💡Alright, so I've got this problem about a magical tree with 60 fruits. Every day, fruits fall from the tree in a specific pattern. On the first day, 1 fruit falls. Starting from the second day, each day, one more fruit falls than the previous day. But here's the catch: if the number of fruits left on the tree is less than the number that should fall on that day, the process resets, and only 1 fruit falls that day, and then it starts increasing again from the next day.Okay, let's break this down. I need to figure out on which day all the fruits will have fallen from the tree.First, let's think about the pattern of fruits falling. On day 1, 1 fruit falls. On day 2, 2 fruits fall. On day 3, 3 fruits fall, and so on. This seems like an arithmetic sequence where each day, the number of fruits falling increases by 1.Now, if there were no reset condition, the total number of fruits fallen after n days would be the sum of the first n natural numbers, which is given by the formula:[ S = frac{n(n + 1)}{2} ]But in this problem, if the number of fruits left is less than the number that should fall on a particular day, the process resets, and only 1 fruit falls that day, and then it starts increasing again from the next day.So, I need to consider this reset condition. Let's see how this plays out.Starting with 60 fruits:- Day 1: 1 fruit falls. Remaining fruits: 60 - 1 = 59- Day 2: 2 fruits fall. Remaining fruits: 59 - 2 = 57- Day 3: 3 fruits fall. Remaining fruits: 57 - 3 = 54- Day 4: 4 fruits fall. Remaining fruits: 54 - 4 = 50- Day 5: 5 fruits fall. Remaining fruits: 50 - 5 = 45- Day 6: 6 fruits fall. Remaining fruits: 45 - 6 = 39- Day 7: 7 fruits fall. Remaining fruits: 39 - 7 = 32- Day 8: 8 fruits fall. Remaining fruits: 32 - 8 = 24- Day 9: 9 fruits fall. Remaining fruits: 24 - 9 = 15- Day 10: 10 fruits fall. Remaining fruits: 15 - 10 = 5Okay, so after 10 days, only 5 fruits remain. Now, on day 11, according to the pattern, we should have 11 fruits fall, but there are only 5 left. So, the process resets, and only 1 fruit falls on day 11.- Day 11: 1 fruit falls. Remaining fruits: 5 - 1 = 4- Day 12: Now, the pattern starts again, so we should have 2 fruits fall, but there are only 4 left. So, 2 fruits fall. Remaining fruits: 4 - 2 = 2- Day 13: Next, we should have 3 fruits fall, but there are only 2 left. So, the process resets again, and only 1 fruit falls. Remaining fruits: 2 - 1 = 1- Day 14: Now, we should have 2 fruits fall, but there's only 1 left. So, the process resets again, and only 1 fruit falls. Remaining fruits: 1 - 1 = 0Wait, that seems a bit off. Let me check that again.After day 10, we have 5 fruits left.- Day 11: Reset, 1 fruit falls. Remaining: 5 - 1 = 4- Day 12: Now, the pattern should increase by 1, so 2 fruits fall. Remaining: 4 - 2 = 2- Day 13: Next, we should have 3 fruits fall, but there are only 2 left. So, reset again, 1 fruit falls. Remaining: 2 - 1 = 1- Day 14: Now, the pattern should increase to 2, but there's only 1 left. So, reset again, 1 fruit falls. Remaining: 1 - 1 = 0So, on day 14, the last fruit falls.But wait, is that correct? Let's see:After day 10: 5 fruits left.Day 11: 1 fruit falls (reset). Remaining: 4Day 12: 2 fruits fall. Remaining: 2Day 13: 3 fruits should fall, but only 2 left. So, reset, 1 fruit falls. Remaining: 1Day 14: 2 fruits should fall, but only 1 left. Reset, 1 fruit falls. Remaining: 0Yes, that seems correct. So, on day 14, the last fruit falls.But let me think again. Is there a more efficient way to calculate this without listing each day?Maybe I can calculate how many fruits fall in each cycle before the reset.The initial cycle is days 1 to 10, where 55 fruits fall in total (sum from 1 to 10).Then, we have 5 fruits left.Now, in the next cycle, starting from day 11, the pattern resets to 1, 2, 3, etc., but we only have 5 fruits left.So, on day 11: 1 fruit falls, remaining: 4Day 12: 2 fruits fall, remaining: 2Day 13: 3 fruits should fall, but only 2 left. So, reset, 1 fruit falls, remaining: 1Day 14: 2 fruits should fall, but only 1 left. Reset, 1 fruit falls, remaining: 0So, it takes 4 more days after day 10 to finish the remaining 5 fruits.Therefore, total days: 10 + 4 = 14 days.Yes, that seems consistent.Alternatively, I could think of it as:Total fruits: 60Sum of first n days: S = n(n + 1)/2We need to find the smallest n such that S >= 60.But considering the reset condition, it's not straightforward.Alternatively, think of it as cycles where each cycle starts with 1 fruit and increases by 1 each day until the number of fruits left is less than the required number, then reset.So, first cycle: days 1-10, sum 55, fruits left: 5Second cycle: days 11-14, sum 1 + 2 + 1 + 1 = 5Total days: 14Yes, that matches.So, the answer is day 14.

💡Okay, so I have this problem about vectors. It says that vectors a and b have magnitudes 1 and 2 respectively, and the angle between them is π/3. I need to find the magnitude of the vector a + b. Hmm, alright, let me think about how to approach this.First, I remember that when you add two vectors, the magnitude of the resultant vector can be found using the law of cosines. I think that's the formula that relates the lengths of the sides of a triangle to the cosine of one of its angles. So, in this case, the vectors a and b can be thought of as two sides of a triangle, and the angle between them is given as π/3.Let me recall the formula for the magnitude of the sum of two vectors. I believe it's |a + b|² = |a|² + |b|² + 2|a||b|cosθ, where θ is the angle between them. Yeah, that sounds right. So, I can plug in the values given into this formula.Given that |a| is 1, |b| is 2, and θ is π/3. Let me write that down:|a + b|² = (1)² + (2)² + 2 * 1 * 2 * cos(π/3)Calculating each part step by step. First, (1)² is 1, and (2)² is 4. So, adding those together gives 1 + 4 = 5.Next, the cross term: 2 * 1 * 2 is 4. And cos(π/3) is... hmm, what's the cosine of π/3? I remember that π/3 is 60 degrees, and the cosine of 60 degrees is 0.5. So, cos(π/3) = 0.5.Therefore, the cross term becomes 4 * 0.5 = 2.Now, adding that to the previous sum: 5 + 2 = 7.So, |a + b|² = 7. To find |a + b|, I need to take the square root of 7. That gives me √7.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the formula:|a + b|² = |a|² + |b|² + 2|a||b|cosθPlugging in the values:1² + 2² + 2*1*2*cos(π/3) = 1 + 4 + 4*0.5 = 5 + 2 = 7Yes, that seems correct. So, taking the square root of 7 gives the magnitude of the resultant vector.I think I got it right. So, the magnitude of a + b is √7.**Final Answer**The magnitude of the vector (vec{a} + vec{b}) is (boxed{sqrt{7}}).

💡Okay, so I have this problem here: E_n is defined as 3 raised to the power of 3^n, and then you add 1. I need to find the units digit of E_{1000}. Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I know that the units digit of a number is just the last digit of that number. So, if I can figure out the last digit of 3^{3^{1000}}, and then add 1, I can find the units digit of E_{1000}. That makes sense.I remember that powers of numbers cycle their units digits in a pattern. For example, 3^1 is 3, 3^2 is 9, 3^3 is 27 (so units digit 7), 3^4 is 81 (units digit 1), and then it repeats: 3, 9, 7, 1, and so on. So, the cycle length for the units digit of powers of 3 is 4. That means every 4 exponents, the units digit cycles back.So, if I can figure out where 3^{1000} falls in this cycle, I can determine the units digit of 3^{3^{1000}}. But wait, 3^{1000} is a huge exponent. How do I handle that?I think I need to use modular arithmetic here. Specifically, since the cycle length is 4, I can find the remainder when 3^{1000} is divided by 4. That will tell me where in the cycle the exponent falls.Let me recall that 3 mod 4 is 3, and 3^2 mod 4 is 1. So, 3^2 ≡ 1 mod 4. That means that any even power of 3 will be congruent to 1 mod 4, and any odd power will be congruent to 3 mod 4. Since 1000 is an even number, 3^{1000} will be congruent to 1 mod 4.So, 3^{1000} ≡ 1 mod 4. That means that 3^{3^{1000}} is equivalent to 3^1 mod 10, because the exponent cycles every 4. So, 3^{3^{1000}} ≡ 3^1 ≡ 3 mod 10.Therefore, the units digit of 3^{3^{1000}} is 3. Then, adding 1 to that, the units digit of E_{1000} is 3 + 1 = 4.Wait, let me double-check that. So, the cycle for 3^n mod 10 is 4: 3, 9, 7, 1. So, to find 3^{3^{1000}} mod 10, I need to find 3^{1000} mod 4 first. Since 3^2 ≡ 1 mod 4, 3^{1000} = (3^2)^500 ≡ 1^500 ≡ 1 mod 4. So, 3^{3^{1000}} ≡ 3^1 ≡ 3 mod 10. Adding 1 gives 4. Yeah, that seems right.I think I got it. The units digit cycles every 4, so by figuring out the exponent modulo 4, I can find the corresponding units digit. Since the exponent 3^{1000} is congruent to 1 mod 4, the units digit is the same as 3^1, which is 3. Adding 1 gives 4. So, the units digit of E_{1000} is 4.

💡Okay, so I have this problem about a right triangle with legs of lengths 1 and √2. The smallest angle is α, and I need to find the exact value of cos 8α. Hmm, let me think about how to approach this.First, since it's a right triangle, I can use the Pythagorean theorem to find the hypotenuse. The legs are 1 and √2, so the hypotenuse c should be √(1² + (√2)²) = √(1 + 2) = √3. Got that down.Now, the smallest angle α is opposite the shorter leg, which is 1. So, in this triangle, the sides are 1, √2, and √3. Let me label them: opposite α is 1, adjacent is √2, and hypotenuse is √3.To find cos α, I remember that cosine is adjacent over hypotenuse. So, cos α = √2 / √3. That simplifies to √(2/3). Okay, so cos α is √(2/3). Maybe I can rationalize that, but I think it's fine for now.Now, I need to find cos 8α. That seems like a lot of doubling. Maybe I can use the double-angle formula multiple times. The double-angle formula for cosine is cos 2θ = 2cos²θ - 1. So, if I can find cos 2α, then cos 4α, and so on, up to cos 8α.Let me start with cos 2α. Using the formula:cos 2α = 2cos²α - 1= 2*(√(2/3))² - 1= 2*(2/3) - 1= 4/3 - 1= 1/3.Okay, so cos 2α is 1/3. That wasn't too bad. Now, let's find cos 4α. Using the same formula:cos 4α = 2cos²2α - 1= 2*(1/3)² - 1= 2*(1/9) - 1= 2/9 - 1= -7/9.Hmm, cos 4α is -7/9. Interesting, it's negative now. Now, moving on to cos 8α. Again, using the double-angle formula:cos 8α = 2cos²4α - 1= 2*(-7/9)² - 1= 2*(49/81) - 1= 98/81 - 1= 98/81 - 81/81= 17/81.Wait, so cos 8α is 17/81? That seems a bit small, but let me check my steps.Starting from cos α = √(2/3). Then cos 2α = 1/3, which makes sense because 2*(2/3) - 1 = 1/3. Then cos 4α = 2*(1/3)^2 - 1 = 2/9 - 1 = -7/9. That also seems correct. Then cos 8α = 2*(-7/9)^2 - 1 = 2*(49/81) - 1 = 98/81 - 1 = 17/81.Yeah, that seems consistent. I don't see any mistakes in the calculations. So, cos 8α is 17/81.I wonder if there's another way to approach this problem, maybe using multiple-angle formulas or something else, but I think using the double-angle formula step by step is the most straightforward method here. It might be tedious, but it works.Another thought: since α is the smallest angle, it must be less than 45 degrees because the triangle has sides 1, √2, and √3. So, α is opposite the side of length 1, which is the shortest side, so it's indeed the smallest angle. That makes sense.Just to make sure, let me calculate α in degrees to get a sense of what 8α would be. If cos α = √(2/3), then α = arccos(√(2/3)). Let me approximate that. √(2/3) is approximately 0.8165. So, arccos(0.8165) is roughly 35 degrees. So, 8α would be around 280 degrees. Cosine of 280 degrees is cosine of (360 - 80) degrees, which is cosine 80 degrees, but since it's in the fourth quadrant, cosine is positive. Cosine 80 degrees is approximately 0.1736, and 17/81 is approximately 0.2099. Hmm, that's a bit off. Wait, maybe my approximation is rough.Wait, 17/81 is approximately 0.2099, and cos 80 degrees is about 0.1736. So, 0.2099 is a bit higher. Maybe my angle approximation is not precise enough. Let me check with a calculator.If α = arccos(√(2/3)), let me compute that more accurately. √(2/3) ≈ 0.8164965809. So, arccos(0.8164965809) ≈ 35.264 degrees. So, 8α ≈ 8 * 35.264 ≈ 282.112 degrees. Cosine of 282.112 degrees is cosine(360 - 77.888) = cosine(77.888 degrees). Cosine 77.888 degrees is approximately 0.2079. Hmm, 17/81 is approximately 0.2099, which is very close. So, that seems consistent. So, 17/81 is the exact value, and the approximate value is about 0.2099, which matches the cosine of 8α.Therefore, I think my answer is correct. So, cos 8α is 17/81.

💡Alright, so I have this problem here about imports and exports for a city from 2017 to 2020. They've given me the total import amounts (x) and export amounts (y) for each year, and they've told me that there's a linear correlation between them, which is given by the equation y = bx - 0.84. My job is to find the value of b, and then use that to estimate the import total if the export is planned to be 6 billion yuan in 2022.Okay, first, I need to understand what a linear correlation is. From what I remember, a linear correlation means that there's a straight-line relationship between two variables. In this case, the import amounts (x) and export amounts (y) are linearly related, so as imports increase, exports also increase in a predictable way, and vice versa.The equation they've given is y = bx - 0.84. This looks like a linear equation in the slope-intercept form, which is usually written as y = mx + b, where m is the slope and b is the y-intercept. But in this case, it's written as y = bx - 0.84, so I guess here, b is acting like the slope, and -0.84 is the y-intercept. So, I need to find the value of b that best fits the data they've given.To find b, I think I need to use the method of least squares, which is a common way to find the best-fitting line for a set of data points. But wait, the problem mentions that the data satisfies a linear correlation, so maybe I don't need to do all that work. Maybe I can just use the averages of x and y to find b.Let me recall: if I have a linear regression line, it always passes through the point (x̄, ȳ), where x̄ is the mean of the x-values and ȳ is the mean of the y-values. So, if I can find the means of x and y, I can plug them into the equation y = bx - 0.84 and solve for b.Alright, let's calculate the means. The x-values are 1.8, 2.2, 2.6, and 3.0. To find the mean x̄, I add them up and divide by 4.x̄ = (1.8 + 2.2 + 2.6 + 3.0) / 4Let me do that addition step by step:1.8 + 2.2 = 4.04.0 + 2.6 = 6.66.6 + 3.0 = 9.6So, x̄ = 9.6 / 4 = 2.4Okay, so the mean of x is 2.4 billion yuan.Now, let's find the mean of y. The y-values are 2.0, 2.8, 3.2, and 4.0.ȳ = (2.0 + 2.8 + 3.2 + 4.0) / 4Again, adding step by step:2.0 + 2.8 = 4.84.8 + 3.2 = 8.08.0 + 4.0 = 12.0So, ȳ = 12.0 / 4 = 3.0Alright, so the mean of y is 3.0 billion yuan.Now, since the regression line passes through (x̄, ȳ), I can plug these values into the equation y = bx - 0.84 to find b.So, substituting x̄ = 2.4 and ȳ = 3.0 into the equation:3.0 = b * 2.4 - 0.84Now, I need to solve for b.First, add 0.84 to both sides to isolate the term with b:3.0 + 0.84 = b * 2.43.84 = 2.4bNow, divide both sides by 2.4 to solve for b:b = 3.84 / 2.4Let me calculate that. 3.84 divided by 2.4.Hmm, 2.4 goes into 3.84 how many times?Well, 2.4 * 1 = 2.42.4 * 1.5 = 3.62.4 * 1.6 = 3.84Oh, perfect! So, b = 1.6Alright, so the slope of the linear correlation is 1.6.Now, the second part of the problem asks: If the plan is to achieve an export total of 6 billion yuan in 2022, the estimated import total for that year is approximately how much?So, they want to find the import amount x when the export amount y is 6.0 billion yuan.We have the equation y = 1.6x - 0.84We need to solve for x when y = 6.0So, substituting y = 6.0 into the equation:6.0 = 1.6x - 0.84First, add 0.84 to both sides:6.0 + 0.84 = 1.6x6.84 = 1.6xNow, divide both sides by 1.6 to solve for x:x = 6.84 / 1.6Let me calculate that.6.84 divided by 1.6.Well, 1.6 goes into 6.84 how many times?1.6 * 4 = 6.4Subtract 6.4 from 6.84: 6.84 - 6.4 = 0.44Now, bring down a zero: 0.4401.6 goes into 0.440 how many times?1.6 * 0.275 = 0.44So, total x = 4 + 0.275 = 4.275So, x = 4.275 billion yuan.Therefore, if the export total is planned to be 6 billion yuan in 2022, the estimated import total would be approximately 4.275 billion yuan.Let me just double-check my calculations to make sure I didn't make any mistakes.First, calculating the means:x̄ = (1.8 + 2.2 + 2.6 + 3.0) / 4 = 9.6 / 4 = 2.4ȳ = (2.0 + 2.8 + 3.2 + 4.0) / 4 = 12.0 / 4 = 3.0That seems correct.Then, plugging into y = bx - 0.84:3.0 = 2.4b - 0.84Adding 0.84: 3.84 = 2.4bDividing: b = 3.84 / 2.4 = 1.6That looks good.Then, for y = 6.0:6.0 = 1.6x - 0.84Adding 0.84: 6.84 = 1.6xDividing: x = 6.84 / 1.6 = 4.275Yes, that seems correct.So, I think my answers are accurate.

💡Okay, so I have this problem about triangle ABC where points D and E are the two trisection points of side AB. They also gave me that vector CA is equal to 3a and vector CB is equal to 2b. I need to express vectors DE, CD, and CE in terms of a and b. Hmm, okay, let's break this down step by step.First, I should probably draw a diagram to visualize the problem. Triangle ABC with points D and E on side AB. Since D and E are trisection points, that means they divide AB into three equal parts. So, AD = DE = EB. That makes sense. So, D is closer to A, and E is closer to B.Now, they've given me vectors CA and CB. Vector CA is 3a, and vector CB is 2b. So, in terms of position vectors, if I consider point C as the origin, then vector CA would be the position vector of point A, and vector CB would be the position vector of point B. But wait, actually, in vector notation, vector CA is from point C to point A, so it's A - C. Similarly, vector CB is B - C. So, if I consider point C as the origin, then vector CA would just be the position vector of A, and vector CB would be the position vector of B. So, maybe I can treat C as the origin for simplicity. Let me assume that.So, if C is the origin, then the position vector of A is 3a, and the position vector of B is 2b. Therefore, vector AB would be B - A, which is 2b - 3a. That seems right.Now, since D and E are the trisection points of AB, I can find their position vectors. Since AD = DE = EB, each segment is one-third of AB. So, starting from A, moving one-third towards B gets me to D, and moving two-thirds towards B gets me to E.So, the position vector of D would be A + (1/3)AB. Similarly, the position vector of E would be A + (2/3)AB.Let me compute that. Vector AB is 2b - 3a, so (1/3)AB is (2b - 3a)/3, which is (2/3)b - a. Similarly, (2/3)AB is (4/3)b - 2a.Therefore, the position vector of D is A + (1/3)AB. Since A is 3a, adding (2/3)b - a gives 3a + (2/3)b - a, which simplifies to 2a + (2/3)b.Similarly, the position vector of E is A + (2/3)AB. So, that's 3a + (4/3)b - 2a, which simplifies to a + (4/3)b.Okay, so now I have the position vectors of D and E. Now, I need to find vectors DE, CD, and CE.Starting with DE. Vector DE is from D to E, so it's E - D. So, subtracting the position vector of D from E.E is a + (4/3)b, and D is 2a + (2/3)b. So, DE = (a + (4/3)b) - (2a + (2/3)b) = a - 2a + (4/3)b - (2/3)b = (-a) + (2/3)b. So, DE is (-a) + (2/3)b, which can also be written as (2/3)b - a.Next, vector CD. Vector CD is from C to D. Since C is the origin, the position vector of D is the same as vector CD. So, CD is 2a + (2/3)b.Similarly, vector CE is from C to E. Again, since C is the origin, the position vector of E is the same as vector CE. So, CE is a + (4/3)b.Wait, let me double-check these calculations to make sure I didn't make any mistakes.Starting with vector AB: B - A = 2b - 3a. Correct.Position vector of D: A + (1/3)AB = 3a + (1/3)(2b - 3a) = 3a + (2/3)b - a = 2a + (2/3)b. Correct.Position vector of E: A + (2/3)AB = 3a + (2/3)(2b - 3a) = 3a + (4/3)b - 2a = a + (4/3)b. Correct.Vector DE: E - D = (a + (4/3)b) - (2a + (2/3)b) = -a + (2/3)b. Correct.Vector CD: Since C is origin, CD is position vector of D: 2a + (2/3)b. Correct.Vector CE: Similarly, position vector of E: a + (4/3)b. Correct.So, all the vectors seem to check out. I think that's it. So, DE is (2/3)b - a, CD is 2a + (2/3)b, and CE is a + (4/3)b.Just to make sure, let me think about the directions. Vector DE goes from D to E, which is towards B, so it should have a component in the direction of b and against a, which matches (2/3)b - a. Vector CD goes from C to D, which is towards A and a bit towards B, so 2a + (2/3)b makes sense. Similarly, CE goes from C to E, which is more towards B, so a + (4/3)b is appropriate.Yeah, I think I've got it right. So, the final answers should be:DE = (2/3)b - aCD = 2a + (2/3)bCE = a + (4/3)b**Final Answer**The vectors are expressed as:[overrightarrow{DE} = boxed{dfrac{2}{3}overrightarrow{b} - overrightarrow{a}}, quadoverrightarrow{CD} = boxed{2overrightarrow{a} + dfrac{2}{3}overrightarrow{b}}, quadoverrightarrow{CE} = boxed{overrightarrow{a} + dfrac{4}{3}overrightarrow{b}}]