answer：To determine the smallest number among 1, 0, -2, and -3, I will compare each number based on their numerical value.First, I recognize that negative numbers are smaller than positive numbers. Therefore, -2 and -3 are smaller than 1 and 0.Next, I compare the two negative numbers: -2 and -3. Since -3 is further to the left on the number line compared to -2, it has a smaller value.Thus, the smallest number among the four is -3.

answer：Okay, so I have this problem here: Among 8 consecutive natural numbers, there are k numbers such that any two of them are relatively prime. I need to find the maximum value of k, and the options are 3, 4, 5, or 6. Hmm, interesting.First, let me make sure I understand the problem correctly. We have 8 consecutive natural numbers, say n, n+1, n+2, ..., n+7. Among these, we need to find the largest possible k such that there are k numbers where every pair is relatively prime. That is, the greatest common divisor (GCD) of any two numbers in this subset is 1.So, my goal is to figure out the maximum k possible. The options go up to 6, so maybe it's 4 or 5? I need to think about how to approach this.I remember that in any set of consecutive numbers, there are certain properties related to divisibility and prime numbers. For example, among any two consecutive numbers, they are always coprime because consecutive integers are coprime. Similarly, numbers that are two apart might share a common divisor, like 2 and 4 share 2, but 3 and 5 don't share any common divisors except 1.So, maybe I can try to construct a subset of these 8 numbers where each number is as far apart as possible to minimize the chance of sharing common factors. But since they are consecutive, they are only 1 apart, so that might not help directly.Wait, another thought: if I can include numbers that are primes, that might help because primes are only divisible by 1 and themselves. So, if I include several primes in my subset, they might be coprime with each other. But primes aren't necessarily coprime with all numbers; for example, 2 and 4 are not coprime, even though 2 is prime.So, perhaps I need to include primes and numbers that are not multiples of any smaller primes. Hmm, but in 8 consecutive numbers, how many primes can I expect? It depends on where the sequence starts. For example, if I start at 2, the numbers are 2,3,4,5,6,7,8,9. Here, primes are 2,3,5,7. So, four primes. But 2 and 4 are not coprime, 2 and 6 are not coprime, etc. So, just taking all primes might not work because some of them might share factors with non-prime numbers in the subset.Wait, actually, in this case, the primes themselves are 2,3,5,7, which are all primes, but 2 is even, so any even number in the subset would conflict with 2. So, if I include 2, I can't include any other even numbers. Similarly, if I include 3, I can't include 6 or 9, which are multiples of 3.So, maybe the strategy is to include as many primes as possible without including their multiples. Let me try with the sequence starting at 2: 2,3,4,5,6,7,8,9.If I include 2, I can't include 4,6,8. So, the remaining numbers are 3,5,7,9. Now, 3 and 9 are multiples of 3, so if I include 3, I can't include 9. So, from 3,5,7,9, I can include 3,5,7 or 5,7,9. But 5 and 7 are primes, so they are coprime with everything. 3 and 5 are coprime, 3 and 7 are coprime, 5 and 7 are coprime. So, if I include 2,3,5,7, that's four numbers, and they are all pairwise coprime.Wait, but 2 and 3 are coprime, 2 and 5 are coprime, 2 and 7 are coprime, 3 and 5 are coprime, 3 and 7 are coprime, 5 and 7 are coprime. So, that's four numbers. So, in this case, k is 4.But is this the maximum? Let me check another sequence. Maybe starting at a higher number where there are more primes.Wait, for example, starting at 14: 14,15,16,17,18,19,20,21.Primes here are 17,19. So, only two primes. So, that's worse. Maybe starting at 1: 1,2,3,4,5,6,7,8.Here, primes are 2,3,5,7. So, same as before. Including 2,3,5,7. So, k=4.Wait, but 1 is a natural number. Is 1 considered? Because 1 is coprime with every number. So, if I include 1, I can include more numbers. Let me see.In the sequence 1,2,3,4,5,6,7,8, if I include 1, then I can include 2,3,5,7 as well, because 1 is coprime with all. So, that's 5 numbers: 1,2,3,5,7. Are they pairwise coprime? Let's check:1 and 2: GCD 11 and 3: GCD 11 and 5: GCD 11 and 7: GCD 12 and 3: GCD 12 and 5: GCD 12 and 7: GCD 13 and 5: GCD 13 and 7: GCD 15 and 7: GCD 1So, yes, all pairs are coprime. So, that's 5 numbers. So, k=5.Wait, so in this case, starting at 1, I can get k=5. But is this always possible? Or does it depend on the starting number?Wait, if I start at 1, I have 1, which is special because it's coprime with everything. So, maybe that allows me to include more numbers.But in the problem statement, it's about "8 consecutive natural numbers." So, does it specify starting from 1? No, it's any 8 consecutive natural numbers. So, depending on where you start, you might have different numbers of coprimes.So, in some sequences, you might have 1, which allows you to include more numbers, but in others, you might not.Wait, but 1 is only in the sequence if the starting number is 1. So, if the sequence is from n to n+7, and n=1, then 1 is included. Otherwise, it's not.So, in the case where n=1, we can have k=5. But in other sequences, like n=2, we can only get k=4.But the question is asking for the maximum value of k among 8 consecutive natural numbers. So, if in some sequences, you can have k=5, then 5 is a possible answer. But is 5 the maximum? Or can you get higher?Wait, let me think. If I have 8 consecutive numbers, is it possible to have 6 numbers that are pairwise coprime?Let me try to see. Let's take a sequence where 1 is included, like 1,2,3,4,5,6,7,8. If I include 1,2,3,5,7, what about 4? 4 is 2 squared, so 4 and 2 share a common factor. 4 and 1 are coprime, but 4 and 2 are not. So, if I include 4, I have to exclude 2. Similarly, 6 is a multiple of 2 and 3, so including 6 would require excluding both 2 and 3.So, if I want to include 1, I can include 1, and then other numbers that are coprime with each other.Wait, let me try to include 1,2,3,5,7, and maybe 4. But 4 and 2 are not coprime, so I can't include both. Alternatively, 1,3,4,5,7, something else? 1,3,4,5,7, maybe 8? 8 is 2^3, so 8 and 4 share a common factor, 8 and 2 share a common factor. So, 8 can't be included with 2 or 4.Alternatively, 1,3,5,7, and then maybe 8? 8 is coprime with 3,5,7, but 8 is even, so if I include 8, I can't include 2 or 4 or 6. But 1 is already included, so 8 can be included as long as we don't include 2,4,6.But 1,3,5,7,8: are these pairwise coprime? Let's check:1 is coprime with all.3 and 5: GCD 13 and 7: GCD 13 and 8: GCD 15 and 7: GCD 15 and 8: GCD 17 and 8: GCD 1So, yes, 1,3,5,7,8 are all pairwise coprime. That's 5 numbers.Wait, so in this case, we have 5 numbers. So, same as before.But can we get 6? Let's see.If I try to include another number, say 9, but 9 is not in the sequence. Wait, the sequence is 1,2,3,4,5,6,7,8. So, 9 is not there. Alternatively, can I include 6? 6 is a multiple of 2 and 3, so if I include 6, I can't include 2 or 3. So, if I include 6, I have to exclude 2 and 3. So, let's see:1,5,6,7,8. Let's check:1 is coprime with all.5 and 6: GCD 15 and 7: GCD 15 and 8: GCD 16 and 7: GCD 16 and 8: GCD 2, which is not 1. So, 6 and 8 share a common factor. So, they are not coprime. So, I can't include both 6 and 8.Alternatively, 1,5,6,7. That's 4 numbers. Not better than before.Alternatively, 1,3,5,7,8, and maybe something else. Wait, 1,3,5,7,8, and 4? 4 and 3: GCD 1, 4 and 5: GCD 1, 4 and 7: GCD 1, 4 and 8: GCD 4, which is not 1. So, can't include both 4 and 8.Alternatively, 1,3,5,7,4, something else. 1,3,4,5,7: 5 numbers. If I include 6, as before, conflicts with 3. If I include 2, conflicts with 4. So, no, can't get 6.Alternatively, maybe another sequence where 1 is not included. Let's take n=2: 2,3,4,5,6,7,8,9.If I try to include 2,3,5,7,9. Let's check:2 and 3: GCD 12 and 5: GCD 12 and 7: GCD 12 and 9: GCD 13 and 5: GCD 13 and 7: GCD 13 and 9: GCD 3, which is not 1. So, can't include both 3 and 9.So, if I include 3, I can't include 9. So, 2,3,5,7: 4 numbers.Alternatively, 2,5,7,9: 4 numbers.Alternatively, 3,5,7,9: 4 numbers.Alternatively, 5,7: only 2 numbers.Alternatively, 2,5,7, and maybe 4? 2 and 4: GCD 2, so no.Alternatively, 3,5,7, and maybe 4? 3 and 4: GCD 1, 5 and 4: GCD 1, 7 and 4: GCD 1. So, 3,4,5,7: 4 numbers.So, same as before.Alternatively, 5,7, and then maybe 8? 5 and 8: GCD 1, 7 and 8: GCD 1, but 8 is even, so if I include 8, I can't include 2 or 4 or 6.So, 5,7,8: 3 numbers.Alternatively, 5,7,8,9: 5 and 9: GCD 1, 7 and 9: GCD 1, 8 and 9: GCD 1. So, 5,7,8,9: 4 numbers.So, again, 4 numbers.So, in this case, starting at 2, the maximum k is 4.Wait, so in the first case, starting at 1, we can get k=5, but starting at 2, we can only get k=4. So, the maximum k is 5.But wait, let me try another sequence where 1 is not included, but maybe there are more primes.Wait, for example, starting at 14: 14,15,16,17,18,19,20,21.Primes here are 17,19. So, only two primes. So, not helpful.Alternatively, starting at 24: 24,25,26,27,28,29,30,31.Primes here: 29,31. So, still only two primes.Alternatively, starting at 11: 11,12,13,14,15,16,17,18.Primes: 11,13,17. So, three primes.So, if I include 11,13,17, and maybe 12? 12 is 2^2*3, so 12 and 11: GCD 1, 12 and 13: GCD 1, 12 and 17: GCD 1. So, 11,12,13,17: 4 numbers.Alternatively, 11,13,17, and 14: 14 is 2*7. 14 and 11: GCD 1, 14 and 13: GCD 1, 14 and 17: GCD 1. So, 11,13,14,17: 4 numbers.Alternatively, 11,13,17, and 15: 15 is 3*5. 15 and 11: GCD 1, 15 and 13: GCD 1, 15 and 17: GCD 1. So, 11,13,15,17: 4 numbers.Alternatively, 11,13,17, and 16: 16 is 2^4. 16 and 11: GCD 1, 16 and 13: GCD 1, 16 and 17: GCD 1. So, 11,13,16,17: 4 numbers.So, in this case, we can get 4 numbers.Alternatively, including 12,14,15,16: but 12 and 14 share 2, 12 and 15 share 3, 14 and 15 share nothing, 14 and 16 share 2, 15 and 16 share nothing. So, not all pairwise coprime.So, seems like 4 is the maximum here.Wait, so in some sequences, starting at 1, we can get k=5, but in others, only k=4. So, is 5 the maximum possible? Or is there a sequence where k can be higher?Wait, let me think about another sequence where 1 is included. For example, starting at 1: 1,2,3,4,5,6,7,8.As I saw earlier, including 1,2,3,5,7 gives 5 numbers. Alternatively, including 1,3,4,5,7,8: let's check.1 is coprime with all.3 and 4: GCD 13 and 5: GCD 13 and 7: GCD 13 and 8: GCD 14 and 5: GCD 14 and 7: GCD 14 and 8: GCD 4, which is not 1. So, 4 and 8 can't both be included.So, if I include 1,3,4,5,7: 5 numbers.Alternatively, 1,3,5,7,8: 5 numbers.So, same as before.Wait, can I include 1,3,5,7, and two more numbers? Let's see.If I include 1,3,5,7, and 2: 2 is coprime with 1,3,5,7. So, 1,2,3,5,7: 5 numbers.Alternatively, 1,3,5,7,4: 4 is coprime with 1,3,5,7. So, 1,3,4,5,7: 5 numbers.Alternatively, 1,3,5,7,6: 6 is 2*3. 6 and 3 share a common factor, so can't include both.Similarly, 1,3,5,7,8: 8 is 2^3, which is coprime with 3,5,7, but not with 2. So, if I include 8, I can't include 2 or 4 or 6.But 1,3,5,7,8: 5 numbers.So, seems like 5 is the maximum in this case.Is it possible to get 6? Let's see.If I include 1,2,3,5,7, and another number.Let's try 1,2,3,5,7,4: 4 and 2 share a common factor. So, can't include both.1,2,3,5,7,6: 6 and 2 share a common factor, 6 and 3 share a common factor. So, can't include both.1,2,3,5,7,8: 8 and 2 share a common factor. So, can't include both.1,2,3,5,7,9: 9 is not in the sequence. Wait, the sequence is up to 8, so 9 is not included.Alternatively, 1,3,5,7,8, and something else. 1,3,5,7,8,4: 4 and 8 share a common factor.1,3,5,7,8,6: 6 and 3 share a common factor.1,3,5,7,8, something else: 1,3,5,7,8, and 10? No, 10 is not in the sequence.Wait, the sequence is only up to 8, so 10 isn't there. So, no.Alternatively, 1,3,5,7,8, and 12? No, 12 is not in the sequence.So, seems like in this case, 5 is the maximum.Wait, but let me think again. Maybe another sequence where 1 is not included, but there are more coprimes.Wait, for example, starting at 15: 15,16,17,18,19,20,21,22.Primes here: 17,19. So, only two primes.Alternatively, starting at 23: 23,24,25,26,27,28,29,30.Primes: 23,29. So, two primes.Alternatively, starting at 16: 16,17,18,19,20,21,22,23.Primes:17,19,23. So, three primes.So, 17,19,23: three primes.Including 16: 16 is 2^4, which is coprime with 17,19,23.Including 18: 18 is 2*3^2. 18 and 16 share 2, so can't include both.Including 20: 20 is 2^2*5. 20 and 16 share 4, so can't include both.Including 22: 22 is 2*11. 22 and 16 share 2, so can't include both.Including 21: 21 is 3*7. 21 and 16: GCD 1, 21 and 17: GCD 1, 21 and 19: GCD 1, 21 and 23: GCD 1. So, 21 can be included.So, 16,17,19,21,23: 5 numbers.Wait, 16 and 21: GCD 1, 16 and 17: GCD 1, 16 and 19: GCD 1, 16 and 23: GCD 1.17 and 19: GCD 1, 17 and 21: GCD 1, 17 and 23: GCD 1.19 and 21: GCD 1, 19 and 23: GCD 1.21 and 23: GCD 1.So, yes, 16,17,19,21,23: 5 numbers.So, that's 5 numbers.Alternatively, 17,19,23, and maybe 18: 18 is 2*3^2. 18 and 17: GCD 1, 18 and 19: GCD 1, 18 and 23: GCD 1. So, 17,18,19,23: 4 numbers.Alternatively, 17,19,23,20: 20 is 2^2*5. 20 and 17: GCD 1, 20 and 19: GCD 1, 20 and 23: GCD 1. So, 17,19,20,23: 4 numbers.Alternatively, 17,19,23,22: 22 is 2*11. 22 and 17: GCD 1, 22 and 19: GCD 1, 22 and 23: GCD 1. So, 17,19,22,23: 4 numbers.Alternatively, 17,19,23,21: 21 is 3*7. 21 and 17: GCD 1, 21 and 19: GCD 1, 21 and 23: GCD 1. So, 17,19,21,23: 4 numbers.So, in this case, the maximum is 5 numbers.Wait, so in this case, starting at 16, we can get 5 numbers. So, same as starting at 1.So, seems like in some sequences, you can get 5 numbers, and in others, only 4.But the question is asking for the maximum value of k among 8 consecutive natural numbers. So, if in some sequences, you can get 5, then 5 is the maximum.But wait, the options are 3,4,5,6. So, 5 is an option, so maybe 5 is the answer.But wait, let me think again. Is 5 always possible? Or is there a case where you can't get 5?Wait, in the sequence starting at 1, we can get 5. In the sequence starting at 16, we can get 5. In the sequence starting at 2, we can only get 4. So, depending on the starting number, the maximum k varies.But the problem is asking for the maximum value of k, so it's the highest possible k across all possible sequences. So, if in some sequences, k can be 5, then 5 is the answer.But wait, wait, wait. Let me think about another sequence where maybe k can be 6.Wait, let's take the sequence starting at 1: 1,2,3,4,5,6,7,8.We saw that we can get 5 numbers: 1,2,3,5,7 or 1,3,5,7,8.But can we get 6?Let me try to include 1,2,3,5,7, and another number.If I include 1,2,3,5,7, and 4: 4 and 2 share a common factor, so can't include both.1,2,3,5,7,6: 6 and 2 share a common factor, 6 and 3 share a common factor.1,2,3,5,7,8: 8 and 2 share a common factor.1,2,3,5,7,9: 9 is not in the sequence.Alternatively, 1,3,5,7,8, and 4: 4 and 8 share a common factor.1,3,5,7,8, and 6: 6 and 3 share a common factor.So, seems like 6 is not possible.Alternatively, maybe another sequence where 1 is not included, but there are more coprimes.Wait, let me think of a sequence where there are more primes.Wait, for example, starting at 23: 23,24,25,26,27,28,29,30.Primes: 23,29. So, only two primes.Alternatively, starting at 11: 11,12,13,14,15,16,17,18.Primes:11,13,17. So, three primes.Including 11,13,17, and maybe 12: 12 is 2^2*3. 12 and 11: GCD 1, 12 and 13: GCD 1, 12 and 17: GCD 1. So, 11,12,13,17: 4 numbers.Alternatively, 11,13,17, and 14: 14 is 2*7. 14 and 11: GCD 1, 14 and 13: GCD 1, 14 and 17: GCD 1. So, 11,13,14,17: 4 numbers.Alternatively, 11,13,17, and 15: 15 is 3*5. 15 and 11: GCD 1, 15 and 13: GCD 1, 15 and 17: GCD 1. So, 11,13,15,17: 4 numbers.Alternatively, 11,13,17, and 16: 16 is 2^4. 16 and 11: GCD 1, 16 and 13: GCD 1, 16 and 17: GCD 1. So, 11,13,16,17: 4 numbers.So, in this case, 4 numbers.Alternatively, 12,14,15,16: but 12 and 14 share 2, 12 and 15 share 3, 14 and 15 share nothing, 14 and 16 share 2, 15 and 16 share nothing. So, not all pairwise coprime.So, seems like 4 is the maximum here.Wait, another thought: maybe if the sequence includes more numbers that are powers of primes, like 4=2^2, 8=2^3, 9=3^2, etc. But those are still not coprime with their bases.Alternatively, maybe numbers that are products of different primes, but in 8 consecutive numbers, it's hard to have too many such numbers.Wait, let me think about the sequence starting at 14: 14,15,16,17,18,19,20,21.Primes:17,19. So, two primes.Including 14:14 is 2*7.15:3*5.16:2^4.17: prime.18:2*3^2.19: prime.20:2^2*5.21:3*7.So, if I try to include 14,15,17,19: 14 and 15: GCD 1, 14 and 17: GCD 1, 14 and 19: GCD 1, 15 and 17: GCD 1, 15 and 19: GCD 1, 17 and 19: GCD 1. So, 14,15,17,19: 4 numbers.Alternatively, 16,17,19,21: 16 and 17: GCD 1, 16 and 19: GCD 1, 16 and 21: GCD 1, 17 and 19: GCD 1, 17 and 21: GCD 1, 19 and 21: GCD 1. So, 16,17,19,21: 4 numbers.Alternatively, 14,15,17,19,21: 14 and 15: GCD 1, 14 and 17: GCD 1, 14 and 19: GCD 1, 14 and 21: GCD 7, which is not 1. So, can't include both 14 and 21.So, seems like 4 is the maximum here.Wait, so in all these sequences, the maximum k is either 4 or 5. So, 5 is higher, so 5 is the maximum.But wait, let me think again. Is there a sequence where k=6 is possible?Wait, let's take a sequence where 1 is included, and there are more coprimes.Wait, 1,2,3,4,5,6,7,8.We saw that 1,2,3,5,7: 5 numbers.Alternatively, 1,3,4,5,7, something else.Wait, 1,3,4,5,7,8: 4 and 8 share a common factor, so can't include both.Alternatively, 1,3,5,7,8, and something else.Wait, 1,3,5,7,8, and 9: 9 is not in the sequence.Alternatively, 1,3,5,7,8, and 10: 10 is not in the sequence.So, seems like 5 is the maximum.Wait, another idea: maybe if the sequence includes more numbers that are coprime with each other, not necessarily primes.For example, in the sequence 1,2,3,4,5,6,7,8, the numbers 1,3,5,7 are primes or 1, and 4,8 are powers of 2, 6 is 2*3.But 1 is special because it's coprime with everything. So, if I include 1, I can include as many numbers as possible that are coprime with each other.But in this case, 1,2,3,5,7: 5 numbers.Alternatively, 1,3,4,5,7: 5 numbers.So, same.Wait, is there a way to include 1 and 5 numbers that are coprime with each other and with 1?Yes, as above.But 6 numbers? Let me see.If I include 1,2,3,5,7, and another number.But 2 is even, so any other even number can't be included.Similarly, 3 is a multiple of 3, so any multiple of 3 can't be included.So, in the sequence, 1,2,3,5,7: 5 numbers.If I include 4, which is even, conflicts with 2.If I include 6, which is multiple of 2 and 3, conflicts with both.If I include 8, which is even, conflicts with 2.So, can't include any more numbers without conflict.So, 5 is the maximum.Wait, but in the sequence starting at 1, we can include 1,2,3,5,7, which are 5 numbers.But wait, 1,2,3,5,7: 2 and 3 are coprime, 2 and 5 are coprime, 2 and 7 are coprime, 3 and 5 are coprime, 3 and 7 are coprime, 5 and 7 are coprime. So, yes, all pairwise coprime.So, 5 numbers.Alternatively, 1,3,4,5,7: 4 and 3 are coprime, 4 and 5 are coprime, 4 and 7 are coprime, 3 and 5 are coprime, 3 and 7 are coprime, 5 and 7 are coprime. So, 5 numbers.So, same.So, seems like 5 is the maximum.But wait, let me think about another sequence where maybe 1 is not included, but there are more coprimes.Wait, for example, starting at 2: 2,3,4,5,6,7,8,9.If I include 2,3,5,7: 4 numbers.Alternatively, 3,5,7,9: 4 numbers.Alternatively, 5,7,8,9: 4 numbers.So, 4 is the maximum here.Alternatively, starting at 3: 3,4,5,6,7,8,9,10.Primes:3,5,7.Including 3,5,7: 3 numbers.Including 4,5,7,9: 4 numbers.Including 5,7,8,9: 4 numbers.So, 4 is the maximum.Wait, another thought: maybe if the sequence includes numbers that are all coprime with each other, but not necessarily primes.For example, 1,2,3,5,7: all coprime.But in 8 consecutive numbers, it's hard to have more than 5 numbers that are pairwise coprime.Wait, let me think about the sequence starting at 1: 1,2,3,4,5,6,7,8.We can include 1,2,3,5,7: 5 numbers.Alternatively, 1,3,4,5,7: 5 numbers.Alternatively, 1,5,6,7,8: 5 numbers.Wait, 1,5,6,7,8: 6 is 2*3, 8 is 2^3.1 is coprime with all.5 and 6: GCD 15 and 7: GCD 15 and 8: GCD 16 and 7: GCD 16 and 8: GCD 2, which is not 1. So, can't include both 6 and 8.So, 1,5,6,7: 4 numbers.Alternatively, 1,5,7,8: 4 numbers.So, same as before.Wait, another idea: maybe include 1,2,5,7, something else.1,2,5,7, and 9: 9 is not in the sequence.Alternatively, 1,2,5,7, and 4: 4 and 2 share a common factor.So, can't include both.So, seems like 5 is the maximum.Wait, but in the sequence starting at 1, we can get 5, but in others, only 4.So, the maximum k is 5.But wait, let me think about another sequence where maybe 1 is not included, but there are more coprimes.Wait, for example, starting at 15: 15,16,17,18,19,20,21,22.Primes:17,19.Including 16: 16 is 2^4, which is coprime with 17,19.Including 15: 15 is 3*5, which is coprime with 16,17,19.Including 21: 21 is 3*7, which is coprime with 16,17,19.Including 22: 22 is 2*11, which is coprime with 15,17,19,21.So, let's see:15,16,17,19,21,22.Check pairwise GCDs:15 and 16: GCD 115 and 17: GCD 115 and 19: GCD 115 and 21: GCD 3, which is not 1. So, can't include both 15 and 21.So, can't include both.Alternatively, 16,17,19,21,22: 16 and 22: GCD 2, which is not 1. So, can't include both.Alternatively, 15,16,17,19,22: 15 and 16: GCD 1, 15 and 17: GCD 1, 15 and 19: GCD 1, 15 and 22: GCD 11, which is not 1. So, can't include both.Alternatively, 16,17,19,21: 16 and 21: GCD 1, 17 and 21: GCD 1, 19 and 21: GCD 1. So, 16,17,19,21: 4 numbers.Alternatively, 15,17,19,21: 15 and 17: GCD 1, 15 and 19: GCD 1, 15 and 21: GCD 3, which is not 1. So, can't include both.So, seems like in this sequence, the maximum is 4 numbers.Wait, another thought: maybe if the sequence includes more numbers that are powers of primes, but not sharing the same base.For example, 4=2^2, 8=2^3, 9=3^2, 25=5^2, etc.But in 8 consecutive numbers, it's hard to have too many such numbers without overlapping prime bases.Wait, for example, starting at 14: 14,15,16,17,18,19,20,21.16 is 2^4, 17 is prime, 19 is prime, 25 is not in the sequence.So, 16,17,19: 3 numbers. Not enough.Alternatively, 14=2*7, 15=3*5, 16=2^4, 17=prime, 18=2*3^2, 19=prime, 20=2^2*5, 21=3*7.So, 14,15,17,19: 4 numbers.So, same as before.Wait, another idea: maybe if the sequence includes numbers that are coprime with each other but not necessarily primes.For example, 1,2,3,5,7: all coprime.But 1 is special.Alternatively, 3,4,5,7: 3 and 4 are coprime, 3 and 5 are coprime, 3 and 7 are coprime, 4 and 5 are coprime, 4 and 7 are coprime, 5 and 7 are coprime. So, 4 numbers.So, 4 numbers.But in the sequence starting at 1, we can get 5.So, seems like 5 is the maximum.Wait, but let me think about another sequence where maybe 1 is not included, but there are more coprimes.Wait, for example, starting at 24: 24,25,26,27,28,29,30,31.Primes:29,31.Including 25:25 is 5^2, which is coprime with 24,26,27,28,29,30,31.Including 26:26 is 2*13, which is coprime with 25,27,28,29,30,31.Including 27:27 is 3^3, which is coprime with 25,26,28,29,30,31.Including 28:28 is 2^2*7, which is coprime with 25,27,29,30,31.Including 30:30 is 2*3*5, which is coprime with 25,29,31.So, let's see:25,26,27,28,29,30,31.Check pairwise GCDs:25 and 26: GCD 125 and 27: GCD 125 and 28: GCD 125 and 29: GCD 125 and 30: GCD 5, which is not 1. So, can't include both 25 and 30.25 and 31: GCD 126 and 27: GCD 126 and 28: GCD 2, which is not 1. So, can't include both 26 and 28.26 and 29: GCD 126 and 30: GCD 2, which is not 1. So, can't include both 26 and 30.26 and 31: GCD 127 and 28: GCD 127 and 29: GCD 127 and 30: GCD 3, which is not 1. So, can't include both 27 and 30.27 and 31: GCD 128 and 29: GCD 128 and 30: GCD 2, which is not 1. So, can't include both 28 and 30.28 and 31: GCD 129 and 30: GCD 129 and 31: GCD 130 and 31: GCD 1So, seems like it's tricky to include more than 4 numbers.For example, 25,26,27,29,31: 5 numbers.Check pairwise:25 and 26: GCD 125 and 27: GCD 125 and 29: GCD 125 and 31: GCD 126 and 27: GCD 126 and 29: GCD 126 and 31: GCD 127 and 29: GCD 127 and 31: GCD 129 and 31: GCD 1So, yes, 25,26,27,29,31: 5 numbers.Wait, so in this sequence, starting at 24, we can get 5 numbers.So, that's another sequence where k=5.So, seems like 5 is achievable in multiple sequences.But is 6 achievable?Wait, let me try to include another number.If I include 25,26,27,29,31, and another number.Let's see:25,26,27,29,31, and 28: 28 and 26 share a common factor 2, so can't include both.25,26,27,29,31, and 30: 30 and 25 share 5, 30 and 26 share 2, 30 and 27 share 3. So, can't include both.25,26,27,29,31, and 24: 24 is 2^3*3, which shares factors with 26,27, etc.So, can't include 24.So, seems like 5 is the maximum here.Alternatively, 25,26,27,29,31, and something else.Wait, 25,26,27,29,31, and 32: 32 is not in the sequence.So, no.So, seems like 5 is the maximum.Wait, another thought: maybe if the sequence includes numbers that are all prime powers, but with different bases.For example, 16=2^4, 25=5^2, 27=3^3, 29=prime, 31=prime.So, 16,25,27,29,31: 5 numbers.All pairwise coprime.Yes, as above.So, 5 numbers.But can we get 6?Wait, let me think of another sequence.Wait, starting at 23: 23,24,25,26,27,28,29,30.Primes:23,29.Including 25=5^2, 27=3^3, 28=2^2*7.So, 23,25,27,29: 4 numbers.Alternatively, 25,27,29,30: 30 is 2*3*5, which shares factors with 25,27.So, can't include both.Alternatively, 25,27,29,28: 28 and 25: GCD 1, 28 and 27: GCD 1, 28 and 29: GCD 1. So, 25,27,28,29: 4 numbers.So, same as before.So, seems like 4 is the maximum here.Wait, another idea: maybe if the sequence includes numbers that are all coprime with each other, but not necessarily primes or prime powers.For example, 1,2,3,5,7,11: but 11 is not in the sequence.Wait, in the sequence starting at 1: 1,2,3,4,5,6,7,8.We can include 1,2,3,5,7: 5 numbers.Alternatively, 1,3,4,5,7: 5 numbers.So, same.Wait, another thought: maybe if the sequence includes numbers that are all coprime with each other, but not necessarily primes or prime powers.For example, 1,2,3,5,7: all coprime.But in 8 consecutive numbers, it's hard to have more than 5 numbers that are pairwise coprime.Wait, let me think about the sequence starting at 1: 1,2,3,4,5,6,7,8.We can include 1,2,3,5,7: 5 numbers.Alternatively, 1,3,4,5,7: 5 numbers.Alternatively, 1,5,6,7,8: 5 numbers.But in all cases, 5 is the maximum.So, seems like 5 is the maximum value of k.But wait, let me think about another sequence where 1 is not included, but there are more coprimes.Wait, for example, starting at 16: 16,17,18,19,20,21,22,23.Primes:17,19,23.Including 16:16 is 2^4, which is coprime with 17,19,23.Including 18:18 is 2*3^2, which shares a factor with 16.Including 20:20 is 2^2*5, which shares a factor with 16.Including 21:21 is 3*7, which is coprime with 16,17,19,23.Including 22:22 is 2*11, which shares a factor with 16.So, let's see:16,17,19,21,23: 5 numbers.Check pairwise:16 and 17: GCD 116 and 19: GCD 116 and 21: GCD 116 and 23: GCD 117 and 19: GCD 117 and 21: GCD 117 and 23: GCD 119 and 21: GCD 119 and 23: GCD 121 and 23: GCD 1So, yes, 5 numbers.So, in this sequence, we can get 5 numbers.So, seems like 5 is achievable in multiple sequences.But is 6 achievable?Wait, let me try to include another number.If I include 16,17,19,21,23, and another number.Let's see:16,17,19,21,23, and 18: 18 and 16 share a factor 2.16,17,19,21,23, and 20: 20 and 16 share a factor 4.16,17,19,21,23, and 22: 22 and 16 share a factor 2.16,17,19,21,23, and 24: 24 is not in the sequence.So, can't include any more numbers without conflict.So, 5 is the maximum here.So, in conclusion, it seems that the maximum value of k is 5.But wait, let me check the options: (A) 3, (B) 4, (C) 5, (D) 6.So, 5 is an option, so the answer should be C) 5.But wait, I need to make sure that in all sequences, 5 is achievable or not. But the problem is asking for the maximum value of k, which is the highest possible k across all possible sequences. So, if in some sequences, k=5 is achievable, then 5 is the answer.But wait, in the sequence starting at 2: 2,3,4,5,6,7,8,9.We can only get k=4.But the problem is asking for the maximum value of k, so it's the highest possible k across all possible sequences.So, since in some sequences, k=5 is achievable, the maximum k is 5.Therefore, the answer is C) 5.

answer：Alright, so I've got this problem here about boys and girls sitting around a table with 30 buns. Let me try to parse it step by step.First, there are several boys and five girls. So, the number of boys is unknown, let's call that number 'b'. The total number of people around the table would then be b + 5.There are 30 buns on a plate. Each girl gave one bun to each boy she knew, and then each boy gave one bun to each girl he didn't know. After all this giving, all the buns were given away. So, the total number of buns given by girls and boys combined is 30.Let me try to model this.First, let's think about the girls giving buns to the boys they know. Each girl gives one bun to each boy she knows. So, if a girl knows 'k' boys, she gives away 'k' buns. Since there are five girls, the total number of buns given by girls is the sum of the number of boys each girl knows.But wait, the problem doesn't specify how many boys each girl knows, so maybe we need another approach.Alternatively, maybe we can think in terms of the total number of buns given by girls and boys.Let me denote:- Let b = number of boys.- Let g = 5 (number of girls).Each girl gives one bun to each boy she knows. So, if a girl knows 'k' boys, she gives 'k' buns. But since each girl could know a different number of boys, maybe we need to think about the total number of buns given by all girls.Similarly, each boy gives one bun to each girl he doesn't know. So, if a boy doesn't know 'm' girls, he gives 'm' buns. But again, each boy could not know a different number of girls.But perhaps there's a way to relate the number of buns given by girls and boys.Wait, maybe we can think about the total number of buns given by girls and the total number given by boys.Let me denote:Total buns given by girls: Let's say each girl gives 'k_i' buns, where i = 1 to 5. So, total buns from girls is k1 + k2 + k3 + k4 + k5.Similarly, each boy gives 'm_j' buns, where j = 1 to b. So, total buns from boys is m1 + m2 + ... + mb.And we know that (k1 + k2 + k3 + k4 + k5) + (m1 + m2 + ... + mb) = 30.But without knowing the individual k_i and m_j, this might not be directly helpful.Wait, perhaps we can think about the relationship between the girls' giving and the boys' giving.Each girl gives a bun to each boy she knows, and each boy gives a bun to each girl he doesn't know.Is there a way to relate the number of buns given by girls and boys?Let me think about the total number of buns given by girls. Each girl gives one bun to each boy she knows. So, the total number of buns given by girls is equal to the total number of "girl knows boy" relationships.Similarly, each boy gives one bun to each girl he doesn't know. So, the total number of buns given by boys is equal to the total number of "boy doesn't know girl" relationships.But the total number of possible relationships between girls and boys is b * g = 5b. Because each of the 5 girls can potentially know or not know each of the b boys.So, the total number of "girl knows boy" relationships plus the total number of "boy doesn't know girl" relationships should equal 5b.But wait, is that correct?Wait, actually, for each girl and each boy, there are two possibilities: the girl knows the boy, or the girl doesn't know the boy. So, for each pair (girl, boy), exactly one of these is true.Therefore, the total number of "girl knows boy" relationships plus the total number of "girl doesn't know boy" relationships is equal to 5b.But in the problem, the girls give buns to the boys they know, and the boys give buns to the girls they don't know. So, the total buns given by girls is equal to the number of "girl knows boy" relationships, and the total buns given by boys is equal to the number of "boy doesn't know girl" relationships.But wait, for each pair (girl, boy), if the girl knows the boy, she gives a bun, and the boy doesn't give a bun to that girl. Conversely, if the girl doesn't know the boy, the boy gives a bun to the girl.But actually, the boy gives a bun to each girl he doesn't know, regardless of whether the girl knows him or not.Wait, no, the problem says each boy gives one bun to each girl he doesn't know. So, for each boy, he gives a bun to each girl he doesn't know, regardless of whether the girl knows him or not.Similarly, each girl gives a bun to each boy she knows, regardless of whether the boy knows her or not.So, the total number of buns given by girls is equal to the number of "girl knows boy" relationships, and the total number of buns given by boys is equal to the number of "boy doesn't know girl" relationships.But the total number of "girl knows boy" relationships plus the number of "boy doesn't know girl" relationships is equal to 5b, because for each of the 5b pairs, either the girl knows the boy or the boy doesn't know the girl (but actually, these are not mutually exclusive in terms of the relationships, because a girl could know a boy, and the boy could still know the girl or not. Wait, no, actually, if a girl knows a boy, it doesn't necessarily mean the boy knows the girl. So, the two relationships are independent.Wait, but in the problem, the girls give buns based on their knowledge, and the boys give buns based on their lack of knowledge.So, the total buns given by girls is the number of times a girl knows a boy, and the total buns given by boys is the number of times a boy doesn't know a girl.But since each pair (girl, boy) can be in one of four states:1. Girl knows boy, boy knows girl.2. Girl knows boy, boy doesn't know girl.3. Girl doesn't know boy, boy knows girl.4. Girl doesn't know boy, boy doesn't know girl.But in the problem, the girls give buns to boys they know, regardless of whether the boys know them. Similarly, the boys give buns to girls they don't know, regardless of whether the girls know them.So, the total buns given by girls is the number of pairs where the girl knows the boy, which is the sum over all girls of the number of boys they know.Similarly, the total buns given by boys is the number of pairs where the boy doesn't know the girl, which is the sum over all boys of the number of girls they don't know.But the total number of pairs is 5b.So, the total buns given by girls plus the total buns given by boys is equal to the number of "girl knows boy" relationships plus the number of "boy doesn't know girl" relationships.But these two sets of relationships are not necessarily complementary, because a pair could be both "girl knows boy" and "boy doesn't know girl", or neither, or both.Wait, no, actually, for each pair (girl, boy), the girl may or may not know the boy, and the boy may or may not know the girl. So, the total number of "girl knows boy" relationships is some number, say K, and the total number of "boy doesn't know girl" relationships is some number, say M.But K and M are not necessarily related in a straightforward way because they are counts over different sets.However, we know that K + M = total buns given, which is 30.But K is the number of times a girl knows a boy, and M is the number of times a boy doesn't know a girl.But wait, for each pair (girl, boy), the girl could know the boy, and the boy could know or not know the girl. Similarly, the girl could not know the boy, and the boy could know or not know the girl.So, the total number of "girl knows boy" relationships is K, and the total number of "boy doesn't know girl" relationships is M.But K and M are independent counts.Wait, but perhaps we can relate them through the total number of pairs.Each pair (girl, boy) can be in one of four states:1. Girl knows boy, boy knows girl.2. Girl knows boy, boy doesn't know girl.3. Girl doesn't know boy, boy knows girl.4. Girl doesn't know boy, boy doesn't know girl.So, for each pair, exactly one of these four is true.Now, the total number of "girl knows boy" relationships is the sum of cases 1 and 2.The total number of "boy doesn't know girl" relationships is the sum of cases 2 and 4.Therefore, K = cases 1 + 2.M = cases 2 + 4.But we don't know the exact distribution between these cases.However, we do know that K + M = (cases 1 + 2) + (cases 2 + 4) = cases 1 + 2*cases 2 + 4.But this doesn't directly help us.Wait, but perhaps we can think about the total number of pairs, which is 5b.So, cases 1 + 2 + 3 + 4 = 5b.But we also know that K + M = 30.So, K + M = 30.But K = cases 1 + 2.M = cases 2 + 4.So, K + M = cases 1 + 2 + 2*cases 2 + 4 = cases 1 + 3*cases 2 + 4.Wait, that seems complicated.Alternatively, maybe we can think about the total number of buns given by girls and boys as K + M = 30.But K is the number of "girl knows boy" relationships, and M is the number of "boy doesn't know girl" relationships.But we can also note that for each boy, the number of girls he doesn't know is equal to the number of girls who didn't give him a bun, because the girls only gave buns to boys they knew.Wait, no, the girls gave buns to boys they knew, so the boys who received buns from a girl are the ones the girl knew. So, for each boy, the number of girls who didn't give him a bun is equal to the number of girls he didn't know, because those girls didn't know him, so they didn't give him a bun.Wait, no, that's not necessarily true. Because a girl might not know a boy, but the boy might know the girl or not. But in the problem, the girls only give buns to boys they know, regardless of whether the boys know them.So, for a given boy, the number of girls who gave him a bun is equal to the number of girls who know him.Therefore, the number of girls who didn't give him a bun is equal to the number of girls who don't know him.But the boy gives a bun to each girl he doesn't know, regardless of whether the girl knows him or not.Wait, so for each boy, the number of buns he gives is equal to the number of girls he doesn't know.But the number of girls he doesn't know is equal to the number of girls who didn't give him a bun, because the girls only gave buns to boys they knew.Wait, no, that's not necessarily the case because a girl might not know a boy, but the boy might know the girl, so the girl didn't give him a bun, but the boy might give her a bun.Wait, maybe it's better to think in terms of the total buns given.Let me denote:Total buns given by girls: K = sum over all girls of the number of boys they know.Total buns given by boys: M = sum over all boys of the number of girls they don't know.We know that K + M = 30.But we also know that for each boy, the number of girls he doesn't know is equal to 5 minus the number of girls who know him.Wait, no, because the girls who know him are the ones who gave him a bun. So, for each boy, the number of girls who know him is equal to the number of buns he received from girls.Similarly, the number of girls he doesn't know is equal to 5 minus the number of girls who know him.But the number of girls who know him is equal to the number of buns he received from girls.So, for each boy, the number of girls he doesn't know is 5 - (number of buns he received from girls).But the number of buns he gives is equal to the number of girls he doesn't know, which is 5 - (number of buns he received from girls).Therefore, for each boy, the number of buns he gives is 5 - (number of buns he received from girls).Therefore, the total number of buns given by boys is sum over all boys of [5 - (number of buns received by each boy)].But the total number of buns received by all boys is equal to K, because that's the total buns given by girls.So, the total number of buns given by boys is b*5 - K.Because each boy gives 5 - (number of buns he received), so summing over all boys, it's 5b - K.Therefore, total buns given by boys is 5b - K.We know that total buns given by girls is K, and total buns given by boys is 5b - K.So, total buns given is K + (5b - K) = 5b.But we are told that the total buns given is 30.Therefore, 5b = 30.So, b = 6.Wait, that seems straightforward. Let me verify.So, if there are b boys, then total buns given by girls is K, and total buns given by boys is 5b - K.Therefore, total buns is K + (5b - K) = 5b.Given that total buns is 30, 5b = 30, so b = 6.Yes, that makes sense.So, the number of boys is 6.

answer：Okay, so I have this problem here: I need to determine the number of terms in the polynomial of ( x ) obtained from expanding ( (sqrt{3}x + sqrt[3]{2})^{100} ) where the coefficient is a rational number. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I remember that when you expand a binomial like ( (a + b)^n ), you can use the binomial theorem. The binomial theorem says that each term in the expansion is of the form ( binom{n}{k} a^{n - k} b^k ), where ( k ) ranges from 0 to ( n ). So, in this case, ( a = sqrt{3}x ) and ( b = sqrt[3]{2} ), and ( n = 100 ).So, each term in the expansion will be ( binom{100}{k} (sqrt{3}x)^{100 - k} (sqrt[3]{2})^k ). I need to figure out for which values of ( k ) the coefficient of ( x ) in this term is a rational number.Let me write out the general term:( T_k = binom{100}{k} (sqrt{3}x)^{100 - k} (sqrt[3]{2})^k )Simplifying this, we get:( T_k = binom{100}{k} (sqrt{3})^{100 - k} x^{100 - k} (2)^{k/3} )So, the coefficient of ( x^{100 - k} ) is ( binom{100}{k} (sqrt{3})^{100 - k} (2)^{k/3} ). We need this coefficient to be a rational number.Now, let's analyze the components of this coefficient:1. ( binom{100}{k} ) is always a rational number because it's a binomial coefficient, which is an integer, and integers are rational.2. ( (sqrt{3})^{100 - k} ) can be written as ( 3^{(100 - k)/2} ). For this to be rational, the exponent ( (100 - k)/2 ) must be an integer because ( 3 ) is a prime number and its irrational unless the exponent is an integer. So, ( (100 - k)/2 ) must be an integer. Let's denote ( m = (100 - k)/2 ), so ( m ) must be an integer. Therefore, ( 100 - k ) must be even, which implies that ( k ) must have the same parity as 100. Since 100 is even, ( k ) must also be even.3. ( (2)^{k/3} ) can be written as ( 2^{k/3} ). For this to be rational, ( k/3 ) must be an integer because 2 is a prime number and its irrational unless the exponent is an integer. Let's denote ( n = k/3 ), so ( n ) must be an integer. Therefore, ( k ) must be a multiple of 3.So, putting points 2 and 3 together, ( k ) must be both even and a multiple of 3. In other words, ( k ) must be a multiple of the least common multiple (LCM) of 2 and 3, which is 6. Therefore, ( k ) must be a multiple of 6.So, ( k ) can be 0, 6, 12, ..., up to 96 or 100? Wait, 100 divided by 6 is approximately 16.666, so the largest multiple of 6 less than or equal to 100 is 96. So, the possible values of ( k ) are 0, 6, 12, ..., 96.Now, let's figure out how many such ( k ) there are. Since ( k ) starts at 0 and increases by 6 each time, the number of terms is given by the number of terms in the arithmetic sequence starting at 0, with a common difference of 6, up to 96.The formula for the number of terms in an arithmetic sequence is:( text{Number of terms} = left( frac{text{Last term} - text{First term}}{text{Common difference}} right) + 1 )Plugging in the values:( text{Number of terms} = left( frac{96 - 0}{6} right) + 1 = 16 + 1 = 17 )Wait, hold on. Let me check that again. If ( k ) starts at 0 and goes up by 6 each time, the terms are 0, 6, 12, ..., 96. So, how many terms is that?Let me think of it as ( k = 6m ), where ( m ) is an integer. Then, ( 6m leq 100 ). So, ( m leq 100/6 approx 16.666 ). So, ( m ) can be 0, 1, 2, ..., 16. That's 17 terms in total. So, yes, 17 terms.But wait, hold on. Let me make sure I didn't make a mistake here. Because 6*16 = 96, which is less than 100, and 6*17 = 102, which is more than 100. So, ( m ) can only go up to 16, which gives 17 terms (including m=0). So, that seems correct.But let me verify with another approach. The number of multiples of 6 up to 100 is equal to the floor of 100 divided by 6, which is 16, plus 1 if we include 0. So, 16 + 1 = 17. Yep, that's consistent.So, does that mean there are 17 terms where the coefficient is rational?Wait, but hold on a second. Let me make sure that for each ( k ) that is a multiple of 6, both ( (sqrt{3})^{100 - k} ) and ( (2)^{k/3} ) are rational. Because even if ( k ) is a multiple of 6, we have to ensure that both components are rational.So, let's take ( k = 6m ), where ( m ) is an integer from 0 to 16.Then, ( 100 - k = 100 - 6m ). For ( (sqrt{3})^{100 - 6m} ) to be rational, ( 100 - 6m ) must be even because ( sqrt{3} ) raised to an even power is rational (since ( (sqrt{3})^{2n} = 3^n )), and raised to an odd power is irrational.So, ( 100 - 6m ) must be even. Since 100 is even and 6m is also even (because 6 is even and m is integer), their difference is even. So, ( 100 - 6m ) is even, which means ( (sqrt{3})^{100 - 6m} ) is rational.Similarly, ( (2)^{k/3} = (2)^{6m / 3} = (2)^{2m} = 4^m ), which is clearly rational because 4 is an integer and any integer raised to an integer power is rational.Therefore, both components are rational, so the entire coefficient is rational.Therefore, each ( k ) that is a multiple of 6 will result in a term with a rational coefficient.So, the number of such terms is 17.Wait, but hold on. Let me check for ( k = 100 ). If ( k = 100 ), then ( 100 - k = 0 ), so ( (sqrt{3})^0 = 1 ), which is rational, and ( (2)^{100/3} ). Wait, 100 divided by 3 is approximately 33.333, which is not an integer. So, ( (2)^{100/3} ) is not rational. Therefore, ( k = 100 ) would not result in a rational coefficient.But in our earlier analysis, ( k ) goes up to 96 because 6*16=96, so 96 is the largest multiple of 6 less than or equal to 100. So, ( k = 96 ) is the last term we consider. Let's check ( k = 96 ):( (sqrt{3})^{100 - 96} = (sqrt{3})^4 = 9 ), which is rational.( (2)^{96/3} = (2)^{32} ), which is rational.So, that term is fine. So, ( k = 96 ) is okay, but ( k = 100 ) is not. So, our count of 17 terms is correct because we only go up to ( k = 96 ).Wait, but hold on again. Let me think about ( k = 0 ). If ( k = 0 ), then the term is ( (sqrt{3}x)^{100} times (sqrt[3]{2})^0 = (sqrt{3})^{100}x^{100} times 1 ). ( (sqrt{3})^{100} = (3)^{50} ), which is rational. So, that term is rational as well. So, ( k = 0 ) is included, which is correct.Therefore, the number of terms with rational coefficients is 17.But wait, hold on one more time. Let me think about ( k = 6 ). Then, ( 100 - 6 = 94 ). ( (sqrt{3})^{94} = 3^{47} ), which is rational. ( (2)^{6/3} = 2^2 = 4 ), which is rational. So, that term is fine.Similarly, ( k = 12 ): ( 100 - 12 = 88 ), ( (sqrt{3})^{88} = 3^{44} ), rational. ( (2)^{12/3} = 2^4 = 16 ), rational.So, yeah, each multiple of 6 for ( k ) gives a term with a rational coefficient.Therefore, the number of such terms is 17.Wait, but hold on. Let me check ( k = 96 ). ( 100 - 96 = 4 ), so ( (sqrt{3})^4 = 9 ), which is rational. ( (2)^{96/3} = 2^{32} ), which is rational. So, that's fine.And ( k = 0 ): ( (sqrt{3})^{100} = 3^{50} ), rational, and ( (2)^0 = 1 ), rational. So, that's also fine.So, I think 17 is the correct number of terms.But just to make sure, let me think about another approach. Maybe using exponents.The exponent of ( sqrt{3} ) is ( 100 - k ), so ( (sqrt{3})^{100 - k} = 3^{(100 - k)/2} ). For this to be rational, ( (100 - k)/2 ) must be an integer, so ( 100 - k ) must be even, so ( k ) must be even.Similarly, the exponent of ( sqrt[3]{2} ) is ( k ), so ( (sqrt[3]{2})^k = 2^{k/3} ). For this to be rational, ( k/3 ) must be an integer, so ( k ) must be a multiple of 3.Therefore, ( k ) must be a multiple of both 2 and 3, so it must be a multiple of 6. So, ( k ) can be 0, 6, 12, ..., 96, which is 17 terms.Yes, that's consistent with my earlier reasoning.So, I think I've double-checked enough. The number of terms with rational coefficients is 17.**Final Answer**The number of terms with rational coefficients is boxed{17}.

answer：Okay, so I have this problem here about finding the number of common tangents between two circles, C₁ and C₂. The equations given are:C₁: x² + y² = 4C₂: x² + y² - 4x + 2y + 4 = 0Hmm, I remember that common tangents are lines that touch both circles without crossing them. Depending on the position of the circles, there can be different numbers of common tangents. If the circles are separate, there are four common tangents; if they touch externally, there are three; if one is inside the other without touching, there are none; and if they intersect, there are two.So, first, I need to figure out the positions of these two circles relative to each other. To do that, I should find their centers and radii.Starting with C₁: x² + y² = 4. That's a standard equation of a circle centered at the origin (0,0) with radius √4 = 2.Now, for C₂: x² + y² - 4x + 2y + 4 = 0. I need to rewrite this in standard form by completing the squares.Let me rearrange the equation:x² - 4x + y² + 2y = -4Now, complete the square for the x terms and the y terms.For x: Take the coefficient of x, which is -4, divide by 2 to get -2, square it to get 4.For y: Take the coefficient of y, which is 2, divide by 2 to get 1, square it to get 1.So, add these squares to both sides:(x² - 4x + 4) + (y² + 2y + 1) = -4 + 4 + 1Simplify:(x - 2)² + (y + 1)² = 1So, the center of C₂ is at (2, -1) and the radius is √1 = 1.Alright, so now I have:C₁: Center (0,0), radius 2C₂: Center (2, -1), radius 1Next step is to find the distance between the centers. Let's compute that.The distance formula is √[(x₂ - x₁)² + (y₂ - y₁)²]Plugging in the centers:Distance = √[(2 - 0)² + (-1 - 0)²] = √[4 + 1] = √5 ≈ 2.236Now, let's see the sum and difference of the radii.Sum of radii: 2 + 1 = 3Difference of radii: 2 - 1 = 1So, the distance between centers is √5 ≈ 2.236.Now, compare this distance with the sum and difference of radii.Since 1 < √5 < 3, which is approximately 1 < 2.236 < 3, this means the circles intersect at two points.Wait, hold on. If the distance between centers is less than the sum of radii and greater than the difference, they intersect at two points. So, in that case, how many common tangents are there?I think when two circles intersect at two points, they have two common tangents. But wait, let me recall. If the circles are separate, four tangents; if they touch externally, three; intersecting, two; internally tangent, one; and one inside the other, none.So, yes, since they intersect at two points, there should be two common tangents.But wait, hold on. Let me make sure. I remember that for intersecting circles, the common tangents are the two lines that pass through the points of intersection? Or is that secants?No, wait, common tangents are lines that touch both circles without crossing them. So, if the circles intersect, the only lines that can touch both without crossing would be the two external tangents.Wait, but actually, when circles intersect, they don't have external or internal tangents because the circles overlap. Hmm, maybe I'm confusing something.Wait, no. Let me think again. For two circles, depending on their positions:1. If one is entirely inside another without touching, no common tangents.2. If they are tangent internally, one common tangent at the point of contact.3. If they intersect at two points, two common tangents.4. If they are separate but not overlapping, four common tangents: two external and two internal.Wait, so in the case of intersecting circles, they have two common tangents. So, in this case, since the distance between centers is √5 ≈ 2.236, which is less than the sum of radii (3) and greater than the difference (1), so they intersect at two points. Therefore, they have two common tangents.But wait, I'm a bit confused because I thought when circles intersect, they don't have external or internal tangents. Maybe I need to visualize this.Let me try to sketch it mentally. C₁ is centered at (0,0) with radius 2, so it's a decently sized circle. C₂ is centered at (2, -1) with radius 1, so it's a smaller circle located to the right and slightly downward from the origin.The distance between centers is √5, which is about 2.236. Since the radius of C₁ is 2, and the radius of C₂ is 1, so the distance between centers is more than the radius of C₁ minus C₂ (which is 1) and less than the sum (3). So, yes, they intersect at two points.In such a case, the two circles intersect, so the only common tangents are the two lines that touch both circles without crossing through the overlapping area. So, these would be the external tangents.Wait, but actually, when circles intersect, they don't have external or internal tangents because the circles overlap. Hmm, maybe I need to clarify.Wait, no. The external tangent is a line that touches both circles and lies outside the area between them. But if the circles intersect, the external tangent still exists, but the internal tangent would cross between the circles, but since they intersect, the internal tangent would actually intersect the circles at two points each, making them secants instead of tangents.Therefore, when two circles intersect, they only have two common external tangents.Wait, but I think actually, when two circles intersect, they don't have any common tangents because any line that touches both circles would have to pass through the overlapping region, making it a secant. But that contradicts what I thought earlier.Wait, maybe I need to think about this more carefully.Let me recall the definitions. A common tangent is a line that touches both circles at exactly one point each. If the circles are separate, you can have external tangents that don't cross the line segment joining the centers, and internal tangents that cross between them. If the circles intersect, then any line that would be an internal tangent would actually intersect both circles at two points, making them secants, not tangents. Therefore, only external tangents can exist when circles intersect.But wait, if the circles intersect, can you still have external tangents? Because the external tangents would lie outside the overlapping area.Wait, let me think about two intersecting circles. Yes, you can still draw two lines that touch both circles without crossing into the overlapping area. These would be the external tangents.So, in that case, two common tangents exist.But I'm still a bit confused because I thought that when circles intersect, they don't have common tangents. Maybe I need to verify this.Wait, let me think about two circles that intersect at two points. If I try to draw a tangent line, it has to touch each circle at exactly one point. Since the circles are overlapping, the only way a line can touch both without crossing is if it's on the outer sides of both circles.So, yes, two such lines exist, each on opposite sides of the line connecting the centers.Therefore, the number of common tangents is two.But wait, hold on. Let me check with the formula or method to calculate the number of common tangents.I remember that the number of common tangents depends on the position:- 4 if the circles are separate and neither intersecting nor touching.- 3 if they are externally tangent (touching at one point from outside).- 2 if they intersect at two points.- 1 if they are internally tangent (touching at one point from inside).- 0 if one is entirely inside the other without touching.So, according to this, since our circles intersect at two points, the number of common tangents is two.Therefore, the answer should be two.But just to make sure, let me try to compute the equations of the common tangents and see if they exist.To find the equations of the common tangents, one method is to use the condition that the distance from the center to the line is equal to the radius.Let me denote the equation of a common tangent as y = mx + c.Then, the distance from center of C₁ (0,0) to this line should be equal to radius of C₁, which is 2.Similarly, the distance from center of C₂ (2, -1) to this line should be equal to radius of C₂, which is 1.So, the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / √(a² + b²).In our case, the line is y = mx + c, which can be rewritten as mx - y + c = 0.So, for C₁: |0*m + 0*(-1) + c| / √(m² + 1) = |c| / √(m² + 1) = 2For C₂: |m*2 + (-1)*(-1) + c| / √(m² + 1) = |2m + 1 + c| / √(m² + 1) = 1So, we have two equations:1. |c| = 2√(m² + 1)2. |2m + 1 + c| = √(m² + 1)Let me drop the absolute values for a moment and consider the equations, keeping in mind that we might have to consider different cases.From equation 1: c = ±2√(m² + 1)Let me substitute c into equation 2.Case 1: c = 2√(m² + 1)Then equation 2 becomes:|2m + 1 + 2√(m² + 1)| = √(m² + 1)Let me denote √(m² + 1) as t, so t = √(m² + 1) ≥ 1Then equation becomes:|2m + 1 + 2t| = tWhich implies:2m + 1 + 2t = t or 2m + 1 + 2t = -tFirst subcase:2m + 1 + 2t = t=> 2m + 1 = -tBut t = √(m² + 1) ≥ 1, so RHS is negative or zero, but LHS is 2m + 1.So, 2m + 1 = -√(m² + 1)Let me square both sides to eliminate the square root:(2m + 1)² = (√(m² + 1))²4m² + 4m + 1 = m² + 1Simplify:4m² + 4m + 1 - m² - 1 = 03m² + 4m = 0m(3m + 4) = 0So, m = 0 or m = -4/3Now, check these solutions in the original equation because squaring can introduce extraneous solutions.First, m = 0:Then, t = √(0 + 1) = 1From 2m + 1 = -t:2*0 + 1 = -1 => 1 = -1, which is false. So, m = 0 is extraneous.Next, m = -4/3:Compute t = √( (16/9) + 1 ) = √(25/9) = 5/3From 2m + 1 = -t:2*(-4/3) + 1 = -5/3=> -8/3 + 3/3 = -5/3=> (-5/3) = (-5/3), which is true.So, m = -4/3 is a valid solution.Compute c:c = 2√(m² + 1) = 2*(5/3) = 10/3So, one tangent line is y = (-4/3)x + 10/3Case 1 gives us one solution.Case 2: c = -2√(m² + 1)Substitute into equation 2:|2m + 1 - 2√(m² + 1)| = √(m² + 1)Again, let t = √(m² + 1)So, |2m + 1 - 2t| = tWhich gives two subcases:2m + 1 - 2t = t or 2m + 1 - 2t = -tFirst subcase:2m + 1 - 2t = t=> 2m + 1 = 3tBut t = √(m² + 1), so:2m + 1 = 3√(m² + 1)Let me square both sides:(2m + 1)² = 9(m² + 1)4m² + 4m + 1 = 9m² + 9Bring all terms to left:4m² + 4m + 1 - 9m² - 9 = 0-5m² + 4m - 8 = 0Multiply both sides by -1:5m² - 4m + 8 = 0Discriminant D = 16 - 160 = -144 < 0No real solutions here.Second subcase:2m + 1 - 2t = -t=> 2m + 1 = tBut t = √(m² + 1), so:2m + 1 = √(m² + 1)Square both sides:(2m + 1)² = m² + 14m² + 4m + 1 = m² + 1Simplify:3m² + 4m = 0m(3m + 4) = 0So, m = 0 or m = -4/3Check these in the original equation.First, m = 0:t = √(0 + 1) = 1From 2m + 1 = t:2*0 + 1 = 1 => 1 = 1, which is true.So, m = 0 is a valid solution.Compute c:c = -2√(m² + 1) = -2*1 = -2So, the tangent line is y = 0x - 2 => y = -2Second, m = -4/3:Compute t = √( (16/9) + 1 ) = √(25/9) = 5/3From 2m + 1 = t:2*(-4/3) + 1 = 5/3=> -8/3 + 3/3 = 5/3=> (-5/3) = 5/3, which is false.So, m = -4/3 is extraneous.Therefore, from Case 2, we get another tangent line: y = -2So, in total, we have two common tangent lines:1. y = (-4/3)x + 10/32. y = -2Wait, that's only two lines. So, does that mean there are only two common tangents?But earlier, I thought that when circles intersect, they have two common tangents, which seems to be confirmed here.But wait, let me check if these lines are actually common tangents.First, y = -2. Let's see if it's a tangent to both circles.For C₁: x² + y² = 4. Substitute y = -2:x² + 4 = 4 => x² = 0 => x = 0So, it touches C₁ at (0, -2). That's a single point, so it's a tangent.For C₂: (x - 2)² + (y + 1)² = 1. Substitute y = -2:(x - 2)² + (-2 + 1)² = (x - 2)² + 1 = 1=> (x - 2)² = 0 => x = 2So, it touches C₂ at (2, -2). Single point, so it's a tangent.Good, so y = -2 is a common tangent.Now, the other line: y = (-4/3)x + 10/3Let me check if it's tangent to both circles.For C₁: x² + y² = 4Substitute y = (-4/3)x + 10/3 into the equation:x² + [(-4/3 x + 10/3)]² = 4Compute [(-4/3 x + 10/3)]²:= (16/9)x² - (80/9)x + 100/9So, equation becomes:x² + (16/9)x² - (80/9)x + 100/9 = 4Multiply all terms by 9 to eliminate denominators:9x² + 16x² - 80x + 100 = 36Combine like terms:25x² - 80x + 100 - 36 = 025x² - 80x + 64 = 0Compute discriminant D = 6400 - 4*25*64 = 6400 - 6400 = 0Since discriminant is zero, the line touches C₁ at exactly one point, so it's a tangent.Similarly, check for C₂: (x - 2)² + (y + 1)² = 1Substitute y = (-4/3)x + 10/3:(x - 2)² + [(-4/3 x + 10/3 + 1)]² = 1Simplify the y-term:(-4/3 x + 10/3 + 3/3) = (-4/3 x + 13/3)So, equation becomes:(x - 2)² + [(-4/3 x + 13/3)]² = 1Compute each term:(x - 2)² = x² - 4x + 4[(-4/3 x + 13/3)]² = (16/9)x² - (104/9)x + 169/9So, equation becomes:x² - 4x + 4 + 16/9 x² - 104/9 x + 169/9 = 1Multiply all terms by 9:9x² - 36x + 36 + 16x² - 104x + 169 = 9Combine like terms:(9x² + 16x²) + (-36x - 104x) + (36 + 169) = 925x² - 140x + 205 = 925x² - 140x + 196 = 0Compute discriminant D = 19600 - 4*25*196 = 19600 - 19600 = 0Again, discriminant is zero, so the line touches C₂ at exactly one point, hence a tangent.Therefore, both lines are indeed common tangents.Wait, but earlier I thought there might be four tangents, but in this case, we only found two. Is that because the circles intersect, so only two common tangents exist?Yes, that seems to be the case. So, the number of common tangents is two.Therefore, the answer is 2.**Final Answer**The number of common tangents is boxed{2}.

answer：First, determine the number of roses Kyle picked last year. He picked a dozen roses, which is 12 roses.This year, he was able to pick half the number of roses, so he picked 6 roses.Kyle wants to give his mother a bouquet with twice as many roses as last year, which means he needs 24 roses.He already has 6 roses from his garden, so he needs to buy an additional 18 roses from the grocery store.Since each rose costs 3, the total cost for 18 roses would be 18 multiplied by 3, which equals 54.