answer：Okay, so I have this problem: Find all natural numbers ( n geq 2 ) such that ( 20^n + 19^n ) is divisible by ( 20^{n-2} + 19^{n-2} ). Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I need to understand what it means for one number to be divisible by another. If ( 20^n + 19^n ) is divisible by ( 20^{n-2} + 19^{n-2} ), that means when I divide ( 20^n + 19^n ) by ( 20^{n-2} + 19^{n-2} ), there should be no remainder. In other words, ( 20^{n-2} + 19^{n-2} ) should be a factor of ( 20^n + 19^n ).Let me write that out mathematically:[20^n + 19^n = k cdot (20^{n-2} + 19^{n-2})]where ( k ) is some integer.I think I can factor out ( 20^{n-2} ) and ( 19^{n-2} ) from both terms. Let me try that.Starting with ( 20^n ), that's the same as ( 20^{n-2} cdot 20^2 ), right? Similarly, ( 19^n ) is ( 19^{n-2} cdot 19^2 ). So, substituting back in:[20^n + 19^n = 20^{n-2} cdot 20^2 + 19^{n-2} cdot 19^2]Which simplifies to:[20^n + 19^n = 20^{n-2} cdot 400 + 19^{n-2} cdot 361]Now, let me factor out ( 20^{n-2} + 19^{n-2} ) from this expression. Hmm, how can I do that?Let me think. If I have ( a cdot x + b cdot y ), and I want to factor out ( x + y ), I need to find some multiple that relates them. Maybe I can write it as:[20^{n-2} cdot 400 + 19^{n-2} cdot 361 = (20^{n-2} + 19^{n-2}) cdot k]So, I need to find ( k ) such that this equation holds. Let me rearrange the equation:[(20^{n-2} + 19^{n-2}) cdot k = 20^{n-2} cdot 400 + 19^{n-2} cdot 361]Let me factor out ( 20^{n-2} ) and ( 19^{n-2} ) on the right side:[(20^{n-2} + 19^{n-2}) cdot k = 20^{n-2} cdot 400 + 19^{n-2} cdot 361]Hmm, so if I factor ( 20^{n-2} ) from the first term and ( 19^{n-2} ) from the second term, I can write:[(20^{n-2} + 19^{n-2}) cdot k = 20^{n-2} cdot (400) + 19^{n-2} cdot (361)]Wait, maybe I can write this as:[(20^{n-2} + 19^{n-2}) cdot k = 20^{n-2} cdot (400) + 19^{n-2} cdot (361)]So, if I divide both sides by ( 20^{n-2} + 19^{n-2} ), I get:[k = frac{20^{n-2} cdot 400 + 19^{n-2} cdot 361}{20^{n-2} + 19^{n-2}}]Hmm, that seems a bit complicated. Maybe I can express this as:[k = frac{20^{n-2} cdot 400 + 19^{n-2} cdot 361}{20^{n-2} + 19^{n-2}} = frac{20^{n-2} cdot (400) + 19^{n-2} cdot (361)}{20^{n-2} + 19^{n-2}}]I wonder if I can split this fraction into two parts. Let me try:[k = frac{20^{n-2} cdot 400}{20^{n-2} + 19^{n-2}} + frac{19^{n-2} cdot 361}{20^{n-2} + 19^{n-2}}]Hmm, not sure if that helps. Maybe I can factor out 39 from somewhere? Wait, 400 - 361 is 39. Let me check:[400 - 361 = 39]Yes, that's correct. Maybe I can use that somehow. Let me try to write the numerator as:[20^{n-2} cdot 400 + 19^{n-2} cdot 361 = 20^{n-2} cdot (361 + 39) + 19^{n-2} cdot 361]Wait, that might not be helpful. Alternatively, maybe I can write the numerator as:[20^{n-2} cdot 400 + 19^{n-2} cdot 361 = 20^{n-2} cdot 400 + 19^{n-2} cdot 361]Hmm, perhaps I can factor out 19^{n-2} from both terms? Let me try:[20^{n-2} cdot 400 + 19^{n-2} cdot 361 = 19^{n-2} cdot (something) + 20^{n-2} cdot (something)]Not sure. Maybe I should consider specific values of ( n ) to see if I can spot a pattern.Let's start with ( n = 2 ):[20^2 + 19^2 = 400 + 361 = 761]And ( 20^{0} + 19^{0} = 1 + 1 = 2 ). So, is 761 divisible by 2? Well, 761 divided by 2 is 380.5, which is not an integer. So, ( n = 2 ) doesn't work.Next, ( n = 3 ):[20^3 + 19^3 = 8000 + 6859 = 14859]And ( 20^{1} + 19^{1} = 20 + 19 = 39 ). Is 14859 divisible by 39? Let me check:[14859 ÷ 39 = 381]Yes, because 39 × 381 = 14859. So, ( n = 3 ) works.Now, ( n = 4 ):[20^4 + 19^4 = 160000 + 130321 = 290321]And ( 20^{2} + 19^{2} = 400 + 361 = 761 ). Is 290321 divisible by 761? Let me check:[290321 ÷ 761 ≈ 381.5]Wait, 761 × 381 = 290, let me calculate 761 × 380 = 761 × 300 = 228,300; 761 × 80 = 60,880; so total is 228,300 + 60,880 = 289,180. Then 761 × 381 = 289,180 + 761 = 290, 941. Wait, that's more than 290,321. Hmm, so 761 × 381 is 290,941, which is more than 290,321. So, 290,321 is not divisible by 761. Therefore, ( n = 4 ) doesn't work.Wait, maybe I made a mistake in calculation. Let me double-check:20^4 is 160,000, 19^4 is 130,321, so total is 160,000 + 130,321 = 290,321. Correct.20^{2} + 19^{2} = 400 + 361 = 761. Correct.Now, 761 × 380 = 761 × 300 = 228,300; 761 × 80 = 60,880; total 228,300 + 60,880 = 289,180. Then 761 × 381 = 289,180 + 761 = 290, 941. So, 290,941 is more than 290,321, so 290,321 is not divisible by 761. So, ( n = 4 ) doesn't work.Next, ( n = 5 ):[20^5 + 19^5 = 3,200,000 + 2,476,099 = 5,676,099]And ( 20^{3} + 19^{3} = 8000 + 6859 = 14,859 ). Is 5,676,099 divisible by 14,859? Let me check:[5,676,099 ÷ 14,859 ≈ 382.5]Hmm, 14,859 × 382 = let's calculate 14,859 × 300 = 4,457,700; 14,859 × 80 = 1,188,720; 14,859 × 2 = 29,718. So total is 4,457,700 + 1,188,720 = 5,646,420 + 29,718 = 5,676,138. That's more than 5,676,099. So, 14,859 × 382 = 5,676,138, which is 39 more than 5,676,099. So, it's not divisible. Therefore, ( n = 5 ) doesn't work.Hmm, so far only ( n = 3 ) works. Let me try ( n = 6 ):[20^6 + 19^6 = 64,000,000 + 47,045,881 = 111,045,881]And ( 20^{4} + 19^{4} = 160,000 + 130,321 = 290,321 ). Is 111,045,881 divisible by 290,321? Let me check:[111,045,881 ÷ 290,321 ≈ 382.5]Wait, 290,321 × 382 = let's calculate 290,321 × 300 = 87,096,300; 290,321 × 80 = 23,225,680; 290,321 × 2 = 580,642. So total is 87,096,300 + 23,225,680 = 110,321,980 + 580,642 = 110,902,622. That's less than 111,045,881. The difference is 111,045,881 - 110,902,622 = 143,259. So, not divisible. Therefore, ( n = 6 ) doesn't work.Hmm, seems like only ( n = 3 ) works so far. Maybe I should try to generalize this.Let me go back to the original expression:[20^n + 19^n = k cdot (20^{n-2} + 19^{n-2})]I can rewrite this as:[20^n + 19^n = k cdot 20^{n-2} + k cdot 19^{n-2}]Let me bring all terms to one side:[20^n - k cdot 20^{n-2} + 19^n - k cdot 19^{n-2} = 0]Factor out ( 20^{n-2} ) and ( 19^{n-2} ):[20^{n-2} (20^2 - k) + 19^{n-2} (19^2 - k) = 0]Which simplifies to:[20^{n-2} (400 - k) + 19^{n-2} (361 - k) = 0]Hmm, so:[20^{n-2} (400 - k) = -19^{n-2} (361 - k)]Or:[20^{n-2} (400 - k) = 19^{n-2} (k - 361)]Let me divide both sides by ( 19^{n-2} ):[left( frac{20}{19} right)^{n-2} (400 - k) = k - 361]Let me denote ( r = frac{20}{19} ), which is approximately 1.0526. So, the equation becomes:[r^{n-2} (400 - k) = k - 361]Let me solve for ( k ):[r^{n-2} (400 - k) = k - 361]Bring all terms to one side:[r^{n-2} cdot 400 - r^{n-2} k - k + 361 = 0]Factor out ( k ):[r^{n-2} cdot 400 + 361 - k (r^{n-2} + 1) = 0]So,[k (r^{n-2} + 1) = r^{n-2} cdot 400 + 361]Therefore,[k = frac{r^{n-2} cdot 400 + 361}{r^{n-2} + 1}]Hmm, that's a bit complicated, but maybe I can simplify it. Let me write ( r^{n-2} ) as ( left( frac{20}{19} right)^{n-2} ). So,[k = frac{left( frac{20}{19} right)^{n-2} cdot 400 + 361}{left( frac{20}{19} right)^{n-2} + 1}]Let me multiply numerator and denominator by ( 19^{n-2} ) to eliminate the fraction:[k = frac{20^{n-2} cdot 400 + 19^{n-2} cdot 361}{20^{n-2} + 19^{n-2}}]Wait, that's the same expression I had earlier. So, I'm back to where I started.Maybe I can consider the ratio ( frac{20^{n} + 19^{n}}{20^{n-2} + 19^{n-2}} ) and see if it's an integer. Let me write this ratio as:[frac{20^{n} + 19^{n}}{20^{n-2} + 19^{n-2}} = frac{20^{n-2} cdot 20^2 + 19^{n-2} cdot 19^2}{20^{n-2} + 19^{n-2}}]Which simplifies to:[frac{20^{n-2} cdot 400 + 19^{n-2} cdot 361}{20^{n-2} + 19^{n-2}}]Let me denote ( a = 20^{n-2} ) and ( b = 19^{n-2} ). Then the ratio becomes:[frac{a cdot 400 + b cdot 361}{a + b}]I can write this as:[frac{400a + 361b}{a + b}]Hmm, maybe I can express this as a weighted average or something. Let me see:[frac{400a + 361b}{a + b} = frac{400a + 361b}{a + b} = frac{400a + 400b - 39b}{a + b} = 400 - frac{39b}{a + b}]Ah, that's interesting. So,[frac{400a + 361b}{a + b} = 400 - frac{39b}{a + b}]Therefore, for the ratio to be an integer, ( frac{39b}{a + b} ) must be an integer. Since ( a = 20^{n-2} ) and ( b = 19^{n-2} ), we have:[frac{39 cdot 19^{n-2}}{20^{n-2} + 19^{n-2}} text{ must be an integer}]So, ( 20^{n-2} + 19^{n-2} ) must divide ( 39 cdot 19^{n-2} ). Let me write that as:[20^{n-2} + 19^{n-2} mid 39 cdot 19^{n-2}]Which means that ( 20^{n-2} + 19^{n-2} ) is a divisor of ( 39 cdot 19^{n-2} ).Now, since ( 20^{n-2} ) and ( 19^{n-2} ) are coprime (because 20 and 19 are coprime), the sum ( 20^{n-2} + 19^{n-2} ) must divide 39. Because ( 20^{n-2} + 19^{n-2} ) divides ( 39 cdot 19^{n-2} ), and since ( 20^{n-2} + 19^{n-2} ) is coprime with ( 19^{n-2} ) (as 20 and 19 are coprime), it must divide 39.So, ( 20^{n-2} + 19^{n-2} ) must be a divisor of 39. Let's list the positive divisors of 39:1, 3, 13, 39.So, ( 20^{n-2} + 19^{n-2} ) must be equal to one of these numbers.Let's check each case:1. ( 20^{n-2} + 19^{n-2} = 1 ) - Since ( 20^{n-2} ) and ( 19^{n-2} ) are both at least 1 (for ( n geq 2 )), their sum is at least 2. So, this case is impossible.2. ( 20^{n-2} + 19^{n-2} = 3 ) - Let's see if there exists ( n ) such that ( 20^{n-2} + 19^{n-2} = 3 ). - For ( n = 2 ): ( 20^{0} + 19^{0} = 1 + 1 = 2 ) (too small) - For ( n = 3 ): ( 20^{1} + 19^{1} = 20 + 19 = 39 ) (too big) - So, no solution here.3. ( 20^{n-2} + 19^{n-2} = 13 ) - Let's check: - For ( n = 2 ): 2 (too small) - For ( n = 3 ): 39 (too big) - For ( n = 4 ): ( 20^{2} + 19^{2} = 400 + 361 = 761 ) (way too big) - So, no solution here either.4. ( 20^{n-2} + 19^{n-2} = 39 ) - Let's check: - For ( n = 3 ): ( 20^{1} + 19^{1} = 20 + 19 = 39 ) (bingo!) - For ( n = 4 ): 761 (too big) - So, only ( n = 3 ) satisfies this condition.Therefore, the only natural number ( n geq 2 ) that satisfies the given condition is ( n = 3 ).

answer：Let's denote the two nonzero real numbers as ( x ) and ( y ). According to the problem, their sum is 8 times their product. This can be written as:[x + y = 8xy]We need to find the sum of the reciprocals of these two numbers, which is:[frac{1}{x} + frac{1}{y}]To find this sum, we can manipulate the original equation. Let's start by dividing both sides of the equation ( x + y = 8xy ) by ( xy ):[frac{x}{xy} + frac{y}{xy} = frac{8xy}{xy}]Simplifying each term, we get:[frac{1}{y} + frac{1}{x} = 8]This shows that the sum of the reciprocals of ( x ) and ( y ) is 8.Therefore, the sum of the reciprocals of these two numbers is (boxed{8}).

answer：First, I need to determine the total number of heads in the caravan. This includes the heads of all the animals and the keepers.Next, I'll calculate the total number of feet. Each type of animal and the keepers have a specific number of feet:- Hens have 2 feet each.- Goats have 4 feet each.- Camels have 4 feet each.- Keepers have 2 feet each.After finding the total number of feet, I'll subtract the total number of heads from the total number of feet to find out how many more feet there are than heads in the caravan.

answer：Okay, so I'm trying to figure out this clock problem. It says that certain positions of the hour and minute hands are possible, like the hour hand halfway between 5 and 6 and the minute hand exactly at 6. But other positions are impossible, like the hour hand exactly at 5 and the minute hand exactly at 6. The question is asking how many different positions are there that would remain possible if the hour and minute hands were switched.Hmm, okay, so first, I need to understand what makes a position possible or impossible. I know that the hour and minute hands move at different speeds. The minute hand moves faster than the hour hand. Specifically, the minute hand makes a full revolution every hour, while the hour hand takes 12 hours to make a full revolution. So, the minute hand is 12 times faster than the hour hand.Now, if we think about the positions of the hands, each position can be represented by the angles of the hour and minute hands from 12 o'clock. Let's denote the time elapsed since 12:00 as ( t ) hours. Then, the angle of the hour hand from 12 o'clock is ( 30t ) degrees (since ( 360/12 = 30 ) degrees per hour). The angle of the minute hand is ( 360t ) degrees (since it makes a full circle every hour).But since the clock is circular, angles are modulo 360 degrees. So, the position of the hour hand is ( 30t mod 360 ) and the minute hand is ( 360t mod 360 ), which simplifies to ( 0 ) degrees because ( 360t ) is a multiple of 360. Wait, that can't be right because the minute hand moves continuously. Maybe I should think in terms of fractions of a circle instead of degrees.Let me try again. Let's represent the position of each hand as a fraction of the circle. For the hour hand, after ( t ) hours, it has moved ( t/12 ) of the circle. For the minute hand, after ( t ) hours, it has moved ( t ) of the circle. So, their positions can be represented as ( t/12 ) and ( t ) modulo 1, since the circle is 1 full revolution.Now, the problem is asking about positions where if we switch the hour and minute hands, the configuration remains possible. That means, if originally the hour hand is at position ( t/12 ) and the minute hand is at position ( t ), after switching, the hour hand should be at ( t ) and the minute hand should be at ( t/12 ), and this new configuration should also be possible for some time ( t' ).So, we need to find all ( t ) such that there exists a ( t' ) where:- The original hour hand position ( t/12 ) equals the new minute hand position ( t'/12 ).- The original minute hand position ( t ) equals the new hour hand position ( t' ).So, setting up the equations:1. ( t/12 = t'/12 )2. ( t = t' )Wait, that seems too straightforward. If ( t = t' ), then substituting into the first equation, ( t/12 = t/12 ), which is always true. So, does that mean every position is possible when switching the hands? That can't be right because the problem states that some positions are impossible when switched.Maybe I'm missing something. Let's think differently. When we switch the hands, the new hour hand position should correspond to some time ( t' ), and the new minute hand position should correspond to the same time ( t' ). So, the new hour hand position is ( t ), and the new minute hand position is ( t/12 ). These should satisfy the relationship that the minute hand is 12 times faster than the hour hand.So, for the new configuration to be possible, the minute hand position ( t/12 ) should be equal to ( 12 times ) the hour hand position ( t ) modulo 1. That is:[ t/12 equiv 12t mod 1 ]Let me write that equation:[ t/12 = 12t + k ]for some integer ( k ).But since both ( t ) and ( t/12 ) are fractions between 0 and 1 (because they represent positions on the clock), ( k ) must be 0. So:[ t/12 = 12t ][ t/12 - 12t = 0 ][ t(1/12 - 12) = 0 ][ t(-143/12) = 0 ]So, ( t = 0 ).But ( t = 0 ) corresponds to 12:00, which is a valid position, but it's just one position. The problem is asking for how many different positions remain possible when switching the hands, not just at 12:00.Wait, maybe I made a mistake in setting up the equation. Let's try again.If we switch the hands, the new hour hand position is ( t ) and the new minute hand position is ( t/12 ). For this new configuration to be possible, the minute hand must be 12 times the hour hand position modulo 1. So:[ t/12 equiv 12 times t mod 1 ][ t/12 equiv 12t mod 1 ]Let me write this as:[ 12t equiv t/12 mod 1 ]Multiply both sides by 12 to eliminate the fraction:[ 144t equiv t mod 12 ][ 144t - t equiv 0 mod 12 ][ 143t equiv 0 mod 12 ]So, ( 143t ) must be an integer multiple of 12. Since 143 and 12 are coprime (their greatest common divisor is 1), this implies that ( t ) must be a multiple of ( 12/143 ).Therefore, ( t = frac{12k}{143} ) for ( k = 0, 1, 2, ldots, 142 ).But since the clock is periodic with a 12-hour cycle, we only need to consider ( t ) in the interval [0, 12). So, ( k ) ranges from 0 to 142, giving 143 distinct values of ( t ).However, we need to check if these positions are distinct when switching the hands. Since each ( t ) gives a unique configuration, and switching the hands corresponds to a different ( t' ), but the positions themselves are unique.Therefore, there are 143 different positions where switching the hour and minute hands results in another valid position.Wait, but earlier I thought it might be 11. Maybe I confused something. Let me double-check.If we consider the equation ( 143t equiv 0 mod 12 ), the solutions are ( t = frac{12k}{143} ) for ( k = 0, 1, ldots, 142 ). Each ( k ) gives a distinct ( t ) in [0, 12). Therefore, there are 143 distinct positions.Yes, that makes sense. So, the answer should be 143.

answer：First, I need to determine how many bananas each person has in their basket after eating some.For Raj:- Raj ate 120 bananas.- He has twice as many bananas remaining in his basket.- Therefore, Raj has 2 * 120 = 240 bananas in his basket.For Asha:- Asha ate 100 bananas.- She has thrice as many bananas remaining in her basket.- Therefore, Asha has 3 * 100 = 300 bananas in her basket.For Vijay:- Vijay ate 80 bananas.- He has four times as many bananas remaining in his basket.- Therefore, Vijay has 4 * 80 = 320 bananas in his basket.Next, I'll add up the bananas each person has in their basket and the ones left on the tree to find the initial number of bananas.Bananas in Raj's basket: 240Bananas in Asha's basket: 300Bananas in Vijay's basket: 320Bananas left on the tree: 430Initial number of bananas = 240 + 300 + 320 + 430 = 1290So, there were initially 1290 bananas on the tree.

answer：Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. First, the problem says it's a positive geometric sequence {a_n}. I remember that in a geometric sequence, each term is multiplied by a common ratio to get the next term. So, if the first term is a_1 and the common ratio is r, then a_n = a_1 * r^(n-1). The problem gives me that a_1008 * a_1009 = 1/100. Hmm, okay. So, let me write that down:a_1008 * a_1009 = 1/100Since it's a geometric sequence, I can express a_1008 and a_1009 in terms of a_1 and r. a_1008 = a_1 * r^(1007)a_1009 = a_1 * r^(1008)So, multiplying them together:a_1008 * a_1009 = (a_1 * r^(1007)) * (a_1 * r^(1008)) = a_1^2 * r^(2015)And this equals 1/100. So,a_1^2 * r^(2015) = 1/100Okay, so that's one equation. Now, the question is asking for the sum of the logarithms of the terms from a_1 to a_2016. Specifically:lg a_1 + lg a_2 + ... + lg a_2016I remember that the sum of logarithms is the logarithm of the product. So, this sum is equal to:lg(a_1 * a_2 * ... * a_2016)So, I need to find the product of all terms from a_1 to a_2016 and then take the logarithm base 10 of that product.Let me think about the product of the terms in a geometric sequence. The product of the terms from a_1 to a_n in a geometric sequence can be expressed as (a_1 * a_n)^(n/2). Wait, is that right? Hmm, actually, that formula is for the sum of the terms, not the product. Wait, no, for the product, maybe it's different. Let me think. If I have a geometric sequence, the product of the terms can be written as (a_1 * a_2 * ... * a_n). Since each term is a multiple of the previous term by r, the product would be a_1^n * r^(1 + 2 + ... + (n-1)).Because each term is a_1, a_1*r, a_1*r^2, ..., a_1*r^(n-1). So, multiplying them all together:Product = a_1 * (a_1*r) * (a_1*r^2) * ... * (a_1*r^(n-1)) = a_1^n * r^(0 + 1 + 2 + ... + (n-1))The exponent on r is the sum of the first (n-1) integers, which is (n-1)*n/2. So, the product is:Product = a_1^n * r^((n(n-1))/2)Okay, so in this problem, n is 2016. So, the product of the terms from a_1 to a_2016 is:Product = a_1^2016 * r^((2016*2015)/2)Hmm, that seems a bit complicated, but maybe I can relate it to the given information.Earlier, I had the equation:a_1^2 * r^(2015) = 1/100Let me write that again:a_1^2 * r^(2015) = 1/100So, if I can express the product in terms of this, maybe I can find the value.Looking back at the product:Product = a_1^2016 * r^(2016*2015/2)Let me factor this expression:Product = (a_1^2)^1008 * (r^(2015))^(1008)Because 2016 is 2*1008, so a_1^2016 is (a_1^2)^1008, and r^(2016*2015/2) is (r^(2015))^(1008).So, Product = (a_1^2 * r^(2015))^1008But from the given equation, a_1^2 * r^(2015) = 1/100. So, substituting that in:Product = (1/100)^1008So, the product of all terms from a_1 to a_2016 is (1/100)^1008.Now, taking the logarithm base 10 of this product:lg(Product) = lg((1/100)^1008) = 1008 * lg(1/100)I know that lg(1/100) is the same as lg(10^(-2)) which is -2. So,lg(Product) = 1008 * (-2) = -2016Therefore, the sum of the logarithms is -2016.Wait, let me double-check that. So, the product is (1/100)^1008, which is 10^(-2016). Then, lg(10^(-2016)) is indeed -2016. Yeah, that makes sense.So, the answer should be -2016, which is option D.I think that's correct. I went through the steps, expressed the product in terms of the given information, used logarithm properties, and everything seems consistent.**Final Answer**boxed{D}