( r = 8 ): ( 2 + (6) times 256 = 2 + 1536 = 1538 )- ( r = 9 ): ( 2 + (7) times 512 = 2 + 3584 = 3586 )So, ( S_{255} = 1538 ) and ( S_{511} = 3586 ). Since 1994 is between 1538 and 3586, the value of ( n ) must be between 256 and 511.Now, let me denote ( n = 255 + m ), where ( m ) is the number of terms beyond 255. Each term from 256 onwards contributes ( lfloor log_2 x rfloor = 8 ) because ( 256 = 2^8 ), and ( x ) is less than ( 2^9 = 512 ).Therefore, the sum ( S_n = S_{255} + 8 times m ).We have:[S_n = 1538 + 8m = 1994]Solving for ( m ):[8m = 1994 - 1538 = 456][m = 456 / 8 = 57]Therefore, ( n = 255 + 57 = 312 ).Wait, let me verify this. If ( n = 312 ), then ( S_{312} = 1538 + 57 times 8 = 1538 + 456 = 1994 ). That seems correct.But I need to make sure that 312 is the largest such ( n ). Let me check ( n = 313 ):( S_{313} = S_{312} + lfloor log_2 313 rfloor = 1994 + 8 = 2002 ), which is greater than 1994. Therefore, 312 is indeed the largest ( n ) such that the sum equals 1994.So, the answer is ( boxed{312} ).**Final Answer**boxed{312}

answer：Okay, so I need to figure out the range of the function y = [f(x)], where f(x) is given as (2^(x+1) - 1)/(2^x + 1). Hmm, I remember that the floor function, or Gauss function, [x], gives the greatest integer less than or equal to x. So, for example, [2.3] is 2, and [-3.7] is -4. First, I think I should simplify f(x) as much as possible to understand its behavior. Let me write down f(x):f(x) = (2^(x+1) - 1)/(2^x + 1)Hmm, 2^(x+1) is the same as 2*2^x, right? So I can rewrite the numerator:f(x) = (2*2^x - 1)/(2^x + 1)Maybe I can factor out a 2 from the numerator:f(x) = 2*(2^x) - 1 / (2^x + 1)Wait, that might not be helpful. Let me try another approach. Maybe I can split the fraction:f(x) = (2*2^x)/(2^x + 1) - 1/(2^x + 1)That gives me two separate terms. Let me see if I can simplify each term. The first term is 2*2^x/(2^x + 1). Maybe I can write this as 2*(2^x)/(2^x + 1). Similarly, the second term is -1/(2^x + 1).Hmm, maybe I can combine these terms differently. Let me try adding and subtracting something to make it easier. Wait, another idea: maybe I can perform polynomial long division or something similar. Let me see.If I let u = 2^x, then f(x) becomes (2u - 1)/(u + 1). Let me write that:f(u) = (2u - 1)/(u + 1)Now, I can perform division on this. Let's divide 2u - 1 by u + 1.Dividing 2u by u gives 2. Then, multiply (u + 1) by 2, which is 2u + 2. Subtracting this from 2u - 1:(2u - 1) - (2u + 2) = -3So, f(u) = 2 - 3/(u + 1)Therefore, f(x) = 2 - 3/(2^x + 1)Okay, that seems simpler. So f(x) is equal to 2 minus 3 divided by (2^x + 1). Now, I need to find the range of y = [f(x)].First, let's understand the behavior of f(x). Since 2^x is always positive for any real x, 2^x + 1 is always greater than 1. Therefore, 3/(2^x + 1) is always between 0 and 3. Specifically, as x approaches infinity, 2^x becomes very large, so 3/(2^x + 1) approaches 0. As x approaches negative infinity, 2^x approaches 0, so 3/(2^x + 1) approaches 3/(0 + 1) = 3.So, 3/(2^x + 1) is in the interval (0, 3). Therefore, f(x) = 2 - 3/(2^x + 1) will be in the interval (2 - 3, 2 - 0) = (-1, 2).Wait, let me double-check that. If 3/(2^x + 1) is between 0 and 3, then subtracting that from 2 would give:When 3/(2^x + 1) is near 0, f(x) is near 2.When 3/(2^x + 1) is near 3, f(x) is near 2 - 3 = -1.So, f(x) is in (-1, 2). That makes sense.Now, since f(x) is in (-1, 2), we need to find the range of y = [f(x)]. The floor function [x] gives the greatest integer less than or equal to x. So, for any x in (-1, 2), [x] can be -1, 0, or 1.Let me verify this. If f(x) is between -1 and 0, then [f(x)] would be -1. If f(x) is between 0 and 1, [f(x)] would be 0. If f(x) is between 1 and 2, [f(x)] would be 1. But wait, does f(x) actually reach 0 or 1? Let's see.First, can f(x) be equal to 0? Let's solve for x when f(x) = 0:0 = 2 - 3/(2^x + 1)So, 3/(2^x + 1) = 2Multiply both sides by (2^x + 1):3 = 2*(2^x + 1)3 = 2*2^x + 2Subtract 2:1 = 2*2^xDivide by 2:1/2 = 2^xSo, 2^x = 1/2, which implies x = -1.So, when x = -1, f(x) = 0. Therefore, f(x) can be 0.Similarly, can f(x) be equal to 1?Set f(x) = 1:1 = 2 - 3/(2^x + 1)So, 3/(2^x + 1) = 1Multiply both sides by (2^x + 1):3 = 2^x + 1Subtract 1:2 = 2^xSo, 2^x = 2, which implies x = 1.Therefore, when x = 1, f(x) = 1.So, f(x) can reach 0 and 1. But can it reach -1 or 2?When x approaches negative infinity, f(x) approaches -1, but never actually reaches -1 because 3/(2^x + 1) approaches 3, so f(x) approaches 2 - 3 = -1. Similarly, as x approaches positive infinity, f(x) approaches 2, but never actually reaches 2 because 3/(2^x + 1) approaches 0, so f(x) approaches 2 - 0 = 2.Therefore, f(x) is in the interval (-1, 2), but does not include -1 and 2.Therefore, the possible values of [f(x)] are:- When f(x) is in (-1, 0), [f(x)] = -1- When f(x) is in [0, 1), [f(x)] = 0- When f(x) is in [1, 2), [f(x)] = 1But wait, earlier we saw that f(x) can actually be 0 and 1. So, at x = -1, f(x) = 0, so [f(x)] = 0. Similarly, at x = 1, f(x) = 1, so [f(x)] = 1.Therefore, the possible integer values that [f(x)] can take are -1, 0, and 1.Hence, the range of y = [f(x)] is {-1, 0, 1}.I think that's it. Let me just recap to make sure I didn't miss anything.1. Simplified f(x) to 2 - 3/(2^x + 1).2. Determined that 3/(2^x + 1) is in (0, 3), so f(x) is in (-1, 2).3. Analyzed the floor function over this interval, which can take values -1, 0, or 1.4. Verified that f(x) can actually attain 0 and 1, so these values are included.5. Concluded that the range is {-1, 0, 1}.Yeah, that seems solid. I don't think I missed any steps or made any calculation errors. So, I'm confident with this answer.

answer：Okay, so I need to simplify the expression tan(π/12) + tan(5π/12). Hmm, I remember that tan can be expressed in terms of sine and cosine, so maybe I can rewrite the expression that way. Let me try that.So, tan(π/12) is sin(π/12)/cos(π/12) and tan(5π/12) is sin(5π/12)/cos(5π/12). If I add these together, I get:sin(π/12)/cos(π/12) + sin(5π/12)/cos(5π/12)To add these two fractions, I need a common denominator. The common denominator would be cos(π/12) * cos(5π/12). So, I can rewrite the expression as:[sin(π/12) * cos(5π/12) + sin(5π/12) * cos(π/12)] / [cos(π/12) * cos(5π/12)]Wait a second, the numerator looks familiar. Isn't there a sine addition formula? Let me recall: sin(A + B) = sin A cos B + cos A sin B. So, the numerator is actually sin(π/12 + 5π/12). Let me compute that angle:π/12 + 5π/12 = (1 + 5)π/12 = 6π/12 = π/2.So, sin(π/2) is 1. That simplifies the numerator to 1. So now the expression becomes:1 / [cos(π/12) * cos(5π/12)]Hmm, okay, so now I have 1 divided by the product of cos(π/12) and cos(5π/12). I need to simplify this further. Maybe I can use some trigonometric identities to simplify the denominator.I remember that there's a product-to-sum identity for cosine. Let me recall it: cos A cos B = [cos(A + B) + cos(A - B)] / 2. Let me apply that here with A = π/12 and B = 5π/12.So, cos(π/12) * cos(5π/12) = [cos(π/12 + 5π/12) + cos(π/12 - 5π/12)] / 2.Calculating the angles inside the cosines:π/12 + 5π/12 = 6π/12 = π/2, and π/12 - 5π/12 = -4π/12 = -π/3.So, cos(π/2) is 0, and cos(-π/3) is the same as cos(π/3) because cosine is an even function. Cos(π/3) is 1/2.Therefore, cos(π/12) * cos(5π/12) = [0 + 1/2] / 2 = (1/2) / 2 = 1/4.So, the denominator is 1/4. Therefore, the expression becomes:1 / (1/4) = 4.Wait, that seems too straightforward. Let me double-check my steps.First, I converted tan to sin/cos, which is correct. Then, I used the sine addition formula on the numerator, which gave me sin(π/2) = 1. That seems right.Then, for the denominator, I used the product-to-sum identity. Let me verify that:cos A cos B = [cos(A + B) + cos(A - B)] / 2.Yes, that's correct. So, plugging in A = π/12 and B = 5π/12, I get:cos(π/12 + 5π/12) = cos(π/2) = 0, and cos(π/12 - 5π/12) = cos(-π/3) = cos(π/3) = 1/2.So, adding those gives 0 + 1/2 = 1/2, and then dividing by 2 gives 1/4. So, the denominator is indeed 1/4.Therefore, 1 divided by 1/4 is 4. So, the simplified expression is 4.Wait, but in the initial problem, the user had written the expression as tan(π/12) + tan(5π/12). So, according to my calculation, that's equal to 4.But let me think again. Maybe I made a mistake in the product-to-sum identity. Let me double-check that.The identity is:cos A cos B = [cos(A + B) + cos(A - B)] / 2.Yes, that's correct. So, with A = π/12 and B = 5π/12, we have:cos(π/12) cos(5π/12) = [cos(6π/12) + cos(-4π/12)] / 2 = [cos(π/2) + cos(-π/3)] / 2.Since cos is even, cos(-π/3) = cos(π/3) = 1/2. And cos(π/2) = 0. So, [0 + 1/2] / 2 = 1/4. That seems correct.Therefore, 1 / (1/4) = 4. So, the expression simplifies to 4.Wait, but I recall that tan(π/12) is 2 - √3 and tan(5π/12) is 2 + √3. Let me check that.Yes, tan(π/12) is indeed 2 - √3, and tan(5π/12) is 2 + √3. So, adding them together: (2 - √3) + (2 + √3) = 4. So, that confirms it.Therefore, the simplified expression is 4.I think that's correct. So, the final answer is 4.

answer：First, I note that there were initially 22 children on the bus.At the bus stop, 60 children got off the bus, leaving 22 - 60 = -38 children on the bus. This negative number indicates that more children got off than were originally on the bus.Let ( x ) represent the number of children who got on the bus at the bus stop.After these children boarded, the total number of children on the bus became 2. Therefore, the equation is:[-38 + x = 2]Solving for ( x ):[x = 2 + 38][x = 40]Thus, 40 children got on the bus at the bus stop.

answer：Okay, so I have this problem where Billy's baseball team has scored a certain number of points over the first 13 games of the season. The scores are listed as: 41, 46, 50, 50, 55, 55, 55, 59, 63, 68, 72, 75, 80. Now, they scored 36 in the 14th game, and I need to figure out which of the given statistics will show an increase. The options are Range, Median, Mean, Mode, and Mid-range.First, I need to understand each of these statistics:1. **Range**: This is the difference between the highest and lowest values in the dataset.2. **Median**: This is the middle value when the data is ordered from least to greatest.3. **Mean**: This is the average of all the numbers.4. **Mode**: This is the number that appears most frequently.5. **Mid-range**: This is the average of the highest and lowest values.Alright, let's start by calculating each of these statistics for the original dataset of 13 games.**Original Dataset:**41, 46, 50, 50, 55, 55, 55, 59, 63, 68, 72, 75, 80**1. Range:**The highest score is 80, and the lowest is 41.Range = 80 - 41 = 39.**2. Median:**Since there are 13 games, the median will be the 7th value when the data is ordered.Counting to the 7th value: 41, 46, 50, 50, 55, 55, 55, 59, 63, 68, 72, 75, 80.So, the median is 55.**3. Mean:**To find the mean, I need to sum all the scores and divide by the number of games.Sum = 41 + 46 + 50 + 50 + 55 + 55 + 55 + 59 + 63 + 68 + 72 + 75 + 80.Let me calculate that step by step:41 + 46 = 8787 + 50 = 137137 + 50 = 187187 + 55 = 242242 + 55 = 297297 + 55 = 352352 + 59 = 411411 + 63 = 474474 + 68 = 542542 + 72 = 614614 + 75 = 689689 + 80 = 769So, the total sum is 769.Mean = 769 / 13 ≈ 59.15.**4. Mode:**Looking at the dataset, the number that appears most frequently is 55, which appears three times.**5. Mid-range:**This is the average of the highest and lowest values.Mid-range = (80 + 41) / 2 = 121 / 2 = 60.5.Now, let's add the 14th game score of 36 to the dataset and recalculate each statistic.**New Dataset:**36, 41, 46, 50, 50, 55, 55, 55, 59, 63, 68, 72, 75, 80**1. New Range:**The highest score is still 80, and the lowest is now 36.Range = 80 - 36 = 44.So, the range has increased from 39 to 44.**2. New Median:**Now there are 14 games, so the median will be the average of the 7th and 8th values.Ordered dataset: 36, 41, 46, 50, 50, 55, 55, 55, 59, 63, 68, 72, 75, 80.7th value: 558th value: 55Median = (55 + 55) / 2 = 55.So, the median remains the same.**3. New Mean:**Sum of the new dataset = previous sum (769) + 36 = 769 + 36 = 805.Mean = 805 / 14 ≈ 57.5.So, the mean has decreased from approximately 59.15 to 57.5.**4. New Mode:**Looking at the new dataset, 55 still appears three times, which is more frequent than any other number.So, the mode remains 55.**5. New Mid-range:**Mid-range = (80 + 36) / 2 = 116 / 2 = 58.So, the mid-range has decreased from 60.5 to 58.From this analysis, only the range has increased. The median stayed the same, the mean and mid-range decreased, and the mode remained unchanged.Therefore, the statistic that shows an increase is the range.

answer：First, I need to determine the amount of solid content in each batch of fresh grapes before drying. Since the water content percentages are given, I can calculate the solid content by subtracting the water percentage from 100%.For Batch A:- Water content: 70%- Solid content: 30%- Total weight: 100 kg- Solid weight: 30% of 100 kg = 30 kgFor Batch B:- Water content: 75%- Solid content: 25%- Total weight: 150 kg- Solid weight: 25% of 150 kg = 37.5 kgFor Batch C:- Water content: 80%- Solid content: 20%- Total weight: 200 kg- Solid weight: 20% of 200 kg = 40 kgNext, I need to calculate the weight of each batch after drying, when the water content is reduced to the specified percentages. The solid content remains the same during drying, so I can use the solid weight to find the new total weight.For Batch A after drying:- Desired water content: 20%- Desired solid content: 80%- Let the dried weight be ( x ) kg- ( 0.80 times x = 30 ) kg- ( x = frac{30}{0.80} = 37.5 ) kgFor Batch B after drying:- Desired water content: 25%- Desired solid content: 75%- Let the dried weight be ( y ) kg- ( 0.75 times y = 37.5 ) kg- ( y = frac{37.5}{0.75} = 50 ) kgFor Batch C after drying:- Desired water content: 30%- Desired solid content: 70%- Let the dried weight be ( z ) kg- ( 0.70 times z = 40 ) kg- ( z = frac{40}{0.70} approx 57.14 ) kgFinally, I add the dried weights of all three batches to find the total combined weight:- Total dried weight = ( 37.5 ) kg + ( 50 ) kg + ( 57.14 ) kg = ( 144.64 ) kg