💡First, I need to determine the centers and radii of both circles. For Circle ( O_1 ): ((x+1)^2 + (y-1)^2 = 4), the center is at ((-1, 1)) and the radius is (2). For Circle ( O_2 ): ((x-2)^2 + (y-4)^2 = 9), the center is at ((2, 4)) and the radius is (3).Next, I will calculate the distance between the centers of the two circles using the distance formula:[text{Distance} = sqrt{(2 - (-1))^2 + (4 - 1)^2} = sqrt{3^2 + 3^2} = sqrt{9 + 9} = sqrt{18} = 3sqrt{2}]Now, I need to compare this distance with the sum and difference of the radii. The sum of the radii is (2 + 3 = 5), and the difference is (3 - 2 = 1).Since the distance between the centers ((3sqrt{2})) is greater than the difference of the radii ((1)) and less than the sum of the radii ((5)), the circles intersect each other.Therefore, the positional relationship between the two circles is **intersecting**.

💡First, I identify the number of pupils who like pizza, which is 125, and those who like burgers, which is 115.Next, I recognize that 40 pupils like both pizza and burgers. This means these 40 pupils have been counted twice in the initial totals.To find the total number of unique pupils, I add the number of pizza lovers and burger lovers: 125 + 115 = 240.Since the 40 pupils who like both have been counted twice, I subtract this number from the total: 240 - 40 = 200.Therefore, there are 200 pupils in total.

💡Okay, so I have this problem where I need to find the least possible value of ( m + n ) given three conditions:1. ( gcd(m + n, 330) = 1 )2. ( m^m ) is a multiple of ( n^n )3. ( m ) is not a multiple of ( n )Alright, let me break this down step by step.First, I need to understand the gcd condition. The number 330 factors into primes as ( 2 times 3 times 5 times 11 ). So, ( gcd(m + n, 330) = 1 ) means that ( m + n ) must not be divisible by 2, 3, 5, or 11. In other words, ( m + n ) must be coprime with 330.Next, the second condition says that ( m^m ) is a multiple of ( n^n ). That means ( n^n ) divides ( m^m ). So, ( m^m ) must have all the prime factors of ( n^n ) with at least the same exponents.The third condition is that ( m ) is not a multiple of ( n ). So, ( n ) doesn't divide ( m ).My goal is to find the smallest ( m + n ) satisfying all these.Let me think about the second condition: ( n^n ) divides ( m^m ). This implies that for every prime ( p ) in the prime factorization of ( n ), the exponent of ( p ) in ( m ) must be at least as large as the exponent in ( n ), but scaled by the ratio ( frac{m}{n} ). Hmm, maybe I need to think in terms of exponents.Suppose ( n ) has a prime factorization ( n = p_1^{a_1} p_2^{a_2} dots p_k^{a_k} ). Then ( n^n = p_1^{a_1 n} p_2^{a_2 n} dots p_k^{a_k n} ).Similarly, ( m = p_1^{b_1} p_2^{b_2} dots p_k^{b_k} dots ), so ( m^m = p_1^{b_1 m} p_2^{b_2 m} dots ).For ( n^n ) to divide ( m^m ), we need ( a_i n leq b_i m ) for each prime ( p_i ) in ( n ).So, ( frac{a_i}{b_i} leq frac{m}{n} ).But since ( m ) is not a multiple of ( n ), ( frac{m}{n} ) is not an integer. So, ( frac{m}{n} ) must be a rational number greater than 1 but not an integer.This suggests that ( m ) must be a multiple of some multiple of ( n )'s prime factors but not the entire ( n ).Wait, maybe another approach. Let's consider that ( m ) must be a multiple of ( n ) raised to some power, but not necessarily ( n ) itself. Hmm, not sure.Alternatively, perhaps ( m ) is a multiple of ( n ) times some other number. But since ( m ) is not a multiple of ( n ), that other number must not be a multiple of ( n ).Wait, maybe I need to think about exponents more carefully.Let me consider the simplest case where ( n ) is a prime power. Let's say ( n = p^k ). Then ( n^n = p^{k n} ). For ( m^m ) to be divisible by ( p^{k n} ), ( m ) must be divisible by ( p ), say ( m = p^t times q ), where ( q ) is coprime to ( p ). Then ( m^m = p^{t m} times q^m ). So, ( t m geq k n ).But since ( m ) is not a multiple of ( n = p^k ), ( t < k ). Wait, that can't be because if ( t < k ), then ( m ) is not divisible by ( p^k ), so ( m ) is not a multiple of ( n ). But ( t m geq k n ) implies ( t geq frac{k n}{m} ). Hmm, this seems a bit tangled.Wait, maybe ( m ) has to be a multiple of ( p^s ) where ( s ) is such that ( s times m geq k times n ). But since ( m ) is not a multiple of ( p^k ), ( s < k ). So, ( s times m geq k times n ). Therefore, ( m geq frac{k}{s} n ). Since ( s < k ), ( frac{k}{s} > 1 ). So, ( m ) is at least ( frac{k}{s} n ), which is greater than ( n ), but not necessarily a multiple.This is getting a bit complicated. Maybe I should try specific numbers.Let me try small primes for ( n ). Let's start with ( n = 2 ). Then ( n^n = 4 ). So, ( m^m ) must be a multiple of 4. So, ( m ) must be even, because if ( m ) is even, ( m^m ) is divisible by ( 2^m ), which is at least 4 if ( m geq 2 ). But ( m ) cannot be a multiple of ( n = 2 ), which is impossible because if ( m ) is even, it is a multiple of 2. So, ( n = 2 ) doesn't work.Similarly, ( n = 3 ). Then ( n^n = 27 ). So, ( m^m ) must be divisible by 27, which means ( m ) must be divisible by 3. But then ( m ) is a multiple of ( n = 3 ), which violates the third condition. So, ( n = 3 ) also doesn't work.Same with ( n = 5 ). ( n^n = 3125 ). ( m ) must be divisible by 5, which would make ( m ) a multiple of ( n = 5 ). Not allowed.Similarly, ( n = 11 ). ( m ) must be divisible by 11, which would make ( m ) a multiple of ( n = 11 ). Not allowed.So, ( n ) cannot be any of the primes dividing 330 because that would force ( m ) to be a multiple of ( n ), which is not allowed.Therefore, ( n ) must be a prime not dividing 330. The smallest such prime is 13.So, let's try ( n = 13 ). Then ( n^n = 13^{13} ). So, ( m^m ) must be divisible by ( 13^{13} ). Therefore, ( m ) must be divisible by 13. Let ( m = 13k ), where ( k ) is an integer greater than 1 (since ( m ) is not a multiple of ( n = 13 ), wait, no. If ( m = 13k ), then ( m ) is a multiple of 13, but ( n = 13 ). So, ( m ) is a multiple of ( n ), which violates the third condition. So, ( n = 13 ) doesn't work either.Wait, maybe ( n ) can be a higher power of 13. Let me try ( n = 13^2 = 169 ). Then ( n^n = 169^{169} = (13^2)^{169} = 13^{338} ). So, ( m^m ) must be divisible by ( 13^{338} ). Therefore, ( m ) must be divisible by 13. Let ( m = 13k ). Then ( m^m = (13k)^{13k} = 13^{13k} times k^{13k} ). So, the exponent of 13 in ( m^m ) is ( 13k ). We need ( 13k geq 338 ), so ( k geq 26 ).But ( m = 13k ) must not be a multiple of ( n = 169 ). Since ( n = 169 = 13^2 ), ( m = 13k ) is a multiple of 13, but not necessarily of ( 13^2 ). So, as long as ( k ) is not a multiple of 13, ( m ) won't be a multiple of ( n = 169 ).So, the smallest ( k ) is 26, but 26 is a multiple of 13 (since 26 = 2×13). So, ( k = 26 ) would make ( m = 13×26 = 338 ), which is ( 2×13^2 ), so ( m = 338 ) is a multiple of ( n = 169 ), which is not allowed. So, we need ( k ) such that ( k ) is not a multiple of 13, and ( k geq 26 ).Wait, no. ( k ) just needs to be such that ( m = 13k ) is not a multiple of ( 13^2 ). So, ( k ) must not be a multiple of 13. So, the smallest ( k ) is 26, but 26 is a multiple of 13, so next is 27, which is not a multiple of 13. So, ( k = 27 ).Wait, but ( k ) needs to satisfy ( 13k geq 338 ). Wait, no, ( 13k ) is the exponent in ( m^m ). Wait, no, ( m = 13k ), and ( m^m = (13k)^{13k} = 13^{13k} times k^{13k} ). So, the exponent of 13 in ( m^m ) is ( 13k ). We need ( 13k geq 338 ), so ( k geq 26 ).So, ( k ) must be at least 26, but ( k ) must not be a multiple of 13 to ensure ( m = 13k ) is not a multiple of ( 13^2 = 169 ).So, the smallest ( k ) is 26, but 26 is a multiple of 13, so next is 27. Let me check ( k = 27 ). Then ( m = 13×27 = 351 ). Is 351 a multiple of 169? 169×2 = 338, 169×3 = 507, so no, 351 is not a multiple of 169. So, ( m = 351 ) is acceptable.Now, check the gcd condition: ( gcd(m + n, 330) = gcd(351 + 169, 330) = gcd(520, 330) ).Let's compute ( gcd(520, 330) ). 520 ÷ 330 = 1 with remainder 190. Then, ( gcd(330, 190) ). 330 ÷ 190 = 1 with remainder 140. ( gcd(190, 140) ). 190 ÷ 140 = 1 with remainder 50. ( gcd(140, 50) ). 140 ÷ 50 = 2 with remainder 40. ( gcd(50, 40) ). 50 ÷ 40 = 1 with remainder 10. ( gcd(40, 10) = 10 ). So, ( gcd(520, 330) = 10 ), which is not 1. So, this doesn't satisfy the first condition.So, ( m = 351 ) and ( n = 169 ) give ( m + n = 520 ), which shares a common factor with 330 (specifically, 10). So, we need to find a larger ( k ) such that ( m + n ) is coprime with 330.Since ( m = 13k ) and ( n = 169 ), ( m + n = 13k + 169 = 13(k + 13) ). So, ( gcd(13(k + 13), 330) = 1 ). Since 13 is a prime not dividing 330, we need ( gcd(k + 13, 330) = 1 ).So, ( k + 13 ) must be coprime with 330. Therefore, ( k + 13 ) must not be divisible by 2, 3, 5, or 11.So, ( k + 13 ) must be an integer greater than or equal to 26 + 13 = 39 (since ( k geq 26 )), and coprime with 330.So, starting from ( k = 26 ), ( k + 13 = 39 ). Is 39 coprime with 330? 39 is divisible by 3 and 13. 330 is divisible by 3, so ( gcd(39, 330) = 3 neq 1 ). So, ( k = 26 ) is invalid.Next, ( k = 27 ), ( k + 13 = 40 ). 40 is divisible by 2 and 5. 330 is divisible by 2 and 5, so ( gcd(40, 330) = 10 neq 1 ). Not good.( k = 28 ), ( k + 13 = 41 ). 41 is a prime number, and 41 doesn't divide 330. So, ( gcd(41, 330) = 1 ). So, ( k = 28 ) works.So, ( m = 13×28 = 364 ), ( n = 169 ). Then, ( m + n = 364 + 169 = 533 ). Check ( gcd(533, 330) ). 533 ÷ 330 = 1 with remainder 203. ( gcd(330, 203) ). 330 ÷ 203 = 1 with remainder 127. ( gcd(203, 127) ). 203 ÷ 127 = 1 with remainder 76. ( gcd(127, 76) ). 127 ÷ 76 = 1 with remainder 51. ( gcd(76, 51) ). 76 ÷ 51 = 1 with remainder 25. ( gcd(51, 25) ). 51 ÷ 25 = 2 with remainder 1. ( gcd(25, 1) = 1 ). So, ( gcd(533, 330) = 1 ). Perfect.Now, check the other conditions. ( m = 364 ), ( n = 169 ). Is ( m^m ) a multiple of ( n^n )? ( n^n = 169^{169} = 13^{338} ). ( m = 364 = 4×91 = 4×7×13 ). So, ( m = 2^2 × 7 × 13 ). Therefore, ( m^m = (2^2 × 7 × 13)^{364} = 2^{728} × 7^{364} × 13^{364} ). The exponent of 13 in ( m^m ) is 364, which is greater than 338, so yes, ( n^n ) divides ( m^m ).Also, ( m = 364 ) is not a multiple of ( n = 169 ), since 364 ÷ 169 = 2.153... which is not an integer.So, ( m = 364 ), ( n = 169 ), ( m + n = 533 ).Wait, but earlier I thought ( k = 27 ) gave ( m = 351 ), but that didn't satisfy the gcd condition. Then ( k = 28 ) gives ( m = 364 ), which works. So, 533 is a candidate.But is this the minimal ( m + n )? Let me see if I can find a smaller ( m + n ).Wait, maybe ( n ) doesn't have to be 169. Maybe a higher prime power? Or perhaps a different prime.Wait, 13 is the smallest prime not dividing 330. If I choose a larger prime, say 17, then ( n = 17^2 = 289 ). Then ( n^n = 289^{289} = 17^{578} ). So, ( m ) must be a multiple of 17, say ( m = 17k ). Then ( m^m = (17k)^{17k} = 17^{17k} times k^{17k} ). So, ( 17k geq 578 ), so ( k geq 34 ). But ( m = 17k ) must not be a multiple of ( n = 289 ), so ( k ) must not be a multiple of 17. So, the smallest ( k ) is 34, but 34 is a multiple of 17, so next is 35.Then, ( m = 17×35 = 595 ), ( n = 289 ). ( m + n = 595 + 289 = 884 ). Check ( gcd(884, 330) ). 884 ÷ 330 = 2 with remainder 224. ( gcd(330, 224) ). 330 ÷ 224 = 1 with remainder 106. ( gcd(224, 106) ). 224 ÷ 106 = 2 with remainder 12. ( gcd(106, 12) ). 106 ÷ 12 = 8 with remainder 10. ( gcd(12, 10) = 2 ). So, ( gcd(884, 330) = 2 neq 1 ). So, invalid.Next, ( k = 36 ). ( m = 17×36 = 612 ). ( m + n = 612 + 289 = 901 ). Check ( gcd(901, 330) ). 901 ÷ 330 = 2 with remainder 241. ( gcd(330, 241) ). 330 ÷ 241 = 1 with remainder 89. ( gcd(241, 89) ). 241 ÷ 89 = 2 with remainder 63. ( gcd(89, 63) ). 89 ÷ 63 = 1 with remainder 26. ( gcd(63, 26) ). 63 ÷ 26 = 2 with remainder 11. ( gcd(26, 11) ). 26 ÷ 11 = 2 with remainder 4. ( gcd(11, 4) ). 11 ÷ 4 = 2 with remainder 3. ( gcd(4, 3) = 1 ). So, ( gcd(901, 330) = 1 ). So, ( m = 612 ), ( n = 289 ), ( m + n = 901 ). But 901 is larger than 533, so 533 is still better.Alternatively, maybe ( n ) can be a composite number not sharing factors with 330. For example, ( n = 13×17 = 221 ). Then ( n^n = 221^{221} ). So, ( m ) must be a multiple of 221, but ( m ) cannot be a multiple of 221. Wait, that's a contradiction. So, ( n ) cannot be composite because then ( m ) would have to be a multiple of ( n )'s prime factors, but not necessarily ( n ) itself. Hmm, maybe.Wait, let's try ( n = 13×17 = 221 ). Then ( n^n = 221^{221} = (13×17)^{221} = 13^{221} × 17^{221} ). So, ( m ) must be divisible by both 13 and 17. Let ( m = 13×17×k = 221k ). Then ( m^m = (221k)^{221k} = 221^{221k} × k^{221k} ). So, the exponents of 13 and 17 in ( m^m ) are ( 221k ). We need ( 221k geq 221 ), so ( k geq 1 ). But ( m = 221k ) must not be a multiple of ( n = 221 ). So, ( k ) must not be a multiple of 1. Wait, that doesn't make sense. If ( k ) is 1, ( m = 221 ), which is a multiple of ( n = 221 ). So, ( k ) must be greater than 1, but ( m = 221k ) would then be a multiple of 221, which is ( n ). So, this approach doesn't work because ( m ) would always be a multiple of ( n ).Therefore, ( n ) cannot be composite because it would force ( m ) to be a multiple of ( n ). So, ( n ) must be a prime power not dividing 330. The smallest such prime is 13, so ( n = 13^2 = 169 ) is the way to go.So, going back to ( n = 169 ), ( m = 364 ), ( m + n = 533 ). But earlier, I thought of ( k = 28 ), but let me check if there's a smaller ( k ) that satisfies ( k + 13 ) coprime with 330.Wait, ( k ) must be at least 26, so ( k + 13 ) is at least 39. The next candidate after ( k = 28 ) (which gives ( k + 13 = 41 )) is ( k = 29 ), ( k + 13 = 42 ). 42 is divisible by 2, 3, 7. So, ( gcd(42, 330) = 6 neq 1 ). Not good.( k = 30 ), ( k + 13 = 43 ). 43 is prime, doesn't divide 330. So, ( gcd(43, 330) = 1 ). So, ( k = 30 ) is another candidate. Then ( m = 13×30 = 390 ), ( n = 169 ), ( m + n = 559 ). Check ( gcd(559, 330) ). 559 ÷ 330 = 1 with remainder 229. ( gcd(330, 229) ). 330 ÷ 229 = 1 with remainder 101. ( gcd(229, 101) ). 229 ÷ 101 = 2 with remainder 27. ( gcd(101, 27) ). 101 ÷ 27 = 3 with remainder 20. ( gcd(27, 20) ). 27 ÷ 20 = 1 with remainder 7. ( gcd(20, 7) ). 20 ÷ 7 = 2 with remainder 6. ( gcd(7, 6) = 1 ). So, ( gcd(559, 330) = 1 ). So, ( m = 390 ), ( n = 169 ), ( m + n = 559 ). But 559 is larger than 533, so 533 is still better.Wait, but earlier I had ( k = 28 ) giving ( m = 364 ), ( n = 169 ), ( m + n = 533 ). Is there a smaller ( k ) between 26 and 28 that satisfies ( k + 13 ) coprime with 330?Wait, ( k = 26 ): ( k + 13 = 39 ), which is divisible by 3. Not good.( k = 27 ): ( k + 13 = 40 ), divisible by 2 and 5. Not good.( k = 28 ): ( k + 13 = 41 ), which is prime and coprime with 330. So, 28 is the smallest ( k ) that works.So, ( m = 364 ), ( n = 169 ), ( m + n = 533 ).But wait, earlier I thought of ( n = 169 ), ( m = 221 ), but that didn't satisfy the gcd condition. Wait, no, ( m = 221 ) would be ( k = 17 ), but ( k = 17 ) is less than 26, so it doesn't satisfy ( 13k geq 338 ). So, ( k = 17 ) is too small.Wait, let me recast this. The exponent condition is ( 13k geq 338 ), so ( k geq 26 ). So, ( k ) must be at least 26. So, ( k = 26 ) is the minimum, but ( k = 26 ) gives ( m = 338 ), which is a multiple of ( n = 169 ), which is not allowed. So, the next possible ( k ) is 27, but ( k + 13 = 40 ), which is not coprime with 330. Then ( k = 28 ), which works.So, ( m = 364 ), ( n = 169 ), ( m + n = 533 ).But wait, is there a way to get a smaller ( m + n ) by choosing a different ( n )?What if ( n ) is a higher power of 13, like ( n = 13^3 = 2197 ). Then ( n^n = 2197^{2197} = 13^{6591} ). So, ( m ) must be a multiple of 13, say ( m = 13k ). Then ( m^m = (13k)^{13k} = 13^{13k} times k^{13k} ). So, ( 13k geq 6591 ), so ( k geq 507 ). That's way too big, so ( m + n ) would be much larger than 533.Alternatively, maybe ( n ) is a prime larger than 13, like 17. But as I saw earlier, ( n = 17^2 = 289 ) gives ( m + n = 901 ), which is larger than 533.Alternatively, maybe ( n ) is a product of two primes not dividing 330, like 13×17=221. But as I saw earlier, that forces ( m ) to be a multiple of 221, which would make ( m ) a multiple of ( n = 221 ), violating the third condition.So, perhaps ( n = 13^2 = 169 ) is indeed the best choice, giving ( m + n = 533 ).Wait, but earlier I thought of ( k = 28 ), ( m = 364 ), but maybe there's a smaller ( k ) that satisfies ( k + 13 ) coprime with 330 and ( k geq 26 ).Wait, ( k = 28 ) is the first ( k geq 26 ) where ( k + 13 ) is coprime with 330. So, 28 is the minimal ( k ).Therefore, the minimal ( m + n ) is 533.Wait, but in the initial problem, the user's thought process led to ( m = 221 ), ( n = 169 ), ( m + n = 390 ). But that was incorrect because ( m = 221 ) is a multiple of ( n = 169 ) only if ( 221 ) is a multiple of ( 169 ), which it's not. Wait, 221 ÷ 169 = 1.307... So, ( m = 221 ) is not a multiple of ( n = 169 ). Wait, but in that case, does ( m = 221 ) satisfy the exponent condition?Wait, ( n = 169 = 13^2 ), so ( n^n = 13^{338} ). ( m = 221 = 13×17 ). So, ( m^m = (13×17)^{221} = 13^{221} × 17^{221} ). The exponent of 13 in ( m^m ) is 221, which is less than 338. So, ( m^m ) is not a multiple of ( n^n ). Therefore, ( m = 221 ) doesn't satisfy the second condition. So, that approach was wrong.So, the correct minimal ( m + n ) is 533.Wait, but in the initial problem, the user's thought process had ( m = 221 ), ( n = 169 ), ( m + n = 390 ), but that was incorrect because ( m^m ) wasn't a multiple of ( n^n ). So, the correct answer is 533.But wait, let me double-check. If ( m = 364 ), ( n = 169 ), then ( m + n = 533 ). Is there a way to get a smaller ( m + n ) by choosing a different ( n )?Wait, what if ( n ) is a higher power of 13, but that would make ( n ) larger, so ( m + n ) would be larger. Alternatively, maybe ( n ) is a prime larger than 13, but that would also make ( n ) larger, leading to a larger ( m + n ).Alternatively, maybe ( n ) is a composite number not sharing factors with 330, but as I saw earlier, that forces ( m ) to be a multiple of ( n ), which is not allowed.So, I think 533 is indeed the minimal ( m + n ).Wait, but let me check if there's a smaller ( n ) that is a prime power not dividing 330. The primes not dividing 330 are 13, 17, 19, etc. So, the smallest prime power is 13^2 = 169, then 17^2 = 289, etc. So, 169 is the smallest possible ( n ).Therefore, the minimal ( m + n ) is 533.Wait, but in the initial problem, the user's thought process had ( m = 221 ), ( n = 169 ), ( m + n = 390 ), but that was incorrect because ( m^m ) wasn't a multiple of ( n^n ). So, the correct answer is 533.But wait, let me check if ( m = 364 ), ( n = 169 ), ( m + n = 533 ), and ( gcd(533, 330) = 1 ). Yes, as computed earlier.Also, ( m^m ) is a multiple of ( n^n ), since ( m = 364 = 2^2 × 7 × 13 ), so ( m^m = (2^2 × 7 × 13)^{364} = 2^{728} × 7^{364} × 13^{364} ), which has ( 13^{364} ), which is more than ( 13^{338} ) in ( n^n ).And ( m = 364 ) is not a multiple of ( n = 169 ), since 364 ÷ 169 = 2.153... which is not an integer.So, all conditions are satisfied, and ( m + n = 533 ) is the minimal possible value.Wait, but I just realized that 533 is quite a large number. Maybe there's a smaller ( n ) that is a prime power not dividing 330, but I thought 13^2 = 169 is the smallest. Let me confirm.The primes not dividing 330 are 13, 17, 19, etc. So, the smallest prime power is 13^2 = 169. So, yes, 169 is the smallest possible ( n ).Therefore, the minimal ( m + n ) is 533.Wait, but in the initial problem, the user's thought process had ( m = 221 ), ( n = 169 ), ( m + n = 390 ), but that was incorrect because ( m^m ) wasn't a multiple of ( n^n ). So, the correct answer is 533.But wait, let me check if there's a way to have ( n ) as a higher power of 13 but with a smaller ( m + n ). For example, ( n = 13^3 = 2197 ), but then ( m ) would have to be much larger, making ( m + n ) larger than 533.Alternatively, maybe ( n ) is a higher power of a larger prime, but that would also make ( n ) larger, leading to a larger ( m + n ).So, I think 533 is indeed the minimal possible value.

💡Okay, so I have this problem here from the American Mathematics Olympiad in 1976. It says:"Let ( a, a_{0}, a_{1}, cdots, a_{n} ) be arbitrary integers. Is the necessary and sufficient condition for the integer ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) to be divisible by ( a^{2}+a+1 ) (or by ( a^{2}-a+1 )) that the integer ( sum_{k=0}^{n}(-1)^{k} a_{k} ) is divisible by ( a^{2}+a+1 ) (or by ( a^{2}-a+1 )?"Hmm, okay. So, the problem is asking whether the divisibility of a certain sum by either ( a^2 + a + 1 ) or ( a^2 - a + 1 ) is equivalent to the divisibility of another sum by the same polynomials. Specifically, it's relating the sum ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) to ( sum_{k=0}^{n}(-1)^{k} a_{k} ).First, I need to understand what each part of the problem is saying. Let's break it down.We have integers ( a, a_0, a_1, ldots, a_n ). The first sum is ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ). That looks like a weighted sum where each term ( a_k ) is multiplied by ( (a^2 + 1)^{3k} ). The question is about the divisibility of this sum by either ( a^2 + a + 1 ) or ( a^2 - a + 1 ).The second sum is ( sum_{k=0}^{n}(-1)^{k} a_{k} ). This is a simpler alternating sum of the coefficients ( a_k ). The question is whether the divisibility of the first sum by ( a^2 + a + 1 ) or ( a^2 - a + 1 ) is equivalent to the divisibility of the second sum by the same polynomials.So, the problem is asking if these two conditions are equivalent. That is, if ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) is divisible by ( a^2 + a + 1 ) if and only if ( sum_{k=0}^{n}(-1)^{k} a_{k} ) is divisible by ( a^2 + a + 1 ), and similarly for ( a^2 - a + 1 ).I think the key here is to analyze the relationship between ( (a^2 + 1)^{3k} ) and the polynomials ( a^2 + a + 1 ) and ( a^2 - a + 1 ). Maybe there's a modular relationship or some kind of periodicity when raising ( a^2 + 1 ) to powers modulo these polynomials.Let me consider ( a^2 + a + 1 ) first. Let me denote ( b = a^2 + a + 1 ). Then, I can try to compute ( (a^2 + 1)^3 ) modulo ( b ).So, ( a^2 + 1 ) modulo ( b ) is equal to ( a^2 + 1 - (a^2 + a + 1) = -a ). Therefore, ( a^2 + 1 equiv -a mod b ).Therefore, ( (a^2 + 1)^3 equiv (-a)^3 = -a^3 mod b ).Now, let's compute ( -a^3 ) modulo ( b ). Since ( b = a^2 + a + 1 ), we can express ( a^3 ) in terms of ( b ).We know that ( a^3 - 1 = (a - 1)(a^2 + a + 1) ). Therefore, ( a^3 = 1 + (a - 1)b ). So, ( a^3 equiv 1 mod b ).Therefore, ( -a^3 equiv -1 mod b ). So, ( (a^2 + 1)^3 equiv -1 mod b ).That's interesting. So, ( (a^2 + 1)^3 equiv -1 mod b ). Therefore, ( (a^2 + 1)^{3k} equiv (-1)^k mod b ).Therefore, the sum ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) modulo ( b ) is equivalent to ( sum_{k=0}^{n} (-1)^k a_k mod b ).Therefore, ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} equiv sum_{k=0}^{n} (-1)^k a_k mod b ).Therefore, the divisibility of the first sum by ( b = a^2 + a + 1 ) is equivalent to the divisibility of the second sum by ( b ).So, that seems to answer the question for ( a^2 + a + 1 ). Now, what about ( a^2 - a + 1 )?Let me denote ( c = a^2 - a + 1 ). I can perform a similar analysis.Compute ( a^2 + 1 ) modulo ( c ). So, ( a^2 + 1 equiv a^2 + 1 - (a^2 - a + 1) = a mod c ).Therefore, ( a^2 + 1 equiv a mod c ). Then, ( (a^2 + 1)^3 equiv a^3 mod c ).Now, let's compute ( a^3 ) modulo ( c ). Since ( c = a^2 - a + 1 ), we can express ( a^3 ) in terms of ( c ).Note that ( a^3 + 1 = (a + 1)(a^2 - a + 1) ). Therefore, ( a^3 = -1 + (a + 1)c ). So, ( a^3 equiv -1 mod c ).Therefore, ( (a^2 + 1)^3 equiv -1 mod c ).Wait, that's the same result as with ( b ). So, ( (a^2 + 1)^3 equiv -1 mod c ).Therefore, ( (a^2 + 1)^{3k} equiv (-1)^k mod c ).Therefore, the sum ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) modulo ( c ) is equivalent to ( sum_{k=0}^{n} (-1)^k a_k mod c ).Hence, the divisibility of the first sum by ( c = a^2 - a + 1 ) is equivalent to the divisibility of the second sum by ( c ).Therefore, in both cases, whether we're considering ( a^2 + a + 1 ) or ( a^2 - a + 1 ), the necessary and sufficient condition for the sum ( sum_{k=0}^{n}left(a^{2}+1right)^{3 k} a_{k} ) to be divisible by these polynomials is that the sum ( sum_{k=0}^{n}(-1)^{k} a_{k} ) is divisible by the same polynomials.So, putting it all together, the answer is yes, the condition is necessary and sufficient.**Final Answer**boxed{text{True}}

💡First, I need to determine the daily wages of a, b, and c based on the given ratios. The ratio of their daily wages is 3:4:5, which means I can express their wages as 3x, 4x, and 5x respectively.Next, I'll calculate the total earnings for each person by multiplying their daily wage by the number of days they worked:- a worked for 6 days, so their total earning is 6 * 3x = 18x.- b worked for 9 days, so their total earning is 9 * 4x = 36x.- c worked for 4 days, so their total earning is 4 * 5x = 20x.Adding these together gives the total earning of all three: 18x + 36x + 20x = 74x.The total earning is given as 1480, so I can set up the equation 74x = 1480 and solve for x:x = 1480 / 74x = 20Now that I have the value of x, I can find the daily wage of c:Daily wage of c = 5x = 5 * 20 = 100Therefore, the daily wage of c is 100.

💡Alright, so I have this problem where I need to determine which of the given functions are "geometric functions." A geometric function is defined such that if you take any geometric sequence {a_n}, applying the function f to each term of the sequence results in another geometric sequence {f(a_n)}. The functions given are:① f(x) = 2^x ② f(x) = log₂|x| ③ f(x) = x² ④ f(x) = ln(2^x)I need to figure out which of these functions satisfy the condition that for any geometric sequence {a_n}, {f(a_n)} is also a geometric sequence.First, I should recall what a geometric sequence is. A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, r. So, if {a_n} is a geometric sequence, then a_{n+1} = r * a_n for all n. This implies that a_{n+2} = r * a_{n+1} = r² * a_n. An important property of geometric sequences is that the square of any middle term is equal to the product of the terms on either side. So, for a geometric sequence {a_n}, we have:a_n * a_{n+2} = (a_{n+1})²This property will be useful because if {f(a_n)} is to be a geometric sequence, then the same relationship must hold for the transformed terms:f(a_n) * f(a_{n+2}) = [f(a_{n+1})]²So, for each function f(x), I need to check whether this equation holds for any geometric sequence {a_n}.Let's analyze each function one by one.**Function ①: f(x) = 2^x**Let's compute f(a_n) * f(a_{n+2}) and see if it equals [f(a_{n+1})]².f(a_n) = 2^{a_n} f(a_{n+2}) = 2^{a_{n+2}} Multiplying these together:f(a_n) * f(a_{n+2}) = 2^{a_n} * 2^{a_{n+2}} = 2^{a_n + a_{n+2}}Now, [f(a_{n+1})]² = (2^{a_{n+1}})² = 2^{2a_{n+1}}So, for f(a_n) * f(a_{n+2}) to equal [f(a_{n+1})]², we need:2^{a_n + a_{n+2}} = 2^{2a_{n+1}} Which implies:a_n + a_{n+2} = 2a_{n+1}But wait, in a geometric sequence, we have:a_{n+2} = r² * a_n a_{n+1} = r * a_nSo, let's substitute these into the equation:a_n + r² * a_n = 2 * (r * a_n) Factor out a_n:a_n (1 + r²) = 2r * a_nSince a_n is non-zero (as the domain is (-∞, 0) ∪ (0, ∞)), we can divide both sides by a_n:1 + r² = 2r Which simplifies to:r² - 2r + 1 = 0 (r - 1)² = 0 So, r = 1This means that the only time when f(a_n) * f(a_{n+2}) = [f(a_{n+1})]² is when the common ratio r is 1. However, a geometric sequence with r = 1 is just a constant sequence, which is a trivial case. Since the problem states "for any given geometric sequence," this function doesn't satisfy the condition for all geometric sequences, only for the trivial ones. Therefore, function ① is not a geometric function.**Function ②: f(x) = log₂|x|**Again, let's compute f(a_n) * f(a_{n+2}) and compare it to [f(a_{n+1})]².f(a_n) = log₂|a_n| f(a_{n+2}) = log₂|a_{n+2}| Multiplying these together:f(a_n) * f(a_{n+2}) = log₂|a_n| * log₂|a_{n+2}| Now, [f(a_{n+1})]² = [log₂|a_{n+1}|]²So, we need:log₂|a_n| * log₂|a_{n+2}| = [log₂|a_{n+1}|]²Let's express a_{n+1} and a_{n+2} in terms of a_n:a_{n+1} = r * a_n a_{n+2} = r² * a_nSo, substituting these into the equation:log₂|a_n| * log₂|r² * a_n| = [log₂|r * a_n|]²Simplify each logarithm:log₂|a_n| * [log₂(r²) + log₂|a_n|] = [log₂r + log₂|a_n|]²Let me denote log₂|a_n| as L and log₂r as R for simplicity.So, the equation becomes:L * (2R + L) = (R + L)²Expanding both sides:Left side: L * 2R + L² = 2RL + L² Right side: R² + 2RL + L²So, setting them equal:2RL + L² = R² + 2RL + L²Subtracting 2RL + L² from both sides:0 = R²Which implies R = 0, so log₂r = 0, meaning r = 1.Again, this only holds when the common ratio r is 1, which is the trivial case. Therefore, function ② is not a geometric function for all geometric sequences.**Function ③: f(x) = x²**Let's compute f(a_n) * f(a_{n+2}) and compare it to [f(a_{n+1})]².f(a_n) = (a_n)² f(a_{n+2}) = (a_{n+2})² Multiplying these together:f(a_n) * f(a_{n+2}) = (a_n)² * (a_{n+2})² = (a_n * a_{n+2})²Now, [f(a_{n+1})]² = [(a_{n+1})²]² = (a_{n+1})⁴So, we need:(a_n * a_{n+2})² = (a_{n+1})⁴Taking square roots on both sides:a_n * a_{n+2} = (a_{n+1})²But this is exactly the property of a geometric sequence! As we know, in a geometric sequence:a_{n+2} = r² * a_n a_{n+1} = r * a_nSo, substituting:a_n * (r² * a_n) = (r * a_n)² Left side: r² * (a_n)² Right side: r² * (a_n)²They are equal. Therefore, for function ③, f(a_n) * f(a_{n+2}) = [f(a_{n+1})]² holds for any geometric sequence {a_n}. Thus, function ③ is a geometric function.**Function ④: f(x) = ln(2^x)**First, let's simplify f(x):f(x) = ln(2^x) = x * ln(2)So, f(x) is a linear function, specifically scaling x by ln(2). Let's compute f(a_n) * f(a_{n+2}) and compare it to [f(a_{n+1})]².f(a_n) = a_n * ln(2) f(a_{n+2}) = a_{n+2} * ln(2) Multiplying these together:f(a_n) * f(a_{n+2}) = (a_n * ln(2)) * (a_{n+2} * ln(2)) = (a_n * a_{n+2}) * (ln(2))²Now, [f(a_{n+1})]² = [a_{n+1} * ln(2)]² = (a_{n+1})² * (ln(2))²So, we need:(a_n * a_{n+2}) * (ln(2))² = (a_{n+1})² * (ln(2))²We can divide both sides by (ln(2))² since ln(2) ≠ 0:a_n * a_{n+2} = (a_{n+1})²Again, this is the property of a geometric sequence. As before, in a geometric sequence:a_{n+2} = r² * a_n a_{n+1} = r * a_nSo, substituting:a_n * (r² * a_n) = (r * a_n)² Left side: r² * (a_n)² Right side: r² * (a_n)²They are equal. Therefore, for function ④, f(a_n) * f(a_{n+2}) = [f(a_{n+1})]² holds for any geometric sequence {a_n}. Thus, function ④ is a geometric function.**Summary:**- Function ①: Not a geometric function.- Function ②: Not a geometric function.- Function ③: Geometric function.- Function ④: Geometric function.Therefore, the serial numbers of the "geometric functions" are ③ and ④.**Final Answer**The serial numbers of the geometric functions are boxed{③④}.