🧠:[<think>Okay, so I need to find the units digit of this big product plus one: (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1) + 1. The options are A:2, B:4, C:6, D:8. Hmm. Let's break it down step by step.First, let me understand the structure of the product. Each term seems to be 2 raised to a power plus 1. The exponents are 1, 2, 4, 6, ..., 32. Wait, hold on. The first term is (2+1), which is 2¹ +1. Then the next term is 2² +1, then 2⁴ +1, then 2⁶ +1? Wait, that seems inconsistent. Let me check the exponents again.The problem says: (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). Let's list the exponents:First term: exponent 1 (2¹ +1)Second term: exponent 2 (2² +1)Third term: exponent 4 (2⁴ +1)Fourth term: exponent 6 (2⁶ +1)...Last term: exponent 32 (2³² +1)Wait, so the exponents start at 1, then go to 2, then 4, 6, ..., 32? That seems a bit irregular. Let me check if there's a pattern here. From 1 to 2 is +1, then 2 to 4 is +2, then 4 to 6 is +2, and so on up to 32? Hmm, maybe it's a typo? Because usually, in such products, the exponents double each time, like 1, 2, 4, 8, 16, 32. That would make sense as a geometric progression. But here, after 4, it goes to 6, which is not doubling. Maybe the problem statement is correct, but let me confirm.The user wrote: (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). So the exponents are 1, 2, 4, 6, 8, ..., 32? Wait, maybe after 4, it increases by 2 each time? Let's see: 1, 2, 4, 6, 8, 10, 12, 14, 16, ..., 32? But 2⁶ is 64, but that's not part of the exponent here. Wait, but the exponents themselves are 1, 2, 4, 6, ..., 32. So the exponents go 1, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32? That's 17 terms? Wait, 1 to 32, increasing by 1 first, then by 2? Hmm, maybe not. Wait, starting from 1, then 2, then 4, then 6... Hmm, maybe there's a typo, but assuming the problem is written correctly, let's proceed.Wait, maybe the exponents are supposed to be 1, 2, 4, 8, 16, 32, which are powers of 2. But in the problem, it's written as 1, 2, 4, 6, 8,...32. Hmm. Let me check the original problem again.Original problem: (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1)+1. So the exponents are 1, 2, 4, 6, ..., 32. So starting from 1, then 2, 4, 6, 8, 10, ..., up to 32? That seems like after the first two terms, the exponents increase by 2 each time. Let me count how many terms there are.From 1 to 32, with steps: 1, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32. Let's count: 1 (1), 2 (2), 4 (3), 6 (4), 8 (5), 10 (6), 12 (7), 14 (8), 16 (9), 18 (10), 20 (11), 22 (12), 24 (13), 26 (14), 28 (15), 30 (16), 32 (17). So 17 terms? Hmm, that seems a lot. But maybe that's correct? Wait, but if we consider exponents 1, 2, 4, 8, 16, 32, that's 6 terms. So 2+1)(2²+1)(2⁴+1)(2⁸+1)...(2³²+1). But the problem has exponents 1, 2, 4, 6, 8,...,32. So perhaps the problem is written differently. Alternatively, maybe the exponents are 1, 2, 4, 6, 16, 32? Wait, that also doesn't make sense. Hmm.Wait, maybe the problem is mistyped? Because in some textbooks, there's a similar problem where you have (2+1)(2²+1)(2⁴+1)...(2^{2^n}+1) +1, which equals 2^{2^{n+1}}} or something like that. But here, the exponents are 1, 2, 4, 6, ...,32. Hmm. Maybe it's supposed to be exponents doubling each time, but with a typo. But assuming the problem is as stated, let's see.Alternatively, maybe the exponents are 1, 2, 4, 8, 16, 32, but the problem wrote 6 instead of 8? Because 2⁶ is 64, which is 2^6, but if the exponent is 6, then that's different. Wait, but the problem has (2⁶ +1). Hmm. Maybe the original problem is correct. Let's proceed as per the given exponents.So, first, let me try to compute the units digit of the product (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1) +1.Since we are only interested in the units digit, we can compute each term modulo 10, multiply them modulo 10, then add 1 and take modulo 10 again.So, first, compute each term (2^k +1) mod 10, then multiply all those remainders together, mod 10, then add 1 and take mod 10.Let's start by finding each term's units digit.First term: (2^1 +1) = 2 +1 = 3. Units digit 3.Second term: (2^2 +1) = 4 +1 = 5. Units digit 5.Third term: (2^4 +1). 2^4 = 16, so 16 +1 =17. Units digit 7.Fourth term: (2^6 +1). 2^6 = 64. 64 +1 =65. Units digit 5.Fifth term: (2^8 +1). 2^8 =256. 256 +1=257. Units digit 7.Wait, but hold on, in the problem, after (2⁴ +1), it's (2⁶ +1), so exponents are 1,2,4,6,8,...32.Wait, so starting from exponent 1, then 2, then 4, then 6, 8, 10, ..., up to 32. Let's confirm the exponents:Yes, the problem says: (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). So the exponents after 4 increase by 2 each time: 6,8,10,...,32. So exponents are 1,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32. That is 17 terms in total. Let's verify:Starting at 1, then 2, then 4. Then from 4 onwards, step of 2 until 32. So 4,6,8,...,32. How many terms is that? Let's compute the number of terms from 4 to 32 with step 2: (32 -4)/2 +1 = (28)/2 +1 =14 +1=15 terms. Then adding the first two terms (1 and 2), total terms:15 +2=17. So 17 terms.But maybe it's easier not to count them, but rather note that each term from exponent 4 onwards, it's every even number starting from 4 up to 32. Wait, no, starting from exponent 1:1,2,4,6,8,...32. So that's 1,2, then 4,6,8,...,32. So from 4 to 32 inclusive, step 2. So (32-4)/2 +1=15 terms. Then 1 and 2 make total 17.But perhaps we can find a pattern in the units digits.First, let's compute each term's units digit:Term 1: exponent 1: 2^1=2, +1=3. Units digit 3.Term 2: exponent 2: 4 +1=5. Units digit 5.Term 3: exponent 4: 16 +1=17. Units digit 7.Term 4: exponent 6: 64 +1=65. Units digit 5.Term 5: exponent 8: 256 +1=257. Units digit 7.Term 6: exponent 10: 2^10=1024. 1024 +1=1025. Units digit 5.Term 7: exponent 12: 2^12=4096. 4096 +1=4097. Units digit 7.Term 8: exponent 14: 2^14=16384. 16384 +1=16385. Units digit 5.Term 9: exponent 16: 2^16=65536. 65536 +1=65537. Units digit 7.Term 10: exponent 18: 2^18=262144. 262144 +1=262145. Units digit 5.Term 11: exponent 20: 2^20=1048576. 1048576 +1=1048577. Units digit 7.Term 12: exponent 22: 2^22=4194304. 4194304 +1=4194305. Units digit 5.Term 13: exponent 24: 2^24=16777216. +1=16777217. Units digit 7.Term 14: exponent 26: 2^26=67108864. +1=67108865. Units digit 5.Term 15: exponent 28: 2^28=268435456. +1=268435457. Units digit 7.Term 16: exponent 30: 2^30=1073741824. +1=1073741825. Units digit 5.Term 17: exponent 32: 2^32=4294967296. +1=4294967297. Units digit 7.So listing all the units digits:Term 1:3Term 2:5Term 3:7Term 4:5Term 5:7Term 6:5Term 7:7Term 8:5Term 9:7Term10:5Term11:7Term12:5Term13:7Term14:5Term15:7Term16:5Term17:7So, units digits of the factors: 3,5,7,5,7,5,7,5,7,5,7,5,7,5,7,5,7.So starting from term 3 onwards (exponent 4), the units digits alternate between 7 and 5, starting with 7 for exponent 4, then 5 for exponent 6, 7 for exponent 8, etc. Wait, but when we list them:Term3:7 (exponent4)Term4:5 (exponent6)Term5:7 (exponent8)Term6:5 (exponent10)So alternating 7,5,7,5,... So from term3 to term17 (15 terms), starting with 7 and alternating. Since 15 terms: starting with 7, the pattern is 7,5,7,5,...15 terms. Since 15 is odd, the last term (term17) is 7.So the units digits are: 3,5, then 7,5,7,5,...7 (total 15 terms after the first two).So let's separate the units digits into:First two terms: 3 and 5.Then the next 15 terms: starting with 7, alternating 5,7,..., ending with 7.So to compute the total product mod 10, we can compute the product of all these units digits modulo 10.Let's first compute the product of the first two terms: 3 * 5 = 15. 15 mod10=5.Now, the remaining 15 terms: 7,5,7,5,...,7. Since each pair (7,5) multiplies to 35, which is 5 mod10. Since there are 15 terms, which is 7.5 pairs? Wait, 15 terms starting with 7. Let's group them:Term3:7Term4:5Term5:7Term6:5...Term17:7So, starting from term3, every two terms form a pair: (7,5). How many such pairs are there?15 terms: each pair is 2 terms, so 15/2=7.5. So 7 full pairs and one remaining term. Since 15 is odd. So:Number of pairs:7 pairs (each 7,5) and one remaining term (7).Each pair (7*5)=35. 35 mod10=5. So each pair contributes a factor of 5 mod10. Then the remaining term is 7.So the total contribution from the 15 terms is (5^7)*7 mod10.So first, compute 5^7 mod10. But 5^n mod10 is always 5 if n>=1. Because 5*5=25 mod10=5, then 5*5=5, etc. So 5^7 mod10=5.Then, multiply by 7: 5*7=35. 35 mod10=5.Therefore, the 15 terms contribute 5 mod10.So now, combining everything:First two terms: 5 mod10.Next 15 terms:5 mod10.Total product:5 *5=25 mod10=5.Then add 1:5 +1=6. So 6 mod10=6.Therefore, the units digit is 6, which is option C.Wait, but let me verify this step by step again to be sure.First, compute the first two terms: (3)*(5)=15. 15 mod10=5.Then, the next 15 terms: as calculated, they contribute 5 mod10.So total product is 5*5=25 mod10=5.Then, adding 1 gives 5+1=6 mod10=6. So answer is C:6.But let me check if my grouping is correct.Alternative approach: compute the entire product step by step, taking modulus 10 at each step.Start with 3 (from first term).Multiply by second term 5:3*5=15. 15 mod10=5.Multiply by third term 7:5*7=35. 35 mod10=5.Multiply by fourth term 5:5*5=25. 25 mod10=5.Multiply by fifth term7:5*7=35. 35 mod10=5.Multiply by sixth term5:5*5=25. 25 mod10=5.Multiply by seventh term7:5*7=35. 35 mod10=5.Multiply by eighth term5:5*5=25. 25 mod10=5.Multiply by ninth term7:5*7=35. 35 mod10=5.Multiply by tenth term5:5*5=25. 25 mod10=5.Multiply by eleventh term7:5*7=35. 35 mod10=5.Multiply by twelfth term5:5*5=25. 25 mod10=5.Multiply by thirteenth term7:5*7=35. 35 mod10=5.Multiply by fourteenth term5:5*5=25. 25 mod10=5.Multiply by fifteenth term7:5*7=35. 35 mod10=5.Multiply by sixteenth term5:5*5=25. 25 mod10=5.Multiply by seventeenth term7:5*7=35. 35 mod10=5.Wait, hold on. Wait, the 15 terms after the first two would be 15 terms: term3 to term17. So starting from term3 (7), term4(5),...,term17(7). So that's 15 terms. If we start with 5 (from first two terms), then multiply by term3(7):5*7=35 mod10=5. Then term4(5):5*5=25 mod10=5. Term5(7):5*7=35 mod10=5. Term6(5):5*5=25 mod10=5. Term7(7):5*7=35 mod10=5. Term8(5):5*5=25 mod10=5. Term9(7):5*7=35 mod10=5. Term10(5):5*5=25 mod10=5. Term11(7):5*7=35 mod10=5. Term12(5):5*5=25 mod10=5. Term13(7):5*7=35 mod10=5. Term14(5):5*5=25 mod10=5. Term15(7):5*7=35 mod10=5. Term16(5):5*5=25 mod10=5. Term17(7):5*7=35 mod10=5.Wait, that gives after each multiplication, the result is alternating between 5 and 5. Wait, no. Wait, if we start with 5, then multiply by 7:5*7=35 mod10=5. Then 5*5=25 mod10=5. Then 5*7=35 mod10=5, and so on. So every multiplication, regardless of multiplying by 5 or 7, when the current product is 5 mod10, multiplying by 5 gives 25 mod10=5, and multiplying by 7 gives 35 mod10=5. So regardless, after each multiplication, the product remains 5 mod10. So after multiplying all 15 terms, the product is still 5 mod10.Therefore, the total product is 5 mod10. Then add 1:5+1=6 mod10=6. Therefore, the answer is C:6.Alternatively, this seems to be correct, but let me check with another approach.Another way to think about this is recognizing that multiplying numbers ending with 5 and 7 alternately, when starting with 5, the result cycles. But in our case, after the first two terms, the product is 5 mod10, and then each subsequent multiplication by 5 or 7 keeps it at 5 mod10. Because 5*5=25 mod10=5, and 5*7=35 mod10=5. So once the product is 5 mod10, multiplying by any number of 5s and 7s will keep it 5 mod10. Therefore, after the first two terms (which give 5 mod10), multiplying all remaining 15 terms (alternating 7 and 5) keeps it at 5 mod10. Then adding 1 gives 6 mod10.Alternatively, perhaps there's a formula or identity that can be applied here. The given product resembles the form of (2 -1)(2 +1)(2² +1)(2⁴ +1)... which is a telescoping product. Wait, but in this problem, the first term is (2 +1), not (2 -1). Let's see:If we have (2 -1)(2 +1)(2² +1)(2⁴ +1)...(2^{2^n} +1) = 2^{2^{n+1}}} -1. But in our problem, the first term is (2 +1), so maybe we can adjust it.But the given product is (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). If we multiply and divide by (2 -1), which is 1, we get:[(2 -1)(2 +1)(2² +1)(2⁴ +1)(2⁶ +1)...(2³² +1)] / (2 -1) = [ (2² -1)(2² +1)(2⁴ +1)(2⁶ +1)...(2³² +1) ] /1 = (2⁴ -1)(2⁴ +1)(2⁶ +1)...(2³² +1) = ... and so on. But this telescoping only works if the exponents are powers of 2. However, in our problem, after 2⁴ +1, we have 2⁶ +1, which breaks the telescoping pattern. Therefore, this approach might not work here.Therefore, the earlier approach of calculating each term's units digit and multiplying step by step modulo10 seems the way to go.Hence, the units digit is 6, so the answer is C.Final Answerboxed{C}</think>To find the units digit of the expression (left(2+1right)left(2^{2}+1right)left(2^{4}+1right)left(2^{6}+1right)ldots left(2^{32}+1right)+1), we need to consider the units digit of each term in the product and then compute the result modulo 10.1. Units digits of each term: - (2^1 + 1 = 3) (units digit 3) - (2^2 + 1 = 5) (units digit 5) - (2^4 + 1 = 17) (units digit 7) - (2^6 + 1 = 65) (units digit 5) - (2^8 + 1 = 257) (units digit 7) - (2^{10} + 1 = 1025) (units digit 5) - (2^{12} + 1 = 4097) (units digit 7) - (2^{14} + 1 = 16385) (units digit 5) - (2^{16} + 1 = 65537) (units digit 7) - (2^{18} + 1 = 262145) (units digit 5) - (2^{20} + 1 = 1048577) (units digit 7) - (2^{22} + 1 = 4194305) (units digit 5) - (2^{24} + 1 = 16777217) (units digit 7) - (2^{26} + 1 = 67108865) (units digit 5) - (2^{28} + 1 = 268435457) (units digit 7) - (2^{30} + 1 = 1073741825) (units digit 5) - (2^{32} + 1 = 4294967297) (units digit 7)2. Product of units digits modulo 10: - The units digits of the terms are: 3, 5, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5, 7. - The first two terms: (3 times 5 = 15) (units digit 5). - The remaining 15 terms alternate between 7 and 5. Multiplying these terms modulo 10: - Each pair (7, 5) contributes (7 times 5 = 35) (units digit 5). - There are 7 such pairs and one remaining 7, contributing (5^7 times 7). - Since (5^n equiv 5 pmod{10}) for any (n geq 1), this simplifies to (5 times 7 = 35) (units digit 5).3. Combining results: - The product of all terms modulo 10 is (5 times 5 = 25) (units digit 5). - Adding 1: (5 + 1 = 6).Thus, the units digit of the given expression is (boxed{C}).,<think>Okay, so I need to find the units digit of this big product plus one. The expression is (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1) + 1. Hmm, let's break it down step by step. First, maybe I can compute the units digit of each term in the product and then multiply them all together, adding 1 at the end. But let's see.Starting with the first term: 2 + 1 = 3. The units digit is 3.Next term: 2² + 1 = 4 + 1 = 5. Units digit is 5.Third term: 2⁴ + 1. 2⁴ is 16, so 16 + 1 = 17. Units digit is 7.Fourth term: 2⁶ + 1. Wait, 2⁶ is 64, so 64 + 1 = 65. Units digit is 5.Wait, hold on, the exponents here: 2, 4, 6, ..., 32. Wait, the first term is (2+1), which is exponent 1, but 2^1 +1? Wait, no. Wait, the problem is written as (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). Let me check the exponents again.Starting from the first term: 2^1 +1, then 2^2 +1, then 2^4 +1, then 2^6 +1... Wait, but 2^1 is 2, so 2 +1 = 3. Then 2² +1 = 5, then 2⁴ +1 = 17, then 2⁶ +1 = 65, then 2^8 +1=257, 2^10 +1=1025, etc., up to 2^32 +1. Wait, but the exponents seem to be 1, 2, 4, 6, 8, ..., 32? Wait, that seems inconsistent. Wait, 1, 2, 4, 6... Hmm, maybe there's a typo in the problem? Because after 2^4, the next exponent is 6, which is 4 + 2. Then 8, 10, etc., up to 32? Let me confirm.Wait, maybe it's supposed to be exponents that are powers of 2? Usually, when you see a product like (2+1)(2²+1)(2⁴+1)... it's exponents doubling each time: 1, 2, 4, 8, 16, 32. But in the problem here, after 2^4 +1, it's 2^6 +1. So maybe it's 1, 2, 4, 6, 8, ..., 32. Wait, 6 is not a power of 2. Hmm. That seems odd. Let me check again.Wait, perhaps the original problem is written with exponents increasing by 2 each time starting from 1? Let's list all the exponents:First term: exponent 1 (2^1 +1)Second term: exponent 2 (2² +1)Third term: exponent 4 (2^4 +1)Fourth term: exponent 6 (2^6 +1)Wait, that's inconsistent. 1, 2, 4, 6... Maybe it's a typo, and it should be 1, 2, 4, 8, 16, 32? That would make sense as each exponent is doubling. But if it's written as 1, 2, 4, 6, 8,... up to 32, then the exponents after 4 are increasing by 2 each time. Let me count how many terms there are. Starting at 1, then 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32. Wait, from 1 to 32, increasing by 1, but the exponents given are 1, 2, 4, 6, ..., 32. Wait, maybe it's 1, 2, 4, 8, 16, 32? That would be exponents as powers of 2. But the problem says (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). The fourth term is 2^6 +1. So, exponent 6. That complicates things. Maybe the problem is correct as stated, and the exponents are 1, 2, 4, 6, 8, ..., 32. Let's see. Starting at 1, then 2, then 4, then 6, 8, ..., up to 32. That would mean the exponents after 2 are increasing by 2 each time. So from 4 to 32, step 2. Wait, 4,6,8,10,12,14,16,18,20,22,24,26,28,30,32. So starting from 4, stepping by 2 up to 32. Let's count how many terms that is. From 4 to 32 inclusive, step 2: (32 - 4)/2 +1 = (28)/2 +1=14 +1=15 terms. Then including the first two terms (1 and 2), total terms would be 15 + 2 =17? Wait, but the problem is written as (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1). So the first term is exponent 1, then 2, then 4, then 6, etc., up to 32. So exponents: 1,2,4,6,8,...,32.This seems a bit odd, but maybe it's intentional. Maybe there is a pattern here. Let's check. Alternatively, maybe the problem was meant to have exponents as powers of two: 1, 2, 4, 8, 16, 32. Then the product would be (2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1). That is a known formula which telescopes when multiplied by (2 -1). But in our problem, the exponents after 4 are 6,8,...32. So not sure. Wait, let me check again.Wait, if the problem is (2+1)(2²+1)(2⁴+1)(2⁶+1)...(2³²+1)+1, then maybe the exponents start at 1 and then go even numbers up to 32? Wait, 1 is odd, then 2 is even, 4 is even, 6 is even... So the exponents are 1, 2, 4, 6, 8, ...,32. That seems inconsistent. Maybe there was a mistake in the exponent sequence. Alternatively, perhaps the problem is written correctly and the exponents are 1,2,4,6,8,...,32. Let's assume that's the case.So, to compute the units digit, we can compute the product modulo 10. Because the units digit is the remainder when divided by 10. So, compute each term modulo 10, multiply them all together modulo 10, then add 1 and take modulo 10 again.Let me list each term's units digit:First term: (2 +1) = 3. Units digit 3.Second term: 2² +1 = 4 +1 =5. Units digit 5.Third term: 2⁴ +1. 2⁴ is 16, so 16 +1=17. Units digit 7.Fourth term: 2⁶ +1. 2⁶ is 64, so 64 +1=65. Units digit 5.Fifth term: 2⁸ +1. 2⁸=256, so 256 +1=257. Units digit 7.Sixth term: 2¹⁰ +1. 2¹⁰=1024, 1024 +1=1025. Units digit 5.Seventh term: 2¹² +1. 2¹²=4096, 4096 +1=4097. Units digit 7.Eighth term: 2¹⁴ +1. 16384 +1=16385. Units digit 5.Ninth term: 2¹⁶ +1. 65536 +1=65537. Units digit 7.Tenth term: 2¹⁸ +1. 262144 +1=262145. Units digit 5.Eleventh term: 2²⁰ +1. 1048576 +1=1048577. Units digit 7.Twelfth term: 2²² +1. 4194304 +1=4194305. Units digit 5.Thirteenth term: 2²⁴ +1. 16777216 +1=16777217. Units digit 7.Fourteenth term: 2²⁶ +1. 67108864 +1=67108865. Units digit 5.Fifteenth term: 2²⁸ +1. 268435456 +1=268435457. Units digit 7.Sixteenth term: 2³⁰ +1. 1073741824 +1=1073741825. Units digit 5.Seventeenth term: 2³² +1. 4294967296 +1=4294967297. Units digit 7.Wait, but let me confirm. The exponents after the first two terms are 4,6,8,...32. So starting from exponent 1, then 2, then 4,6,8,...32. So total terms: exponent 1,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32. Let's count: 1,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32. That's 17 terms. So 17 terms in the product. Each term's units digit alternates between 3,5,7,5,7,...,5,7. Wait, the first term is 3, then starting from the second term, the units digits go 5,7,5,7,... up to the last term, which is exponent 32. Let's list all units digits:Term 1: 3Term 2: 5Term 3: 7Term 4: 5Term 5: 7...Continuing this pattern up to term 17. Let's check the pattern.From term 2 onwards (exponents 2,4,6,...32):Exponent even number. 2^(2k) +1. Let's find the units digit of 2^(2k) +1.The units digit of 2^n cycles every 4:2^1=2, 2^2=4, 2^3=8, 2^4=6, 2^5=2, etc. So cycle: 2,4,8,6,2,4,8,6,...So for exponent 2k, which is even. Let's let n = 2k. Then n is even, so n mod 4 is either 0 or 2.If n ≡ 0 mod 4, then 2^n ends with 6. So 2^n +1 ends with 7.If n ≡ 2 mod 4, then 2^n ends with 4. So 2^n +1 ends with 5.Therefore, for exponents 2k:If k is odd (so n=2k ≡ 2 mod 4), units digit is 5.If k is even (so n=2k ≡ 0 mod 4), units digit is 7.Wait, let's verify with exponents:Exponent 2: 2^2=4. 4 +1=5. Correct.Exponent 4: 2^4=16. 16 +1=17. Units digit 7.Exponent 6: 2^6=64. 64 +1=65. Units digit 5.Exponent 8: 2^8=256. 256 +1=257. Units digit 7.Exponent 10: 2^10=1024. Units digit 4. 4 +1=5.So yes, alternating 5,7,5,7,... starting from exponent 2: when exponent ≡ 2 mod 4 (2,6,10,...), units digit is 5. When exponent ≡ 0 mod 4 (4,8,12,...), units digit is 7.So starting from term 2 (exponent 2), the units digits alternate 5,7,5,7,... Since exponents from 2 onwards go 2,4,6,8,10,...,32, which in terms of mod 4:Exponent 2: 2 mod 4 ⇒ 5Exponent 4: 0 mod 4 ⇒7Exponent 6: 2 mod 4 ⇒5Exponent 8: 0 mod 4 ⇒7And so on. Therefore, starting from term 2 (exponent 2), the pattern is 5,7,5,7,... alternately.So term 2:5, term3:7, term4:5, term5:7,..., term17: exponent 32.Exponent 32: 32 mod 4=0, so units digit 7.Therefore, the units digits of the terms are:Term1:3,Term2:5,Term3:7,Term4:5,Term5:7,Term6:5,Term7:7,Term8:5,Term9:7,Term10:5,Term11:7,Term12:5,Term13:7,Term14:5,Term15:7,Term16:5,Term17:7.So total terms: 1 (3), followed by 16 terms alternating 5 and7, starting with 5 and ending with7 (since 16 terms: 8 pairs of 5 and7, so 8*2=16). Wait, term2 to term17:16 terms.Therefore, units digits:3,5,7,5,7,...,5,7.So let's compute the product modulo 10. Since multiplying numbers modulo 10, we can take the units digit of each term, multiply them step by step, taking modulo 10 at each step.Let's proceed step by step:Start with term1: 3.Multiply by term2: 3 * 5 = 15. Units digit 5.Multiply by term3: 5 * 7 = 35. Units digit 5.Multiply by term4: 5 * 5 =25. Units digit 5.Multiply by term5: 5 * 7=35. Units digit 5.Multiply by term6:5 *5=25. Units digit5.Multiply by term7:5*7=35. Units digit5.Multiply by term8:5*5=25. Units digit5.Multiply by term9:5*7=35. Units digit5.Multiply by term10:5*5=25. Units digit5.Multiply by term11:5*7=35. Units digit5.Multiply by term12:5*5=25. Units digit5.Multiply by term13:5*7=35. Units digit5.Multiply by term14:5*5=25. Units digit5.Multiply by term15:5*7=35. Units digit5.Multiply by term16:5*5=25. Units digit5.Multiply by term17:5*7=35. Units digit5.So after multiplying all terms, the product modulo10 is5. Then add1:5 +1=6. So the units digit is6.Wait, that's the conclusion? Let me verify the steps again, because this seems a bit repetitive.Starting with 3.After multiplying by term2 (5): 3*5=15 → units digit5.Then term3 (7):5*7=35 → units digit5.Term4 (5):5*5=25→5.Term5 (7):5*7=35→5.Term6 (5):5*5=25→5.Term7 (7):5*7=35→5.Term8 (5):5*5=25→5.Term9 (7):5*7=35→5.Term10 (5):5*5=25→5.Term11 (7):5*7=35→5.Term12 (5):5*5=25→5.Term13 (7):5*7=35→5.Term14 (5):5*5=25→5.Term15 (7):5*7=35→5.Term16 (5):5*5=25→5.Term17 (7):5*7=35→5.So after multiplying all 17 terms, the units digit is5. Then add1:5 +1=6. Therefore, the answer should beC:6.But wait, that seems straightforward, but let me check if there's a different approach.Alternatively, maybe there's a formula for such products. For example, products of the form (2 +1)(2² +1)(2⁴ +1)...(2^{2^n} +1) = 2^{2^{n+1}} -1. But in this problem, the exponents are 1,2,4,6,8,...,32, which doesn't follow the doubling exponents. However, if we assume that the exponents after 4 are increasing by 2 each time, which is different from the usual telescoping product.Alternatively, maybe if we can find a pattern in the product modulo10. Let's see:Alternatively, perhaps observe that after a few terms, the product modulo10 cycles or stabilizes. Let's test:First term:3.After term1:3.After term2:3*5=15→5.After term3:5*7=35→5.After term4:5*5=25→5.After term5:5*7=35→5.Wait, it's always 5 from term2 onwards. Wait, starting from term2, each subsequent multiplication either by5 or7, but since 5*5=25≡5 mod10, and 5*7=35≡5 mod10. Therefore, once the product becomes5, multiplying by5 or7 will keep it at5. Hence, once we multiply the first two terms (3*5=15≡5 mod10), then all subsequent multiplications by whatever (5 or7) will keep the product as5. Then, adding1 gives6. Therefore, regardless of the number of terms after the first two, the product modulo10 is5, then +1=6.Therefore, the answer is6, which is optionC.But let me confirm this reasoning. Once the product becomes5 mod10, multiplying by5 gives5*5=25≡5, and multiplying by7 gives5*7=35≡5. So regardless of multiplying by5 or7, the result remains5 mod10. Therefore, once we have a units digit of5, multiplying by any number of terms with units digits5 or7 will keep it at5. Therefore, after the first two terms (3*5=15≡5), the rest of the multiplications don't change the units digit. Therefore, the product is5, then +1=6. Therefore, the units digit is6. So the answer isC.But let's check with a smaller example. Suppose we have (2+1)(2²+1) +1 =3*5 +1=15 +1=16. Units digit6. Which matches. Then, if we add another term, say (2+1)(2²+1)(2^4 +1) +1=3*5*17 +1=255 +1=256. Units digit6. Wait, 255 +1=256. Units digit6. Then adding another term, (2+1)(2²+1)(2^4 +1)(2^6 +1)+1=3*5*17*65 +1=3*5=15, 15*17=255, 255*65=16575, 16575 +1=16576. Units digit6. So indeed, the units digit remains6 after the first two terms. Wait, but in our previous step-by-step calculation, after multiplying the first two terms, we had15→5, then multiply by17→5*7=35→5, then multiply by65→5*5=25→5, etc., leading to the same result. So even though the actual product's units digit cycles between5, when you add1, it becomes6. So regardless of the number of terms beyond the first two, the product modulo10 is5, so +1 gives6. Therefore, answer isC:6.Final Answerboxed{C}</think>To find the units digit of the expression (left(2+1right)left(2^{2}+1right)left(2^{4}+1right)left(2^{6}+1right)ldots left(2^{32}+1right)+1), we analyze the units digit of each term in the product and then compute the result modulo 10.1. The units digits of each term are calculated as follows: - (2 + 1 = 3) (units digit 3) - (2^2 + 1 = 5) (units digit 5) - (2^4 + 1 = 17) (units digit 7) - (2^6 + 1 = 65) (units digit 5) - (2^8 + 1 = 257) (units digit 7) - Subsequent terms alternate between units digits 5 and 7 depending on the exponent modulo 4. For even exponents: - If (2k equiv 2 mod 4), the units digit is 5. - If (2k equiv 0 mod 4), the units digit is 7.2. The units digits of the terms are: 3, 5, 7, 5, 7, 5, ..., 7 (alternating 5 and 7 starting from the third term).3. Multiplying these units digits step-by-step modulo 10: - Start with 3. - Multiply by 5: (3 times 5 = 15 equiv 5 mod 10). - Subsequent multiplications by 5 or 7 will keep the product as 5 modulo 10 because: - (5 times 5 = 25 equiv 5 mod 10) - (5 times 7 = 35 equiv 5 mod 10)4. After multiplying all terms, the product modulo 10 is 5. Adding 1 gives: - (5 + 1 = 6 mod 10)Thus, the units digit of the given expression is (boxed{C}).]

🔑:To solve for the units digit of the expression left(2+1right)left(2^{2}+1right)left(2^{4}+1right)left(2^{6}+1right)ldots left(2^{32}+1right)+1, we can simplify the expression step by step:1. First, notice that left(2+1right)left(2^{2}+1right)left(2^{4}+1right)left(2^{6}+1right)ldots left(2^{32}+1right)+1 can be rewritten by substituting the first term left(2+1right) with left(2^{2}-1right), which gives us: [ left(2^{2}-1right)left(2^{2}+1right)left(2^{4}+1right)ldots left(2^{32}+1right)+2 ]2. This expression simplifies by using the difference of squares formula, a^2 - b^2 = (a-b)(a+b), which applies to the first two terms, giving us: [ left(2^{4}-1right)left(2^{4}+1right)ldots left(2^{32}+1right)+2-1 ]3. Continuing this pattern of simplification, we eventually reach: [ 2^{64}-1+1 ]4. Simplifying the above expression, we get: [ 2^{64} ]5. To find the units digit of 2^{64}, we observe the pattern of the units digits in powers of 2: - The units digit of 2^1 is 2. - The units digit of 2^2 is 4. - The units digit of 2^3 is 8. - The units digit of 2^4 is 6.6. Since the units digits of powers of 2 repeat every 4 powers, and 64 is a multiple of 4, the units digit of 2^{64} will be the same as the units digit of 2^4, which is 6.Therefore, the units digit of the given expression is boxed{6}, which corresponds to choice boxed{C}.

🧠:[<think>Okay, so I need to find the length of side AC in triangle ABC where BC is 1 unit, angle B is 60 degrees, and the area of the triangle is √3. Hmm, let me visualize the triangle first. Let me sketch it out in my mind. Triangle ABC with points A, B, and C. BC is 1, angle at B is 60 degrees, and the area is √3. So, angle B is between sides AB and BC, right? Since BC is given as 1, and angle B is 60°, maybe I can use the formula for the area of a triangle involving two sides and the included angle. The formula is (1/2)*ab*sinθ, where a and b are the sides and θ is the included angle. Wait, in this case, angle B is between sides AB and BC. So, the two sides forming the angle B are AB and BC. BC is given as 1, but AB isn't. So, if I let AB be, let's say, x, then the area would be (1/2)*AB*BC*sin(60°). Plugging in the known values, that would be (1/2)*x*1*sin(60°) = √3. So, solving for x would give me AB. Then once I have AB, I can use the Law of Cosines to find AC. Let me check if that makes sense.First, let's compute sin(60°). Sin of 60 degrees is √3/2. So, substituting into the area formula:Area = (1/2)*x*1*(√3/2) = √3. Let's compute that left side:(1/2)*x*(√3/2) = (x√3)/4. So, (x√3)/4 = √3. Solving for x:Multiply both sides by 4: x√3 = 4√3Then divide both sides by √3: x = 4. So, AB is 4 units. Okay, now that I have AB = 4, BC = 1, angle B = 60°, I can use the Law of Cosines to find AC. The Law of Cosines states that in any triangle, c² = a² + b² - 2ab*cosθ, where θ is the angle opposite side c. Wait, actually, no. Let me recall: Law of Cosines is c² = a² + b² - 2ab*cos(C), where C is the angle opposite side c. So, in our case, angle B is 60°, and sides AB and BC are adjacent to angle B, so the side opposite angle B is AC. Wait, is that right? Let me confirm. In triangle ABC, angle B is at vertex B, so sides AB and BC are adjacent to angle B. The side opposite angle B is AC. So, to apply the Law of Cosines to find AC, which is opposite angle B, the formula should be:AC² = AB² + BC² - 2*AB*BC*cos(angle B)Yes, that seems right. So, plugging in the values:AC² = 4² + 1² - 2*4*1*cos(60°)Compute each term:4² = 161² = 12*4*1 = 8cos(60°) = 0.5So, AC² = 16 + 1 - 8*0.5 = 17 - 4 = 13Therefore, AC = √13. Wait, but hold on. Let me check the calculations again step by step to make sure I didn't make a mistake.First, AB = 4, BC = 1. Angle B = 60°, so Law of Cosines:AC² = AB² + BC² - 2*AB*BC*cos(angle B)So, 4² is 16, 1² is 1, sum is 17. Then subtract 2*4*1*cos(60°). 2*4*1 is 8, cos(60°) is 0.5, so 8*0.5 is 4. Therefore, 17 - 4 = 13. So AC² is 13, so AC is √13. That seems correct.But let me think again. Is there another way to verify this? Maybe using the area formula with Heron's formula or something else? Let me see.Alternatively, once we have all sides, we can check the area. Wait, we found AB = 4, BC = 1, angle B = 60°, and AC = √13. Then, let's compute the area again using the formula (1/2)*AB*BC*sin(angle B). So that's (1/2)*4*1*(√3/2) = (4*√3)/4 = √3. Which matches the given area. So that checks out. So, the answer must be √13. Wait, but let me try another approach to confirm. Maybe coordinate geometry. Let's place point B at the origin (0,0), side BC along the x-axis, so point C is at (1,0). Angle at B is 60°, so point A will be somewhere in the plane. Let me compute coordinates of point A.Since AB is 4 units, and angle at B is 60°, so from point B at (0,0), making a 60° angle with the x-axis (since BC is along the x-axis). So, coordinates of point A can be found using trigonometry. The x-coordinate of A is AB*cos(60°) = 4*(1/2) = 2. The y-coordinate is AB*sin(60°) = 4*(√3/2) = 2√3. So, point A is at (2, 2√3). Then, point C is at (1,0). Therefore, the distance AC can be calculated using the distance formula:AC = √[(2 - 1)² + (2√3 - 0)²] = √[1 + (4*3)] = √[1 + 12] = √13. Yep, same result. So that confirms it.Alternatively, if I use the coordinates, the area can be computed using the shoelace formula. Coordinates of B (0,0), C (1,0), A (2, 2√3). The area is |(0*(0 - 2√3) + 1*(2√3 - 0) + 2*(0 - 0))/2| = |0 + 2√3 + 0|/2 = 2√3/2 = √3. Which matches the given area. So all checks out.Therefore, the length of AC is √13. So, the answer is √13.Wait, but just to make sure there isn't another possible configuration. Since in the problem, they didn't specify if the triangle is acute or obtuse, but given that angle B is 60°, which is acute, and BC is 1, AB is 4, which seems quite long. But when we calculated using coordinates, everything worked out, so I think that's the correct answer. Is there a possibility of another triangle with the same area and angle but different side lengths? Hmm. Let me think.Suppose we fix BC = 1, angle B = 60°, area √3. The area formula gives us AB = 4 as we found. So, AB is uniquely determined by the area and the angle. Therefore, there's only one possible length for AB, which is 4. Then, once AB is fixed, AC is determined by the Law of Cosines, giving √13. So, no, there's no other possible configuration. Therefore, the answer must be √13.I think that's solid. All the methods lead to the same answer, and cross-checked with coordinate geometry. So, I'm confident that AC = √13.Final AnswerThe length of AC is boxed{sqrt{13}}.</think>To find the length of side ( AC ) in (triangle ABC) where ( BC = 1 ), (angle B = 60^circ), and the area is (sqrt{3}), we start by using the area formula for a triangle involving two sides and the included angle:[text{Area} = frac{1}{2} times AB times BC times sin(angle B)]Given the area is (sqrt{3}), we substitute the known values:[sqrt{3} = frac{1}{2} times AB times 1 times sin(60^circ)]Since (sin(60^circ) = frac{sqrt{3}}{2}), we have:[sqrt{3} = frac{1}{2} times AB times frac{sqrt{3}}{2}]Solving for ( AB ):[sqrt{3} = frac{AB times sqrt{3}}{4} implies AB = 4]Next, we use the Law of Cosines to find ( AC ). The Law of Cosines states:[AC^2 = AB^2 + BC^2 - 2 times AB times BC times cos(angle B)]Substituting ( AB = 4 ), ( BC = 1 ), and (cos(60^circ) = 0.5 ):[AC^2 = 4^2 + 1^2 - 2 times 4 times 1 times 0.5][AC^2 = 16 + 1 - 4 = 13]Thus, ( AC = sqrt{13} ).The length of ( AC ) is (boxed{sqrt{13}}).,<think>Okay, let's see. I need to find the length of side AC in triangle ABC where BC is 1, angle B is 60 degrees, and the area is √3. Hmm, okay. Let me start by drawing a rough sketch of the triangle. So, triangle ABC, with angle B at vertex B. BC is 1 unit long. The area is given as √3. I need to find AC. First, let me recall the formula for the area of a triangle when two sides and the included angle are known. The formula is (1/2)*ab*sinθ, where a and b are the sides and θ is the included angle. In this case, angle B is 60 degrees, and sides adjacent to angle B are AB and BC. Wait, BC is given as 1. Let me note that down. BC = 1, angle B = 60°, area = √3. Let me denote the sides as per standard triangle notation: side opposite angle A is BC, which would be a, side opposite angle B is AC, which is b, and side opposite angle C is AB, which is c. Wait, maybe that's confusing. Alternatively, maybe just label the sides as AB = c, BC = a = 1, and AC = b, which is the one we need to find. Wait, perhaps better to use coordinates. Maybe coordinate geometry can help here. Let me think.Alternatively, let's use the area formula. The area is (1/2)*AB*BC*sin(angle B). Since BC is 1, angle B is 60°, so sin(60°) is √3/2. Let me write that down. Area = (1/2)*AB*1*(√3/2) = (1/2)*AB*(√3/2) = AB*(√3)/4. But the area is given as √3. So, set that equal to √3:AB*(√3)/4 = √3Then, solving for AB:AB = (√3) * 4 / √3 = 4So AB is 4. Wait, that seems straightforward. So AB is 4 units. Then, now that I know AB and BC and angle B, can I use the Law of Cosines to find AC? Yes. The Law of Cosines states that AC² = AB² + BC² - 2*AB*BC*cos(angle B). Let's plug in the values:AC² = 4² + 1² - 2*4*1*cos(60°)Calculating that:AC² = 16 + 1 - 8*(0.5) because cos(60°) is 0.5.So AC² = 17 - 4 = 13Therefore, AC = √13.Wait, but let me check if this makes sense. Let me verify the steps again. First, area formula: area = (1/2)*AB*BC*sin(angle B). Given that BC = 1, angle B = 60°, so:√3 = (1/2)*AB*1*(√3/2)Simplify the right side: (1/2)*(√3/2)*AB = (√3/4)*ABSet equal to √3: (√3/4)*AB = √3Multiply both sides by 4: √3*AB = 4√3Divide both sides by √3: AB = 4. That's correct. So AB is indeed 4. Then, using the Law of Cosines:AC² = AB² + BC² - 2*AB*BC*cos(angle B)Plugging in AB = 4, BC = 1, angle B = 60°:AC² = 16 + 1 - 2*4*1*0.5Calculate term by term:16 + 1 = 172*4*1 = 8, 8*0.5 = 4So, 17 - 4 = 13. Thus, AC² = 13, so AC = √13. That seems right.Wait, but let me think again. Alternatively, maybe using coordinates. Let me try that approach to confirm. Let me place point B at the origin (0,0). Since angle B is 60°, and BC is 1, let me place point C at (1,0). Then, point A is somewhere in the plane. The coordinates of A can be determined using the length AB = 4 and angle at B is 60°. Wait, if angle at B is 60°, then the angle between BA and BC is 60°, so BA makes 60° angle with BC. Since BC is along the x-axis from B(0,0) to C(1,0), then BA would be a line from B(0,0) making 60° with BC. So coordinates of A can be found using polar coordinates. If AB is 4, then coordinates of A would be (AB*cos(60°), AB*sin(60°)) = (4*(0.5), 4*(√3/2)) = (2, 2√3). Then, AC is the distance between point A(2, 2√3) and C(1,0). Let's compute that:AC = sqrt[(2 - 1)^2 + (2√3 - 0)^2] = sqrt[1 + (4*3)] = sqrt[1 + 12] = sqrt[13]. Yep, that's the same as before. So AC is indeed √13.So both methods agree. Therefore, the answer should be √13. Let me check once more if I didn't make any calculation errors. First method: area gives AB=4, Law of Cosines gives AC=√13. Second method: coordinates, compute AC as √13. All steps seem correct. Therefore, AC is √13.Final Answerboxed{sqrt{13}}</think>Given triangle ABC with BC = 1, angle B = 60^circ, and the area of triangle ABC is sqrt{3}, we need to find the length of AC.First, we use the formula for the area of a triangle with two sides and the included angle:[text{Area} = frac{1}{2} times AB times BC times sin(angle B)]Given BC = 1, angle B = 60^circ, and the area is sqrt{3}, we substitute these values into the formula:[sqrt{3} = frac{1}{2} times AB times 1 times sin(60^circ)]Since sin(60^circ) = frac{sqrt{3}}{2}, we have:[sqrt{3} = frac{1}{2} times AB times frac{sqrt{3}}{2}]Simplifying the right-hand side:[sqrt{3} = frac{sqrt{3}}{4} times AB]Solving for AB:[AB = frac{sqrt{3} times 4}{sqrt{3}} = 4]Next, we use the Law of Cosines to find AC:[AC^2 = AB^2 + BC^2 - 2 times AB times BC times cos(angle B)]Substituting AB = 4, BC = 1, and cos(60^circ) = 0.5:[AC^2 = 4^2 + 1^2 - 2 times 4 times 1 times 0.5]Calculating each term:[AC^2 = 16 + 1 - 4 = 13]Thus, the length of AC is:[AC = sqrt{13}][boxed{sqrt{13}}]]

🔑:Given that the area of triangle ABC is S= dfrac {1}{2} times BC times AB times sin B = sqrt {3},Substituting the given values, we get dfrac {1}{2} times 1 times AB times dfrac { sqrt {3}}{2} = sqrt {3}. Solving for AB, we get AB=4.By using the cosine rule, we can find AC = sqrt {4^{2}+1^{2}-2 times 4 times 1 times dfrac {1}{2}} = sqrt {13}.Hence, the answer is boxed{sqrt {13}}.This problem involves using the sine and cosine rules to solve for sides of a triangle, and also involves the formula for the area of a triangle. It is a basic problem.

🧠:[<think>Okay, let's see. I need to factorize the expression ( x^{2} + 4y^{2} - 4xy ). Hmm, factorization. Alright, where do I start? Maybe I should look for common factors first. But in this case, the terms are ( x^2 ), ( 4y^2 ), and ( -4xy ). I don't see a common factor in all three terms. The coefficients are 1, 4, and -4. Hmmm. So maybe common factors aren't the way to go here. Alternatively, maybe this is a quadratic in terms of one variable. Let's see. If I treat it as a quadratic in x, then it's ( x^2 -4xy +4y^2 ). Yeah, that's the standard quadratic form ( ax^2 + bx + c ), where here, the coefficients would be: a = 1, b = -4y, and c = 4y². Maybe I can use the quadratic formula to factor this? Wait, but since it's a quadratic in x, maybe it factors into two binomials. Let me try that.If it's factorable, it would look like (x + m)(x + n), but since the coefficient of x² is 1, right? But wait, actually, since there are y terms involved, maybe the factors are binomials involving both x and y. Let me think. Suppose it factors into something like (x - ay)(x - by). Let's test that.Expanding (x - ay)(x - by) would give x² - (a + b)xy + aby². Comparing this to our original expression: x² -4xy +4y². So that would require:1. The coefficient of xy is -(a + b) = -4. So a + b = 4.2. The constant term is ab = 4.So we need two numbers a and b such that their sum is 4 and their product is 4. Let's see. Factors of 4 are 1 & 4, 2 & 2. Let's check if 2 and 2 add up to 4. Yes, 2 + 2 = 4 and 2*2=4. Perfect. So that means a = 2 and b = 2. Therefore, the factorization is (x - 2y)(x - 2y) which is (x - 2y)². Let me check that:Expand (x - 2y)²: x² -4xy +4y². Yes, that's exactly the original expression. So the factorization is (x - 2y) squared. Alternatively, maybe recognizing that the original expression is a perfect square trinomial. Let's recall that a perfect square trinomial is of the form a² - 2ab + b² = (a - b)². Comparing to our expression: x² +4y² -4xy. Let me rearrange the terms: x² -4xy +4y². If I consider x as 'a' and 2y as 'b', since (2y)² = 4y². Then, substituting into the formula: a² - 2ab + b² = (a - b)². Here, a = x, b = 2y. Then, x² - 2*(x)*(2y) + (2y)² = x² -4xy +4y², which is exactly the given expression. Therefore, it factors to (x - 2y)².Another way to approach this might be to recognize that the expression is similar to (x - 2y)^2. Let me verify:(x - 2y)^2 = x^2 - 2*x*2y + (2y)^2 = x² -4xy +4y². Exactly. So that's the same result.Alternatively, if I didn't recognize it as a perfect square, perhaps I could try grouping terms. Let's see. The expression is x² -4xy +4y². Let me group the first two terms and the last term: (x² -4xy) +4y². Then, factor out an x from the first group: x(x -4y) +4y². Hmm, but that doesn't seem helpful. Alternatively, maybe group differently. But since there are only three terms, grouping might not be straightforward here. So the perfect square approach is more effective.Wait, another thought: maybe completing the square. But since it's a quadratic in two variables, maybe that complicates things, but in this case, since it's a trinomial that we already saw is a perfect square, that seems redundant. But just for thoroughness, let's try that.Starting with x² -4xy +4y². Let's treat this as a quadratic in x. So x² -4xy +4y². To complete the square, we can take the coefficient of x, which is -4y, divide by 2, which gives -2y, then square it: (-2y)^2 = 4y². So the expression becomes:x² -4xy +4y² = (x² -4xy +4y²) = (x -2y)^2. Yep, same result. So completing the square also leads us to the same factorization.Therefore, regardless of the method—whether factoring as a quadratic in x, recognizing the perfect square trinomial, or completing the square—we arrive at the factorization (x - 2y)².Let me double-check by expanding (x - 2y)^2 to ensure there are no mistakes. Expansion:First term: x squared.Outer term: x times -2y, which is -2xy.Inner term: -2y times x, which is another -2xy.Last term: -2y times -2y, which is +4y².Adding them up: x² -2xy -2xy +4y² = x² -4xy +4y². Perfect, matches the original expression. So the factorization is correct.Therefore, the answer should be (x - 2y) squared, or (x - 2y)^2. In boxed form, that's boxed{(x - 2y)^2}.Final AnswerThe factorization of ( x^{2} + 4y^{2} - 4xy ) is boxed{(x - 2y)^2}.</think>To factorize the expression ( x^{2} + 4y^{2} - 4xy ):1. Recognize the expression as a quadratic in ( x ): ( x^2 - 4xy + 4y^2 ).2. Notice that it resembles the form of a perfect square trinomial ( a^2 - 2ab + b^2 ), where ( a = x ) and ( b = 2y ).3. Verify by expanding ( (x - 2y)^2 ): [ (x - 2y)^2 = x^2 - 2 cdot x cdot 2y + (2y)^2 = x^2 - 4xy + 4y^2 ]4. Confirm that the expanded form matches the original expression.Thus, the factorization of ( x^{2} + 4y^{2} - 4xy ) is (boxed{(x - 2y)^2}).,<think>Okay, let's see. The problem is to factorize the expression ( x^{2} + 4y^{2} - 4xy ). Hmm, factorization. Alright, so first, I need to recall different factorization techniques. Maybe this is a quadratic in terms of x or y? Let me check.Looking at the expression, it's ( x^2 ) plus ( 4y^2 ) minus ( 4xy ). Let me rearrange the terms to see if it becomes more familiar. If I write it as ( x^2 - 4xy + 4y^2 ), does that look like a standard form? Hmm, yes, that seems like a quadratic in x. The standard quadratic form is ( ax^2 + bx + c ). Here, the coefficient of ( x^2 ) is 1, the coefficient of x is -4y, and the constant term is ( 4y^2 ).Alternatively, since the expression has both x and y, maybe it's a perfect square trinomial. Let me recall that ( (a - b)^2 = a^2 - 2ab + b^2 ). Let's see if the given expression fits this pattern.Comparing ( x^2 - 4xy + 4y^2 ) with ( a^2 - 2ab + b^2 ):- The first term is ( x^2 ), so a would be x.- The last term is ( 4y^2 ), which can be written as ( (2y)^2 ), so b would be 2y.- The middle term should then be -2ab, which would be -2 * x * 2y = -4xy. Yes! The middle term is indeed -4xy, which matches. So that means the expression can be written as ( (a - b)^2 ), which is ( (x - 2y)^2 ).Let me double-check by expanding ( (x - 2y)^2 ):( (x - 2y)^2 = x^2 - 4xy + 4y^2 ), which is exactly the original expression. So that works.Alternatively, maybe there's another way to factor this? Let me think. If I group terms differently. Let's see:Original expression: ( x^2 + 4y^2 - 4xy ). Group x^2 - 4xy first, then add 4y^2. So that would be ( x(x - 4y) + 4y^2 ). Hmm, not sure if that helps. Maybe completing the square?But since we already saw that it's a perfect square, I think that's the simplest factorization here.Wait, another thought: sometimes expressions can be factored as a difference of squares. But here, we have all plus and minus signs. Let me check if that applies here.Difference of squares is ( a^2 - b^2 = (a - b)(a + b) ). But our expression is ( x^2 + 4y^2 - 4xy ), which doesn't look like a difference of two squares. It has three terms. So that approach might not work here.Therefore, the most straightforward factorization is recognizing the perfect square trinomial. So the answer should be ( (x - 2y)^2 ).I think that's it. Let me just confirm once again by expanding:( (x - 2y)^2 = x^2 - 4xy + 4y^2 ). Yes, exactly matches. So the factorization is correct.Final AnswerThe factorized form is boxed{(x - 2y)^2}.</think>The given expression is ( x^2 + 4y^2 - 4xy ). To factorize this, we recognize that it is a quadratic in terms of ( x ). Rewriting the expression as ( x^2 - 4xy + 4y^2 ), we notice that it fits the form of a perfect square trinomial ( (a - b)^2 ), where ( a = x ) and ( b = 2y ).Comparing the terms:- The first term ( x^2 ) corresponds to ( a^2 ).- The last term ( 4y^2 ) corresponds to ( (2y)^2 ).- The middle term ( -4xy ) corresponds to ( -2ab ).Thus, the expression ( x^2 - 4xy + 4y^2 ) can be written as ( (x - 2y)^2 ).Expanding ( (x - 2y)^2 ) confirms this:[ (x - 2y)^2 = x^2 - 4xy + 4y^2 ]Therefore, the factorized form is (boxed{(x - 2y)^2}).]

🔑:To factorize the given expression x^{2}+4y^{2}-4xy, we follow these steps:1. Rearrange the terms to follow a recognizable pattern: x^{2}-4xy+4y^{2}.2. Recognize that this is a perfect square trinomial, which can be factored into the square of a binomial.3. The pattern for a perfect square trinomial is a^{2}-2ab+b^{2}=(a-b)^{2}.4. Comparing, we identify a=x and b=2y, thus applying the pattern: (x-2y)^{2}.Therefore, the factorized form of the given expression is boxed{(x-2y)^{2}}.

🧠:[<think>Okay, so I have this problem here about a regular hexagon with a point inside it. The distances from this point to three consecutive vertices are given as 1, 1, and 2. I need to find the side length of the hexagon. Hmm, regular hexagons have all sides equal and all internal angles equal, right? Let me recall some properties of regular hexagons.First, a regular hexagon can be divided into six equilateral triangles, each with side length equal to the hexagon's side. The center of the hexagon is equidistant from all vertices, and that distance is equal to the side length. So, the radius of the circumscribed circle around the hexagon is equal to the side length. Let's denote the side length as 's'. Now, there's a point inside the hexagon, not necessarily the center, from which the distances to three consecutive vertices are 1, 1, and 2. Let me try to visualize this. Let's label the hexagon's vertices as A, B, C, D, E, F in order. Suppose the point P is inside the hexagon, and the distances from P to A, B, and C are 1, 1, and 2, respectively. So PA = 1, PB = 1, PC = 2. Wait, three consecutive vertices. So maybe A, B, C, with distances 1, 1, 2. Or maybe another set? But since the problem says consecutive, it's three in a row. So A, B, C or B, C, D, etc. But since the hexagon is regular, the configuration should be symmetric, so it doesn't matter which three consecutive vertices we take. So let's fix A, B, C for simplicity.Now, I need to relate these distances to the side length. Since the hexagon is regular, coordinates might help here. Let me place the hexagon on a coordinate system with its center at the origin. Let's assume vertex A is at (s, 0), vertex B at (s/2, (s√3)/2), vertex C at (-s/2, (s√3)/2), and so on. Wait, no. Wait, in a regular hexagon, each vertex is at 60-degree increments. Let me recall the coordinates for a regular hexagon centered at the origin.If the hexagon is centered at the origin with a vertex at (s, 0), then the coordinates of the six vertices can be given by (s*cos(0°), s*sin(0°)), (s*cos(60°), s*sin(60°)), (s*cos(120°), s*sin(120°)), (s*cos(180°), s*sin(180°)), (s*cos(240°), s*sin(240°)), and (s*cos(300°), s*sin(300°)). So yes, that would be (s, 0), (s/2, (s√3)/2), (-s/2, (s√3)/2), (-s, 0), (-s/2, -(s√3)/2), (s/2, -(s√3)/2). So vertices A, B, C, D, E, F correspond to these coordinates respectively.So, point P is somewhere inside this hexagon, and its distances to A, B, and C are 1, 1, and 2. Let me assign coordinates to point P. Let me denote the coordinates of P as (x, y). Then, the distances from P to A, B, and C can be written using the distance formula.First, distance from P to A: A is at (s, 0). So,PA² = (x - s)² + (y - 0)² = (x - s)² + y² = 1² = 1.Similarly, distance from P to B: B is at (s/2, (s√3)/2). So,PB² = (x - s/2)² + (y - (s√3)/2)² = 1² = 1.Distance from P to C: C is at (-s/2, (s√3)/2). So,PC² = (x + s/2)² + (y - (s√3)/2)² = 2² = 4.So now we have three equations:1. (x - s)² + y² = 1. (Equation 1)2. (x - s/2)² + (y - (s√3)/2)² = 1. (Equation 2)3. (x + s/2)² + (y - (s√3)/2)² = 4. (Equation 3)Now, the goal is to solve these three equations to find the value of s, which is the side length of the hexagon. Since there are three equations and three unknowns (x, y, s), this should be possible. Let's proceed step by step.First, let's subtract Equation 1 from Equation 2. That might eliminate some terms.Equation 2 - Equation 1:[(x - s/2)² + (y - (s√3)/2)²] - [(x - s)² + y²] = 1 - 1 = 0.Expanding both squares:First term: (x - s/2)² = x² - s x + (s²)/4Second term: (y - (s√3)/2)² = y² - s√3 y + (3 s²)/4Third term: (x - s)² = x² - 2 s x + s²Fourth term: y² remains y².Putting it all together:[x² - s x + (s²)/4 + y² - s√3 y + (3 s²)/4] - [x² - 2 s x + s² + y²] = 0Simplify term by term:x² - s x + (s²)/4 + y² - s√3 y + (3 s²)/4 - x² + 2 s x - s² - y² = 0Simplify:x² cancels with -x²y² cancels with -y²-s x + 2 s x = s x(s²)/4 + (3 s²)/4 - s² = (s²)/4 + 3 s²/4 - s² = (4 s²)/4 - s² = s² - s² = 0Then the remaining term is -s√3 ySo overall:s x - s√3 y = 0Divide both sides by s (assuming s ≠ 0, which it can't be as it's a side length):x - √3 y = 0 ⇒ x = √3 y. (Equation 4)Okay, so we have a relation between x and y: x is √3 times y. Let's note that.Now, let's subtract Equation 2 from Equation 3. Let's see:Equation 3 - Equation 2:[(x + s/2)² + (y - (s√3)/2)²] - [(x - s/2)² + (y - (s√3)/2)²] = 4 - 1 = 3.Again, expanding the squares:First term: (x + s/2)² = x² + s x + (s²)/4Second term: (y - (s√3)/2)² = same as before, y² - s√3 y + (3 s²)/4Third term: (x - s/2)² = x² - s x + (s²)/4Fourth term: same second term.So subtract:[x² + s x + (s²)/4 + y² - s√3 y + (3 s²)/4] - [x² - s x + (s²)/4 + y² - s√3 y + (3 s²)/4] = 3Simplify term by term:x² - x² = 0s x - (-s x) = 2 s x(s²)/4 - (s²)/4 = 0y² - y² = 0-s√3 y - (-s√3 y) = 0(3 s²)/4 - (3 s²)/4 = 0So, all terms cancel except 2 s x.Therefore:2 s x = 3 ⇒ x = 3/(2 s). (Equation 5)From Equation 4, we have x = √3 y, so substituting into Equation 5:√3 y = 3/(2 s) ⇒ y = 3/(2 s √3) = √3/(2 s). (Equation 6)So now we have expressions for x and y in terms of s. Let's plug these back into one of the original equations. Let's choose Equation 1, which is (x - s)^2 + y^2 = 1.Substituting x = 3/(2 s) and y = √3/(2 s):[ (3/(2 s) - s ) ]^2 + [ √3/(2 s) ]^2 = 1.Let me compute each term.First term: (3/(2 s) - s) = (3 - 2 s²)/(2 s). So squared: (3 - 2 s²)^2 / (4 s²)Second term: [ √3/(2 s) ]^2 = (3)/(4 s²)Therefore:[ (3 - 2 s²)^2 / (4 s²) ] + [ 3 / (4 s²) ] = 1Combine terms:[ (3 - 2 s²)^2 + 3 ] / (4 s²) = 1Multiply both sides by 4 s²:(3 - 2 s²)^2 + 3 = 4 s²Expand (3 - 2 s²)^2:= 9 - 12 s² + 4 s^4So:9 - 12 s² + 4 s^4 + 3 = 4 s²Combine like terms:4 s^4 - 12 s² + 12 = 4 s²Bring all terms to left side:4 s^4 - 12 s² + 12 - 4 s² = 0 ⇒ 4 s^4 - 16 s² + 12 = 0Divide both sides by 4:s^4 - 4 s² + 3 = 0Let me set u = s². Then equation becomes:u² - 4 u + 3 = 0Quadratic equation: u = [4 ± √(16 - 12)] / 2 = [4 ± √4]/2 = [4 ± 2]/2 ⇒ u = 3 or u = 1.Therefore, s² = 3 or s² = 1 ⇒ s = √3 or s = 1.But s is the side length, which must be positive. So s = 1 or √3. But we need to check which one is valid.Wait, if s = 1, let's check if the coordinates x and y are inside the hexagon. Let's compute x and y:From Equation 5, x = 3/(2 s). If s = 1, x = 3/2. But the hexagon with side length 1 has vertices at (1, 0), (0.5, √3/2 ≈ 0.866), etc. The x-coordinate of the point P is 3/2 = 1.5, which is outside the hexagon, since the maximum x-coordinate in the hexagon is 1 (vertex A at (1, 0)). So x = 1.5 would be outside. Therefore, s = 1 is invalid.Therefore, s must be √3. Let's check that. If s = √3, then x = 3/(2 * √3) = (3)/(2√3) = (√3)/2 ≈ 0.866. Similarly, y = √3/(2 * √3) = 1/2. So coordinates (√3/2, 1/2). Let's check if this is inside the hexagon.The hexagon with side length √3 has vertices at (√3, 0), (√3/2, (√3 * √3)/2) = (√3/2, 3/2), (-√3/2, 3/2), etc. The coordinates (√3/2 ≈ 0.866, 1/2 = 0.5) should be inside the hexagon. Let's see: the rightmost point is (√3, 0), so x = √3/2 is less than √3, so that's fine. The top point in that area would be at (√3/2, 3/2), so y = 0.5 is below 3/2 = 1.5, so yes, it's inside. So s = √3 is valid.Therefore, the side length is √3. Wait, but let's confirm by plugging back into the original equations.Let me check s = √3, x = √3/2, y = 1/2.First equation: (x - s)^2 + y^2 = ( (√3/2 - √3) )^2 + (1/2)^2 = ( (-√3/2) )^2 + 1/4 = (3/4) + 1/4 = 1. Correct.Second equation: (x - s/2)^2 + (y - (s√3)/2)^2 = ( (√3/2 - √3/2 ) )^2 + (1/2 - (√3 * √3)/2 )^2 = 0 + (1/2 - 3/2)^2 = (-1)^2 = 1. Correct.Third equation: (x + s/2)^2 + (y - (s√3)/2)^2 = ( (√3/2 + √3/2 ) )^2 + (1/2 - 3/2 )^2 = ( √3 )^2 + (-1)^2 = 3 + 1 = 4. Correct.So all three equations are satisfied. Therefore, s = √3 is indeed the solution.Therefore, the side length of the hexagon is √3.Final AnswerThe side length of the hexagon is boxed{sqrt{3}}.</think>Given a regular hexagon with a point inside it, the distances from this point to three consecutive vertices are 1, 1, and 2. We need to find the side length of the hexagon.1. Coordinates of Vertices: - The vertices of the hexagon are labeled as (A), (B), (C), (D), (E), (F). - Coordinates of vertices (A), (B), and (C) are ((s, 0)), (left(frac{s}{2}, frac{ssqrt{3}}{2}right)), and (left(-frac{s}{2}, frac{ssqrt{3}}{2}right)) respectively.2. Equations for Distances: - Let the coordinates of the point (P) be ((x, y)). - Using the distance formula, we set up the following equations: [ (x - s)^2 + y^2 = 1 quad text{(Equation 1)} ] [ left(x - frac{s}{2}right)^2 + left(y - frac{ssqrt{3}}{2}right)^2 = 1 quad text{(Equation 2)} ] [ left(x + frac{s}{2}right)^2 + left(y - frac{ssqrt{3}}{2}right)^2 = 4 quad text{(Equation 3)} ]3. Subtracting Equations: - Subtracting Equation 1 from Equation 2: [ s x - ssqrt{3} y = 0 implies x = sqrt{3} y quad text{(Equation 4)} ] - Subtracting Equation 2 from Equation 3: [ 2 s x = 3 implies x = frac{3}{2s} quad text{(Equation 5)} ]4. Solving for (x) and (y): - Substituting (x = sqrt{3} y) into Equation 5: [ sqrt{3} y = frac{3}{2s} implies y = frac{sqrt{3}}{2s} ] - Substituting (x = frac{3}{2s}) and (y = frac{sqrt{3}}{2s}) into Equation 1: [ left(frac{3}{2s} - sright)^2 + left(frac{sqrt{3}}{2s}right)^2 = 1 ] Simplifying this equation leads to solving a quadratic equation in terms of (s^2): [ s^4 - 4s^2 + 3 = 0 ] Solving this, we get (s^2 = 3) or (s^2 = 1), leading to (s = sqrt{3}) or (s = 1).5. Validation: - Checking (s = 1) results in coordinates outside the hexagon, hence invalid. - Checking (s = sqrt{3}) results in valid coordinates inside the hexagon.Thus, the side length of the hexagon is (boxed{sqrt{3}}).,<think>Okay, so I need to find the side length of a regular hexagon where there's a point inside it. The distances from this point to three consecutive vertices are 1, 1, and 2. Hmm, let's see. First, I should recall some properties of regular hexagons. A regular hexagon can be divided into six equilateral triangles, all with sides equal to the side length of the hexagon. So, if the side length is, say, 's', then each of those triangles has sides of length 's'. The center of the hexagon is equidistant from all vertices, and that distance is equal to the side length.But in this problem, the point isn't necessarily the center. It's some interior point, and the distances to three consecutive vertices are 1, 1, and 2. Let me try to visualize this. Let's label the regular hexagon as ABCDEF, with each vertex labeled in order. Let's say the point P is inside the hexagon, and the distances from P to vertices A, B, and C are 1, 1, and 2, respectively. So, PA = 1, PB = 1, PC = 2. Since these are consecutive vertices, A, B, C are adjacent to each other.Since it's a regular hexagon, all sides are equal, and all internal angles are 120 degrees. The distance between non-consecutive vertices can be calculated using the law of cosines. For example, the distance between A and C would be 2s sin(60°) = s√3, but maybe that's not directly helpful here.Wait, maybe coordinate geometry can help here. Let me place the hexagon on a coordinate system to make things easier. Let's center the hexagon at the origin, with vertex A at (s, 0). Then the coordinates of the vertices can be determined using the 60° angles.So, vertex A is at (s, 0). Vertex B would be at (s/2, (s√3)/2). Vertex C is at (-s/2, (s√3)/2). Wait, hold on. If the hexagon is centered at the origin, then the coordinates of the six vertices can be given by (s cos(0°), s sin(0°)), (s cos(60°), s sin(60°)), (s cos(120°), s sin(120°)), and so on. So yes:A: (s, 0)B: (s/2, (s√3)/2)C: (-s/2, (s√3)/2)D: (-s, 0)E: (-s/2, -(s√3)/2)F: (s/2, -(s√3)/2)But in this case, the point P is inside the hexagon, and PA = 1, PB = 1, PC = 2. So, if I can assign coordinates to point P, say (x, y), then the distances from (x, y) to A, B, and C would be 1, 1, and 2 respectively. Therefore, I can set up equations based on the distance formula.So, the coordinates:PA = 1: distance from (x, y) to A(s, 0) is 1:√[(x - s)^2 + (y - 0)^2] = 1 → (x - s)^2 + y^2 = 1. (1)PB = 1: distance from (x, y) to B(s/2, (s√3)/2) is 1:√[(x - s/2)^2 + (y - (s√3)/2)^2] = 1 → (x - s/2)^2 + (y - (s√3)/2)^2 = 1. (2)PC = 2: distance from (x, y) to C(-s/2, (s√3)/2) is 2:√[(x + s/2)^2 + (y - (s√3)/2)^2] = 2 → (x + s/2)^2 + (y - (s√3)/2)^2 = 4. (3)So, equations (1), (2), and (3) are three equations with two variables (x, y) and the parameter s. The goal is to solve for s.But since there are three equations and two variables, this system might be overdetermined. However, since the point P exists inside the hexagon, there should be a solution. So, we need to solve these equations for x, y, and s.Let me try subtracting equations (1) and (2) to eliminate the quadratic terms. Let's write them out:Equation (1): (x - s)^2 + y^2 = 1.Equation (2): (x - s/2)^2 + (y - (s√3)/2)^2 = 1.Subtract equation (1) from equation (2):[(x - s/2)^2 + (y - (s√3)/2)^2] - [(x - s)^2 + y^2] = 1 - 1 = 0.Expanding both squares:First, (x - s/2)^2 = x² - s x + (s²)/4(y - (s√3)/2)^2 = y² - s√3 y + (3 s²)/4So, adding those: x² - s x + (s²)/4 + y² - s√3 y + (3 s²)/4 = x² + y² - s x - s√3 y + s²Similarly, equation (1): (x - s)^2 + y² = x² - 2 s x + s² + y²Therefore, subtracting equation (1) from equation (2):[ x² + y² - s x - s√3 y + s² ] - [ x² - 2 s x + s² + y² ] = 0Simplify term by term:x² - x² + y² - y² -s x + 2 s x - s√3 y + s² - s² = 0Which simplifies to:s x - s√3 y = 0Divide both sides by s (assuming s ≠ 0, which it is since it's a hexagon):x - √3 y = 0 → x = √3 y. (4)So, from equations (1) and (2), we find that x = √3 y. Now, let's use this in equation (1) to express x in terms of y and s.From equation (1):(x - s)^2 + y² = 1Substitute x = √3 y:(√3 y - s)^2 + y² = 1Expand (√3 y - s)^2:3 y² - 2 s √3 y + s² + y² = 1Combine like terms:4 y² - 2 s √3 y + s² - 1 = 0 (Equation 5)Now, let's use equation (3) with x = √3 y.Equation (3): (x + s/2)^2 + (y - (s√3)/2)^2 = 4Substitute x = √3 y:(√3 y + s/2)^2 + (y - (s√3)/2)^2 = 4Let's expand both squares.First square: (√3 y + s/2)^2 = 3 y² + √3 s y + (s²)/4Second square: (y - (s√3)/2)^2 = y² - s√3 y + (3 s²)/4Add them together:3 y² + √3 s y + s²/4 + y² - s√3 y + 3 s²/4 = 4Combine like terms:(3 y² + y²) + (√3 s y - √3 s y) + (s²/4 + 3 s²/4) = 4Which simplifies to:4 y² + s² = 4 (Equation 6)Now, from equation 5: 4 y² - 2 s √3 y + s² - 1 = 0But equation 6: 4 y² + s² = 4 → 4 y² = 4 - s². Let's substitute 4 y² from equation 6 into equation 5.From equation 5:(4 - s²) - 2 s √3 y + s² - 1 = 0Simplify:4 - s² - 2 s √3 y + s² - 1 = (4 - 1) + (-s² + s²) - 2 s √3 y = 3 - 2 s √3 y = 0Therefore:3 - 2 s √3 y = 0 → 2 s √3 y = 3 → y = 3 / (2 s √3) = √3 / (2 s) (rationalizing the denominator)So, y = √3 / (2 s). Then, since x = √3 y, x = √3 * (√3 / (2 s)) ) = 3 / (2 s).So, x = 3/(2 s), y = √3/(2 s). Now, we can substitute these into equation 6 to solve for s.From equation 6: 4 y² + s² = 4Substitute y:4 * ( (√3)/(2 s) )² + s² = 4Compute:4 * (3)/(4 s²) + s² = 4Simplify:(3)/(s²) + s² = 4Multiply both sides by s² to eliminate the denominator:3 + s⁴ = 4 s²Bring all terms to left side:s⁴ - 4 s² + 3 = 0Let me set u = s², so the equation becomes:u² - 4 u + 3 = 0Solve quadratic equation:u = [4 ± √(16 - 12)] / 2 = [4 ± √4]/2 = [4 ± 2]/2So, u = (4 + 2)/2 = 3 or u = (4 - 2)/2 = 1Thus, u = 3 or u = 1, so s² = 3 or s² = 1, hence s = √3 or s = 1. But since s is the side length of the hexagon, which must be positive.But we need to check which solution is valid. Let's check s = 1 first.If s = 1, then from above, x = 3/(2*1) = 3/2, y = √3/(2*1) = √3/2.But in a regular hexagon with side length 1, the coordinates of vertex B are (1/2, (√3)/2). So, point P would be at (3/2, √3/2). But wait, in a hexagon with side length 1, the maximum x-coordinate is 1 (at vertex A). So, x = 3/2 would be outside the hexagon. Hence, s = 1 cannot be valid because point P would be outside the hexagon.Therefore, the valid solution is s = √3. Let's check that.If s = √3, then x = 3/(2*√3) = (√3)/2, y = √3/(2*√3) = 1/2.So, coordinates of P are (√3/2, 1/2). Let's check if this is inside the hexagon.The hexagon with side length √3 has vertices at:A: (√3, 0)B: (√3/2, (√3 * √3)/2) = (√3/2, 3/2)C: (-√3/2, 3/2)D: (-√3, 0)E: (-√3/2, -3/2)F: (√3/2, -3/2)So, the point P is at (√3/2, 1/2). Let's see if this is inside the hexagon.The hexagon can be thought of as having six sides. The rightmost vertex is at (√3, 0), so the x-coordinate of P, which is √3/2 ≈ 0.866, is less than √3 ≈ 1.732, so that's fine. The y-coordinate is 1/2. Let's check if this point is below the line connecting B to C.The line from B to C goes from (√3/2, 3/2) to (-√3/2, 3/2). That's a horizontal line at y = 3/2. Since 1/2 < 3/2, the point is below that line.Similarly, the other sides: the line from A to B is from (√3, 0) to (√3/2, 3/2). Let's find the equation of that line.The slope between A and B is (3/2 - 0)/(√3/2 - √3) = (3/2)/(-√3/2) = -3/√3 = -√3. So, the equation is y - 0 = -√3(x - √3). So, y = -√3 x + 3. Let's see if the point (√3/2, 1/2) is below this line.Plug x = √3/2 into the equation: y = -√3*(√3/2) + 3 = -3/2 + 3 = 3/2. The y-coordinate of the point is 1/2, which is less than 3/2, so it's below the line.Similarly, check the line from C to D: from (-√3/2, 3/2) to (-√3, 0). But since the point is in the right half, we don't need to check all sides. The key is whether the point is inside all the boundaries.Alternatively, since all the coordinates of P are less than the maximum coordinates of the hexagon, and given the hexagon is convex, the point is inside if it satisfies all the inequalities defining the hexagon.Alternatively, maybe there's another way to verify.Alternatively, perhaps better to check the distance from P to the center. The center is at (0,0). The distance from P (√3/2, 1/2) to the center is √( (√3/2)^2 + (1/2)^2 ) = √( 3/4 + 1/4 ) = √1 = 1. So, the point is at distance 1 from the center. In a regular hexagon, the distance from the center to any vertex is equal to the side length. Wait, if the side length is s, then the distance from center to vertices is s. But in our case, s = √3, so the distance from center to vertex is √3. The point P is at distance 1 from the center, which is less than √3, so it's inside.Therefore, s = √3 is a valid solution.But let's also check if equations (1), (2), (3) hold for s = √3 and point (√3/2, 1/2).Equation (1): distance from P to A: A is at (√3, 0). So, distance is √( (√3/2 - √3)^2 + (1/2 - 0)^2 ) = √( (-√3/2)^2 + (1/2)^2 ) = √( 3/4 + 1/4 ) = √1 = 1. Correct.Equation (2): distance from P to B: B is at (√3/2, 3/2). Distance is √( (√3/2 - √3/2)^2 + (1/2 - 3/2)^2 ) = √(0 + (-1)^2) = √1 = 1. Correct.Equation (3): distance from P to C: C is at (-√3/2, 3/2). Distance is √( (√3/2 + √3/2)^2 + (1/2 - 3/2)^2 ) = √( (√3)^2 + (-1)^2 ) = √(3 + 1) = √4 = 2. Correct.So, all distances check out. Therefore, the side length is indeed √3. So, the answer should be √3, which is boxed{sqrt{3}}.Final Answerboxed{sqrt{3}}</think>Given a regular hexagon with a point inside it, the distances from this point to three consecutive vertices are 1, 1, and 2. We need to find the side length of the hexagon.1. Labeling the Hexagon and Setting Up Coordinates: - The hexagon is centered at the origin with vertices labeled (A, B, C, D, E, F). - Coordinates of vertices: - (A: (s, 0)) - (B: left(frac{s}{2}, frac{ssqrt{3}}{2}right)) - (C: left(-frac{s}{2}, frac{ssqrt{3}}{2}right))2. Setting Up Equations Using Distance Formula: - Let the coordinates of the point (P) be ((x, y)). - Distances from (P) to (A), (B), and (C) are 1, 1, and 2 respectively: - ((x - s)^2 + y^2 = 1) - (left(x - frac{s}{2}right)^2 + left(y - frac{ssqrt{3}}{2}right)^2 = 1) - (left(x + frac{s}{2}right)^2 + left(y - frac{ssqrt{3}}{2}right)^2 = 4)3. Solving the System of Equations: - Subtracting the first equation from the second: - (s x - ssqrt{3} y = 0 implies x = sqrt{3} y) - Substituting (x = sqrt{3} y) into the first equation: - ((sqrt{3} y - s)^2 + y^2 = 1 implies 4 y^2 - 2 s sqrt{3} y + s^2 = 1) - Using the third equation: - (left(sqrt{3} y + frac{s}{2}right)^2 + left(y - frac{ssqrt{3}}{2}right)^2 = 4 implies 4 y^2 + s^2 = 4) - Substituting (4 y^2 = 4 - s^2) into the previous equation: - (3 - 2 s sqrt{3} y = 0 implies y = frac{sqrt{3}}{2 s}) - Substituting (y) back into the equation (4 y^2 + s^2 = 4): - (s^4 - 4 s^2 + 3 = 0 implies (s^2 - 1)(s^2 - 3) = 0) - Solutions: (s = 1) or (s = sqrt{3}). Checking validity, (s = sqrt{3}) is valid as it places (P) inside the hexagon.4. Verification: - Coordinates of (P) for (s = sqrt{3}) are (left(frac{sqrt{3}}{2}, frac{1}{2}right)), which satisfy all distance conditions.Thus, the side length of the hexagon is (boxed{sqrt{3}}).]

🔑:Let A, B, C be consecutive vertices of a hexagon, 0 be a point inside it, and O = 2.First method. Consider the other neighboring vertex F of A. Then FABC is an isosceles trapezoid (see left figure). The point O lies on the common perpendicular bisector of its bases FC and AB, so OF = OC = 2. Since FC = 2AB, triangles AOB and FOC are similar with a ratio of 2. Since point O is the center of homothety that transforms one of these triangles into the other, it lies on the diagonal AC (similarly, it lies on the diagonal BF). Then angle OBC = 120^circ - 30^circ = 90^circ, and BC = sqrt{2^2 - 1^2} = sqrt{3}.Second method. The point O lies on the perpendicular bisector of side AB. Construct an equilateral triangle ABD outside the hexagon (see right figure). Clearly, then OD is the perpendicular bisector of segment AB. Moreover, since DB : DC = 1 : 2 = OB : OC, OD is the angle bisector of angle BAO = angle ABO = angle BOE = angle EOC, points A, O, C lie on the same line. In triangle ADC, angle D = 60^circ, DC = 2DA, so angle angle DAC is a right angle, and AB = AD = AC cot angle D = 3 cot 60^circ = sqrt{3}.## Answersqrt{3}.Author: Yurran A. Yu.Trapezoid ABCD is inscribed in a circle. Its base AB is three times the length of base CD. Tangents to the circumscribed circle at points A and C intersect at point K. Prove that angle KDA is a right angle.## SolutionLet DH be the height of the trapezoid, then ADCH is a parallelogram (see left figure). Let M be its center. Then KM is the perpendicular bisector of diagonal AC. By the theorem on the angle between a chord and a tangent, (angle KAD = angle ABD = angle BAC = angle KMD) (the last two angles are angles with mutually perpendicular sides). Therefore, points A, K, D, and M lie on the same circle, from which (angle KDA = angle KMA = 90^circ).Submit a comment

🧠:[<think>Okay, so I need to find the remainder when the sum of the squares of all odd numbers from 1 to 99 is divided by 1000. Hmm, let me break this down step by step. First, the problem is asking for the sum: 1² + 3² + 5² + ... + 99². Then, once I compute that sum, I have to find what the remainder is when it's divided by 1000. Alright, so maybe I can start by figuring out a formula for the sum of squares of the first n odd numbers. I remember there's a formula for the sum of squares of the first n numbers, which is n(n + 1)(2n + 1)/6. But this is for all numbers, not just odd ones. So how do I adjust that formula for only odd numbers?Let me think. The odd numbers can be represented as 2k - 1, where k starts from 1. So the first odd number is 2(1) - 1 = 1, the second is 2(2) - 1 = 3, and so on, up to 2k - 1 = 99. Let me solve for k here. 2k - 1 = 99 → 2k = 100 → k = 50. So there are 50 terms in this sum. That makes sense because from 1 to 99, every other number is odd, and 99 is the 50th odd number.So, the sum we're looking at is the sum from k=1 to k=50 of (2k - 1)². Let me expand that expression. (2k - 1)² = 4k² - 4k + 1. Therefore, the sum becomes the sum from k=1 to 50 of (4k² - 4k + 1). I can split this into three separate sums: 4*sum(k²) - 4*sum(k) + sum(1). Now, let's recall the formulas for these sums. Sum of k² from 1 to n is n(n + 1)(2n + 1)/6. Sum of k from 1 to n is n(n + 1)/2. Sum of 1 from 1 to n is just n. So plugging in n = 50:First term: 4 * [50*51*101)/6]. Let me compute that. Let's calculate the sum inside first. 50*51 = 2550, then 2550*101. Hmm, 2550*100 = 255,000, plus 2550*1 = 2,550. So total is 255,000 + 2,550 = 257,550. Then divide by 6: 257,550 / 6. Let's see, 257,550 divided by 6. 6*42,000 = 252,000. Subtract 252,000 from 257,550: 5,550. Then 5,550 /6 = 925. So total is 42,000 + 925 = 42,925. Then multiply by 4: 42,925 *4. Let me compute that. 40,000*4=160,000, 2,925*4=11,700. So total is 160,000 +11,700 = 171,700.Second term: -4 * [50*51)/2]. Compute that. 50*51=2550. Divided by 2 is 1275. Multiply by 4: 1275*4=5,100. But since it's negative, it's -5,100.Third term: sum(1) from 1 to 50 is 50. So adding those three terms together: 171,700 -5,100 +50. Let's do the subtraction first: 171,700 -5,100 = 166,600. Then add 50: 166,600 +50 = 166,650.So the total sum of the squares of the odd numbers from 1 to 99 is 166,650. Now, we need to find the remainder when this number is divided by 1000. So, essentially, we need to compute 166,650 mod 1000. To find the remainder when divided by 1000, we can look at the last three digits of the number. 166,650. The last three digits are 650. Therefore, the remainder should be 650. But wait, let me verify this. Let's divide 166,650 by 1000. 1000*166 = 166,000. Subtract that from 166,650: 166,650 -166,000 = 650. So yes, the remainder is 650. Hmm, so is that the answer? But let me just cross-validate. Maybe I made a mistake in calculating the sum. Let's check the individual components again.First term: sum of squares of first 50 odd numbers. Let's verify the formula again. The formula I used was expanding (2k -1)^2 and splitting into three sums. Let me check if that's correct.Yes, (2k -1)^2 =4k² -4k +1. Correct. Then sum from k=1 to 50: 4Σk² -4Σk + Σ1. Correct. Then substituting the formulas.Sum of k² from 1 to 50: 50*51*101/6. Let me confirm that. 50*51=2550, 2550*101=257,550. Divided by 6: 257,550 /6=42,925. Multiply by 4: 42,925*4=171,700. That seems correct.Sum of k from 1 to 50: 50*51/2=1275. Multiply by -4: -5,100. Correct.Sum of 1:50. Correct. So 171,700 -5,100 +50=166,650. Yes. Then mod 1000: 650. That seems right.But let me just check with a small example to see if the formula works. For example, sum of squares of first few odd numbers. Let's say 1² +3²=1+9=10. Using the formula:Sum from k=1 to 2. 4*(2*3*5)/6 -4*(2*3)/2 +2. Let's compute:First term: 4*(2*3*5)/6 =4*(30)/6=4*5=20Second term: -4*(2*3)/2= -4*3= -12Third term: 2Total:20 -12 +2=10. Which matches. So formula works here. Another check: 1² +3² +5²=1+9+25=35. Let's use formula for n=3:4*(3*4*7)/6 -4*(3*4)/2 +3First term:4*(84)/6=4*14=56Second term: -4*(12)/2= -24Third term:3Total:56 -24 +3=35. Correct. So formula works. Therefore, trusting the formula.Thus, answer is 650. Wait, but before finalizing, let me compute 166,650 divided by 1000. 1000*166=166,000. 166,650 -166,000=650. So remainder is 650. Therefore, the remainder is 650.Alternatively, another way: Since 1000=8*125, perhaps compute modulo 8 and modulo 125, then use Chinese remainder theorem? But maybe overcomplicating. But let's see.But since the number is 166,650. Let's compute modulo 8 and modulo 125.But maybe the initial method is correct, as we are directly calculating the sum and then taking modulo 1000. The problem is, 166,650 is a big number. Did I compute the sum correctly? Let's cross-validate with another approach.Alternatively, the sum of the squares of the first n odd numbers can be given by the formula: n(2n +1)(2n -1)/3. Wait, is that correct? Let me check with n=1: 1(2*1 +1)(2*1 -1)/3=1*3*1/3=1. Correct. For n=2: 2(5)(3)/3=2*5=10. Which matches. For n=3:3(7)(5)/3=35. Correct. So formula seems n(2n +1)(2n -1)/3. Let me compute that.For n=50: 50*(101)*(99)/3. Compute step by step. 50/3=50/3≈16.666..., but let's do exact calculation.50*101=5050. Then 5050*99. Compute 5050*100=505,000 minus 5050*1=5050. So 505,000 -5,050=499,950. Then divide by 3: 499,950 /3=166,650. Which matches the previous sum. So that's a good check. Therefore, the formula works here. Therefore, the sum is indeed 166,650, and mod 1000 gives 650. So answer is 650. So I think that's correct.But just to be thorough, maybe I should compute the sum in another way. Let's compute the sum of squares of odd numbers up to 99.Alternatively, note that sum of squares of all numbers up to 99 is equal to sum of squares of even numbers up to 98 plus sum of squares of odd numbers up to 99. So if we compute sum of squares up to 99, subtract sum of squares of even numbers up to 98, we get the desired sum.Let's try this method.Sum of squares from 1 to 99: 99*100*199 /6. Formula is n(n +1)(2n +1)/6. So n=99. So 99*100*199 /6. Compute that.First, 99*100=9900, then 9900*199. Let's compute 9900*200=1,980,000, subtract 9900*1=9900: 1,980,000 -9,900=1,970,100. Then divide by 6:1,970,100 /6. Let's compute that. 6*328,350=1,970,100. Therefore, the sum is 328,350.Sum of squares of even numbers up to 98: 2² +4² +...+98². Factor out 4: 4(1² +2² +...+49²). So sum is 4*(49*50*99)/6. Let's compute that. 49*50=2450. 2450*99=2450*(100 -1)=245,000 -2,450=242,550. Then divide by 6:242,550 /6=40,425. Multiply by 4:40,425*4=161,700.So sum of squares of odd numbers up to 99 is total sum minus even sum:328,350 -161,700=166,650. Same as before. Therefore, that's a cross-check. So sum is indeed 166,650. Therefore, mod 1000:650. So the answer is 650. Therefore, I think that's correct.Alternatively, maybe I can compute the sum modulo 1000 directly, without calculating the entire sum. Because 166,650 is a big number, but perhaps we can compute the sum modulo 1000 term by term, which might be easier. Let's see. Because if we add each term modulo 1000, then we can keep the running total under control.But since the problem is already solved, maybe not necessary. But just to verify again, let's try.Compute the sum 1² +3² +5² +...+99² mod 1000. Each term is (2k -1)^2 where k from 1 to 50. So compute each (2k -1)^2 mod 1000, sum them up, then take mod 1000.But 50 terms is manageable. Let me see.But 50 terms is a bit time-consuming, but maybe we can note that (2k -1)^2 =4k² -4k +1. Therefore, sum is 4Σk² -4Σk +Σ1. So sum modulo 1000 can be computed as (4*S2 -4*S1 +S0) mod 1000, where S2=Σk² mod (1000/ gcd(4,1000)), but since we need to compute 4*S2 mod 1000, perhaps we can compute S2 first.Wait, maybe let's compute S2 = sum_{k=1 to 50} k². Then compute 4*S2 mod 1000. Similarly, S1 = sum_{k=1 to50}k, compute -4*S1 mod 1000. Then S0=50. Then sum them all mod 1000.Compute each part:First, compute S1 = sum_{k=1 to50}k =50*51/2=1275. So -4*S1= -4*1275= -5100. Then mod 1000. Since -5100 /1000= -5*1000 -100, so -5100 ≡ -100 mod 1000. But -100 mod 1000 is 900. So -5100 ≡ 900 mod 1000.Next, compute S2 = sum_{k=1 to50}k²=50*51*101/6=50*51*101/6. Let me compute this modulo (1000 / gcd(4,1000)). Since we need to compute 4*S2 mod 1000. gcd(4,1000)=4, so 1000/4=250. Therefore, S2 modulo 250 is sufficient because 4*(S2 mod 250) ≡4*S2 mod 1000. Therefore, compute S2 mod 250.So compute S2=50*51*101/6 mod 250.First, compute numerator:50*51*101. Let's compute each step mod 250*6=1500, but maybe breaking down.First, note that 50/6 is not integer, but 50*51*101 divided by 6.But perhaps compute S2 = (50*51*101)/6.Let me factor 50 and 6. 50=5*10, 6=2*3. So 50/6= (5*10)/(2*3)= (5/3)*(10/2)= (5/3)*5. Hmm, not helpful. Maybe compute step by step:50*51=2550. Then 2550*101=2550*(100 +1)=2550*100 +2550*1=255,000 +2,550=257,550. Then divide by 6:257,550 ÷6=42,925. So S2=42,925. Therefore, 42,925 mod 250. 250*171=42,750. 42,925 -42,750=175. So S2 ≡175 mod 250. Therefore, 4*S2 ≡4*175=700 mod 1000.Then S0=50. So adding up all parts: 4*S2 -4*S1 +S0 ≡700 +900 +50 mod 1000. 700 +900=1600. 1600 +50=1650. 1650 mod 1000=650. So same result. So that's a good check. So using modular arithmetic step by step also gives 650. So that confirms the answer.Therefore, the remainder is 650 when the sum is divided by 1000. I think that's solid.Final Answerboxed{650}</think>To find the remainder when (1^2 + 3^2 + 5^2 + cdots + 99^2) is divided by 1000, we start by recognizing that this is the sum of the squares of the first 50 odd numbers. The sum of the squares of the first (n) odd numbers can be calculated using the formula:[sum_{k=1}^{n} (2k - 1)^2 = frac{n(2n + 1)(2n - 1)}{3}]For (n = 50):[sum_{k=1}^{50} (2k - 1)^2 = frac{50 cdot 101 cdot 99}{3}]Calculating the numerator:[50 cdot 101 = 5050][5050 cdot 99 = 5050 cdot (100 - 1) = 505000 - 5050 = 499950]Dividing by 3:[frac{499950}{3} = 166650]Next, we find the remainder when 166650 is divided by 1000. This is done by looking at the last three digits of 166650, which are 650. Therefore, the remainder is:[166650 mod 1000 = 650]Thus, the remainder when (1^2 + 3^2 + 5^2 + cdots + 99^2) is divided by 1000 is (boxed{650}).,<think>Okay, let me try to figure out how to find the remainder when the sum of the squares of all odd numbers from 1 to 99 is divided by 1000. Hmm, so the problem is asking for (1^2 + 3^2 + 5^2 + dots + 99^2 mod 1000). Alright, first step, I need to understand how many terms are in this sequence and maybe find a formula for the sum of squares of odd numbers.Let me recall that the sum of the squares of the first n natural numbers is given by (frac{n(n+1)(2n+1)}{6}). But here, we only want the squares of the odd numbers. So maybe I can find a way to relate this to the sum of all squares and subtract the sum of the even squares?Wait, the sum of the squares of the first n odd numbers. Let me check. If I consider the first m odd numbers, they are 1, 3, 5, ..., (2m-1). So the sum of their squares would be (sum_{k=1}^{m} (2k - 1)^2). Let me compute that.Expanding ((2k - 1)^2) gives (4k^2 - 4k + 1). So the sum becomes (sum_{k=1}^{m} (4k^2 - 4k + 1) = 4sum_{k=1}^{m} k^2 - 4sum_{k=1}^{m} k + sum_{k=1}^{m} 1).I can compute each of these sums separately. The sum of (k^2) from 1 to m is (frac{m(m+1)(2m+1)}{6}), the sum of k from 1 to m is (frac{m(m+1)}{2}), and the sum of 1 m times is m. So substituting these into the equation:(4 cdot frac{m(m+1)(2m+1)}{6} - 4 cdot frac{m(m+1)}{2} + m)Simplify each term:First term: (4 cdot frac{m(m+1)(2m+1)}{6} = frac{4}{6}m(m+1)(2m+1) = frac{2}{3}m(m+1)(2m+1))Second term: (-4 cdot frac{m(m+1)}{2} = -2m(m+1))Third term: (+m)Combine all terms:(frac{2}{3}m(m+1)(2m+1) - 2m(m+1) + m)Let me factor out m from each term:(m left[ frac{2}{3}(m+1)(2m+1) - 2(m+1) + 1 right])Maybe factor out (m+1) from the first two terms:(m left[ (m+1)left( frac{2}{3}(2m+1) - 2 right) + 1 right])Compute the expression inside the brackets:First, inside the (m+1) factor:(frac{2}{3}(2m + 1) - 2 = frac{4m + 2}{3} - 2 = frac{4m + 2 - 6}{3} = frac{4m - 4}{3} = frac{4(m - 1)}{3})So now, substituting back:(m left[ (m+1)cdot frac{4(m - 1)}{3} + 1 right])Simplify:(m left[ frac{4(m+1)(m - 1)}{3} + 1 right] = m left[ frac{4(m^2 - 1)}{3} + 1 right])Combine the terms inside:( frac{4m^2 - 4}{3} + 1 = frac{4m^2 - 4 + 3}{3} = frac{4m^2 - 1}{3} )Therefore, the entire expression becomes:(m cdot frac{4m^2 - 1}{3} = frac{m(4m^2 - 1)}{3})So the formula for the sum of the squares of the first m odd numbers is (frac{m(4m^2 - 1)}{3}). Let me check this formula with a small example to make sure.Take m = 1: sum is 1² = 1. Formula gives (1*(4*1 -1))/3 = (1*3)/3 = 1. Correct.m = 2: sum is 1² + 3² = 1 + 9 = 10. Formula: (2*(4*4 -1))/3 = 2*(16 -1)/3 = 2*15/3 = 10. Correct.m = 3: 1 + 9 + 25 = 35. Formula: 3*(4*9 -1)/3 = (3*35)/3 = 35. Correct. Okay, seems good.So now, how many terms do we have in the sum from 1² to 99² of odd numbers? The numbers are 1, 3, 5, ..., 99. This is an arithmetic sequence with first term 1, common difference 2. The nth term is 1 + (n-1)*2 = 2n -1. So 2n -1 = 99 => 2n = 100 => n = 50. So there are 50 terms. Therefore, m = 50.So substituting m = 50 into the formula:Sum = (frac{50(4*50^2 - 1)}{3})Compute this. First compute 50^2 = 2500. Then 4*2500 = 10,000. Then 10,000 -1 = 9,999. Then 50*9,999 = 50*(10,000 -1) = 500,000 - 50 = 499,950. Then divide by 3: 499,950 /3.Let me compute that. 3 into 499,950: 3*166,650 = 499,950. So sum is 166,650. So the total sum is 166,650.But wait, we need the remainder when this sum is divided by 1000. So compute 166,650 mod 1000.But 166,650 divided by 1000 is 166.650, so the remainder is the last three digits. 166,650. The last three digits are 650. So the remainder is 650. But wait, wait, is this correct?Hold on, let me verify my steps again. Because 166,650 divided by 1000 is 166 with a remainder of 650. So the remainder is 650. But I need to check whether my calculation of the sum is correct.Wait, let's double-check the formula. The formula is for the sum of the squares of the first m odd numbers, which is (frac{m(4m^2 -1)}{3}). So with m=50, sum is 50*(4*(50)^2 -1)/3.Compute 4*50^2: 4*2500=10,000. 10,000 -1=9,999. 50*9,999= 499,950. Then divided by 3: 499,950 ÷3. Let me compute 3*166,650= 499,950. So yes, 166,650. Then 166,650 mod 1000 is 650. So the answer is 650. Hmm.But wait, let me cross-verify this with another approach to make sure. Maybe calculate the sum using another method.Alternatively, the sum of the squares of the first n odd numbers can also be calculated as the sum of all squares minus the sum of even squares. Let's try this.The sum of squares from 1 to 99: but we have to be careful here. Wait, the numbers go up to 99, but since we are dealing with odd numbers up to 99, the even numbers go up to 98. Let me see.The sum of all squares from 1 to 99 is (sum_{k=1}^{99} k^2). The sum of even squares is (sum_{k=1}^{49} (2k)^2) because the even numbers from 2 to 98 are 2,4,...,98, which is 49 terms (since 98/2=49). So sum of even squares is 4*(sum_{k=1}^{49} k^2). Then the sum of odd squares is sum of all squares up to 99 minus sum of even squares up to 98.Compute sum of all squares up to 99: (frac{99*100*199}{6}). Let's compute that.First, 99*100=9900. 9900*199. Let's compute 9900*200 = 1,980,000. Subtract 9900: 1,980,000 -9,900 = 1,970,100. Then divide by 6: 1,970,100 /6 = 328,350.Sum of even squares: 4* sum of squares from 1 to 49. Sum of squares from 1 to 49 is (frac{49*50*99}{6}). Compute that: 49*50=2450. 2450*99=2450*(100-1)=245,000 -2,450=242,550. Divide by 6: 242,550 /6=40,425. Then multiply by 4: 40,425*4=161,700.Therefore, sum of odd squares is 328,350 -161,700=166,650. Which matches the previous result. So that's good.Therefore, the sum is indeed 166,650. Then 166,650 divided by 1000 is 166 with remainder 650. So the remainder is 650. Therefore, the answer is 650. But wait, let me check once more.Alternatively, perhaps compute the sum modulo 1000 directly. Since we are dealing with a sum of numbers, each squared, and we can compute each square modulo 1000, then add them up modulo 1000. However, since there are 50 terms, that might take a while, but maybe there's a pattern or periodicity in the squares modulo 1000. Let's see.But given that the total sum is 166,650, and 166,650 mod 1000 is 650, that seems straightforward, but I want to be absolutely sure. Let me compute 166,650 divided by 1000. 1000*166=166,000. Subtract 166,000 from 166,650: 650. So remainder is 650. So yes, 650. Therefore, the answer is 650. But wait, the problem is from 1² to 99² of odd numbers. Wait, but in the first approach, when we use m=50, we get sum as 166,650, which is correct. Then 166,650 mod 1000 is 650.Alternatively, maybe the user is expecting the answer to be 650, but let me just check once again.Alternatively, perhaps compute the sum modulo 1000 term by term. Let's see. Since 1000 is 8*125, and modulus 8 and 125, then use Chinese Remainder Theorem. But perhaps that's more complicated. Alternatively, compute the sum mod 8 and mod 125, then combine.But maybe that's not necessary here. Alternatively, note that 166,650 divided by 1000 is 166*1000 + 650, so the remainder is 650. So the remainder is indeed 650.But wait, let me check with a smaller example. Let's take sum up to 5² (1² + 3² +5²)=1+9+25=35. 35 mod 1000 is 35. Using the formula m=3: 3*(4*9 -1)/3= (3*35)/3=35. Correct. So the formula works. So confidence is high.Therefore, the remainder is 650. However, let me check once more.Wait, 166,650 divided by 1000. 1000 * 166 = 166,000. 166,650 - 166,000 = 650. So remainder 650. So yes.Alternatively, let me compute the sum modulo 1000 step by step. Maybe the formula gives 166,650 which is 650 mod 1000, but perhaps if I compute each term modulo 1000 and sum up, I might get a different result. Wait, that can't be. If the total sum is 166,650, then modulo 1000 must be 650.But perhaps the formula is incorrect? Wait, we checked with small numbers. It works. So unless my arithmetic is wrong here.Wait, 50*9999= 50*(10,000 -1)=500,000 -50=499,950. 499,950 divided by 3 is 166,650. Correct.Yes, seems correct.Alternatively, if I compute the sum as follows: each odd number from 1 to 99. Let's note that modulo 1000, squares repeat every certain number. However, since 1000 is 8*125, maybe compute modulo 8 and modulo 125, then combine.But perhaps that's overcomplicating. Alternatively, note that:Sum = 1² +3² +5² +…+99².If we group terms in pairs: (1² +99²), (3² +97²), ..., up to (49² +51²). There are 25 such pairs. Each pair is ( (2k-1)^2 + (101 - 2k)^2 ). Let me compute (2k-1)^2 + (100 - (2k -1))^2. Wait, 99 = 100 -1, 97=100-3, etc. So each pair is (2k -1)^2 + (100 - (2k -1))^2.Compute (2k -1)^2 + (100 - (2k -1))^2 = (2k -1)^2 + (101 -2k)^2.Expanding both terms:First term: (2k -1)^2 =4k² -4k +1Second term: (101 -2k)^2= (101)^2 -2*101*2k + (2k)^2 = 10,201 -404k +4k²Adding both terms: 4k² -4k +1 +10,201 -404k +4k² =8k² -408k +10,202Therefore, each pair sums to 8k² -408k +10,202. Hmm, but this might not lead to an easy formula.Alternatively, maybe compute each pair modulo 1000. Since 100 - (2k -1) = 101 - 2k. Let's compute (2k -1)^2 + (101 -2k)^2 modulo 1000.Let me denote x = 2k -1, then 101 -2k = 101 - (x +1) = 100 - x. So each pair is x² + (100 -x)^2.Compute x² + (100 -x)^2 = x² + 10,000 -200x +x² =2x² -200x +10,000.Thus, modulo 1000, this becomes (2x² -200x +10,000) mod 1000.But 10,000 mod 1000 is 0. So the expression becomes (2x² -200x) mod 1000. Factor out 2: 2(x² -100x) mod 1000. Hmm.Alternatively, note that x is an odd number from 1 to 49, since the pairs go up to 49 and 51. Wait, if x is from 1 to 49 (since the pairs are (1,99), (3,97), ..., (49,51)), so x is 1,3,...,49. So x ranges from 1 to 49, step 2.So for each pair, compute 2(x² -100x) mod 1000. Then sum over all pairs. Plus, if there is an unpaired term. Wait, since there are 50 terms, which is even, 25 pairs. So no unpaired term.Therefore, total sum modulo 1000 is 25 pairs, each contributing 2(x² -100x) mod 1000. Then sum all these 25 terms and take modulo 1000.Wait, but x varies over 1,3,...,49. Let me denote x=2k-1 where k=1 to 25.Wait, maybe this approach is getting complicated. Alternatively, note that x is odd, so x=2m-1 where m=1 to 25. Wait, but x=1,3,...,49, so m=1 to 25.But this might not help much. Alternatively, compute each pair's contribution modulo 1000.Take x=1: pair (1,99). Contribution: 2(1 -100*1) =2(1 -100)=2(-99)= -198 mod 1000. But we can compute (1² +99²) mod 1000. 1²=1, 99²=9801. 1 +9801=9802. 9802 mod 1000=802. So contribution is 802.But according to the formula 2(x² -100x) mod 1000 when x=1: 2(1 -100)=2(-99)= -198 mod 1000=802. Yes, matches.Similarly, x=3: 3² +97²=9 +9409=9418 mod1000=418. Formula:2(9 -300)=2(-291)= -582 mod1000=418. Correct.Similarly, x=5: 25 + 9025=9050 mod1000=50. Formula:2(25 -500)=2(-475)= -950 mod1000=50. Correct.So each pair's contribution is 2(x² -100x) mod1000. So, the total sum modulo 1000 is the sum over x=1,3,...,49 of [2(x² -100x)] mod1000.But since modulus is 1000, we can compute each term mod1000 and sum them up. Then take the total mod1000.Therefore, compute Sum_{x=1,3,...,49} [2x² -200x] mod1000.Factor out 2: 2 Sum_{x=1,3,...,49} (x² -100x) mod1000.Let me compute Sum_{x=1,3,...,49} (x² -100x). Let's split it into Sum x² - 100 Sum x.Compute Sum x² where x is 1,3,...,49. This is the sum of squares of first 25 odd numbers? Wait, x here is 1,3,...,49. Wait, but 49 is the 25th odd number. Wait, no. Wait, 1 is the 1st odd number, 3 the 2nd, ..., 49 is the 25th odd number. Wait, 25th odd number is 2*25 -1=49. Yes. So Sum x² where x=1,3,...,49 is the sum of squares of first 25 odd numbers. Using the formula earlier, this would be (frac{25(4*25^2 -1)}{3}).Compute 25^2=625, 4*625=2500, 2500 -1=2499. 25*2499=62,475. Divide by 3:62,475 /3=20,825. So Sum x²=20,825.Sum x where x=1,3,...,49. This is the sum of first 25 odd numbers. The sum of first m odd numbers is m². So m=25, sum=25²=625.Therefore, Sum (x² -100x)=20,825 -100*625=20,825 -62,500= -41,675.Multiply by 2:2*(-41,675)= -83,350.Now compute -83,350 mod1000. First, divide 83,350 by1000: 83*1000=83,000, 83,350-83,000=350. So -83,350 mod1000= -350 mod1000=650.Therefore, the total sum mod1000 is 650. Which matches the previous result. So this confirms that the answer is 650.Alternatively, since we have Sum = 166,650, which is 166*1000 + 650, so mod1000=650. Therefore, the remainder is 650.But to be absolutely thorough, let me compute a few pairs manually and add them up modulo 1000 to see if they align.First pair:1² +99²=1 +9801=9802. 9802 mod1000=802.Second pair:3² +97²=9 +9409=9418. 9418 mod1000=418.Third pair:5² +95²=25 +9025=9050. 9050 mod1000=50.Fourth pair:7² +93²=49 +8649=8698. 8698 mod1000=698.Fifth pair:9² +91²=81 +8281=8362. 8362 mod1000=362.Sixth pair:11² +89²=121 +7921=8042. 8042 mod1000=42.Seventh pair:13² +87²=169 +7569=7738. 7738 mod1000=738.Eighth pair:15² +85²=225 +7225=7450. 7450 mod1000=450.Ninth pair:17² +83²=289 +6889=7178. 7178 mod1000=178.Tenth pair:19² +81²=361 +6561=6922. 6922 mod1000=922.Eleventh pair:21² +79²=441 +6241=6682. 6682 mod1000=682.Twelfth pair:23² +77²=529 +5929=6458. 6458 mod1000=458.Thirteenth pair:25² +75²=625 +5625=6250. 6250 mod1000=250.Fourteenth pair:27² +73²=729 +5329=6058. 6058 mod1000=58.Fifteenth pair:29² +71²=841 +5041=5882. 5882 mod1000=882.Sixteenth pair:31² +69²=961 +4761=5722. 5722 mod1000=722.Seventeenth pair:33² +67²=1089 +4489=5578. 5578 mod1000=578.Eighteenth pair:35² +65²=1225 +4225=5450. 5450 mod1000=450.Nineteenth pair:37² +63²=1369 +3969=5338. 5338 mod1000=338.Twentieth pair:39² +61²=1521 +3721=5242. 5242 mod1000=242.Twenty-first pair:41² +59²=1681 +3481=5162. 5162 mod1000=162.Twenty-second pair:43² +57²=1849 +3249=5098. 5098 mod1000=98.Twenty-third pair:45² +55²=2025 +3025=5050. 5050 mod1000=50.Twenty-fourth pair:47² +53²=2209 +2809=5018. 5018 mod1000=18.Twenty-fifth pair:49² +51²=2401 +2601=5002. 5002 mod1000=2.Now, sum all these remainders:802 +418 +50 +698 +362 +42 +738 +450 +178 +922 +682 +458 +250 +58 +882 +722 +578 +450 +338 +242 +162 +98 +50 +18 +2.Let me compute this step by step.Start with 802.802 +418 =12201220 +50=12701270 +698=19681968 +362=23302330 +42=23722372 +738=31103110 +450=35603560 +178=37383738 +922=46604660 +682=53425342 +458=58005800 +250=60506050 +58=61086108 +882=69906990 +722=77127712 +578=82908290 +450=87408740 +338=90789078 +242=93209320 +162=94829482 +98=95809580 +50=96309630 +18=96489648 +2=9650.Now, 9650 mod1000=650.Which matches the earlier result. So even when calculating each pair's contribution modulo1000 and summing them up, we get 9650, which modulo1000 is 650. Therefore, the remainder is indeed 650.Therefore, the answer is boxed{650}.Final Answerboxed{650}</think>To find the remainder when (1^2 + 3^2 + 5^2 + cdots + 99^2) is divided by 1000, we start by using the formula for the sum of the squares of the first (m) odd numbers, which is (frac{m(4m^2 - 1)}{3}).1. Determine the number of terms: The sequence of odd numbers from 1 to 99 forms an arithmetic sequence with the first term 1 and common difference 2. The (n)-th term is given by (2n - 1). Solving (2n - 1 = 99) gives (n = 50). Therefore, there are 50 terms.2. Apply the formula: Substituting (m = 50) into the formula: [ text{Sum} = frac{50(4 cdot 50^2 - 1)}{3} ] Calculate the values step-by-step: [ 50^2 = 2500 ] [ 4 cdot 2500 = 10000 ] [ 10000 - 1 = 9999 ] [ 50 cdot 9999 = 499950 ] [ frac{499950}{3} = 166650 ]3. Find the remainder modulo 1000: The sum (166650) divided by 1000 gives a quotient of 166 and a remainder of 650. Therefore, the remainder when (166650) is divided by 1000 is: [ 166650 mod 1000 = 650 ]4. Verification: To ensure correctness, we verified the result by calculating the sum modulo 1000 using pairs of terms and summing their contributions modulo 1000. This method also confirmed that the remainder is 650.Thus, the remainder when (1^2 + 3^2 + 5^2 + cdots + 99^2) is divided by 1000 is (boxed{650}).]

🔑:Answer: 650 We have S=sum_{i=0}^{49}(2 i+1)^{2}=sum_{i=0}^{49} 4 i^{2}+4 i+1=4 cdot frac{49 cdot 50 cdot 99}{6}+4 cdot frac{49 cdot 50}{2}+50 equiv 700+900+50(bmod 1000) equiv 650(bmod 1000)

🧠:[<think>Okay, let's try to figure out this problem step by step. So we have two points, A(3,5) and B(5,7), and we need to find the inclination angle of a line l whose slope is √3 times the slope of line AB. Alright, let's break this down.First, I remember that the slope of a line between two points (x1, y1) and (x2, y2) is calculated by (y2 - y1)/(x2 - x1). So for points A and B, let's compute that. Point A is (3,5) and point B is (5,7). So the difference in y-coordinates is 7 - 5 = 2, and the difference in x-coordinates is 5 - 3 = 2. So the slope of AB is 2/2 = 1. That seems straightforward. So the slope of AB is 1.Now, the problem says that the slope of line l is √3 times the slope of AB. Since the slope of AB is 1, multiplying that by √3 gives a slope of √3 for line l. So slope of l is √3.Now, we need to find the inclination angle of line l. I recall that the inclination angle θ of a line with slope m is given by the relation m = tanθ. So if the slope is √3, then we have tanθ = √3. So now we need to find θ such that tanθ = √3. From trigonometry, I remember that tan(60°) = √3. Since 60 degrees is π/3 radians, the inclination angle should be 60 degrees or π/3 radians. But let me make sure there's no catch here.Wait, inclination angle is defined as the angle measured from the positive direction of the x-axis to the line in the counterclockwise direction, right? And this angle should be between 0° and 180°, or 0 and π radians. Since the slope is positive √3, the line is going upwards from left to right, so the angle should be in the first quadrant. So 60° makes sense here. Let me just double-check the calculations. Slope of AB: (7-5)/(5-3) = 2/2 = 1. That's correct. Then slope of l is √3 * 1 = √3. Then arctangent of √3 is indeed 60°, because tan(60°) = opposite/adjacent = √3/1 in a 30-60-90 triangle. Yeah, that's right. So the inclination angle of line l is 60 degrees, which is π/3 radians. Depending on what unit the answer is expected, but since the question doesn't specify, but in math problems like this, sometimes radians are preferred. But maybe not necessarily. Wait, let me check the original problem again. It just says "the inclination angle of line l is ______." So it might need to be in degrees or radians. Hmm. But in Chinese textbooks, it's often degrees, but in calculus, it's radians. Wait, the problem didn't specify. Hmm. But since the answer is a standard angle, both 60° and π/3 are correct. But let's see what's the standard answer for inclination angle. Looking back, in some contexts, inclination angles are given in degrees, but mathematically, especially in higher mathematics, radians are more common. However, since the slope is a simple √3, which corresponds to 60°, which is a commonly known angle. Also, π/3 radians is equivalent to 60°. So maybe both are acceptable, but I need to check if there's a convention here. Wait, in the problem statement, the other slope was 1, which corresponds to an angle of 45°, so if they wanted the answer in degrees, it would be 60°, if in radians, π/3. The problem doesn't specify, so perhaps either is acceptable, but since it's a fill-in-the-blank, probably either is okay, but maybe the question expects degrees? Wait, the original problem says "inclination angle"—I think in many geometry contexts, that's often in degrees unless specified otherwise. Wait, but actually, in calculus and vectors, angles are in radians. Hmm. Wait, but the problem is about coordinate geometry, so maybe degrees. But I'm not entirely sure. Wait, perhaps I should check some references.Wait, but maybe in the original problem's source, they use degrees. But since the user hasn't specified, perhaps it's safer to give both? Wait, no, the user says "put your final answer within boxed{}". So they probably expect one answer. Hmm.Wait, let me think. When you take arctangent of √3 on a calculator, if the calculator is in degree mode, you get 60°, if in radians, you get π/3. Since the problem doesn't specify, but in most mathematical contexts where calculus is involved, radians are the default. But in problems involving just geometry, especially if from a pre-calculus course, degrees might be expected.But since the problem mentions "inclination angle", let me recall the exact definition. The inclination angle of a line is the smallest non-negative angle formed between the positive x-axis and the line, measured counterclockwise. This angle is typically expressed in degrees in many geometry courses, but it's not incorrect to use radians. However, given that the answer is likely expecting a common angle, and √3 corresponds to 60°, which is a standard degree measure, but π/3 is also standard. Wait, but the problem is presented in Chinese, probably translated. Wait, no, the user wrote the problem in English. So maybe the answer is expected in degrees.Alternatively, maybe the problem is from an SAT or ACT or GRE Math subject test, which usually use degrees for angles unless specified. Hmm. But in the absence of specific information, I might need to make a call. Wait, but in my own studies, I recall that sometimes inclination angles are in radians, but also in degrees. Wait, the term "inclination angle" in mathematics is often associated with slope, and slope is rise over run, which is tangent of the angle. So tan(theta) = m. The angle theta is between 0 and 180 degrees (or 0 and pi radians). So if m is positive, theta is between 0 and 90 degrees (or 0 and pi/2 radians). If m is negative, theta is between 90 and 180 degrees (or pi/2 and pi radians). So in this case, since m is sqrt(3) which is positive, theta is arctangent(sqrt(3)), which is 60 degrees or pi/3 radians.Given that pi/3 is approximately 1.047 radians, but since the problem is asking for an exact value, either 60 degrees or pi/3 radians is acceptable. However, in Chinese math education, angles are typically expressed in degrees unless specified otherwise. But since the user hasn't specified, but the problem is written in English, perhaps the answer is expected in radians. Wait, but in many English math problems, especially at the high school level, angles are still often in degrees. Hmm. This is a bit confusing.Alternatively, maybe check common standard angles. Tan(theta) = sqrt(3) corresponds to 60 degrees, which is a standard angle. So in the absence of specific units, the answer is likely 60 degrees. But I need to confirm if there's a standard convention here.Wait, in mathematics, when angles are involved in calculus or higher-level math, radians are standard. But in geometry problems, especially those not involving calculus, degrees are common. Since this problem is about slopes and inclination angles, which is a geometry topic, and not involving calculus, maybe degrees are expected here. Also, since pi/3 is an exact value, but 60 degrees is also exact. So both are correct. Hmm.Wait, actually, in the study of slopes and inclination angles, the angle is often given in degrees in many textbooks. For example, a slope of 1 corresponds to 45 degrees, which is more intuitive for most people. Similarly, a slope of sqrt(3) corresponds to 60 degrees. So I think the answer is 60 degrees. Therefore, the inclination angle is 60 degrees. But to be thorough, let me verify once more.If we have tan(theta) = sqrt(3), solving for theta gives theta = arctan(sqrt(3)). The principal value of arctan(sqrt(3)) is pi/3 radians, which is 60 degrees. Therefore, both 60 degrees and pi/3 radians are correct. However, since the problem does not specify the unit, but given that the slope is presented as sqrt(3), which is a number without units, the angle could be presented in radians. Wait, but the arctangent of sqrt(3) is a common angle, known to be 60 degrees or pi/3. So in the absence of units, it's ambiguous.Wait, but in mathematical literature, when angles are given without units, they are typically assumed to be in radians. However, in many geometry problems, especially at the high school level, angles are given in degrees. Since this problem involves basic coordinate geometry (slope between two points), which is a high school topic, and inclination angle is a term used there, maybe the answer is expected in degrees. But to resolve this, perhaps we can look at the problem's origin.But since we don't have more context, maybe the answer is expected in degrees. Alternatively, if the answer is to be given in radians, it's pi/3. But since the problem mentions "the inclination angle of line l is ______", and in the Chinese education system, angles are usually in degrees unless specified. But the user is presenting the problem in English. Hmm.Wait, let's check another angle. Suppose the slope was 1, which would be 45 degrees or pi/4 radians. If the question had been about that, the answer would probably be 45 degrees, as it's more intuitive. Similarly, sqrt(3) is 60 degrees, so maybe they expect 60 degrees here.Alternatively, maybe check the definition of inclination angle. According to some sources, the inclination angle is typically expressed in degrees between 0° and 180°, but in calculus contexts, it's in radians. However, given that the problem is purely algebraic/geometric, perhaps degrees are intended. But since the problem didn't specify units, but in mathematical notation, if they want radians, they usually put it in terms of pi. Since 60 degrees is pi/3 radians, maybe writing pi/3 is better here because it's a more mathematical answer. Wait, but maybe not. If the question is from a geometry course, they might prefer degrees. Alternatively, since the answer is to be boxed, and mathematical notation often uses radians without the degree symbol, but in many cases, if it's a standard angle like 30°, 45°, 60°, they can be written either way. But in this case, pi/3 is a more precise answer in a mathematical context. Hmm.Wait, let me check some references. For example, in the textbook "Calculus" by James Stewart, angles in the context of slope and arctangent are usually in radians. However, in a geometry textbook like "Geometry" by Jurgensen, Brown, and Jurgensen, angles are in degrees. Since this problem is about the slope and inclination angle, which is covered in both geometry and calculus, but here it's more of a geometry problem.Given that, I think the answer is 60 degrees. But to be safe, maybe check both. Wait, but the problem is given in English, and the answer is to be boxed. If the answer is 60 degrees, then boxed{60^circ}, if radians, then boxed{dfrac{pi}{3}}. But since the problem didn't specify, perhaps both are acceptable. Wait, but let me check the initial problem statement again.Wait, the original problem says: "The inclination angle of line l is ______." There's no unit specified. Hmm. In such cases, if the angle is a common one in both degrees and radians, like 30°, 45°, 60°, 90°, which correspond to pi/6, pi/4, pi/3, pi/2, then perhaps the answer is expected in radians since it's more mathematically concise. However, in many problems, especially in geometry, degrees are used. Hmm.Alternatively, maybe the problem comes from a source where the answer is expected in radians. For example, in higher-level math, angles are in radians. Since the multiplication factor is sqrt(3), which is an irrational number, but the angle itself is a common angle in radians. So pi/3 radians is 60 degrees. Hmm. So maybe the answer is pi/3. But how to decide?Wait, let's see. If I recall correctly, in the SAT or ACT, angles are in degrees, but in AP Calculus or higher, they are in radians. Since this problem is about the slope and inclination angle, which is a concept that's covered in pre-calculus. In pre-calculus, sometimes both units are used, but when dealing with inverse trigonometric functions, radians are more common. However, the term "inclination angle" might be more of a geometry term, where degrees are standard.But since the problem involves a slope that is sqrt(3) times another slope, and the arctangent of sqrt(3) is a standard angle, the answer is likely 60 degrees. Therefore, the inclination angle is 60 degrees, so the answer is boxed{60^circ}.Alternatively, if the answer is in radians, it's pi over 3. But given that the problem is presented without any specific context, and considering that 60 degrees is more intuitive for a geometry problem, I think 60 degrees is the answer expected here. However, to be thorough, let's check the calculation once again.Slope of AB: (7-5)/(5-3) = 2/2 = 1. Correct. Then slope of line l is sqrt(3) * 1 = sqrt(3). Then tan(theta) = sqrt(3). The angle whose tangent is sqrt(3) is 60 degrees, as tan(60°) = sqrt(3). Therefore, the inclination angle is 60 degrees. So I think that's solid.Yes, I think that's the answer. So the answer is 60 degrees, boxed as boxed{60^circ}. Alternatively, if radians, boxed{dfrac{pi}{3}}. But since the problem doesn't specify, but given the presence of sqrt(3), which relates to 60 degrees, which is a common angle, I think 60 degrees is the answer they expect.Final AnswerThe inclination angle of line ( l ) is boxed{60^circ}.</think>Given points ( A(3,5) ) and ( B(5,7) ), we first calculate the slope of line ( AB ). The slope is found using the formula:[text{slope of } AB = frac{7 - 5}{5 - 3} = frac{2}{2} = 1]The slope of line ( l ) is given as ( sqrt{3} ) times the slope of line ( AB ). Therefore, the slope of line ( l ) is:[sqrt{3} times 1 = sqrt{3}]The inclination angle ( theta ) of a line with slope ( m ) is given by ( tan(theta) = m ). For line ( l ), we have:[tan(theta) = sqrt{3}]The angle ( theta ) whose tangent is ( sqrt{3} ) is ( 60^circ ) (or ( frac{pi}{3} ) radians). Since the problem does not specify the unit, but the angle is a common standard angle in geometry, we conclude the inclination angle is:[boxed{60^circ}],<think>Okay, so I need to find the inclination angle of line ( l ). The problem says that the slope of line ( l ) is ( sqrt{3} ) times the slope of line ( AB ). First, I should probably figure out what the slope of line ( AB ) is. The points given are ( A(3,5) ) and ( B(5,7) ). Alright, the formula for slope between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is ( m = frac{y_2 - y_1}{x_2 - x_1} ). Let me plug in the coordinates of points A and B here. So, substituting, the slope ( m_{AB} ) would be ( frac{7 - 5}{5 - 3} ), which simplifies to ( frac{2}{2} = 1 ). So the slope of line ( AB ) is 1. Then, the slope of line ( l ) is ( sqrt{3} ) times that, so ( m_l = sqrt{3} times 1 = sqrt{3} ). Therefore, the slope of line ( l ) is ( sqrt{3} ). Now, the question is asking for the inclination angle of line ( l ). I recall that the inclination angle ( theta ) of a line with slope ( m ) is given by ( theta = arctan(m) ). So, if the slope is ( sqrt{3} ), then the inclination angle should be ( arctan(sqrt{3}) ). Hmm, arctangent of ( sqrt{3} ). Let me think. I remember from trigonometry that ( tan(60^circ) = sqrt{3} ), because in a 30-60-90 triangle, the sides are in the ratio ( 1 : sqrt{3} : 2 ). So, the tangent of 60 degrees is ( sqrt{3} ). Therefore, ( arctan(sqrt{3}) = 60^circ ). Wait, but is that in radians or degrees? The problem doesn't specify, but inclination angles are usually measured in degrees unless stated otherwise. Let me check. In some contexts, angles in calculus are in radians, but since the problem is about geometry, maybe degrees. However, both are possible. Let me see. Wait, the problem just says "the inclination angle of line ( l ) is ______." It doesn't specify units, but in math problems like this, unless told otherwise, angles in trigonometry are often in degrees when common angles like 30°, 45°, 60° are involved. So, 60 degrees makes sense here. But let me make sure. Alternatively, if they want radians, 60 degrees is ( pi/3 ) radians. So, maybe they want ( pi/3 ). But the problem is in Chinese maybe? Wait, no, the problem is in English. Let me check the original problem again. It says "the inclination angle of line ( l ) is ______." Probably expects degrees here. But I can check. Alternatively, in some contexts, inclination angle is defined as the angle measured counterclockwise from the positive x-axis to the line, and it's typically between 0° and 180°, or in radians, between 0 and ( pi ). But regardless, since the slope is positive ( sqrt{3} ), the angle is acute, between 0° and 90°, and since ( tan(theta) = sqrt{3} ), the angle is 60°, which is ( pi/3 ) radians. But since the problem is written without specifying units, and given that the answer is likely a common angle, perhaps both 60° and ( pi/3 ) are acceptable. However, since the original problem uses numbers like 3, 5, 7, which are integers, and ( sqrt{3} ), which is common in trigonometry, it's more probable that the answer is expected in degrees. Also, in many geometry problems, angles are given in degrees unless stated otherwise. Therefore, the inclination angle of line ( l ) is 60 degrees. Let me just recap the steps to make sure I didn't make a mistake. First, slope of AB: points A(3,5) and B(5,7). Difference in y is 7-5=2, difference in x is 5-3=2, so slope is 2/2=1. Then, slope of l is ( sqrt{3} times 1 = sqrt{3} ). Then, arctangent of ( sqrt{3} ) is 60 degrees. That all checks out. Wait, just to confirm, let me verify using another approach. If the slope is ( sqrt{3} ), then for a right triangle, the rise over run would be ( sqrt{3}/1 ), so if we imagine a right triangle where the opposite side is ( sqrt{3} ) and the adjacent side is 1, then the angle ( theta ) would satisfy ( tan(theta) = sqrt{3}/1 ), which as I thought earlier is 60 degrees. Alternatively, if the slope was 1, that would correspond to 45 degrees, which is a familiar result. Since ( sqrt{3} ) is a steeper slope than 1, the angle should be greater than 45 degrees, which 60 degrees is. That makes sense. Therefore, I think I'm confident that the inclination angle is 60 degrees. But just to make sure, let me check if there's any other detail I might have missed. For example, is there any possibility that the inclination angle could be in a different quadrant? But inclination angles are defined as the smallest positive angle between the x-axis and the line, measured counterclockwise. Since the slope is positive, the line is going up to the right, so the angle should be between 0 and 90 degrees. So 60 degrees is correct. Another check: if the slope was negative, the inclination angle would be 180° minus the arctangent of the absolute value of the slope. But in this case, the slope is positive, so no need for that. So yeah, conclusion: 60 degrees. Therefore, the answer is 60°, which in a box would be boxed{60^circ} or boxed{dfrac{pi}{3}} if in radians. But considering the problem context, 60 degrees is more likely. Wait, the problem says "the inclination angle of line ( l ) is ______." It doesn't specify, but in Chinese math problems, often degrees are used. Wait, maybe check standard definitions. Looking up, in mathematics, the angle of inclination is typically measured in degrees unless specified otherwise. For example, common angles like 30°, 45°, 60° are usually expressed in degrees. So I think the answer is 60 degrees. Alternatively, if the problem is from a calculus or advanced math context, maybe radians. But given the problem is about basic slope and inclination, probably degrees. But just to be thorough, let me confirm. If we consider the arctangent of ( sqrt{3} ), in radians that is ( pi/3 ). So 60 degrees is ( pi/3 ) radians. But again, unless specified, it's safer to assume degrees. But wait, maybe the problem expects radians. Wait, if I check examples in textbooks, often angles of inclination are given in degrees. For instance, a slope of 1 corresponds to 45°, slope ( sqrt{3} ) is 60°, slope ( 1/sqrt{3} ) is 30°, which are all standard. So yes, 60° is the answer. Therefore, the answer should be boxed{60^circ}. However, if the problem expects radians, it would be boxed{dfrac{pi}{3}}. Hmm. The problem is given in Chinese or English? The user instruction is in English, so likely the problem is in English. In English math problems, unless stated otherwise, angles in such contexts are often in degrees. But sometimes, in higher-level math, radians are default. Wait, the problem is about inclination angle. Let me check the definition. According to standard math definitions, the angle of inclination is the angle formed between the positive x-axis and the line, measured counterclockwise, and this angle is typically between 0° and 180°, and can be measured in degrees or radians depending on the context. But since the problem gives slopes involving sqrt(3), which relates to 60°, which is a commonly known angle in degrees, I think the answer is 60 degrees. Moreover, if they wanted radians, they would probably mention it. Therefore, I think the answer is 60 degrees, so boxed{60^circ}. Final Answerboxed{60^circ}</think>Given points ( A(3,5) ) and ( B(5,7) ), we first calculate the slope of line ( AB ). Using the formula for slope between two points ((x_1, y_1)) and ((x_2, y_2)):[m_{AB} = frac{7 - 5}{5 - 3} = frac{2}{2} = 1]The slope of line ( l ) is (sqrt{3}) times the slope of line ( AB ):[m_l = sqrt{3} times 1 = sqrt{3}]The inclination angle (theta) of a line with slope ( m ) is given by (theta = arctan(m)). For slope (sqrt{3}):[theta = arctan(sqrt{3})]We know that (tan(60^circ) = sqrt{3}), so the inclination angle is:[theta = 60^circ]Thus, the inclination angle of line ( l ) is (boxed{60^circ}).]

🔑:According to the problem, let the inclination angle of line l be theta, then the slope of line l is k_l = tan theta.Given A(3,5) and B(5,7), we have k_{AB} = 1.Since the slope of line l is sqrt{3} times the slope of line AB,we have k_l = tan theta = sqrt{3}.Therefore, theta = 60^{circ}.Hence, the answer is boxed{60^{circ}}.This problem examines the relationship between the slope of a line and the inclination angle of the line, involving the tangent formula of double angles, and is considered a basic question.