🧠:[<think>Okay, so I need to figure out which of the cones listed (A to E) can be formed from a 252-degree sector of a circle with radius 10 by aligning the two straight sides. Hmm, let's start by recalling how creating a cone from a sector works.First, when you roll up a sector of a circle into a cone, the radius of the original sector becomes the slant height of the cone. That makes sense because the straight sides of the sector become the slant height when you bring them together. So in this case, the slant height (let's denote it as ( l )) is 10. Got that.Next, the arc length of the sector becomes the circumference of the base of the cone. The original sector has a central angle of 252 degrees. The circumference of the full circle would be ( 2pi r ), but since we have a sector, the arc length is a portion of that. Let me write that down.The arc length ( L ) of a sector with central angle ( theta ) degrees and radius ( r ) is given by:[L = frac{theta}{360} times 2pi r]So substituting the given values:[L = frac{252}{360} times 2pi times 10]Calculating that:First, simplify ( frac{252}{360} ). Both are divisible by 36? 252 ÷ 36 = 7, 360 ÷ 36 = 10. So that's ( frac{7}{10} ).Therefore:[L = frac{7}{10} times 20pi = 14pi]So the arc length is ( 14pi ). This arc length becomes the circumference of the base of the cone. The circumference of the base of the cone is ( 2pi r ), where ( r ) is the radius of the base of the cone. So:[2pi r = 14pi implies r = 7]Alright, so the radius of the base of the cone should be 7 units.Now, we know the slant height ( l = 10 ), and the radius ( r = 7 ). To find the height ( h ) of the cone, we can use the Pythagorean theorem in the right triangle formed by the radius, height, and slant height:[l^2 = r^2 + h^2]Plugging in the known values:[10^2 = 7^2 + h^2 implies 100 = 49 + h^2 implies h^2 = 51 implies h = sqrt{51}]Hmm, so the height of the cone should be ( sqrt{51} ), which is approximately 7.14. But looking at the answer choices, the cones are labeled with integer radii and heights. The options given are:A) Radius 6, Slant height 10 (height not labeled)B) Radius 6, Height 10C) Radius 7, Slant height 10D) Radius 7, Height 10E) Radius 8, Slant height 10Wait, from the Asymptote code, it seems each cone is labeled with radius, height, and slant height. For example, pic1 is labeled (6, 0, 10), meaning radius 6, height 0, and slant height 10? Wait, that can't be. If the height is 0, it's not a cone. Maybe the Asymptote code uses parameters (r, h, sh), where sh is the slant height. So if they have pic1 as (6, 0, 10), that might mean radius 6, height 0 (which doesn't make sense) and slant height 10. But probably, maybe in the images, some labels are omitted. Let me check the description again.The Asymptote code shows labels for r, h, sh. So for example, pic1 is conePic(pic1, 6, 0, 10), which would be radius 6, height 0 (which is probably not labeled, since h=0), and slant height 10. Similarly, pic2 is (6, 10, 0), so radius 6, height 10, slant height 0? That also seems wrong. Maybe the parameters are different. Let me look at the code.Looking at the Asymptote code, there's a function conePic with parameters (real r, real h, real sh). Then in the code, they label (string) r on the radius, (string) h on the height, and (string) sh on the slant height. But if h or sh is 0, perhaps they don't label them? For example, pic1 is conePic(pic1, 6, 0, 10), so labels radius 6, slant height 10, and height 0 (which is impossible). Wait, that's confusing. Maybe the labels are either radius and height or radius and slant height? Wait, looking at the code, for pic1, they have:label(pic, (string) r, center--radPt, dir(270)); which is radius. Then:if (h != 0) { label(pic, (string) h, heightPt--center, dir(0)); }if (sh != 0) { label(pic, (string) sh, heightPt--radPt, dir(0)); }So, if h is non-zero, label the height; if sh is non-zero, label the slant height. So in pic1 (6, 0, 10), h is 0, so only label the radius 6 and slant height 10. Similarly, pic2 (6, 10, 0): h=10, sh=0, so labels radius 6 and height 10. So, each picture either labels radius and slant height (if sh is non-zero) or radius and height (if h is non-zero). So the options are:A: Radius 6, Slant height 10B: Radius 6, Height 10C: Radius 7, Slant height 10D: Radius 7, Height 10E: Radius 8, Slant height 10Therefore, the possible options have either radius and slant height or radius and height. Since from our calculation, the correct cone should have radius 7 and slant height 10 (since slant height is 10, given by the original sector's radius), and height sqrt(51) ≈7.14. So among the options, option C is radius 7 and slant height 10, which matches our calculated radius and the given slant height. Option D is radius 7 and height 10, but if the radius is 7 and slant height 10, the height can't be 10. Let's check that.Wait, if someone claims the height is 10, but we know that with radius 7 and slant height 10, the height must be sqrt(10^2 -7^2) = sqrt(51) ≈7.14. Therefore, if the height were 10, then with radius 7, the slant height would be sqrt(7^2 +10^2) = sqrt(149) ≈12.21, which is not 10. So option D is invalid.Similarly, option C has radius 7 and slant height 10, which is exactly what we have. So option C must be the correct answer. Let me just verify once more.Original sector: radius 10 (becomes slant height), arc length 252/360 *2π*10 =14π, which is the circumference of the base. Therefore, base radius is 14π/(2π)=7. So radius 7, slant height 10. So yes, option C is the answer. The other options have different radii or heights, which don't align with these calculations. Therefore, the answer is C.Final Answerboxed{C}</think>To determine which cone can be formed from a (252^circ) sector of a circle with radius 10, we start by noting that the radius of the sector becomes the slant height ((l)) of the cone, which is 10. The arc length of the sector becomes the circumference of the base of the cone.The arc length (L) of the sector is calculated as:[L = frac{252}{360} times 2pi times 10 = frac{7}{10} times 20pi = 14pi]This arc length is the circumference of the base of the cone, so we solve for the radius (r) of the base:[2pi r = 14pi implies r = 7]Using the Pythagorean theorem to find the height (h) of the cone with slant height (l = 10) and radius (r = 7):[l^2 = r^2 + h^2 implies 10^2 = 7^2 + h^2 implies 100 = 49 + h^2 implies h^2 = 51 implies h = sqrt{51}]The options provided include cones with different radii and heights. The correct cone must have a radius of 7 and a slant height of 10. Among the options, option C (radius 7, slant height 10) matches these values.Thus, the correct answer is (boxed{C}).,<think>Okay, so I need to figure out which of the cones (A to E) can be formed from a 252-degree sector of a circle with radius 10. The Asymptote code draws these cones with different radii and heights, so I have to determine which one matches the cone made from the sector.First, let me recall how constructing a cone from a sector works. When you roll up a sector of a circle into a cone, the radius of the original sector becomes the slant height of the cone. The length of the arc of the sector becomes the circumference of the base of the cone. Given the sector has a radius of 10, that means the slant height (l) of the cone is 10. The angle of the sector is 252 degrees. The arc length of this sector will be the circumference of the base of the cone. Let me write that down.The arc length (L) of a sector with angle θ (in degrees) and radius r is given by:L = (θ/360) * 2πrSo plugging in θ = 252° and r = 10:L = (252/360) * 2π * 10Simplify the fraction 252/360. Let's divide numerator and denominator by 36: 252 ÷ 36 = 7, 360 ÷ 36 = 10. So 7/10. Therefore:L = (7/10) * 2π * 10The 10s cancel out:L = 7 * 2π = 14πSo the arc length is 14π. This arc length becomes the circumference of the base of the cone. The circumference of a circle is 2πr, so the radius of the base of the cone (let's call it R) can be found by:2πR = 14πDivide both sides by 2π:R = 7So the radius of the base of the cone should be 7. Then, we need to find the height (h) of the cone. Since we know the slant height (l = 10) and the radius (R = 7), we can use the Pythagorean theorem for the right triangle formed by the radius, height, and slant height:l² = R² + h²Solving for h:h² = l² - R²h² = 10² - 7²h² = 100 - 49h² = 51h = √51Now, √51 is approximately 7.141, but we can leave it as √51. However, looking at the answer choices, the cones have options with radius 6,7,8 and heights 0,10, or slant height 10. Wait, in the Asymptote code, the cones are labeled with r (radius), h (height), and sh (slant height). But looking at the code for the pictures:For example, picture pic1 is conePic(pic1, 6, 0, 10). So the parameters are r, h, sh. So in pic1, radius is 6, height is 0, slant height is 10. Wait, but if height is 0, that would be a flat circle, which doesn't make sense. Maybe the parameters are (r, h, sh), but if h is zero, then maybe sh is slant height? But the labels are r as the radius of the base, h as the height from base to tip, and sh as the slant height?Wait, let me check the code again. The code says:label(pic, (string) r, center--radPt, dir(270));So label r is from center to radPt, which is the radius of the base. Then:if (h != 0) label h from center to heightPt, which is the vertical line from center (base center) to heightPt, which is the apex. So h is the height. Then sh is from heightPt to radPt, which would be the slant height. So in the picture, the labels are r (base radius), h (height), sh (slant height). So the parameters are (r, h, sh). So for example, pic1 is (6,0,10). So base radius 6, height 0, slant height 10. But if height is 0, then slant height would equal the radius, but here slant height is 10. Wait, that seems inconsistent. If the height is 0, then the slant height is the same as the radius? Wait, but in that case, the slant height is the distance from the apex to the edge of the base. If the height is 0, the apex is at the center of the base, so the slant height would be equal to the radius. So if the slant height is 10, but the radius is 6, that's impossible. So maybe there's a mistake in the parameters. Wait, perhaps the parameters are (r, h, sh), but if h is given, then sh is not labeled, and vice versa? Wait, in the code, the function conePic takes r, h, sh. Then, if h is non-zero, it labels h, otherwise if sh is non-zero, it labels sh. Wait, let's look:In the conePic function: if (h != 0) { label(pic, (string) h, heightPt--center, dir(0)); } if (sh != 0) { label(pic, (string) sh, heightPt--radPt, dir(0)); }So, if h is non-zero, it labels the height. If sh is non-zero, it labels the slant height. So in pic1, which is conePic(pic1, 6, 0, 10), since h is 0, it doesn't label h, but labels sh (10). So in pic1, the slant height is 10, the radius is 6. Then, using Pythagorean theorem, height can be calculated. Similarly, in pic2: (6,10,0). Here, h is 10, sh is 0. Wait, but if h is 10, then the slant height can be calculated. But if sh is 0, that doesn't make sense. Wait, maybe when sh is 0, they don't label it. So in pic2, the labels are radius 6 and height 10, so we can compute slant height. But the labels are based on which parameters are non-zero.So in the options, for each cone, either the slant height is labeled (sh) or the height (h). For the cones:A: (6, 0, 10) → r=6, sh=10. So slant height is 10, radius 6. Then, height is sqrt(10² -6²)=sqrt(64)=8.B: (6,10,0) → r=6, h=10. Then, slant height is sqrt(6² +10²)=sqrt(136)=2*sqrt(34)≈11.6619, which is not 10.C: (7,0,10) → r=7, sh=10. So slant height 10, radius 7. Then height is sqrt(10² -7²)=sqrt(51)≈7.141, which is the correct answer we found earlier.D: (7,10,0) → r=7, h=10. Then slant height is sqrt(7² +10²)=sqrt(149)≈12.2066.E: (8,0,10) → r=8, sh=10. Then height is sqrt(10² -8²)=sqrt(36)=6.So the cone formed from the sector should have radius 7, slant height 10, and height sqrt(51). Looking at the options, option C is (7,0,10), which is radius 7 and slant height 10. So even though the height isn't labeled (since h is 0 in the parameters), the actual height is sqrt(51). But in the Asymptote code, since h is set to 0, they only label the slant height. So option C is the correct one.Wait, but just to confirm, when you form the cone from the sector, the slant height is the original radius of the sector, which is 10. The circumference of the base is the arc length of the sector, which we found to be 14π, leading to a base radius of 7. Then, the height is sqrt(10^2 -7^2)=sqrt(51). Therefore, the cone should have radius 7, slant height 10, and height sqrt(51). So in the options, the cone with radius 7 and slant height 10 is option C, which is labeled (7,0,10). But wait, in option C, the height is 0? But that's not the case. The parameters passed to conePic for option C are (7,0,10), meaning that h=0 and sh=10, so in the picture, they show the radius 7 and slant height 10. The height is not zero, but since h is given as 0, they don't label the height. Wait, maybe there's an error here. The code says if h !=0, label h, else if sh !=0, label sh. So in option C, h is 0, so they label sh=10. But the actual height is sqrt(51)≈7.14, which is not 0. But the problem is asking which cone can be formed from the sector. The answer is the cone with radius 7 and slant height 10. So that's option C. Even though in the Asymptote code it's given as (7,0,10), that probably means they just show radius and slant height, and the height is calculated. So the answer is C. Let me check the other options:A: r=6, sh=10. Then h=8. But we need radius 7.B: r=6, h=10. Slant height would be sqrt(6²+10²)=sqrt(136), which is not 10.C: Correct.D: r=7, h=10. Then slant height is sqrt(149), not 10.E: r=8, sh=10. h=6. Not 7 radius.So yeah, option C is the correct answer.Final Answerboxed{C}</think>To determine which cone can be formed from a (252^circ) sector of a circle with radius 10, we start by noting that the radius of the sector becomes the slant height ((l)) of the cone, which is 10. The arc length of the sector becomes the circumference of the base of the cone.First, calculate the arc length ((L)) of the (252^circ) sector:[L = left(frac{252}{360}right) times 2pi times 10 = left(frac{7}{10}right) times 20pi = 14pi]This arc length is the circumference of the base of the cone, so we solve for the radius ((R)) of the base:[2pi R = 14pi implies R = 7]Next, we use the Pythagorean theorem to find the height ((h)) of the cone, where the slant height ((l)) is 10 and the radius ((R)) is 7:[l^2 = R^2 + h^2 implies 10^2 = 7^2 + h^2 implies 100 = 49 + h^2 implies h^2 = 51 implies h = sqrt{51}]The cone formed has a radius of 7, a slant height of 10, and a height of (sqrt{51}). Among the given options, the cone with radius 7 and slant height 10 is labeled as option C. Thus, the correct answer is (boxed{C}).]

🔑:1. Determine the arc length of the sector: The sector of the circle has an angle of (252^circ). The circumference of the full circle with radius (10) is given by: [ C = 2pi times 10 = 20pi ] The arc length (L) of the sector is a fraction of the full circumference, specifically (frac{252}{360}) of it: [ L = frac{252}{360} times 20pi = frac{7}{10} times 20pi = 14pi ]2. Form the cone: When the sector is formed into a cone by aligning the two straight sides, the arc length (L) becomes the circumference of the base of the cone. Let (r) be the radius of the base of the cone. Then: [ 2pi r = 14pi ] Solving for (r): [ r = frac{14pi}{2pi} = 7 ]3. Determine the slant height: The radius of the original circle is the slant height (s) of the cone. Given that the radius of the original circle is (10), the slant height (s) of the cone is: [ s = 10 ]4. Identify the correct cone: We need to find the cone with a base radius of (7) and a slant height of (10). From the given options, we see that option (textbf{C}) has a base radius of (7) and a slant height of (10).Conclusion:[boxed{C}]

🧠:[<think>Okay, so I need to find the smallest positive integer n such that 31 divides 5^n + n. In other words, 5^n + n ≡ 0 mod 31. That means 5^n ≡ -n mod 31. My goal is to find the minimal n where this congruence holds.First, since 31 is a prime number, maybe I can use some properties from number theory related to primes. I remember Fermat's Little Theorem states that if p is a prime and a is not divisible by p, then a^(p-1) ≡ 1 mod p. Here, 5 and 31 are coprime, so 5^30 ≡ 1 mod 31. That tells me that the powers of 5 modulo 31 repeat every 30. So the order of 5 modulo 31 divides 30. Maybe I need to find the order of 5 modulo 31 first? Let's see.To find the order, I need the smallest positive integer k such that 5^k ≡ 1 mod 31. Let's check the divisors of 30 (since the order must divide φ(31) = 30). The divisors are 1,2,3,5,6,10,15,30.Check k=1: 5^1=5≠1 mod31.k=2: 5^2=25≠1 mod31.k=3:5^3=125≡125-4*31=125-124=1 mod31. Wait, 5^3=125, 31*4=124, 125-124=1. So 5^3≡1 mod31. Oh, so the order of 5 mod31 is 3. That's useful! So 5^3 ≡1 mod31, so 5^k cycles every 3 exponents. So 5^n mod31 cycles every 3. So the possible residues of 5^n mod31 are 5^0=1, 5^1=5, 5^2=25, 5^3=1, etc. So cycle is 1,5,25,1,5,25,...Therefore, 5^n mod31 is periodic with period 3, and the residue depends on n mod3. So for n≡0 mod3: 5^n≡1, n≡1 mod3:5, n≡2 mod3:25.So 5^n ≡ -n mod31. Let's write that as:Depending on n mod3:If n ≡0 mod3: 1 ≡ -n mod31 ⇒ n≡-1 mod31. But n≡0 mod3. So solve n≡-1 mod31 and n≡0 mod3.If n≡1 mod3:5≡-n mod31 ⇒n≡-5 mod31. And n≡1 mod3.If n≡2 mod3:25≡-n mod31 ⇒n≡-25 mod31. And n≡2 mod3.So we have three separate congruences to solve, depending on the case. Then, for each case, solve n ≡ a mod31 and n≡b mod3, then find the minimal positive solution.So let's handle each case:Case 1: n ≡0 mod3 and n≡-1 mod31. So n≡30 mod31 (since -1 mod31 is 30). So need to solve n≡30 mod31 and n≡0 mod3.Use Chinese Remainder Theorem. Let n=31k +30. Then 31k +30 ≡0 mod3. Compute 31 mod3=1, so 1*k +30≡k +0≡0 mod3. So k≡0 mod3. So k=3m. Hence n=31*(3m)+30=93m +30. The minimal positive n here is 30, but check if 30≡0 mod3? 30/3=10, yes. So n=30 is a solution here. Wait, but n=30, let's check if 5^30 +30 ≡0 mod31. Since 5^3≡1 mod31, so 5^30=(5^3)^10≡1^10=1 mod31. So 1 +30=31≡0 mod31. So yes, n=30 is a solution here.Case 2: n≡1 mod3 and n≡-5 mod31. -5 mod31 is 26. So n≡26 mod31 and n≡1 mod3.So set n=31k +26. Then 31k +26 ≡1 mod3. 31 mod3=1, so k +26 mod3= k + (26 mod3)=k + 2≡1 mod3. Hence k≡1 -2= -1≡2 mod3. So k=3m +2. Then n=31*(3m +2)+26=93m +62 +26=93m +62 +26=93m +88. The minimal n here is 88. Let's check if n=88: Check 5^88 +88 mod31.But 5^3≡1, so 5^88=5^(3*29 +1)= (5^3)^29 *5^1≡1^29 *5=5 mod31. Then 5 +88=93. 93 divided by31 is 3, so 93≡0 mod31. So 5^88 +88≡0 mod31. Also, n=88≡88 mod3. 88/3=29*3 +1. So 88≡1 mod3. So yes, satisfies both congruences. So 88 is another solution.Case3: n≡2 mod3 and n≡-25 mod31. -25 mod31 is 6. So n≡6 mod31 and n≡2 mod3.Set n=31k +6. Then 31k +6≡2 mod3. 31 mod3=1, so k +6 mod3=k +0≡2 mod3. So k≡2 mod3. Therefore, k=3m +2. Thus n=31*(3m +2) +6=93m +62 +6=93m +68. The minimal n here is 68. Check if n=68: 5^68 +68 mod31. 5^3≡1, so 5^68=5^(3*22 +2)= (5^3)^22 *5^2≡1^22 *25=25 mod31. Then 25 +68=93≡0 mod31. Also, 68 mod3: 68=3*22 +2, so 68≡2 mod3. Correct. So n=68 is a solution here.So now we have three possible solutions: 30,68,88. The minimal among them is 30. Wait, but before I conclude that, let me check n=30.n=30: 5^30 +30 mod31. Since 5^3≡1 mod31, 5^30=(5^3)^10≡1^10=1 mod31. So 1 +30=31≡0 mod31. So yes, that works. So 30 is a solution.But wait, why isn't 30 the answer? The problem says "positive integer n". 30 is positive. Is there a smaller n?Wait, but hold on. When I considered the first case, n≡0 mod3 and n≡30 mod31. The minimal solution is 30, but is there a smaller n? For example, maybe 30 is the minimal, but maybe some smaller n that is not covered in the cases? Wait, but the logic was: 5^n mod31 cycles every 3, so depending on n mod3, 5^n is either 1,5,25. Then in each case, we set 1≡-n, 5≡-n, 25≡-n mod31 and solve for n. So this should cover all possibilities. Therefore, the minimal n is 30. But wait, let's check if there's a smaller n.Wait, let's try n=6. Let's compute 5^6 +6 mod31.5^1=55^2=255^3=125≡15^4=55^5=255^6=1. So 5^6=1. Then 1 +6=7≠0 mod31.n=6:7≡7 mod31≠0.n=5:5^5=25. 25 +5=30≡30 mod31≠0.n=7:5^7=5^(6+1)=5^6*5=1*5=5. 5 +7=12≠0.n=3:5^3=1. 1 +3=4≠0.n=4:5^4=5. 5+4=9≠0.n=8:5^8=5^(6+2)=1*25=25. 25 +8=33≡2 mod31≠0.n=9:5^9=5^(6+3)=1*1=1. 1 +9=10≠0.n=10:5^10=5^(9+1)=1*5=5. 5 +10=15≠0.n=12:5^12=1. 1 +12=13≠0.n=15:5^15=1. 1 +15=16≠0.n=18:5^18=1. 1 +18=19≠0.n=21:5^21=1. 1 +21=22≠0.n=24:5^24=1. 1 +24=25≠0.n=27:5^27=1. 1 +27=28≠0.n=30:5^30=1. 1 +30=31≡0 mod31. So yes, n=30 works.Wait, so 30 is the minimal solution. But according to the three cases, we had solutions at 30,68,88, so 30 is the smallest. But the problem says "positive integer n", so 30 is positive. So why does the question say "the minimum value of n is"? Maybe 30 is the answer. Wait, but hold on, let me check the problem statement again: "Let the positive integer n satisfy 31 ∣(5^{n}+n). Then the minimum value of n is ." So yes, 30. But in the answer above, when the user says "put your final answer within boxed{}", but in the problem statement, in Chinese forums, sometimes similar problems have smaller solutions. Wait, maybe my analysis is wrong.Wait, let's check n=6. Wait, n=6:5^6=1, 1+6=7≠0. n=5:25+5=30≠0. n=25:5^25. Since 5^3=1, 25=3*8 +1, so 5^25=(5^3)^8 *5^1=1^8*5=5. 5 +25=30≡30 mod31≠0. Hmm. So n=30 is the first one that works.Wait, but let's check n=30. 5^30 +30≡1 +30=31≡0 mod31. So yes, correct. So 30 is a solution. Is there a smaller n?Wait, let's check n=68. 5^68=25, 25 +68=93≡0 mod31. But 68 is larger than 30, so not minimal.Similarly, n=88: 5^88=5, 5 +88=93≡0 mod31. Also larger.So 30 is the minimal. But wait, wait, but according to the initial problem, in the problem statement, it's possible that maybe 30 is not the minimal. Let me check some intermediate n between 1 and30.Wait, let's check n=25: As above, 5^25=5, 5+25=30≡30≠0.n=6:5^6=1, 1 +6=7≠0.n=9:1+9=10≠0.n=12:1 +12=13≠0.n=15:1 +15=16≠0.n=18:1 +18=19≠0.n=21:1 +21=22≠0.n=24:1 +24=25≠0.n=27:1 +27=28≠0.n=30:1 +30=31≡0.So yeah, 30 is the first one where it works.But wait, maybe there's a smaller n which is not ≡0 mod3. Let's check.For example, in case 2: n≡1 mod3, solution is n=88. But maybe there is a smaller n≡26 mod31 and ≡1 mod3?Wait, n=26: 26 mod3=26-3*8=26-24=2≡2 mod3. So 26≡2 mod3. So n=26 is not ≡1 mod3. The next number congruent to 26 mod31 is 26+31=57. 57 mod3=57/3=19, remainder 0. So 57≡0 mod3. Then 57+31=88. 88 mod3=88-3*29=88-87=1. So 88≡1 mod3. So the first solution in case2 is 88. So indeed, 88 is the first solution for case2.Similarly for case3: n=6. 6 mod3=0, so n=6≡0 mod3, but we need n≡2 mod3. Next is 6+31=37. 37 mod3=1. Then 37+31=68. 68 mod3=2. So 68 is the first solution for case3.So the three minimal solutions are 30,68,88. Therefore, the minimal n is 30.But wait, let's check n=5. 5^5=25, 25+5=30≡30 mod31≠0.n=10:5^10=5^(3*3 +1)=1^3*5=5. 5 +10=15≡15≠0.n=15:5^15=1^5=1, 1 +15=16≠0.n=20:5^20=5^(3*6 +2)=1^6*25=25. 25 +20=45≡45-31=14≠0.n=25:5^25=5^(3*8 +1)=1^8*5=5. 5 +25=30≠0.n=29:5^29=5^(3*9 +2)=1^9*25=25. 25 +29=54≡54-31=23≠0.n=30:1 +30=31≡0.So yes, n=30 is the minimal. Therefore, the answer is 30. So why does the problem seem to suggest that maybe I have to do more? Maybe I made a mistake in assuming that the cases split only on n mod3. Wait, let's think again.Wait, 5^n mod31 cycles every 3, so 5^n ≡1,5,25 depending on n≡0,1,2 mod3. So we have:If 5^n ≡1, then 1≡-n mod31 ⇒n≡-1≡30 mod31. Also, n≡0 mod3. So solutions are n≡30 mod31 and n≡0 mod3. Minimal n=30.Similarly, if 5^n≡5 mod31, then 5≡-n mod31 ⇒n≡-5≡26 mod31. Also, n≡1 mod3. Solve n≡26 mod31 and n≡1 mod3. The minimal n is 88.Third case: 5^n≡25 mod31, then 25≡-n mod31 ⇒n≡-25≡6 mod31. Also, n≡2 mod3. Minimal n=68.Therefore, 30 is the minimal. So I think 30 is correct.Wait, but maybe I need to check if 30 is actually the minimal. Let me check n=30:Compute 5^30 mod31. Since 5^3=125≡1, then 5^30=(5^3)^10≡1^10=1 mod31. Then 1 +30=31≡0 mod31. So yes, works.Therefore, the minimal n is 30. So the answer is 30. But wait, I found some solutions in each case, but maybe there are overlaps. But 30 is the smallest.Alternatively, maybe there's a mistake in the initial assumption that 5^3≡1 mod31. Let me verify.Compute 5^1=5 mod315^2=25 mod315^3=125 mod31. 31*4=124. 125-124=1. So 5^3=1 mod31. Yes, correct.Therefore, 5^3≡1. So the order is 3. Therefore, exponents cycle every 3. So yes, the analysis is correct.Thus, the minimal n is 30. Therefore, the answer is boxed{30}.Final Answerboxed{30}</think>To find the smallest positive integer ( n ) such that ( 31 mid (5^n + n) ), we need to solve the congruence ( 5^n + n equiv 0 pmod{31} ). This simplifies to ( 5^n equiv -n pmod{31} ).Using Fermat's Little Theorem, we know ( 5^{30} equiv 1 pmod{31} ). However, we find that the order of 5 modulo 31 is 3 because ( 5^3 equiv 1 pmod{31} ). This means the powers of 5 modulo 31 repeat every 3 exponents: ( 5^0 equiv 1 ), ( 5^1 equiv 5 ), ( 5^2 equiv 25 ), ( 5^3 equiv 1 ), etc.We analyze three cases based on ( n mod 3 ):1. Case 1: ( n equiv 0 pmod{3} ) - ( 5^n equiv 1 pmod{31} ) - Solve ( 1 equiv -n pmod{31} ) which gives ( n equiv 30 pmod{31} ) - Combine ( n equiv 0 pmod{3} ) and ( n equiv 30 pmod{31} ) using the Chinese Remainder Theorem: - ( n = 31k + 30 ) - ( 31k + 30 equiv 0 pmod{3} ) gives ( k equiv 0 pmod{3} ) - Minimal solution: ( k = 0 ) gives ( n = 30 )2. Case 2: ( n equiv 1 pmod{3} ) - ( 5^n equiv 5 pmod{31} ) - Solve ( 5 equiv -n pmod{31} ) which gives ( n equiv 26 pmod{31} ) - Combine ( n equiv 1 pmod{3} ) and ( n equiv 26 pmod{31} ) using the Chinese Remainder Theorem: - ( n = 31k + 26 ) - ( 31k + 26 equiv 1 pmod{3} ) gives ( k equiv 2 pmod{3} ) - Minimal solution: ( k = 2 ) gives ( n = 88 )3. Case 3: ( n equiv 2 pmod{3} ) - ( 5^n equiv 25 pmod{31} ) - Solve ( 25 equiv -n pmod{31} ) which gives ( n equiv 6 pmod{31} ) - Combine ( n equiv 2 pmod{3} ) and ( n equiv 6 pmod{31} ) using the Chinese Remainder Theorem: - ( n = 31k + 6 ) - ( 31k + 6 equiv 2 pmod{3} ) gives ( k equiv 2 pmod{3} ) - Minimal solution: ( k = 2 ) gives ( n = 68 )The minimal solutions from each case are 30, 68, and 88. The smallest value is ( n = 30 ).Verification:- ( 5^{30} equiv 1 pmod{31} ) and ( 1 + 30 = 31 equiv 0 pmod{31} )Thus, the minimum value of ( n ) is (boxed{30}).,<think>Okay, so I need to find the smallest positive integer n such that 31 divides 5^n + n. That means 5^n + n ≡ 0 mod 31. Hmm, modular arithmetic. Let me break this down step by step.First, let's restate the problem: Find the minimal n > 0 where 5^n ≡ -n mod 31. So 5^n ≡ (-n) mod 31. So, equivalently, 5^n + n ≡ 0 mod 31. Got it.I think I need to compute 5^n modulo 31 for different values of n and see when it is congruent to -n modulo 31. Since 31 is a prime number, maybe Fermat's little theorem applies here. Fermat's little theorem says that if p is prime and a not divisible by p, then a^(p-1) ≡ 1 mod p. So here, 5^30 ≡ 1 mod 31. So the order of 5 modulo 31 divides 30. Let me check the order. Maybe the order is less than 30? If so, that could help me find a cycle for 5^n mod 31.Let's find the order of 5 modulo 31. The order is the smallest k such that 5^k ≡ 1 mod 31. Let's check divisors of 30: 1, 2, 3, 5, 6, 10, 15, 30.Check k=1: 5^1 = 5 ≡ 5 ≠1 mod31.k=2: 5^2=25≡25≠1 mod31.k=3:5^3=125. 125 divided by 31: 31*4=124, so 125≡1 mod31. Wait, 5^3 ≡1 mod31? Wait, 5^3=125, 125-124=1. Yes, 5^3≡1 mod31. So the order of 5 modulo31 is 3. That's interesting. So 5^3 ≡1, so 5^(3k) ≡1 mod31.Therefore, 5^n ≡5^(n mod3) mod31. Because 5^3≡1, so 5^3k + r ≡ (5^3)^k *5^r ≡1^k *5^r ≡5^r mod31. So the cycle repeats every 3 exponents. So the possible residues for 5^n mod31 are 5^0=1, 5^1=5, 5^2=25, then 5^3=1 again, so cycle is 1,5,25,1,5,25,...So 5^n mod31 cycles every 3 terms: 1,5,25. So depending on n mod3, 5^n is either 1,5,25.Therefore, 5^n ≡ (1 if n≡0 mod3, 5 if n≡1 mod3, 25 if n≡2 mod3) mod31.So 5^n ≡ -n mod31 implies that:Case1: If n ≡0 mod3, then 1 ≡ -n mod31 → n ≡ -1 mod31. But also n ≡0 mod3. So we need n ≡30 mod31 (since -1 mod31 is 30) and n ≡0 mod3. So solve n ≡30 mod31 and n ≡0 mod3. So n =31k +30. Then 31k +30 ≡(31k mod3)+(30 mod3)≡(1k mod3)+(0 mod3)=k mod3. So 31k +30 ≡k mod3. So set k ≡0 mod3. Therefore, the smallest k=0: n=30, but check if n is positive integer. If k=0, n=30. But 30 is positive. Wait, but 31 divides 5^30 +30? Let me check that. But 5^3=125≡1 mod31, so 5^30=(5^3)^10≡1^10=1 mod31. So 5^30 +30≡1 +30=31≡0 mod31. So yes, n=30 works. But maybe there is a smaller n. Wait, maybe k=3? Wait, if k=0 gives n=30, which is already less than 31. So n=30 is possible. But let's see other cases.Case2: If n≡1 mod3, then 5≡-n mod31 → n≡-5 mod31. So n≡26 mod31 (since -5 mod31 is 26). So n=31k +26. Then n must ≡1 mod3. So 31k +26 ≡(31k mod3) + (26 mod3)= (1k mod3)+(2 mod3)=k +2 mod3≡1 mod3. So k +2 ≡1 mod3 →k≡-1≡2 mod3. So k=3m +2. Therefore, n=31*(3m +2) +26=93m +62 +26=93m +88. The smallest positive n here is 88 when m=0. So n=88. But 88 is bigger than 30, so n=30 is smaller.Case3: If n≡2 mod3, then 25≡-n mod31 →n≡-25 mod31≡6 mod31. So n=31k +6. Then n must ≡2 mod3. So 31k +6≡(31k mod3)+(6 mod3)= (1k mod3) +0≡k mod3≡2 mod3. Therefore, k≡2 mod3. So k=3m +2. Therefore, n=31*(3m +2)+6=93m +62 +6=93m +68. The smallest positive n here is 68 when m=0. So n=68. But 68 is larger than 30, so 30 is still smaller.So according to this, the minimal n is 30. But wait, the problem says "positive integer n", and 30 is positive. So why is the answer 30? But the question is from a competition or something, and usually, such answers are smaller. Maybe I made a mistake here. Let me check if n=30 is the minimal one.Wait, let's check n=5. Maybe there is a smaller n. Let me check n=5: 5^5 +5=3125 +5=3130. 3130/31=101. So 31*101=3131. Wait, 3130-3131=-1. So 3130≡-1 mod31. So 5^5 +5≡-1 mod31, not 0. So not valid.Check n=10: 5^10. Let's compute 5^10 mod31. Since 5^3≡1 mod31, 5^3=1, so 5^10=5^(3*3 +1)= (5^3)^3 *5^1=1^3 *5=5 mod31. So 5^10≡5 mod31. Then 5^10 +10≡5 +10=15≡15 mod31≠0.n=15: 5^15= (5^3)^5≡1^5=1 mod31. So 5^15 +15≡1 +15=16≡16 mod31≠0.n=20: 5^20= (5^3)^6 *5^2=1^6 *25=25 mod31. So 25 +20=45≡45-31=14≡14 mod31≠0.n=25: 5^25= (5^3)^8 *5^1=1^8*5=5 mod31. 5 +25=30≡30 mod31≠0.n=30: 5^30=(5^3)^10≡1^10=1 mod31. 1 +30=31≡0 mod31. So yes, n=30 works. So according to this, n=30 is the minimal. But is there a smaller n?Wait, let's check n=6. Let's compute 5^6. Since 5^3=1, so 5^6=(5^3)^2=1^2=1 mod31. So 5^6 +6≡1 +6=7 mod31≠0.n=9: 5^9=(5^3)^3=1^3=1 mod31. 1 +9=10≡10 mod31≠0.n=12:5^12=(5^3)^4=1^4=1. 1 +12=13≠0.n=18:5^18=(5^3)^6=1^6=1. 1 +18=19≠0.n=21:5^21=(5^3)^7=1^7=1. 1 +21=22≠0.n=24:5^24=(5^3)^8=1^8=1. 1 +24=25≠0.n=27:5^27=(5^3)^9=1^9=1. 1 +27=28≠0.n=28: Let's check n=28. 5^28=5^(3*9 +1)=1^9*5=5. So 5 +28=33≡2 mod31≠0.n=29:5^29=5^(3*9 +2)=1^9*25=25. 25 +29=54≡54-31=23≠0.n=30:5^30=1. 1 +30=31≡0 mod31. So n=30 is the first one. Wait, but maybe I missed some n in between.Wait, maybe n=25? Wait, n=25:5^25=5^(3*8 +1)=1^8*5=5. 5 +25=30≡30 mod31. Not 0.n=26:5^26=5^(3*8 +2)=1^8*25=25. 25 +26=51≡51-31=20≡20 mod31≠0.n=27:5^27=1. 1 +27=28≡28≠0.n=28:5^28=5. 5 +28=33≡2 mod31.n=29:25 +29=54≡23 mod31.n=30:1 +30=31≡0. So yes, seems like 30 is the minimal.But wait, 30 is kind of a large number, maybe there's a smaller n. Let me check n=4: 5^4=625. 625 mod31. 31*20=620, so 625-620=5. So 5^4≡5 mod31. Then 5 +4=9≡9 mod31≠0.n=7:5^7=5^(3*2 +1)=1^2*5=5. 5 +7=12≡12≠0.n=8:5^8=5^3*5^5=1*5^5. 5^5=5^3*5^2=1*25=25. So 5^8=25. 25 +8=33≡2≠0.n=10: 5^10=5^(3*3 +1)=1*5=5. 5 +10=15≡15≠0.n=11:5^11=5^(3*3 +2)=1*25=25. 25 +11=36≡5≡5≠0.n=12:5^12=1. 1 +12=13≡13≠0.n=13:5^13=5. 5 +13=18≡18≠0.n=14:5^14=25. 25 +14=39≡8≡8≠0.n=15:1 +15=16≡16≠0.n=16:5 +16=21≡21≠0.n=17:25 +17=42≡11≡11≠0.n=18:1 +18=19≡19≠0.n=19:5 +19=24≡24≠0.n=20:25 +20=45≡14≡14≠0.n=21:1 +21=22≡22≠0.n=22:5 +22=27≡27≠0.n=23:25 +23=48≡17≡17≠0.n=24:1 +24=25≡25≠0.n=25:5 +25=30≡30≠0.n=26:25 +26=51≡20≡20≠0.n=27:1 +27=28≡28≠0.n=28:5 +28=33≡2≡2≠0.n=29:25 +29=54≡23≡23≠0.n=30:1 +30=31≡0. Yes, so n=30 is the first one that works.But wait, I thought maybe there's a smaller n. Let me check n=30. Maybe the answer is 30.But I just want to confirm. Let me check 5^30 +30.5^30 mod31 is 1, as per Fermat's little theorem. 1 +30=31≡0 mod31. So yes, it works.But since the order of 5 modulo31 is 3, then 5^n cycles every 3. So the residues for 5^n mod31 are 1,5,25. So 5^n +n ≡0 mod31 means that:If 5^n ≡1 mod31, then 1 +n ≡0 mod31 →n≡30 mod31. But also n must be ≡0 mod3 (since 5^n ≡1 when n≡0 mod3). So n must ≡30 mod31 and ≡0 mod3. The minimal such n is 30, as 30≡0 mod3 (30 divided by3 is10) and 30≡30 mod31.Similarly, if 5^n≡5 mod31, then 5 +n≡0 mod31→n≡26 mod31. And n≡1 mod3. So solve n≡26 mod31 and n≡1 mod3. The minimal n here is26 +31k where 26 +31k≡1 mod3. 26≡2 mod3, 31≡1 mod3. So 2 +k ≡1 mod3→k≡-1≡2 mod3. So k=2 gives n=26 +62=88. Which is higher than30.If 5^n≡25 mod31, then25 +n≡0 mod31→n≡6 mod31. And n≡2 mod3. So n≡6 mod31 and n≡2 mod3. 6≡0 mod3, so need 0≡2 mod3? Not possible. Wait, but n≡6 mod31, which is≡0 mod3, and we need n≡2 mod3. Contradiction. So no solution in this case. Hence, only possible solutions come from first two cases, where minimal n is30.Thus, the minimal positive integer n is30. But wait, the problem says "positive integer n". So 30 is a positive integer, so that's acceptable.Wait, but maybe there's a mistake here. Let me check for n=5: 5^5 +5. 5^5=3125. 3125 +5=3130. 3130 divided by31: 31*100=3100, 3130-3100=30, so 3130=31*100 +30. So 3130≡30 mod31≡-1 mod31. Not 0. So n=5 doesn't work.n=6:5^6=15625. 15625 mod31. Since 5^3=1, 5^6=(5^3)^2=1^2=1. 1 +6=7≡7 mod31≠0.n=7:5^7=5^6*5=1*5=5. 5 +7=12≡12≠0.n=8:5^8=5^7*5=5*5=25. 25 +8=33≡2≡2≠0.n=9:5^9=5^8*5=25*5=125≡1 mod31. 1 +9=10≡10≠0.n=10:5^10=5^9*5=1*5=5. 5 +10=15≡15≠0.Continuing up to n=30, as before, we see that none of the numbers before n=30 satisfy the condition. Therefore, the minimal n is indeed30.But just to make sure, let me check n=30 again:5^30 mod31: Since the order is3, 30 is a multiple of3 (30=3*10), so 5^30=(5^3)^10=1^10=1 mod31. Then 1 +30=31≡0 mod31. So yes, correct.Therefore, the minimal n is30. But wait, wait. In the initial analysis, the case1 gives n=30, which is the smallest. But maybe there's a different approach where n is smaller.Alternatively, perhaps using the equation 5^n ≡ -n mod31.We can try to solve for n such that 5^n +n ≡0 mod31.Since 5 and31 are coprime, maybe we can write this as n ≡-5^n mod31.So n ≡ -5^n mod31.So we need to find n where n ≡-5^n mod31. Let's tabulate n from1 upwards, compute 5^n mod31, then check if n ≡-5^n mod31.Let me make a table:n | 5^n mod31 | -5^n mod31 | Check if n ≡-5^n mod31---|---------|----------|---------1 |5 |-5≡26 |1≡26? No2 |25 |-25≡6 |2≡6? No3 |1 |-1≡30 |3≡30? No4 |5 |-5≡26 |4≡26? No5 |25 |-25≡6 |5≡6? No6 |1 |-1≡30 |6≡30? No7 |5 |-5≡26 |7≡26? No8 |25 |-25≡6 |8≡6? No9 |1 |-1≡30 |9≡30? No10|5 |-5≡26 |10≡26? No11|25 |-25≡6 |11≡6? No12|1 |-1≡30 |12≡30? No13|5 |-5≡26 |13≡26? No14|25 |-25≡6 |14≡6? No15|1 |-1≡30 |15≡30? No16|5 |-5≡26 |16≡26? No17|25 |-25≡6 |17≡6? No18|1 |-1≡30 |18≡30? No19|5 |-5≡26 |19≡26? No20|25 |-25≡6 |20≡6? No21|1 |-1≡30 |21≡30? No22|5 |-5≡26 |22≡26? No23|25 |-25≡6 |23≡6? No24|1 |-1≡30 |24≡30? No25|5 |-5≡26 |25≡26? No26|25 |-25≡6 |26≡6? Yes! Wait, n=26: check if 26≡6 mod31? No, 26 is not ≡6. Wait, the third column is -5^n mod31. For n=26, 5^26 mod31. Since 5^3=1, 26=3*8 +2, so 5^26=5^(3*8 +2)= (5^3)^8 *5^2=1^8*25=25. So -25≡6 mod31. Then check if n=26≡6 mod31? 26≡26 mod31, 6≡6. 26≠6 mod31. So no.Wait, maybe I made a mistake here.Wait for n=26, the second column is 5^26 mod31=25. Then third column is -25 mod31=6. So check if n=26≡6 mod31? 26≡26, not 6. So no. So n=26 does not work.Wait, let's check n=6. 5^6=1. -1 mod31=30. So check if 6≡30 mod31? 6≠30. So no.Wait, maybe there's a mistake in the table.Let me go step by step:n=1:5^1=5. -5 mod31=26. Check if 1≡26 mod31? No.n=2:5^2=25. -25 mod31=6. Check if 2≡6 mod31? No.n=3:5^3=125≡1 mod31. -1 mod31=30. Check if3≡30 mod31? No.n=4:5^4=5^3*5=1*5=5. -5 mod31=26. Check if4≡26 mod31? No.n=5:5^5=5^4*5=5*5=25. -25 mod31=6. Check if5≡6 mod31? No.n=6:5^6=5^5*5=25*5=125≡1 mod31. -1 mod31=30. Check if6≡30 mod31? No.n=7:5^7=5^6*5=1*5=5. -5=26. Check7≡26? No.n=8:5^8=25. -25=6. Check8≡6? No.n=9:1. -1=30. Check9≡30? No.n=10:5. -5=26. Check10≡26? No.n=11:25. -25=6. Check11≡6? No.n=12:1. -1=30. Check12≡30? No.n=13:5. -5=26. Check13≡26? No.n=14:25. -25=6. Check14≡6? No.n=15:1. -1=30. Check15≡30? No.n=16:5. -5=26. Check16≡26? No.n=17:25. -25=6. Check17≡6? No.n=18:1. -1=30. Check18≡30? No.n=19:5. -5=26. Check19≡26? No.n=20:25. -25=6. Check20≡6? No.n=21:1. -1=30. Check21≡30? No.n=22:5. -5=26. Check22≡26? No.n=23:25. -25=6. Check23≡6? No.n=24:1. -1=30. Check24≡30? No.n=25:5. -5=26. Check25≡26? No.n=26:25. -25=6. Check26≡6? No.n=27:1. -1=30. Check27≡30? No.n=28:5. -5=26. Check28≡26? 28-26=2≡2 mod31. No.n=29:25. -25=6. Check29≡6? No.n=30:1. -1=30. Check30≡30 mod31. Yes! Finally, n=30≡30 mod31. So yes, n=30 satisfies the condition.Therefore, even by brute-forcing, n=30 is the minimal solution. Hence, the answer should be 30.Final AnswerThe minimum value of ( n ) is boxed{30}.</think>To find the smallest positive integer ( n ) such that ( 31 mid (5^n + n) ), we need ( 5^n + n equiv 0 pmod{31} ). This translates to finding ( n ) such that ( 5^n equiv -n pmod{31} ).Using Fermat's Little Theorem, we know ( 5^{30} equiv 1 pmod{31} ). However, we found that the order of 5 modulo 31 is 3, meaning ( 5^3 equiv 1 pmod{31} ). Therefore, ( 5^n ) cycles every 3 terms: ( 5^0 equiv 1 ), ( 5^1 equiv 5 ), ( 5^2 equiv 25 ), and then repeats.We analyze the cases based on ( n mod 3 ):1. Case 1: ( n equiv 0 pmod{3} ) - ( 5^n equiv 1 pmod{31} ) - We need ( 1 equiv -n pmod{31} ) ⟹ ( n equiv -1 pmod{31} ) - Solving ( n equiv 0 pmod{3} ) and ( n equiv 30 pmod{31} ), the smallest solution is ( n = 30 ).2. Case 2: ( n equiv 1 pmod{3} ) - ( 5^n equiv 5 pmod{31} ) - We need ( 5 equiv -n pmod{31} ) ⟹ ( n equiv -5 pmod{31} ) - Solving ( n equiv 1 pmod{3} ) and ( n equiv 26 pmod{31} ), the smallest solution is ( n = 88 ).3. Case 3: ( n equiv 2 pmod{3} ) - ( 5^n equiv 25 pmod{31} ) - We need ( 25 equiv -n pmod{31} ) ⟹ ( n equiv -25 pmod{31} ) - This case has no solution because ( n equiv 2 pmod{3} ) and ( n equiv 6 pmod{31} ) are contradictory.Checking the smallest solution from Case 1, ( n = 30 ) works as ( 5^{30} equiv 1 pmod{31} ) and ( 1 + 30 equiv 0 pmod{31} ).Thus, the minimum value of ( n ) is (boxed{30}).,<think>Okay, so I need to find the smallest positive integer n such that 31 divides 5^n + n. In other words, 5^n + n ≡ 0 mod 31. Hmm, let's break this down.First, since 31 is a prime number, maybe Fermat's Little Theorem could help here. Fermat's Little Theorem says that if p is prime and a is not divisible by p, then a^(p-1) ≡ 1 mod p. Here, p is 31, so 5^30 ≡ 1 mod 31. That means the powers of 5 modulo 31 repeat every 30. So the exponent cycles with period 30. Maybe n is related to this cycle?But n is also part of the equation 5^n + n ≡ 0 mod 31. So we need to solve 5^n ≡ -n mod 31. Since both 5^n and n are involved, this might not be straightforward. Let me think.Perhaps I can check small values of n first to see if any satisfy the condition. Let's start from n = 1 and go upwards until I find one that works.n = 1: 5^1 + 1 = 5 + 1 = 6 ≡ 6 mod 31 ≠ 0n = 2: 25 + 2 = 27 ≡ 27 mod 31 ≠ 0n = 3: 125 + 3 = 128. 128 ÷ 31 is 4*31=124, so 128 - 124 = 4 ≡ 4 mod 31 ≠ 0n = 4: 5^4 = 625. 625 ÷ 31 is 20*31=620, so 625 - 620 = 5. Then 5 + 4 = 9 ≡ 9 mod 31 ≠ 0n = 5: 5^5 = 3125. 3125 ÷ 31: 31*100 = 3100, so 3125 - 3100 = 25. 25 + 5 = 30 ≡ 30 mod 31 ≠ 0n = 6: 5^6 = 15625. Let's divide by 31. 31*500 = 15500, 15625 - 15500 = 125. Then 125 ÷ 31 is 4*31=124, so 125 - 124 = 1. So 5^6 ≡ 1 mod 31. Then 1 + 6 = 7 ≡ 7 mod 31 ≠ 0n = 7: 5^7 = 5^6 * 5 ≡ 1*5 = 5 mod 31. So 5 + 7 = 12 ≡ 12 mod 31 ≠ 0n = 8: 5^8 = 5^7 *5 ≡ 5*5=25 mod 31. 25 + 8 = 33 ≡ 2 mod 31 ≠ 0n = 9: 5^9 = 5^8 *5 ≡25*5=125≡125-4*31=125-124=1 mod 31. 1 + 9 = 10 ≡10 mod 31 ≠0n = 10: 5^10 ≡5^9 *5≡1*5=5 mod 31. 5 +10=15≡15 mod31≠0n=11:5^11≡5*5=25 mod31. 25 +11=36≡5 mod31≠0n=12:5^12≡25*5=125≡1 mod31. 1 +12=13≡13≠0n=13:5^13≡1*5=5 mod31. 5 +13=18≡18≠0n=14:5^14≡5*5=25. 25 +14=39≡8 mod31≠0n=15:5^15≡25*5=125≡1. 1 +15=16≡16≠0n=16:5^16≡1*5=5. 5 +16=21≡21≠0n=17:5^17≡5*5=25. 25 +17=42≡11 mod31≠0n=18:5^18≡25*5=125≡1. 1 +18=19≡19≠0n=19:5^19≡1*5=5. 5 +19=24≡24≠0n=20:5^20≡5*5=25. 25 +20=45≡14 mod31≠0n=21:5^21≡25*5=125≡1. 1 +21=22≡22≠0n=22:5^22≡1*5=5. 5 +22=27≡27≠0n=23:5^23≡5*5=25. 25 +23=48≡17 mod31≠0n=24:5^24≡25*5=125≡1. 1 +24=25≡25≠0n=25:5^25≡1*5=5. 5 +25=30≡30 mod31. Close, but 30 ≡ -1 mod31. So 30 ≡ -1 mod31. So 5^25 +25≡-1 mod31. Not 0. So n=25 gives -1 mod31. Hmm.n=26:5^26≡5*5=25. 25 +26=51. 51-31=20≡20 mod31≠0n=27:5^27≡25*5=125≡1. 1 +27=28≡28≠0n=28:5^28≡1*5=5. 5 +28=33≡2 mod31≠0n=29:5^29≡5*5=25. 25 +29=54≡54-31=23≡23≠0n=30:5^30≡1 mod31. 1 +30=31≡0 mod31. Wait, n=30: 5^30 +30 ≡1 +30=31≡0 mod31. So n=30 works. But is there a smaller n?Wait, I went all the way up to n=30 and found that it works. But maybe there's a smaller n? Let me check again.Wait, when n=25, 5^25 +25≡5 +25=30≡-1 mod31. Hmm. Then n=25 is close but not zero.Similarly, n=5: 5^5 +5=3125 +5=3130. 3130 divided by 31: 31*100=3100, 3130-3100=30. So 30≡-1 mod31. Hmm, same as n=25.Wait, n=5 and n=25 both give 5^n +n ≡-1 mod31. Interesting. Then n=30 works. Maybe 30 is the minimal?But let me check my calculations again. When I computed n=6: 5^6=15625. 15625 divided by 31. Let me do that step again. 31*500=15500. 15625-15500=125. Then 125 divided by 31: 31*4=124. 125-124=1. So 5^6≡1 mod31. Then 1 +6=7 mod31. Correct.Similarly, n=12: 5^12. Since 5^6≡1, then 5^12=(5^6)^2≡1^2=1 mod31. So 1 +12=13. Correct.n=18: same logic, 5^18=(5^6)^3≡1^3=1. 1 +18=19. Correct.n=24: same as above. 5^24=(5^6)^4≡1. 1 +24=25.n=30:5^30≡1 by Fermat's Little Theorem. 1 +30=31≡0 mod31. So yes, n=30 works.But is there a smaller n?Wait, in my first pass from n=1 to n=30, n=30 was the first one that worked. So maybe 30 is the minimal. But perhaps there's a mistake here. Let me double-check.Alternatively, perhaps I can approach this problem more systematically.We need 5^n ≡ -n mod31.Let me denote 5^n mod31 as a function of n. Since 5 has order 5 modulo31? Wait, no. Wait, the multiplicative order of 5 modulo31.The multiplicative order is the smallest k>0 such that 5^k ≡1 mod31.Since 31 is prime, the multiplicative order of 5 must divide 30 (since 5^30≡1 mod31 by Fermat). Let's find the order.Compute 5^k mod31:5^1=55^2=255^3=125≡1 mod31 (since 125-4*31=125-124=1). Wait, 5^3≡1 mod31? Then the order of 5 modulo31 is 3? Wait, that can't be. Wait, 5^3=125. 125 divided by31: 31*4=124, 125-124=1. So yes, 5^3≡1 mod31. So the order is 3. That's surprising. Wait, then 5 has order 3 modulo31. Then the powers of 5 cycle every 3. So 5^1=5, 5^2=25, 5^3=1, 5^4=5, etc. So the cycle is 5,25,1,5,25,1,...Wait, but that would mean 5^3≡1, 5^6≡1, etc. But earlier when I calculated 5^6, I found 5^6≡1 mod31, which matches. Similarly, 5^9≡1, etc. But in that case, the cycle length is 3, not 30. So 5 has order 3 modulo31. That seems correct based on the calculation. So 5^3≡1 mod31.Therefore, the multiplicative order of 5 modulo31 is 3. Therefore, 5^n mod31 cycles with period 3. Therefore, 5^n mod31 is:n ≡1 mod3: 5n≡2 mod3:25n≡0 mod3:1Therefore, we can write 5^n mod31 as:If n ≡1 mod3, then 5If n≡2 mod3, then25If n≡0 mod3, then1Therefore, 5^n ≡ -n mod31. So, depending on n mod3, we can write equations.Case1: n ≡1 mod3. Then 5^n≡5 mod31. So equation is 5 ≡ -n mod31 ⇒ n≡-5 mod31. So n≡26 mod31. But n≡1 mod3. So we need to solve n≡26 mod31 and n≡1 mod3. Find the smallest n that satisfies both.Case2: n≡2 mod3. Then 5^n≡25 mod31. Equation:25≡-n mod31 ⇒n≡-25 mod31⇒n≡6 mod31. Also n≡2 mod3. Solve n≡6 mod31 and n≡2 mod3.Case3: n≡0 mod3. Then5^n≡1 mod31. Equation:1≡-n mod31 ⇒n≡-1 mod31⇒n≡30 mod31. And n≡0 mod3. So solve n≡30 mod31 and n≡0 mod3. Since 30≡0 mod3, so n≡30 mod31 is already congruent to 0 mod3, so solutions are n≡30 mod93? Wait, no. LCM(31,3)=93. Wait, 31 and3 are coprime. So the solutions are n≡30 mod31 and n≡0 mod3. So n≡30 mod31, and since 30≡0 mod3, then n=30 +31k. But 30 +31k ≡0 +k*1 mod3 (since 31≡1 mod3). So 30 +31k≡k mod3. We need k≡0 mod3. So k=3m. So n=30 +31*3m=30 +93m. So the minimal n in this case is30.So in each case, the solutions are:Case1: n≡26 mod31 and n≡1 mod3. Find minimal n.Case2: n≡6 mod31 and n≡2 mod3. Find minimal n.Case3: n≡30 mod31 and n≡0 mod3. Minimal n=30.So now, need to check if in cases1 and2, there exists a smaller n than30.Let me check Case2 first: n≡6 mod31 and n≡2 mod3. Find the smallest n.So n=6 +31k. Let's check when 6 +31k ≡2 mod3. 31≡1 mod3, so31k≡k mod3. So6 +k≡2 mod3. 6≡0 mod3, so k≡2 mod3. So k=2 +3m. Then n=6 +31*(2 +3m)=6 +62 +93m=68 +93m. So minimal n=68. Which is bigger than30, so not smaller.Case1: n≡26 mod31 and n≡1 mod3. So n=26 +31k. Check congruence mod3. 31≡1 mod3, so31k≡k mod3. 26≡26-24=2 mod3. So 26 +k≡2 +k≡1 mod3. So k≡-1 mod3⇒k≡2 mod3. So k=2 +3m. Then n=26 +31*(2 +3m)=26 +62 +93m=88 +93m. So minimal n=88. Also bigger than30.Therefore, the minimal solution is30. So the answer is30. But wait, in the earlier check when n=30, it works. However, I might have made a mistake in the initial check up to n=25. Let me confirm.Wait, when I checked n=30, 5^30≡1 mod31 (by FLT), and 1 +30=31≡0 mod31. So yes, n=30 works. And according to the case analysis, it's the minimal. But in the initial check from n=1 to n=30, it was the first one. So maybe 30 is indeed the minimal. Wait, but let's check if n=5 works. Wait, n=5: 5^5=3125. 3125 mod31. Let's compute 31*100=3100, 3125-3100=25. So 5^5≡25 mod31. Then 25 +5=30≡-1 mod31. Not zero. So n=5 doesn't work. Similarly, n=25:5^25. Since 5^3≡1, 5^25=5^(3*8 +1)= (5^3)^8 *5^1≡1^8 *5=5 mod31. So 5 +25=30≡-1 mod31. Not zero. So n=25 also doesn't work. So yeah, n=30 is the first one that works.Therefore, the minimal n is30. Wait, but maybe there's an error in the case analysis. Let me check another approach.Alternatively, suppose that we model the equation 5^n ≡ -n mod31. Let's try to express n in terms of the exponent. Since 5 has order3, as established, then 5^n cycles every3. So n can be expressed as:If n ≡1 mod3, then 5^n≡5 mod31, so 5 ≡ -n mod31 ⇒n≡-5≡26 mod31.But n also needs to be≡1 mod3. So find n≡26 mod31 and≡1 mod3. As before, solve n=26 +31k. 26≡2 mod3, 31k≡k mod3. So 2 +k ≡1 mod3⇒k≡-1≡2 mod3. So k=2 gives n=26 +62=88.Similarly, for n≡2 mod3, 5^n≡25 mod31⇒25≡-n mod31⇒n≡6 mod31. And n≡2 mod3. So n=6 +31k. 6≡0 mod3, 31k≡k mod3⇒0 +k≡2 mod3⇒k≡2 mod3. So k=2 gives n=6 +62=68.For n≡0 mod3, 5^n≡1 mod31⇒1≡-n mod31⇒n≡30 mod31. Since30≡0 mod3, that's n=30 +31k. The minimal n=30.So yes, case3 gives n=30, which is the smallest.Therefore, the minimal positive integer n is30. But wait, when I checked up to n=30 manually, n=30 was the first one that worked, so that's correct. Therefore, the answer is30.But wait, is there a possibility that for some n <30, even though it doesn't fit into the cases above, maybe 5^n and n combine in such a way that their sum is divisible by31? For example, maybe n=15: 5^15≡1, 1 +15=16≡16 mod31≠0. n=10:5 +10=15≠0. Hmm, no. So seems like n=30 is indeed the minimal. Therefore, the answer is30.Final Answerboxed{30}</think>To find the smallest positive integer ( n ) such that ( 31 mid (5^n + n) ), we need to solve ( 5^n + n equiv 0 pmod{31} ). This translates to finding ( n ) such that ( 5^n equiv -n pmod{31} ).Using Fermat's Little Theorem, we know ( 5^{30} equiv 1 pmod{31} ). However, further analysis shows that the order of 5 modulo 31 is 3, since ( 5^3 equiv 1 pmod{31} ). This means the powers of 5 modulo 31 repeat every 3 terms: ( 5, 25, 1, 5, 25, 1, ldots ).We consider three cases based on ( n mod 3 ):1. Case 1: ( n equiv 1 pmod{3} ) - ( 5^n equiv 5 pmod{31} ) - Solve ( 5 equiv -n pmod{31} ) ⟹ ( n equiv -5 equiv 26 pmod{31} ) - Combine ( n equiv 26 pmod{31} ) and ( n equiv 1 pmod{3} ): The smallest solution is ( n = 88 ).2. Case 2: ( n equiv 2 pmod{3} ) - ( 5^n equiv 25 pmod{31} ) - Solve ( 25 equiv -n pmod{31} ) ⟹ ( n equiv -25 equiv 6 pmod{31} ) - Combine ( n equiv 6 pmod{31} ) and ( n equiv 2 pmod{3} ): The smallest solution is ( n = 68 ).3. Case 3: ( n equiv 0 pmod{3} ) - ( 5^n equiv 1 pmod{31} ) - Solve ( 1 equiv -n pmod{31} ) ⟹ ( n equiv -1 equiv 30 pmod{31} ) - Combine ( n equiv 30 pmod{31} ) and ( n equiv 0 pmod{3} ): The smallest solution is ( n = 30 ).Checking manually for ( n ) from 1 to 30, we find that ( n = 30 ) is the smallest value satisfying ( 5^{30} + 30 equiv 0 pmod{31} ).Thus, the minimum value of ( n ) is (boxed{30}).,<think>Okay, so I need to find the smallest positive integer n such that 31 divides 5^n + n. In other words, 5^n + n ≡ 0 mod 31. That means 5^n ≡ -n mod 31. Hmm, modular arithmetic again. Alright, let's break this down step by step.First, since we're dealing with modulus 31, which is a prime number, maybe Fermat's Little Theorem could be useful here. Fermat's Little Theorem says that for any integer a not divisible by a prime p, a^(p-1) ≡ 1 mod p. Here, p is 31, so 5^30 ≡ 1 mod 31. That might help in reducing the exponent modulo 30? Because exponents cycle every φ(31) = 30. So, 5^n mod 31 will have a period that divides 30. Let me note that.So, 5^n mod 31 cycles every 30. Therefore, 5^n ≡ 5^(n mod 30) mod 31. So, if I can write n as 30k + r, where r is the remainder when n is divided by 30, then 5^n ≡ 5^r mod 31. But how does that help me? Because n itself is part of the equation. The equation is 5^n ≡ -n mod 31. So, even if I express n as 30k + r, then we have 5^r ≡ -(30k + r) mod 31. Hmm, not sure if that helps immediately. Maybe another approach.Alternatively, perhaps I can compute 5^n mod 31 for n = 1, 2, ..., up to some point and see when 5^n ≡ -n mod 31. Since 31 isn't too large, maybe the answer is within manageable numbers.Let me try computing 5^n mod 31 for n from 1 upwards and see what happens.Compute 5^1 mod 31 = 55^2 = 25 mod 31 = 255^3 = 5*25 = 125 mod 31. 31*4=124, so 125 -124=1. So 5^3 ≡1 mod 31.Wait, 5^3 ≡1 mod31? That's interesting. Then 5^3 ≡1, so the order of 5 modulo 31 is 3? Because 5^3 ≡1, but 5^1 and 5^2 aren't 1. So that would mean the multiplicative order of 5 modulo 31 is 3. Let me check that again.5^1 =5 mod315^2=25 mod315^3=125 mod31=125-4*31=125-124=1. Yep, 1. So indeed, the order is 3. Therefore, 5^n mod31 cycles every 3. So 5^n ≡5^(n mod3) mod31. Therefore, the cycle is 3. Therefore, 5^n is periodic modulo31 with period 3. So 5^1=5, 5^2=25, 5^3=1, 5^4=5, 5^5=25, 5^6=1, etc.Therefore, for n ≡1 mod3, 5^n≡5 mod31For n≡2 mod3, 5^n≡25 mod31For n≡0 mod3, 5^n≡1 mod31Therefore, we can write 5^n ≡ [5,25,1] depending on n mod3.So, the equation 5^n +n ≡0 mod31 can be broken down into three cases, based on n mod3.Case1: n ≡1 mod3Then 5^n ≡5 mod31. So 5 +n ≡0 mod31 => n≡-5 mod31. But n also ≡1 mod3. So solve n≡-5 mod31 and n≡1 mod3.Find the smallest positive integer n such that n≡26 mod31 (since -5 mod31 is 26) and n≡1 mod3.So set n=31k +26. Then 31k +26 ≡1 mod3. 31≡1 mod3, so 31k≡k mod3, and 26≡2 mod3. Therefore, k +2 ≡1 mod3 => k ≡-1≡2 mod3. Therefore, k=3m +2. Then n=31*(3m +2)+26=93m +62 +26=93m +88. So the smallest such n is 88 when m=0. Wait, but 31*2 +26=62 +26=88. Check if 88≡1 mod3: 88 divided by3 is29*3=87, remainder1. Yes, 88≡1 mod3. Then 5^88 +88≡5 +88=93≡0 mod31? 31*3=93, so yes, 93≡0 mod31. So n=88 is a solution in this case.Case2: n ≡2 mod3Then 5^n≡25 mod31. So 25 +n≡0 mod31 => n≡-25 mod31 => n≡6 mod31 (since -25 +31=6). Also, n≡2 mod3. So solve n≡6 mod31 and n≡2 mod3.Set n=31k +6. Then 31k +6≡k +6≡2 mod3 => k≡2 -6≡2 -0≡2 mod3? Wait, 31k≡k mod3, 6≡0 mod3. So k +0≡2 mod3 => k≡2 mod3. So k=3m +2. Then n=31*(3m +2)+6=93m +62 +6=93m +68. The smallest n here is 68. Check: 68≡2 mod3? 68 divided by3 is22*3=66, remainder2. Yes. Then 5^68 +68≡25 +68=93≡0 mod31. Correct. So n=68 is another solution.Case3: n≡0 mod3Then 5^n≡1 mod31. So 1 +n≡0 mod31 =>n≡-1 mod31 =>n≡30 mod31. Also, n≡0 mod3. So solve n≡30 mod31 and n≡0 mod3.Set n=31k +30. Then 31k +30≡k +30≡0 mod3 =>k≡-30≡-0≡0 mod3. So k=3m. Then n=31*(3m) +30=93m +30. The smallest n here is30. Check:30≡0 mod3. 5^30≡1 mod31. 1 +30=31≡0 mod31. So yes, n=30 is a solution here.Wait, but n=30 is a solution? Let me check.Compute 5^30 mod31. Since the order is3, 5^3=1, so 5^30=(5^3)^10=1^10=1 mod31. Then 1 +30=31≡0 mod31. Yes, so n=30 is indeed a solution.So in each case, the minimal solutions are 30, 68, and88. The smallest among these is30. Wait, but the problem says "positive integer n". 30 is positive. So is the minimal n 30? But wait, let's check for smaller n. Maybe there is a smaller n that I missed?Wait, my approach was breaking it into cases based on n mod3. But maybe there's a solution with n less than30? Let me check manually for n from1 up to30.Compute 5^n +n mod31 for n=1 to30.n=1:5 +1=6 mod31=6≠0n=2:25 +2=27≠0n=3:1 +3=4≠0n=4:5 +4=9≠0n=5:25 +5=30≠0n=6:1 +6=7≠0n=7:5 +7=12≠0n=8:25 +8=33≡2≠0n=9:1 +9=10≠0n=10:5 +10=15≠0n=11:25 +11=36≡5≠0n=12:1 +12=13≠0n=13:5 +13=18≠0n=14:25 +14=39≡8≠0n=15:1 +15=16≠0n=16:5 +16=21≠0n=17:25 +17=42≡11≠0n=18:1 +18=19≠0n=19:5 +19=24≠0n=20:25 +20=45≡14≠0n=21:1 +21=22≠0n=22:5 +22=27≠0n=23:25 +23=48≡17≠0n=24:1 +24=25≠0n=25:5 +25=30≠0n=26:25 +26=51≡20≠0n=27:1 +27=28≠0n=28:5 +28=33≡2≠0n=29:25 +29=54≡23≠0n=30:1 +30=31≡0. So here, n=30 works.So indeed, the smallest n is30. But wait, in the earlier analysis, in case3, n=30 is a solution. But when I did the case analysis, I considered three cases based on n mod3. Then the minimal solution was n=30. So why does the case1 and case2 give larger solutions? Because when you solve the congruences, the minimal n in those cases are 68 and88, but n=30 is the minimal overall.Therefore, the minimal n is30. But wait, hold on. Wait, in case3, n=30 is the minimal solution. But let me check if n=30 is the minimal. Since when I checked manually, n=30 works, and there is no smaller n from1 to29 that works, so n=30 is indeed the minimal.But wait, let's cross-verify this. The problem statement says "positive integer n", so n=30 is positive. So maybe the answer is30. But let me check once again.Wait, but in my case3, n≡0 mod3, and n≡30 mod31. So n=30 is the smallest such. However, in the cases where n≡1 mod3 and n≡2 mod3, the minimal solutions are88 and68. So 30 is the smallest. Therefore, the answer is30.But hold on, the problem statement says "positive integer n". So30 is acceptable. But I recall that sometimes in such problems, there might be a smaller solution. Wait, in this case, from1 to30, only n=30 works. So 30 is indeed the minimal.But let me check again my manual computation. For n=30:5^30≡1 mod31, so 1 +30=31≡0 mod31. Correct. So that's correct.But wait, let me check with another approach. Suppose we want to solve 5^n ≡ -n mod31.Since 5^3 ≡1 mod31, so 5^n ≡5^{n mod3} mod31. So:If n≡0 mod3: 5^n≡1 mod31, so 1≡-n mod31 =>n≡-1 mod31. So n≡30 mod31, and since n≡0 mod3, the solution is n≡30 mod93. The minimal n is30.If n≡1 mod3:5^n≡5 mod31, so 5≡-n mod31 =>n≡-5 mod31 =>n≡26 mod31. But n≡1 mod3. So solving n≡26 mod31 and n≡1 mod3. As before, n=26 +31k. 26 mod3=26-24=2, so 2 +31k ≡1 mod3. 31≡1 mod3, so k≡(1 -2)≡-1≡2 mod3. So k=2 +3m. Then n=26 +31*(2 +3m)=26 +62 +93m=88 +93m. So minimal n=88.If n≡2 mod3:5^n≡25 mod31, so 25≡-n mod31 =>n≡-25 mod31 =>n≡6 mod31. And n≡2 mod3. So n=6 +31k. 6 mod3=0, so 0 +31k ≡2 mod3 =>k≡2 mod3. So k=2 +3m. Then n=6 +31*(2 +3m)=6 +62 +93m=68 +93m. Minimal n=68.Thus, the minimal solutions in each case are30,68,88,... so 30 is the smallest.But just to be thorough, let me check n=30 in the original equation:5^30 +30.Since 5^3≡1 mod31, then5^30=(5^3)^10≡1^10=1 mod31. So 1 +30=31≡0 mod31. Correct.So the minimal value of n is30.But wait, the answer seems straightforward, but let me check some references or similar problems.Wait, actually, sometimes when the exponent and the linear term are both variables, the minimal solution could be smaller. But in this case, since we checked all n up to30 manually and only n=30 worked, so that's indeed the minimal.But I just want to make sure I didn't make a mistake in the manual computation. Let me check a few more n's in the 20s.n=25:5^25. Since5^3=1, 25=3*8 +1, so5^25=(5^3)^8*5^1≡1^8*5=5 mod31. 5 +25=30≡30 mod31≠0.n=27:5^27≡(5^3)^9=1^9=1 mod31. 1 +27=28≡28≠0.n=28:5^28≡5^(27+1)=1*5=5 mod31.5 +28=33≡2≠0.n=29:5^29≡5^(27+2)=1*25=25 mod31.25 +29=54≡54-31=23≠0.n=30:1 +30=31≡0. Correct. So yeah, n=30 is the first one that works. Therefore, the answer is30.But hold on, the problem says "positive integer n". Is30 positive? Yes. So it's acceptable.But wait, let me check again the problem statement: "Let the positive integer n satisfy 31 ∣ (5^{n} + n). Then the minimum value of n is...."Yes, n=30 is the minimal. Therefore, the answer is30.However, I just realized that in my initial case analysis, when I considered n≡0 mod3, the minimal solution was30, but in the problem statement, it's possible that there is a smaller n. However, since we checked n=1 to30, and only n=30 works, so that's correct.Therefore, the minimal n is30. So the answer should be boxed{30}.But wait, wait a second. Wait, in the original problem statement, the user wrote "31 | (5^n + n)". So 31 divides 5^n +n. Therefore, 5^n +n ≡0 mod31. We found n=30 is the minimal solution.But let me check again with n=30.Compute5^30 mod31. Since5^3=125≡1 mod31, so5^3=1. Then5^30=(5^3)^10=1^10=1. Then5^30 +30=1 +30=31≡0 mod31. Correct. So yes, it's correct.Therefore, the minimal value of n is30. So the answer is30.But wait, but in some references, for example, similar problems, sometimes the minimal solution is larger. But in this case, since the modulus is31 and the exponent base5 has a small order, which is3, maybe the solution is indeed30.Alternatively, maybe the problem is designed such that the answer is larger? Wait, but our manual check shows30 works.Alternatively, perhaps I made a mistake in assuming the order of5 modulo31 is3. Let's confirm.Compute5^1 mod31=55^2=255^3=125. 31*4=124, so125-124=1. Therefore,5^3=1 mod31. Therefore, order is indeed3.Thus, exponent cycles every3. Therefore,5^n mod31 cycles with period3. Therefore, our analysis is correct.So why do I have this hesitation? Because maybe I'm missing something. But given that the manual check from1 to30 only gives30 as a solution, and the case analysis also leads to30,68,88, etc., with30 being the smallest, I think30 is indeed the correct answer.But let me see. Let me check with another approach.Suppose we let n =31k +r, where r=0,1,...,30. Then we can write5^{31k +r} +31k +r ≡0 mod31.But5^{31k +r} ≡(5^{31})^k *5^r mod31. But since5^3≡1 mod31, then5^{31}=5^{3*10 +1}=(5^3)^10 *5^1≡1^10*5=5 mod31.Therefore,5^{31k +r}≡5^k *5^r mod31. Therefore, the equation becomes5^{k +r} +31k +r≡0 mod31.But this seems more complex. Wait, perhaps not helpful. Alternatively, since31 is prime, and5 has order3, as established.Alternatively, note that5^n ≡-n mod31.Given that5^n cycles every3, then for each residue r=0,1,2 mod3, we can set5^n ≡c mod31, and solve for n≡-c mod31. But n must also be ≡r mod3. So that's similar to our original case analysis.But since5^n ≡1,5,25 mod31 depending on n mod3, we have three equations:If n≡0 mod3:1≡-n mod31=>n≡30 mod31. Since n≡0 mod3, we solve n≡30 mod31 and n≡0 mod3. The minimal solution is30.If n≡1 mod3:5≡-n mod31=>n≡26 mod31. Solve n≡26 mod31 and n≡1 mod3. As before, minimal n=88.If n≡2 mod3:25≡-n mod31=>n≡6 mod31. Solve n≡6 mod31 and n≡2 mod3. Minimal n=68.So, indeed,30 is the minimal.Therefore, after thorough analysis and checking, the minimal positive integer n is30. Hence, the answer is boxed{30}.Final Answerboxed{30}</think>To find the smallest positive integer ( n ) such that ( 31 mid (5^n + n) ), we need to solve the congruence ( 5^n + n equiv 0 pmod{31} ). This can be rewritten as ( 5^n equiv -n pmod{31} ).Using Fermat's Little Theorem, we know that ( 5^{30} equiv 1 pmod{31} ). However, we found that the order of 5 modulo 31 is 3, since ( 5^3 equiv 1 pmod{31} ). This means ( 5^n mod 31 ) cycles every 3. Therefore, we consider three cases based on ( n mod 3 ):1. Case 1: ( n equiv 0 pmod{3} ) Here, ( 5^n equiv 1 pmod{31} ). The congruence becomes ( 1 + n equiv 0 pmod{31} ), which simplifies to ( n equiv -1 pmod{31} ). The smallest positive solution is ( n equiv 30 pmod{31} ). Since ( n equiv 0 pmod{3} ), the smallest ( n ) is 30.2. Case 2: ( n equiv 1 pmod{3} ) Here, ( 5^n equiv 5 pmod{31} ). The congruence becomes ( 5 + n equiv 0 pmod{31} ), which simplifies to ( n equiv -5 pmod{31} ). The smallest positive solution is ( n equiv 26 pmod{31} ). Solving ( 26 + 31k equiv 1 pmod{3} ), we find ( k = 2 ), leading to ( n = 88 ).3. Case 3: ( n equiv 2 pmod{3} ) Here, ( 5^n equiv 25 pmod{31} ). The congruence becomes ( 25 + n equiv 0 pmod{31} ), which simplifies to ( n equiv -25 pmod{31} ). The smallest positive solution is ( n equiv 6 pmod{31} ). Solving ( 6 + 31k equiv 2 pmod{3} ), we find ( k = 2 ), leading to ( n = 68 ).Checking all values from 1 to 30 manually, we find that ( n = 30 ) is the smallest solution. Verifying, ( 5^{30} equiv 1 pmod{31} ) and ( 1 + 30 = 31 equiv 0 pmod{31} ).Thus, the minimum value of ( n ) is (boxed{30}).]

🔑:9. 30 .Given 5^{3} equiv 1(bmod 31), when n=3 kleft(k in mathbf{Z}_{+}right), 5^{n}+n equiv 1+n equiv 0(bmod 31), at this time, the minimum value of n is n_{text {min }}=30;When n=3 k+1left(k in mathbf{Z}_{+}right),5^{n}+n equiv 5+n equiv 0(bmod 31),at this time, the minimum value of n is n_{text {min }}=88;When n=3 k+2left(k in mathbf{Z}_{+}right),5^{n}+n equiv 25+n equiv 0(bmod 31),at this time, the minimum value of n is n_{text {min }}=68.Therefore, the minimum value of n that satisfies the condition is n_{min }=30.

🧠:[<think>Okay, so I have this problem here about negating a proposition. The original proposition is "For all x in real numbers, 2x² + 1 > 0." And the question is asking for its negation, with four options provided. Let me try to work through this step by step.First, I remember that the negation of a universal statement ("for all") is an existential statement ("there exists") such that the original predicate does not hold. So, if the original statement is "For all x, P(x)," then the negation would be "There exists an x such that not P(x)." That makes sense. So in this case, the original predicate P(x) is "2x² + 1 > 0." Therefore, the negation should be "There exists an x in real numbers such that 2x² + 1 is not greater than 0," which translates to "There exists an x in real numbers such that 2x² + 1 ≤ 0."Looking at the options, let me check:A: For all x in R, 2x² +1 ≤0. That's not the negation; it's just another universal statement with the inequality flipped. So that's incorrect because the negation should be existential, not universal.B: There exists an x0 in R such that 2x0² +1 >0. Wait, that's actually the same as the original predicate, but stated existentially. But since the original was universal, the negation can't be something that still claims the same inequality. If the original is true, then this statement B is also true because if it's true for all x, then there certainly exists an x. So B can't be the negation. The negation should be a statement that is true when the original is false.C: There exists an x0 in R such that 2x0² +1 ≤0. That seems right because it's the existential statement with the negation of the original inequality. So this should be the correct answer.D: There exists an x0 in R such that 2x0² +1 <0. Hmm, this is similar to C but with a strict inequality. Let me check if this is equivalent. If there exists an x0 such that 2x0² +1 is less than 0, then certainly it would also be less than or equal to 0. But does the negation of ">0" necessarily lead to "<0"? Wait, no. The negation of ">0" is "≤0," which includes both "less than 0" and "equal to 0." So C is the correct negation because it covers both possibilities. D is only part of the negation, not the entire negation. So even though D might be implied if the expression can't be zero, we still have to check if 2x² +1 can ever be zero or negative.Let me think about the actual expression 2x² +1. Since x is a real number, x² is always non-negative. Multiplying by 2 keeps it non-negative, and then adding 1 makes it strictly positive. So 2x² +1 is always at least 1, right? So 2x² +1 is always greater than 0. Therefore, the original statement is true. Then, its negation should be false. But looking at the options, C is "There exists an x0 such that 2x0² +1 ≤0." Since there is no such x0, C is false, which is correct because the original is true. Similarly, D is "There exists an x0 such that 2x0² +1 <0," which is also false. However, even though both C and D are false, the correct negation is still C because the logical negation of "for all x, P(x)" is "there exists x such that not P(x)," and "not (2x² +1 >0)" is "2x² +1 ≤0," not necessarily "2x² +1 <0."So even though in reality, 2x² +1 is never equal to zero either (since 2x² +1 =0 would imply x² = -1/2, which has no real solutions), the logical negation is still about ≤0. So even though both C and D are false in reality, the correct answer must be C because the negation requires the non-strict inequality. The question is about the logical negation, not about whether the negation is true or false in reality. So regardless of whether such an x0 exists, the negation is formed by flipping the quantifier and negating the predicate.Therefore, the answer is C.Final Answerboxed{C}</think>The original proposition is "For all (x in mathbb{R}), (2x^2 + 1 > 0)." To find its negation, we need to change the universal quantifier to an existential quantifier and negate the predicate. The negation of "For all (x in mathbb{R}), (2x^2 + 1 > 0)" is "There exists an (x_0 in mathbb{R}) such that (2x_0^2 + 1 leq 0)."- Option A is incorrect because it is a universal statement, not an existential one.- Option B is incorrect because it restates the original predicate existentially, not the negation.- Option C is correct because it correctly negates the predicate and uses an existential quantifier.- Option D is incorrect because it only considers the strict inequality, not the non-strict inequality required by the logical negation.Thus, the correct answer is (boxed{C}).,<think>Okay, so I need to find the negation of the proposition "For all (x in mathbb{R}), (2x^{2} + 1 > 0)." The options are A, B, C, D. Let me think step by step.First, I remember that the negation of a universal quantifier ("for all") is an existential quantifier ("there exists") with the negation of the original statement. So, if the original statement is (forall x P(x)), the negation should be (exists x neg P(x)). In this case, the original statement is "For all real numbers x, 2x squared plus 1 is greater than 0." So symbolically, that's (forall x in mathbb{R}, 2x^2 + 1 > 0). To negate this, I should switch the "for all" to "there exists" and negate the inequality. So the negation would be "There exists a real number x such that (2x^2 + 1) is not greater than 0." Now, "not greater than 0" is the same as "less than or equal to 0." So the negation should be "There exists (x_0 in mathbb{R}) such that (2x_0^2 + 1 leq 0)." Looking at the options: A: For all x, (2x^2 +1 leq 0). That's not the negation; the negation should have an existential quantifier, not universal. So A is incorrect.B: There exists an x such that (2x^2 +1 > 0). Wait, the original statement was that all x satisfy this inequality. The negation isn't saying that there exists one that does satisfy it—since if all x satisfy it, then there certainly exists some x that do. So B is actually a consequence of the original statement, not its negation. So B is not the answer.C: There exists an x such that (2x^2 +1 leq 0). That's exactly what I came up with earlier. So C seems correct.D: There exists an x such that (2x^2 +1 < 0). Hmm, this is similar to C but without the equality part. Let me think. The original statement's negation requires that at least one x makes (2x^2 +1) not greater than 0, which includes both less than and equal to. So C includes cases where it's equal to 0 or less than 0, while D only includes strictly less than 0. But is there any real x where (2x^2 +1 = 0)? Let's check. If (2x^2 +1 =0), then (2x^2 = -1), so (x^2 = -0.5). But x squared can't be negative in real numbers. Therefore, the equation (2x^2 +1 =0) has no real solutions. So in reality, the negation would require (2x^2 +1 leq 0), but since (2x^2 +1) is always at least 1 (since x² is non-negative, 2x² is non-negative, plus 1 makes it at least 1), actually (2x^2 +1) is always greater than 0. So both the original statement and its negation would be... Wait, the original statement is actually true, right? Because 2x² is non-negative, adding 1 makes it at least 1, which is always positive. So the original statement is a true statement, so its negation should be false. But regardless of the truth value, we just need to find the correct form of the negation. So even if there are no such x, the negation is still "there exists x such that...", which is option C. However, since (2x^2 +1) can never be less than or equal to zero in real numbers, C is a false statement, but it's still the correct negation. D is a stronger statement; it's saying there exists x where it's strictly less than zero. But even though the negation only requires it to be less than or equal, even if the equal part is impossible. So the correct logical negation is C, even if in reality, no such x exists. Therefore, the answer should be C. Let me check again. The original statement is (forall x, P(x)), negation is (exists x, neg P(x)). Here, (P(x)) is (2x^2 +1 >0), so (neg P(x)) is (2x^2 +1 leq 0). So yes, the negation is C. Options D is not the correct negation because it only covers the strict inequality, but the negation should cover all cases where the original statement fails, which includes both < and =, even if the = case is impossible. But logically, we still have to include it. Hence, C is the correct answer.Final Answerboxed{C}</think>To find the negation of the proposition "For all (x in mathbb{R}), (2x^2 + 1 > 0)", we start by understanding that the negation of a universal quantifier ((forall)) is an existential quantifier ((exists)) with the negation of the original statement. The original statement is (forall x in mathbb{R}, 2x^2 + 1 > 0). The negation of this statement is (exists x in mathbb{R}, neg (2x^2 + 1 > 0)), which simplifies to (exists x in mathbb{R}, 2x^2 + 1 leq 0).Analyzing the options:- A is incorrect because it uses a universal quantifier instead of an existential one.- B is incorrect because it states the original proposition's consequence, not its negation.- C is correct as it correctly uses an existential quantifier and the negation of the inequality.- D is incorrect because it only covers the strict inequality ( < 0 ), whereas the negation should include both ( < 0 ) and ( = 0 ).Even though (2x^2 + 1 leq 0) has no real solutions, the logical negation is still correctly formed as option C. Thus, the final answer is (boxed{C}).]

🔑:Since the proposition "For all (x in mathbb{R}), (2x^{2}+1 > 0)" is a universal proposition, the negation of a universal proposition is an existential proposition. Therefore, the negation of the proposition is: "There exists an (x_{0} in mathbb{R}), such that (2x_{0}^{2}+1 leqslant 0)". Hence, the correct choice is: boxed{C}. This conclusion can be reached by understanding that the negation of a universal proposition is an existential proposition. This question mainly tests the negation of propositions containing quantifiers, requiring an understanding that the negation of an existential proposition is a universal proposition, and vice versa, which is quite basic.

🧠:[<think>Okay, let's see. I need to solve this inequality: √(3x - 7) - √(3x² - 13x + 13) ≥ 3x² - 16x + 20. Hmm, that looks a bit complicated with square roots and a quadratic on the right side. Let me break it down step by step.First, I should figure out the domain of the inequality because there are square roots involved. For the square roots to be real numbers, the expressions inside them must be non-negative.Starting with the first square root, √(3x - 7). The expression inside here is 3x - 7, so we need 3x - 7 ≥ 0. Solving that gives 3x ≥ 7 → x ≥ 7/3 ≈ 2.333.Next, the second square root is √(3x² - 13x + 13). The expression inside here is 3x² - 13x + 13. I need this to be greater than or equal to zero. Let's solve 3x² - 13x + 13 ≥ 0.To find where this quadratic is non-negative, I can find its roots and check intervals. The quadratic equation is 3x² -13x +13 =0. Let's compute the discriminant: D = (-13)² - 4*3*13 = 169 - 156 = 13. So the roots are [13 ± √13]/6. Let me calculate approximate values: √13 ≈ 3.606, so the roots are (13 + 3.606)/6 ≈ 16.606/6 ≈ 2.768 and (13 - 3.606)/6 ≈ 9.394/6 ≈ 1.566. So the quadratic 3x² -13x +13 is positive outside the interval [1.566, 2.768]. Since our first square root requires x ≥ 7/3 ≈ 2.333, which is within the interval [1.566, 2.768], so in the domain x ≥ 7/3, the quadratic 3x² -13x +13 is actually negative or zero? Wait, no. Wait, between the roots, the quadratic is negative. So if x is between 1.566 and 2.768, the quadratic is negative. But our first condition is x ≥ 7/3 ≈ 2.333. So x is in [2.333, 2.768] where the quadratic is negative, and x ≥ 2.768 where it's positive. Wait, so there's a conflict here. Because the second square root requires 3x² -13x +13 ≥0, so x ≤1.566 or x ≥2.768. But the first square root requires x ≥2.333. So the intersection of these two domains is x ≥2.768. So the domain is x ≥ (13 + √13)/6 ≈2.768. Therefore, the valid x must be greater than or equal to approximately 2.768. Let me write that as x ≥ (13 + √13)/6. Let me check if that's exact. The roots of the quadratic are [13 ± √13]/6, so the intervals where the quadratic is non-negative are x ≤ [13 - √13]/6 ≈1.566 or x ≥ [13 + √13]/6 ≈2.768. Therefore, combining with the first condition x ≥7/3 ≈2.333, the domain is x ≥ [13 + √13]/6 ≈2.768. Therefore, the domain is x ≥ (13 + √13)/6. So that's the first step.Therefore, x must be ≥ approximately 2.768. Now, moving on to solving the inequality.The inequality is √(3x -7) - √(3x² -13x +13) ≥ 3x² -16x +20.Let me first note that the right-hand side (RHS) is a quadratic: 3x² -16x +20. Let me check if this quadratic can be factored. Let's compute discriminant: D = (-16)^2 -4*3*20 = 256 - 240 =16. So roots are [16 ±4]/6. That is, (16 +4)/6=20/6=10/3≈3.333, and (16-4)/6=12/6=2. So the quadratic factors as 3x² -16x +20=3(x -10/3)(x -2). But written as factors with integer coefficients, maybe not straightforward. Alternatively, we can write it as (3x -10)(x -2). Let me check: (3x -10)(x -2)=3x² -6x -10x +20=3x² -16x +20. Yes, that's correct. So RHS is (3x -10)(x -2). Let me note that.So, the inequality is √(3x -7) - √(3x² -13x +13) ≥ (3x -10)(x -2).Now, since the domain is x ≥ (13 + √13)/6 ≈2.768. Let's see the roots of the RHS quadratic: x=10/3≈3.333 and x=2. So in our domain x≥2.768, the RHS (3x -10)(x -2) is evaluated for x≥2.768.At x=10/3≈3.333, RHS is zero. Let's analyze the sign of RHS. For x >10/3, (3x -10) is positive and (x -2) is positive, so RHS is positive. For x between 2.768 and 10/3≈3.333, (3x -10) is negative and (x -2) is positive, so RHS is negative. Therefore, RHS is negative when 2.768 < x <3.333 and positive when x ≥3.333.So the inequality has different behavior depending on the interval. Let me split the domain into two parts: [(13 + √13)/6, 10/3) and [10/3, ∞). Let me verify if (13 + √13)/6 is less than 10/3. Let's compute (13 + √13)/6. √13≈3.606, so 13 +3.606≈16.606, divided by6≈2.768. 10/3≈3.333. So yes, the first interval is approximately [2.768, 3.333), and the second is [3.333, ∞).Now, in the first interval, RHS is negative, so the inequality becomes √(3x -7) - √(3x² -13x +13) ≥ negative number. Since the left-hand side (LHS) is √(3x -7) - √(3x² -13x +13). Let's see if LHS is positive or negative here.Let me take a sample value in the first interval, say x=3. Let's compute LHS and RHS.At x=3:First square root: √(3*3 -7)=√(9-7)=√2≈1.414.Second square root: √(3*9 -13*3 +13)=√(27 -39 +13)=√1=1.So LHS is √2 -1≈1.414 -1≈0.414.RHS is 3*9 -16*3 +20=27 -48 +20= -1. So the inequality is 0.414 ≥ -1, which is true.So in this interval, the inequality holds because RHS is negative and LHS is positive (or even if LHS is negative but greater than RHS). Wait, but LHS could be negative. Let's check another point.Take x approaching (13 + √13)/6≈2.768. Let's pick x=2.8.Compute 3x -7: 3*2.8 -7=8.4 -7=1.4, so √1.4≈1.183.Compute 3x² -13x +13. Let's compute 3*(2.8)^2 -13*2.8 +13. 2.8 squared is 7.84, times 3 is 23.52. Then 13*2.8=36.4. So 23.52 -36.4 +13=23.52 -36.4= -12.88 +13=0.12. So √0.12≈0.346.Thus LHS≈1.183 -0.346≈0.837.RHS: 3*(2.8)^2 -16*(2.8) +20. 3*7.84=23.52, 16*2.8=44.8, so 23.52 -44.8 +20= (23.52 +20) -44.8=43.52 -44.8≈-1.28. So inequality 0.837 ≥ -1.28, which is true.Therefore, in the first interval [(13 + √13)/6,10/3), the RHS is negative, and LHS is positive (since √(3x -7) is greater than √(3x² -13x +13) here? Wait, not sure. Wait, but when x increases, 3x² -13x +13 increases as well. Let's see.But in the first interval, 3x² -13x +13 is positive (since we are in the domain where x≥(13 +√13)/6≈2.768, which is the upper root, so quadratic is positive here). Wait, actually earlier, the domain is x≥(13 +√13)/6≈2.768, which is where 3x² -13x +13≥0. So in the entire domain, the second square root is valid.But in this first interval, even though RHS is negative, the LHS is √(3x -7) - √(3x² -13x +13). Let's see if LHS can be negative here. Let's take x=3. Then LHS≈0.414 as above. But if x is approaching 10/3≈3.333 from below, say x=3.3.Compute LHS: √(3*3.3 -7)=√(9.9 -7)=√2.9≈1.702.Second square root: √(3*(3.3)^2 -13*3.3 +13). Compute 3*(10.89)=32.67; 13*3.3=42.9. So 32.67 -42.9 +13= (32.67 +13) -42.9=45.67 -42.9≈2.77. √2.77≈1.664. So LHS≈1.702 -1.664≈0.038.RHS at x=3.3: 3*(3.3)^2 -16*(3.3) +20. 3*10.89=32.67; 16*3.3=52.8. So 32.67 -52.8 +20≈(32.67 +20) -52.8≈52.67 -52.8≈-0.13. So inequality is 0.038 ≥ -0.13, which is true. So even as we approach 10/3, LHS is still positive (though getting close to zero) and RHS is approaching zero from below.Wait, but what if x is such that LHS is negative? Let's check x=3.333 (10/3). Wait, x=10/3 is in the next interval. Let me check x=3.4, which is in the second interval.But perhaps in the first interval, LHS is always positive. Let's test x=3.2.LHS: √(3*3.2 -7)=√(9.6 -7)=√2.6≈1.612.Second square root: √(3*(3.2)^2 -13*3.2 +13). 3*(10.24)=30.72; 13*3.2=41.6. 30.72 -41.6 +13= (30.72 +13) -41.6=43.72 -41.6=2.12. √2.12≈1.456. So LHS≈1.612 -1.456≈0.156.RHS: 3*(3.2)^2 -16*3.2 +20. 3*10.24=30.72; 16*3.2=51.2. 30.72 -51.2 +20≈(30.72 +20) -51.2≈50.72 -51.2≈-0.48. So inequality 0.156 ≥ -0.48: true.So in the first interval [(13 + √13)/6,10/3), RHS is negative, and LHS seems to be positive. Therefore, the inequality holds for all x in this interval. Because a positive number is always greater than a negative number.Therefore, in the first interval, the inequality is always true.Now, moving to the second interval: x ≥10/3≈3.333. Here, RHS is non-negative because (3x -10)(x -2) is positive when x>10/3. So in this interval, the inequality is √(3x -7) - √(3x² -13x +13) ≥ positive number. Therefore, LHS must be ≥ positive number. Let's analyze if this is possible.First, let's check the behavior of LHS and RHS as x increases. Let's compute LHS and RHS at x=4.At x=4:First square root: √(3*4 -7)=√(12 -7)=√5≈2.236.Second square root: √(3*16 -13*4 +13)=√(48 -52 +13)=√9=3. So LHS=√5 -3≈2.236 -3≈-0.764.RHS: 3*16 -16*4 +20=48 -64 +20=4. So inequality is -0.764 ≥4, which is false.So at x=4, inequality does not hold.At x=10/3≈3.333:First, check x=10/3=3.333...First square root: √(3*(10/3) -7)=√(10 -7)=√3≈1.732.Second square root: √(3*(100/9) -13*(10/3) +13). Let's compute:3*(100/9)=100/3≈33.333,13*(10/3)=130/3≈43.333,So 33.333 -43.333 +13= (33.333 +13) -43.333≈46.333 -43.333=3. So √3≈1.732.Thus LHS=√3 - √3=0.RHS: 3*(10/3)^2 -16*(10/3) +20=3*(100/9) -160/3 +20=100/3 -160/3 +20=(-60/3)+20=-20 +20=0.So at x=10/3, LHS=0 and RHS=0. Therefore, 0≥0, which is true.So at x=10/3, inequality holds. Now, check just above x=10/3, say x=3.4.At x=3.4:First sqrt: √(3*3.4 -7)=√(10.2 -7)=√3.2≈1.788.Second sqrt: √(3*(3.4)^2 -13*3.4 +13). Compute:3.4^2=11.56, times 3=34.68,13*3.4=44.2,34.68 -44.2 +13=34.68 -44.2= -9.52 +13=3.48,√3.48≈1.865.Thus LHS≈1.788 -1.865≈-0.077.RHS: 3*(3.4)^2 -16*3.4 +20=3*11.56=34.68, 16*3.4=54.4, so 34.68 -54.4 +20= (34.68 +20) -54.4=54.68 -54.4≈0.28.So inequality: -0.077 ≥0.28? No, false.Thus, just above 10/3, the inequality fails. So at x=10/3, it's equal, but slightly above, it's false. Let's check another point. Maybe x=3.3335.But perhaps we can analyze the functions.Let’s consider x ≥10/3. Let’s denote f(x) = √(3x -7) - √(3x² -13x +13) and g(x)=3x² -16x +20. We need to find where f(x) ≥g(x).At x=10/3, f(x)=0, g(x)=0. For x>10/3, we saw that f(x) becomes negative while g(x) becomes positive. So, is there any x >10/3 where f(x) ≥g(x)? Let's check another point. Let's try x=3.5.At x=3.5:First sqrt: √(3*3.5 -7)=√(10.5 -7)=√3.5≈1.87.Second sqrt: √(3*(3.5)^2 -13*3.5 +13)=√(3*12.25 -45.5 +13)=√(36.75 -45.5 +13)=√(4.25)≈2.06.So LHS≈1.87 -2.06≈-0.19.RHS: 3*(3.5)^2 -16*3.5 +20=3*12.25=36.75; 16*3.5=56. So 36.75 -56 +20= (36.75 +20) -56=56.75 -56≈0.75.So inequality: -0.19 ≥0.75? No.What about x=3.333... (exactly 10/3 + ε). Let's take x=10/3 +0.1=3.433...But I think as x increases beyond 10/3, the RHS increases quadratically, while LHS becomes negative and its magnitude might increase, but need to see.Alternatively, let's try to solve the equation f(x)=g(x) in x ≥10/3. If we can find solutions there.So √(3x -7) - √(3x² -13x +13) =3x² -16x +20.But given that at x=10/3, both sides are zero. Let's check if this is the only solution.Suppose x >10/3. Let's denote y = x -10/3, so y >0. Then we can try to see if for y>0, the equation holds. But this might not help directly.Alternatively, we can square both sides to eliminate the square roots, but since we have two square roots, it might get complicated. Let me consider moving one square root to the other side.Starting with the original inequality for x ≥10/3:√(3x -7) - √(3x² -13x +13) ≥3x² -16x +20.But since in this interval, the RHS is positive, and LHS is negative (as we saw in x=3.4 and x=4), so LHS - RHS would be negative - positive, which is more negative. Therefore, the inequality would not hold. However, this is a bit hand-wavy. Let's try to formalize it.Wait, actually, let's consider the original inequality again:√(3x -7) - √(3x² -13x +13) ≥3x² -16x +20.Let me rearrange it as:√(3x -7) -3x² +16x -20 ≥√(3x² -13x +13}.But since in x ≥10/3, the RHS √(3x² -13x +13) is positive, and the left side is √(3x -7) minus a quadratic. But this might not be helpful.Alternatively, let's set A = √(3x -7), B = √(3x² -13x +13). Then the inequality is A - B ≥3x² -16x +20.But in x ≥10/3, 3x² -16x +20 is positive. So we have A - B ≥ positive. Which would require A - B to be positive. But from x=3.4, A≈1.87, B≈2.06, so A -B≈-0.19. Similarly, as x increases, 3x -7 increases linearly, but 3x² -13x +13 increases quadratically, so B increases faster than A. Therefore, A - B becomes more negative as x increases. Therefore, in x ≥10/3, A - B is negative or decreasing? Let's see.Take x=5:A=√(15-7)=√8≈2.828.B=√(75 -65 +13)=√23≈4.796.A -B≈2.828 -4.796≈-1.968.RHS=75 -80 +20=15. So inequality: -1.968 ≥15? No.Thus, as x increases, LHS becomes more negative, and RHS increases. So there's no solution in x >10/3 except at x=10/3 where both sides are zero. Hence, the only solution in x ≥10/3 is x=10/3.Therefore, overall, the solutions to the inequality are [(13 + √13)/6,10/3]. But wait, we need to check if at x=(13 + √13)/6, does the inequality hold?Wait, (13 + √13)/6≈2.768. Let's check x=2.768.Compute LHS and RHS.First square root: √(3x -7). Let's compute 3x=3*(13 + √13)/6=(13 + √13)/2. Then 3x -7=(13 + √13)/2 -7=(13 + √13 -14)/2=(√13 -1)/2. Which is positive? Since √13≈3.606, √13 -1≈2.606, so divided by 2≈1.303. So √(1.303)≈1.141.Second square root: √(3x² -13x +13). At x=(13 + √13)/6, the expression inside the sqrt is exactly zero because that's the root. So √0=0.Thus, LHS=√(3x -7) -0≈1.141.RHS:3x² -16x +20. Let's compute x=(13 + √13)/6.Compute x²=[(13 + √13)/6]^2=(169 +26√13 +13)/36=(182 +26√13)/36=(91 +13√13)/18.So 3x²=3*(91 +13√13)/18=(91 +13√13)/6≈(91 +13*3.606)/6≈(91 +46.878)/6≈137.878/6≈22.979.-16x= -16*(13 + √13)/6= -8*(13 + √13)/3≈-8*(13 +3.606)/3≈-8*(16.606)/3≈-132.848/3≈-44.283.+20.Thus, RHS≈22.979 -44.283 +20≈(22.979 +20) -44.283≈42.979 -44.283≈-1.304.Therefore, the inequality is 1.141 ≥ -1.304, which is true. So at the left endpoint x=(13 + √13)/6≈2.768, the inequality holds.Therefore, combining both intervals, the solution set is [(13 + √13)/6,10/3]. However, we need to verify if there are any points in [(13 + √13)/6,10/3] where the inequality might not hold. Wait, but in the first interval, we saw that RHS is negative and LHS is positive, so inequality holds. At the left endpoint, LHS is positive, RHS is negative. At the right endpoint (x=10/3), both sides are zero, so equality holds. Therefore, the entire interval [(13 + √13)/6,10/3] is the solution.Therefore, all real numbers x between (13 + √13)/6 and 10/3 (inclusive) satisfy the inequality.But the problem asks for the sum of all integer values of x that satisfy the inequality. So we need to find all integers x such that (13 + √13)/6 ≤x ≤10/3.First, let's compute the numerical values:(13 + √13)/6≈(13 +3.6055)/6≈16.6055/6≈2.7676.10/3≈3.3333.So the interval is approximately [2.7676,3.3333]. The integers in this interval are x=3.Because x must be an integer. 2.7676 to3.3333 includes integer 3. Let's confirm:At x=3, does the inequality hold?Compute LHS: √(3*3 -7)=√2≈1.414.Compute second sqrt: √(3*9 -13*3 +13)=√(27 -39 +13)=√1=1.So LHS=1.414 -1≈0.414.RHS=3*9 -16*3 +20=27 -48 +20= -1.So 0.414 ≥ -1: True.At x=3, the inequality holds.Next integer x=4 is outside the interval (since 3.3333 <4), so x=4 is not included. The only integer is x=3.Therefore, the sum of all integer values is 3.Wait, but let me check x=2.7676 to3.3333. Is there another integer? x=3 is the only integer in that interval. So the answer is 3.Hence, the sum is 3.Final AnswerThe sum of all integer values of ( x ) that satisfy the inequality is (boxed{3}).</think>To solve the inequality (sqrt{3x - 7} - sqrt{3x^2 - 13x + 13} geq 3x^2 - 16x + 20), we first determine the domain where the square roots are defined.1. Domain Analysis: - For (sqrt{3x - 7}) to be defined, (3x - 7 geq 0 implies x geq frac{7}{3} approx 2.333). - For (sqrt{3x^2 - 13x + 13}) to be defined, (3x^2 - 13x + 13 geq 0). Solving this quadratic inequality, we find the roots (frac{13 pm sqrt{13}}{6}). The intervals where the quadratic is non-negative are (x leq frac{13 - sqrt{13}}{6} approx 1.566) or (x geq frac{13 + sqrt{13}}{6} approx 2.768). Combining these results, the domain is (x geq frac{13 + sqrt{13}}{6} approx 2.768).2. Analyzing the Inequality: - The right-hand side (RHS) (3x^2 - 16x + 20) factors as ((3x - 10)(x - 2)). The roots are (x = frac{10}{3} approx 3.333) and (x = 2). - For (x geq frac{13 + sqrt{13}}{6} approx 2.768), we split the domain into two intervals: - (left[frac{13 + sqrt{13}}{6}, frac{10}{3}right)) - (left[frac{10}{3}, inftyright))3. Interval Analysis: - In (left[frac{13 + sqrt{13}}{6}, frac{10}{3}right)), the RHS is negative, and the left-hand side (LHS) is positive, so the inequality holds. - In (left[frac{10}{3}, inftyright)), the RHS is positive, but the LHS becomes negative, so the inequality does not hold except at (x = frac{10}{3}) where both sides are zero.4. Integer Solutions: - The interval (left[frac{13 + sqrt{13}}{6}, frac{10}{3}right]) approximately translates to ([2.768, 3.333]). - The only integer in this interval is (x = 3).Thus, the sum of all integer values of (x) that satisfy the inequality is (boxed{3}).,<think>Okay, so I need to solve this inequality: sqrt(3x - 7) - sqrt(3x² - 13x + 13) ≥ 3x² - 16x + 20. Hmm, that looks a bit intimidating with the square roots and the quadratic terms. Let me break it down step by step.First, I know that whenever there are square roots involved, the expressions inside them must be non-negative. So I need to start by finding the domain of the inequality. Let's look at each square root separately.For sqrt(3x - 7) to be real, the expression inside must be ≥ 0. So:3x - 7 ≥ 0 ⇒ 3x ≥ 7 ⇒ x ≥ 7/3 ≈ 2.333...Next, for sqrt(3x² - 13x + 13), the expression inside must also be ≥ 0:3x² - 13x + 13 ≥ 0.Hmm, solving this quadratic inequality. Let me find the discriminant first:D = (-13)² - 4*3*13 = 169 - 156 = 13.Since the discriminant is positive, the quadratic has two real roots. The roots are:x = [13 ± sqrt(13)] / (2*3) ≈ [13 ± 3.6055]/6.Calculating the approximate values:First root: (13 + 3.6055)/6 ≈ 16.6055/6 ≈ 2.7676Second root: (13 - 3.6055)/6 ≈ 9.3945/6 ≈ 1.5658Since the coefficient of x² is positive (3), the parabola opens upwards. Therefore, the quadratic is ≥ 0 outside the interval between the roots. So:x ≤ 1.5658 or x ≥ 2.7676.But from the first square root, we already have x ≥ 7/3 ≈ 2.333. So combining both domain conditions, the domain is x ≥ 2.7676, since 2.7676 is greater than 7/3. So x must be ≥ approximately 2.7676.But let's write the exact roots to be precise. The roots are (13 ± sqrt(13))/6. So the exact domain is x ≤ (13 - sqrt(13))/6 or x ≥ (13 + sqrt(13))/6. But since x must also be ≥ 7/3, which is 7/3 ≈ 2.333, and (13 + sqrt(13))/6 is approximately (13 + 3.605)/6 ≈ 16.605/6 ≈ 2.7675. So 7/3 is approximately 2.333, which is less than 2.7675. Therefore, the domain is x ≥ (13 + sqrt(13))/6. So that's the exact domain.Therefore, any solution must satisfy x ≥ (13 + sqrt(13))/6.Alright, so moving on to the inequality itself:sqrt(3x - 7) - sqrt(3x² - 13x + 13) ≥ 3x² - 16x + 20.This looks complicated. Let me see. The left side is sqrt(3x - 7) minus another square root, and the right side is a quadratic. Since we have square roots on the left and a quadratic on the right, maybe I can rearrange the inequality to isolate the square roots?Alternatively, perhaps consider substituting variables to simplify. Let me think.Let me denote:Let A = sqrt(3x - 7)Let B = sqrt(3x² - 13x + 13)So the inequality becomes A - B ≥ 3x² - 16x + 20.But I need to relate A and B. Let's see if I can express B in terms of A or something else.Alternatively, perhaps square both sides? But squaring can be tricky because the left side is a difference of square roots, and the right side is a quadratic. Also, the direction of the inequality might change when squaring, depending on the sign of both sides. However, first, I need to check the sign of the right side.Let me analyze the right side: 3x² - 16x + 20. Let's see if this quadratic is always positive or not. Let's compute its discriminant:D = (-16)² - 4*3*20 = 256 - 240 = 16.Since D = 16, which is positive, the quadratic has two real roots. The roots are:x = [16 ± sqrt(16)]/(2*3) = [16 ± 4]/6.So:x = (16 + 4)/6 = 20/6 = 10/3 ≈ 3.333...x = (16 - 4)/6 = 12/6 = 2.So the quadratic 3x² -16x +20 factors as 3(x - 10/3)(x - 2). Since the coefficient of x² is positive, it opens upwards. Therefore, the quadratic is ≥ 0 when x ≤ 2 or x ≥ 10/3 ≈ 3.333...But in our domain, x ≥ (13 + sqrt(13))/6 ≈ 2.7675. So in the domain, the quadratic 3x² -16x +20 is negative between 2.7675 and 10/3 ≈ 3.333, and positive for x ≥ 10/3. Wait, let's check:Wait, the quadratic is positive when x ≤2 or x ≥10/3. So in the domain x ≥ ~2.7675, the quadratic is negative between 2.7675 and 3.333..., and positive for x ≥ 3.333...Therefore, the right-hand side (RHS) of the inequality 3x² -16x +20 is negative in part of the domain and positive in another part. Therefore, when moving terms across the inequality, we have to be careful about the sign.But the original inequality is A - B ≥ RHS. Let me consider moving all terms to the left side:sqrt(3x -7) - sqrt(3x² -13x +13) - 3x² +16x -20 ≥ 0.But I don't know if that helps. Alternatively, maybe move the square roots to the right:sqrt(3x -7) - sqrt(3x² -13x +13) ≥ 3x² -16x +20.But given that the RHS can be negative or positive, depending on x, it complicates things.Alternatively, perhaps consider the entire inequality:sqrt(3x -7) - sqrt(3x² -13x +13) ≥ 3x² -16x +20.Since we have square roots on the left, maybe try to bound them? Alternatively, test possible integer solutions within the domain.First, let me note that the domain is x ≥ (13 + sqrt(13))/6 ≈ 2.7675. So the possible integer x's are starting from x = 3 upwards. Let's check x=3, x=4, etc., and see if they satisfy the inequality. Then sum those integer solutions.But before jumping into testing, let's see if we can manipulate the inequality.First, note that sqrt(3x -7) is present. Let me see if 3x² -13x +13 can be related to (3x -7) somehow. Let me check:Suppose I write 3x² -13x +13. Maybe factor it? Let's try:Looking for factors of 3*13=39 that add up to -13. Hmm, -3 and -10? No. Doesn't seem to factor nicely. So perhaps not factorable. So maybe not helpful.Alternatively, let me see if 3x² -13x +13 can be expressed as (sqrt(3x -7))² + something. Let's see:We have 3x² -13x +13. Let's compute (sqrt(3x -7))² = 3x -7. Then:3x² -13x +13 = 3x² -13x +13. Hmm, perhaps if we subtract 3x -7 from both sides:3x² -13x +13 - (3x -7) = 3x² -16x +20. Wait, that's interesting! Because 3x² -13x +13 - (3x -7) = 3x² -16x +20.So, 3x² -16x +20 = (3x² -13x +13) - (3x -7).Therefore, the original inequality can be written as:sqrt(3x -7) - sqrt(3x² -13x +13) ≥ (3x² -13x +13) - (3x -7).Let me denote again A = sqrt(3x -7) and B = sqrt(3x² -13x +13). Then the inequality becomes:A - B ≥ B² - A².Wait, because (3x² -13x +13) is B², and (3x -7) is A². Therefore:A - B ≥ B² - A².Rearranging terms:A - B + A² - B² ≥ 0.Factor A² - B² as (A - B)(A + B):(A - B) + (A - B)(A + B) ≥ 0.Factor out (A - B):(A - B)(1 + A + B) ≥ 0.So, the inequality simplifies to:(A - B)(1 + A + B) ≥ 0.Interesting! So the original inequality can be factored as (sqrt(3x -7) - sqrt(3x² -13x +13))*(1 + sqrt(3x -7) + sqrt(3x² -13x +13)) ≥ 0.Now, note that 1 + A + B is always positive. Because A and B are square roots, so they are non-negative. So 1 + A + B ≥ 1 > 0.Therefore, the sign of the entire expression (A - B)(1 + A + B) is determined solely by (A - B). Because 1 + A + B is always positive, the inequality (A - B)(1 + A + B) ≥ 0 is equivalent to (A - B) ≥ 0.Therefore, the original inequality simplifies to A - B ≥ 0, which is sqrt(3x -7) ≥ sqrt(3x² -13x +13).So, in other words, the original complicated inequality reduces to sqrt(3x -7) ≥ sqrt(3x² -13x +13). That simplifies things!So now, let's solve sqrt(3x -7) ≥ sqrt(3x² -13x +13). Since both sides are square roots, they are non-negative, so we can square both sides without flipping the inequality (as both sides are non-negative):(3x -7) ≥ (3x² -13x +13).Simplify:3x -7 ≥ 3x² -13x +13.Bring all terms to one side:0 ≥ 3x² -13x +13 -3x +7.Simplify:0 ≥ 3x² -16x +20.Which is equivalent to:3x² -16x +20 ≤ 0.But earlier, we found that the quadratic 3x² -16x +20 has roots at x = 2 and x = 10/3 ≈ 3.333. Since the quadratic opens upwards, it is ≤ 0 between its roots: 2 ≤ x ≤ 10/3.But remember the domain of the original inequality: x ≥ (13 + sqrt(13))/6 ≈ 2.7675. So the intersection of 2 ≤ x ≤ 10/3 and x ≥ 2.7675 is 2.7675 ≤ x ≤ 10/3 ≈ 3.333.Therefore, the solution to the inequality is x in [ (13 + sqrt(13))/6 , 10/3 ].But we need to check if squaring both sides introduced any extraneous solutions. Let me verify.When we squared both sides, we had sqrt(a) ≥ sqrt(b) => a ≥ b, provided that a and b are non-negative, which they are in the domain. So squaring preserves the inequality here because both sides are non-negative. Therefore, the solution set [ (13 + sqrt(13))/6 , 10/3 ] is valid.Now, converting the bounds to decimals for clarity:(13 + sqrt(13))/6 ≈ (13 + 3.6055)/6 ≈ 16.6055/6 ≈ 2.767610/3 ≈ 3.3333.Therefore, the solution is x ∈ [2.7676, 3.3333].Now, the problem asks for the sum of all integer values of x that satisfy the inequality. Since x must be an integer within [2.7676, 3.3333], the only integer is x=3.But wait, let me check if x=3 is indeed a solution.Testing x=3:Left side: sqrt(3*3 -7) - sqrt(3*(3)^2 -13*3 +13) = sqrt(9 -7) - sqrt(27 -39 +13) = sqrt(2) - sqrt(1) = sqrt(2) -1 ≈ 1.4142 -1 = 0.4142.Right side: 3*(3)^2 -16*3 +20 = 27 -48 +20 = -1.So the inequality is 0.4142 ≥ -1, which is true. Therefore, x=3 is a solution.Wait, but according to our earlier analysis, the inequality reduces to sqrt(3x -7) ≥ sqrt(3x² -13x +13), but when x=3, sqrt(2) ≥ sqrt(1), which is true. So x=3 is indeed in the solution set.Is there another integer? The upper bound is 10/3 ≈3.3333, so the next integer is x=4, but 4 is outside the upper bound. So x=4 would be 4 ≈ 4.0, which is greater than 10/3 ≈3.333, so it's not included. Therefore, the only integer solution is x=3.Therefore, the sum of all integer values is 3.But let me just check x=3.3333 (10/3). Let's see if that's a solution. x=10/3:Left side: sqrt(3*(10/3) -7) - sqrt(3*(10/3)^2 -13*(10/3) +13)Simplify:sqrt(10 -7) - sqrt(3*(100/9) -130/3 +13)= sqrt(3) - sqrt(100/3 -130/3 +13)= sqrt(3) - sqrt( (-30/3 ) +13 )= sqrt(3) - sqrt( -10 +13 )= sqrt(3) - sqrt(3) = 0.Right side: 3*(10/3)^2 -16*(10/3) +20= 3*(100/9) -160/3 +20= 100/3 -160/3 +60/3= (100 -160 +60)/3= 0/3 = 0.Therefore, at x=10/3, the inequality is 0 ≥ 0, which is true. So x=10/3 is included. But it's not an integer. Similarly, x=(13 + sqrt(13))/6 ≈2.7676 is not an integer.Therefore, the only integer in the interval is x=3.So the sum is 3.But wait, hold on. The original inequality was sqrt(3x -7) - sqrt(3x² -13x +13) ≥ 3x² -16x +20.But when I tested x=3, the left side was approximately 0.4142, and the right side was -1, so 0.4142 ≥ -1 is true. However, according to the reduced inequality sqrt(3x -7) ≥ sqrt(3x² -13x +13), which at x=3 is sqrt(2) ≥ sqrt(1), which is true. However, when the original inequality was transformed, we considered that (A - B)(1 + A + B) ≥0, and since 1 + A + B >0, then A - B ≥0. So in effect, the original inequality reduces to A - B ≥0. However, in the original problem, the RHS was 3x² -16x +20, which for x=3 is -1. So even though A - B is 0.4142, which is greater than -1. But in the transformed inequality, we ended up with A - B ≥0, which is a stricter condition. Wait, but why?Wait, perhaps the steps are correct? Let me recap.Original inequality:A - B ≥ RHS, where RHS = B² - A² (since RHS = 3x² -16x +20 = B² - A²).Then moving all terms to left:A - B - (B² - A²) ≥0 ⇒ A - B - B² + A² ≥0 ⇒ A² - B² + A - B ≥0 ⇒ (A - B)(A + B) + (A - B) ≥0 ⇒ (A - B)(A + B +1) ≥0.Since (A + B +1) >0, then (A - B) ≥0.Therefore, yes, the original inequality reduces to A - B ≥0. So in effect, the original inequality is equivalent to A - B ≥0, given that (A + B +1) is positive.But in testing x=3, A - B ≈0.4142 ≥0 is true, so satisfies the reduced inequality. But the original RHS is -1, so 0.4142 ≥ -1 is also true. So why does the original inequality include x=3, but when transformed, it reduces to A - B ≥0, which is more restrictive? Wait, but in reality, when we transformed the original inequality, we used algebraic manipulation, which shows that the original inequality is equivalent to A - B ≥0. Therefore, the original inequality is equivalent to A - B ≥0, and since A - B ≥0 is more restrictive than the original inequality (since 0.4142 ≥ -1 is true, but the equivalence only holds because the RHS was expressed as B² - A², leading to (A - B)(something positive) ≥0 ⇒ A - B ≥0). Therefore, the solution set is actually the intersection of the domain and A - B ≥0, which is the interval [ (13 + sqrt(13))/6, 10/3 ].But in this case, even though for some x in that interval, the original inequality A - B ≥ RHS (which is negative) is true because A - B is positive and RHS is negative, but when we manipulate the inequality, we actually get that the solution is exactly the interval where A - B ≥0. Therefore, the transformation is valid and the solution set is correct.Therefore, the only integer solution is x=3, so the sum is 3.But just to be thorough, let's check another point in the interval, say x=3. Let's check the original inequality:sqrt(3*3 -7) - sqrt(3*9 -13*3 +13) = sqrt(2) - sqrt(27 -39 +13) = sqrt(2) - sqrt(1) ≈1.414 -1 ≈0.414.RHS: 3*9 -16*3 +20 =27 -48 +20 = -1.So 0.414 ≥ -1, which is true. So x=3 is indeed a solution.Another test at x=10/3 ≈3.333:Left side: sqrt(3*(10/3) -7) - sqrt(3*(10/3)^2 -13*(10/3) +13) = sqrt(10 -7) - sqrt(3*(100/9) -130/3 +13) = sqrt(3) - sqrt(100/3 -130/3 +13) = sqrt(3) - sqrt(-30/3 +13) = sqrt(3) - sqrt(-10 +13) = sqrt(3) - sqrt(3) = 0.RHS: 3*(10/3)^2 -16*(10/3) +20 = 3*(100/9) -160/3 +20 = 100/3 -160/3 +60/3 = 0/3 =0.Therefore, 0 ≥0, which is true.What about a point just above (13 + sqrt(13))/6, say x=2.8:First, check domain: (13 + sqrt(13))/6 ≈2.7675, so x=2.8 is in the domain.Left side: sqrt(3*2.8 -7) - sqrt(3*(2.8)^2 -13*2.8 +13)Compute 3*2.8 =8.4, 8.4 -7=1.4, sqrt(1.4)≈1.1832.3*(2.8)^2=3*7.84=23.52; 13*2.8=36.4; 23.52 -36.4 +13=23.52 -36.4= -12.88 +13=0.12. sqrt(0.12)≈0.3464.Therefore, left side≈1.1832 -0.3464≈0.8368.RHS:3*(2.8)^2 -16*2.8 +20=23.52 -44.8 +20= (23.52 +20) -44.8=43.52 -44.8≈-1.28.Therefore, inequality:0.8368 ≥ -1.28, which is true.But according to the reduced inequality, sqrt(3x -7) ≥ sqrt(3x² -13x +13). At x=2.8:sqrt(3*2.8 -7)=sqrt(1.4)≈1.1832sqrt(3*(2.8)^2 -13*2.8 +13)=sqrt(0.12)≈0.3464So 1.1832 ≥0.3464, which is true. Therefore, x=2.8 is a solution.But since we're only asked for integer solutions, x=3 is the only integer in the interval. Therefore, the answer is 3.Final AnswerThe sum of all integer values of ( x ) that satisfy the inequality is boxed{3}.</think>To solve the inequality [sqrt{3x - 7} - sqrt{3x^2 - 13x + 13} geq 3x^2 - 16x + 20,]we first determine the domain. The expressions under the square roots must be non-negative:1. (3x - 7 geq 0 implies x geq frac{7}{3} approx 2.333).2. (3x^2 - 13x + 13 geq 0). Solving this quadratic inequality, we find the roots are (frac{13 pm sqrt{13}}{6}). The solution is (x leq frac{13 - sqrt{13}}{6}) or (x geq frac{13 + sqrt{13}}{6} approx 2.7675).Combining these, the domain is (x geq frac{13 + sqrt{13}}{6}).Next, we rewrite the inequality using substitutions (A = sqrt{3x - 7}) and (B = sqrt{3x^2 - 13x + 13}):[A - B geq 3x^2 - 16x + 20.]Recognizing that (3x^2 - 16x + 20 = B^2 - A^2), we rewrite the inequality as:[A - B geq B^2 - A^2.]Rearranging terms, we factorize:[(A - B)(1 + A + B) geq 0.]Since (1 + A + B > 0), the inequality simplifies to (A geq B). Squaring both sides (valid as both sides are non-negative):[3x - 7 geq 3x^2 - 13x + 13.]Simplifying, we get:[3x^2 - 16x + 20 leq 0.]Solving this quadratic inequality, we find the roots (x = 2) and (x = frac{10}{3} approx 3.333). The solution is (2 leq x leq frac{10}{3}). Intersecting this with the domain, we get:[frac{13 + sqrt{13}}{6} leq x leq frac{10}{3}.]The only integer within this interval is (x = 3). Verifying, we find (x = 3) satisfies the original inequality. Thus, the sum of all integer solutions is:[boxed{3}]]

🔑:Solution. As a result of the substitution v=sqrt{3 x-7}, u=sqrt{3 x^{2}-13 x+13}, we obtain the equivalent inequality u leqslant v. Therefore, x satisfies the inequality 3 x^{2}-13 x+13 leqslant 3 x-7 Longleftrightarrow 2 leqslant x leqslant 10 / 3. Out of the two integer values x=2 and x=3, only x=3 falls within the domain of valid values.Answer: (B) 3.Answer options.

🧠:[<think>Okay, let's see. So the problem is about Xiao Hu in an archery competition. He has 50 arrows total. When he still has 7 arrows left to shoot, meaning he's already shot 50 - 7 = 43 arrows. At this point, his score is over 150 points. We need to find the minimum number of arrows that must have hit area 1. First, let's recap the scoring rules: area 1 gives 4 points, area 2 gives 3 points, and other areas give 1 point. The goal is to find the least number of area 1 hits (let's call that x) such that the total score from 43 arrows is more than 150. To find the minimum x, we need to consider the scenario where the remaining arrows (other than the x area 1 hits) contribute the least possible points. That way, we maximize the number of low-scoring arrows, which would require more high-scoring arrows (area 1) to reach the total score. So, if x is the number of area 1 hits, then the remaining (43 - x) arrows must be distributed between area 2 and other areas. To minimize the total points from these (43 - x) arrows, we should maximize the number of arrows that hit other areas (1 point each) because that gives the least points. However, we can only do that if there's no constraint on the number of hits in each area. Wait, the problem doesn't mention any limits on how many times each area can be hit, right? So Xiao Hu could have all his remaining arrows hitting other areas. Therefore, to minimize the total score from the remaining arrows, we assume that all non-area 1 hits are in other areas (1 point each). Therefore, the total score would be 4x + 1*(43 - x) > 150. Let's write that equation:4x + (43 - x) > 150Simplify this:4x + 43 - x > 150Combine like terms:3x + 43 > 150Subtract 43 from both sides:3x > 107Divide both sides by 3:x > 107/3Calculate that:107 divided by 3 is approximately 35.666...Since x must be an integer (number of arrows can't be a fraction), we round up to the next whole number, which is 36. So x must be at least 36. But wait, let me verify this. If x is 36, then the total score would be 36*4 + (43 - 36)*1 = 144 + 7 = 151. Which is indeed greater than 150. If x is 35, then the total score would be 35*4 + (43 - 35)*1 = 140 + 8 = 148. Which is less than 150. So 35 is not enough, so 36 is the minimum. Therefore, the answer is 36.But hold on, let me check if there's another possibility where some of the remaining arrows hit area 2. If some of them are area 2, which gives 3 points, then the total points would be higher than if all were 1 point. But since we are trying to find the minimal number of area 1 hits, which requires the remaining arrows to contribute as little as possible, meaning we should assume all non-area 1 arrows are 1 point. So our initial reasoning holds.Alternatively, if we allowed some area 2 hits, the total points would be higher, which would mean that even with fewer area 1 hits, the total could still exceed 150. But that's the opposite of what we want. Since we need to find the minimum number of area 1 hits required to exceed 150, we have to assume the worst-case scenario where the other arrows contribute the least, hence all other arrows are 1 point. Therefore, our calculation is correct.So the answer is 36.Final AnswerThe minimum number of arrows hitting area 1 is boxed{36}.</think>Xiao Hu has 50 arrows in total and has already shot 43 arrows when he still has 7 arrows left. His score at this point exceeds 150 points. We need to determine the minimum number of arrows hitting area 1.The scoring rules are:- Area 1: 4 points- Area 2: 3 points- Other areas: 1 pointTo find the minimum number of arrows hitting area 1 (denoted as ( x )), we assume the remaining arrows (other than those hitting area 1) contribute the least points, i.e., all hit other areas (1 point each). The total score from 43 arrows must be greater than 150 points:[4x + (43 - x) > 150]Simplifying this inequality:[4x + 43 - x > 150 3x + 43 > 150 3x > 107 x > frac{107}{3} approx 35.666...]Since ( x ) must be an integer, we round up to the next whole number, which is 36. Verifying:- If ( x = 36 ), the score is ( 36 times 4 + (43 - 36) times 1 = 144 + 7 = 151 ), which exceeds 150.- If ( x = 35 ), the score is ( 35 times 4 + (43 - 35) times 1 = 140 + 8 = 148 ), which is less than 150.Thus, the minimum number of arrows hitting area 1 is (boxed{36}).,<think>Okay, let's see. So the problem is about Xiao Hu in an archery competition. He has 50 arrows total. When he still has 7 arrows left to shoot, his score exceeds 150 points. We need to find the least number of arrows he must have hit in area 1 up to that point. First, let me parse the problem again. He has 50 arrows. When he still has 7 left, meaning he has already shot 50 - 7 = 43 arrows. At this point, his score is over 150. We need to find the minimum number of area 1 hits he must have had in those 43 arrows. Area 1 gives 4 points, area 2 gives 3 points, and other areas give 1 point. So, we need to figure out the minimum number of 4-point arrows (area 1) such that the total score from 43 arrows is more than 150. Let me think. To minimize the number of high-scoring arrows (area 1), we need to maximize the number of lower-scoring arrows. But the total has to be over 150. So, we need to set up an inequality where the total points from 43 arrows > 150. Let x be the number of area 1 hits, y be the number of area 2 hits, and z be the number of other areas. Then x + y + z = 43, and 4x + 3y + z > 150. We need to find the minimal x such that this holds.But maybe it's easier to approach this by considering that to minimize x, we need to maximize the other points. So, for the remaining arrows (43 - x), we can assign as many as possible to the next highest scoring, which is area 2 (3 points), and then the rest to other areas (1 point). Wait, but actually, if we want to maximize the total score given a certain number of arrows, we would maximize the number of higher-scoring ones. But here, since we need to minimize the number of area 1 hits, we need to maximize the score from the remaining arrows (43 - x). Therefore, to maximize their score, those remaining arrows should be as many area 2 hits as possible, and then the rest as other areas. Wait, no. Wait, no. Wait, if we are trying to find the minimal x (area 1 hits) needed to reach over 150, then given x area 1 hits, the maximum score from the remaining (43 - x) arrows would be if all of them are area 2 hits, which give 3 points each. So, total score would be 4x + 3*(43 - x). If that total is greater than 150, then x is sufficient. So we can set up the inequality:4x + 3*(43 - x) > 150But wait, is that correct? Because in reality, some of the remaining arrows might hit other areas (1 point). So if we assume that the maximum possible score from the remaining (43 - x) arrows is 3*(43 - x), but if we use that to compute the inequality, then if even the maximum possible score (all area 2) plus 4x is just over 150, then x would be the minimal. Because if in reality some of them are 1 point, the total score would be lower. So in order for the total score to exceed 150, even if all non-area 1 arrows are area 2 (highest possible), the total must still be over 150. Therefore, by setting up the inequality with maximum possible points from non-area 1 arrows, we can find the minimal x required.Therefore, solving:4x + 3*(43 - x) > 150Simplify:4x + 129 - 3x > 150x + 129 > 150x > 150 - 129x > 21Therefore, x must be at least 22. But wait, let's check if x=22 suffices. Let's compute 4*22 + 3*(43 -22) = 88 + 3*21 = 88 + 63 = 151. 151 is indeed over 150. If x=21, then 4*21 + 3*(43-21) = 84 + 3*22 = 84 + 66 = 150. 150 is not over 150, it's exactly 150. But the problem states that his score exceeds 150 points. Therefore, 150 is not enough. So x must be at least 22. Therefore, the minimal number is 22.Wait, but hold on. The problem says "hitting other areas scores 1 point." So in reality, if not all remaining arrows (after x area 1 hits) are area 2, but some are other areas, then the total score would be less. So if we assume that as many as possible are area 2, giving the maximum possible score, then the minimal x is 22. However, if in reality some are other areas, then the total would be lower. Therefore, the calculation with all remaining arrows as area 2 gives the minimal x required. Because if even in the best case (all remaining arrows are area 2), you need 22 area 1 hits, then in reality, if some are lower, you might need more. Wait, but the problem says that his score already exceeds 150. Therefore, regardless of how the remaining arrows (43 - x) are distributed between area 2 and other areas, as long as the total score exceeds 150. So in order to find the minimal x, we need to consider the scenario where the remaining arrows (43 - x) give the maximum possible score, which is 3*(43 -x). Because if the maximum possible score with x area 1 hits is still not over 150, then even if you have some area 2s, you still can't reach over 150. Therefore, to guarantee that the total exceeds 150, x must be such that even if all remaining arrows are area 2, the total is over 150. Therefore, the minimal x is 22. Therefore, the answer is 22.But let me verify this again. If x=22, and all remaining 21 arrows are area 2, total score is 22*4 + 21*3 = 88 + 63 = 151. Which is over 150. If x=21, and all remaining 22 arrows are area 2, total score is 21*4 + 22*3 = 84 + 66 = 150. Which is exactly 150, not exceeding. Therefore, x must be at least 22. Therefore, the answer is 22. So the minimal number is 22.But wait, maybe there's another way. Suppose some of the remaining arrows are area 2 and some are other areas. Let's see. For example, if x=21, and 22 arrows are split into y area 2 and z others. Then total score is 21*4 + 3y + z. But y + z = 22. So total score = 84 + 3y + (22 - y) = 84 + 2y + 22 = 106 + 2y. To exceed 150, 106 + 2y > 150. Then 2y > 44, so y > 22. But y can't exceed 22, since y + z = 22. So maximum y is 22. Then 106 + 2*22 = 106 + 44 = 150. Therefore, even if all 22 remaining arrows are area 2, score is 150. So with x=21, it's impossible to exceed 150. Therefore, x must be at least 22. So yes, 22 is the minimal.Alternatively, if we have x=22, then the remaining 21 arrows can be split into y and z. Total score is 88 + 3y + z. With y + z =21. So 88 + 3y + (21 - y) = 88 + 2y +21 = 109 + 2y. To exceed 150, 2y > 41, so y > 20.5. So y must be at least 21. But y can be at most 21, since there are 21 arrows. Therefore, y=21, z=0. Then total score is 109 + 42 = 151. Which is over 150. So even if we have x=22, and all remaining arrows are area 2, it's 151. If some are others, the total would be less, but still 151 is already over. So as long as x=22, even if all remaining are area 2, which gives 151, which is over. Therefore, the minimal x is 22.Therefore, the answer is 22. So I think the answer is boxed{22}.Final Answerboxed{22}</think>Xiao Hu has 50 arrows in total and has already shot 43 arrows when he still has 7 left. His score at this point exceeds 150 points. We need to determine the minimum number of arrows that hit area 1.To minimize the number of area 1 hits (which score 4 points), we maximize the score from the remaining arrows. The remaining arrows (43 - x) should be considered as hitting area 2 (3 points each) to maximize the score. We set up the inequality:[4x + 3(43 - x) > 150]Simplifying this inequality:[4x + 129 - 3x > 150][x + 129 > 150][x > 21]Thus, (x) must be at least 22. Checking if (x = 22) suffices:[4 times 22 + 3 times (43 - 22) = 88 + 63 = 151]This score is 151, which exceeds 150. If (x = 21):[4 times 21 + 3 times (43 - 21) = 84 + 66 = 150]This score is exactly 150, which does not exceed 150. Therefore, the minimum number of arrows hitting area 1 is (boxed{22}).]

🔑:22

🧠:[<think>Okay, let me try to figure out how to solve this problem. The question is about finding the length of the shortest chord passing through the point (3,5) on the circle given by the equation (x^{2}+y^{2}-6x-8y=0). Hmm, chords passing through a specific point... I remember that the shortest chord through a point on a circle is the one that is perpendicular to the radius at that point. But wait, in this case, is the point (3,5) actually on the circle? Let me check that first.To verify if (3,5) is on the circle, I can substitute x=3 and y=5 into the equation and see if it satisfies. Plugging in:(3^2 + 5^2 -6*3 -8*5 = 9 +25 -18 -40 = 34 - 58 = -24). Hmm, that's not equal to zero. So the point (3,5) is not on the circle. That changes things. So the chord passes through a point inside the circle? Because the chord has to pass through (3,5), which is inside the circle. Wait, but how do I know if it's inside? Let me find the center and radius of the circle first.The given equation is (x^{2}+y^{2}-6x-8y=0). To find the center and radius, I need to rewrite it in the standard form ((x - h)^2 + (y - k)^2 = r^2). So, completing the squares for both x and y terms.Starting with the x terms: (x^2 -6x). To complete the square, take half of -6, which is -3, square it to get 9. Similarly for the y terms: (y^2 -8y). Half of -8 is -4, square it to get 16. So adding and subtracting these numbers:(x^2 -6x +9 -9 + y^2 -8y +16 -16 =0)Which simplifies to:((x -3)^2 + (y -4)^2 -25 =0)So the equation becomes:((x -3)^2 + (y -4)^2 =25)Therefore, the center of the circle is at (3,4) and the radius is 5. Okay, so the center is (3,4), radius 5. Now, the point (3,5) is given. Let's see where this point is relative to the circle.First, the center is (3,4), and the point (3,5) is one unit above the center. So the distance from the center to the point (3,5) is sqrt[(3-3)^2 + (5-4)^2] = sqrt[0 +1] =1. Since the radius is 5, which is much larger than 1, the point (3,5) is inside the circle. So any chord passing through (3,5) is a chord of the circle passing through an interior point.Now, the question is asking for the length of the shortest such chord. I remember that the shortest chord through a point inside a circle is the one that is perpendicular to the line joining the center to that point. Is that right? Let me think. If we have a point inside the circle, then the chord perpendicular to the radius at that point would be the shortest. Wait, but actually, since the point isn't the center, the radius isn't going through that point. Wait, no. The chord that is perpendicular to the line connecting the center to the point (3,5) should be the shortest chord through (3,5). Yes, that seems familiar.Let me recall the formula for the length of a chord passing through a point inside a circle. If a chord passes through a point at distance d from the center, then the length of the chord is (2sqrt{r^2 - d^2}). But wait, that's when the chord is perpendicular to the line joining the center to the point. Wait, actually, if the chord is at distance d from the center, then the length is (2sqrt{r^2 - d^2}). But in this case, the point (3,5) is at distance 1 from the center (3,4). So is the length of the shortest chord (2sqrt{5^2 -1^2} = 2sqrt{24} = 4sqrt{6}). Wait, but that doesn't seem right. Wait, hold on.Wait, maybe I'm confusing some concepts here. Let me clarify. The formula (2sqrt{r^2 - d^2}) gives the length of the chord that is at distance d from the center. But in this case, the point (3,5) is not the distance from the center, but the distance from the center to the chord. Wait, perhaps I need to find the distance from the center to the chord, then compute the length.Alternatively, here's another approach. Let's denote the center of the circle as O(3,4), and the point inside the circle as P(3,5). We need to find the shortest chord passing through P.The shortest chord through P is the one perpendicular to the line OP. Since OP is the line from the center to the point P, which in this case is a vertical line (since both O and P have x-coordinate 3). So the line OP is vertical. Therefore, the chord perpendicular to OP would be horizontal.Wait, if OP is vertical, then the chord perpendicular to it would be horizontal. So the shortest chord passing through P(3,5) is a horizontal line passing through (3,5). Let me verify that.So the chord is horizontal, so its equation is y =5. Let's find the intersection points of this line with the circle.Substitute y=5 into the circle equation:((x -3)^2 + (5 -4)^2 =25)Simplifies to:((x -3)^2 +1 =25)((x -3)^2 =24)So x -3 = ±√24 = ±2√6Therefore, x=3±2√6, y=5.Therefore, the two intersection points are (3+2√6,5) and (3-2√6,5). The distance between these two points is the length of the chord. Since the x-coordinates differ by 4√6 and the y-coordinates are the same, the distance is 4√6. But the problem states that this is the shortest chord. But earlier, when I thought the length was 4√6, but I thought maybe I made a mistake. But according to this calculation, the chord length is 4√6. However, let me check if this is indeed the shortest chord.Alternatively, perhaps there's a chord passing through (3,5) that's shorter? Let me think. Wait, another way to compute the length of the chord is using the formula. If a chord passes through a point P inside the circle, then the length of the chord depends on the angle of the chord with respect to OP. The minimum length occurs when the chord is perpendicular to OP, which we just calculated. So in this case, the chord perpendicular to OP is horizontal, length 4√6, which is the shortest. Therefore, the answer should be 4√6. Wait, but let me check again.Wait, OP is the line from center O(3,4) to point P(3,5). The distance from O to P is 1, as calculated before. The length of the chord perpendicular to OP at point P is 2√(r² - d²), where d is the distance from O to P. So substituting, 2√(25 -1) = 2√24 = 4√6. So yes, that's correct. Therefore, the shortest chord length is 4√6. Hmm, but let me make sure.Alternatively, maybe there's a different approach. Let's suppose we take a general chord passing through (3,5). Let me parametrize such a chord. Let's say the chord has a slope m. Then the equation of the chord is y -5 = m(x -3). We can find the points where this line intersects the circle, then compute the distance between them, and find the minimum distance.Let me try this approach. The circle equation is (x² + y² -6x -8y =0). The line is y = m(x -3) +5. Substitute into the circle equation:x² + [m(x -3) +5]^2 -6x -8[m(x -3) +5] =0Expanding the terms:x² + m²(x -3)^2 +10m(x -3) +25 -6x -8m(x -3) -40 =0Let me compute each term:First, expand [m(x -3) +5]^2:= m²(x -3)^2 + 10m(x -3) +25Then subtract 6x and 8[m(x -3) +5]:So total equation:x² + m²(x² -6x +9) +10m(x -3) +25 -6x -8m(x -3) -40 =0Simplify term by term:x² + m²x² -6m²x +9m² +10mx -30m +25 -6x -8mx +24m -40=0Combine like terms:x² + m²x²: x²(1 + m²)-6m²x +10mx -8mx -6x: x terms= x[-6m² + (10m -8m) -6] = x[-6m² +2m -6]Then constants:9m² -30m +25 +24m -40 =9m² -6m -15So the equation becomes:(1 + m²)x² + (-6m² +2m -6)x + (9m² -6m -15) =0This is a quadratic in x. Let me denote coefficients as A, B, C:A =1 +m²B = -6m² +2m -6C=9m² -6m -15Now, for the line to intersect the circle, this quadratic has two real solutions, which it will since the line passes through the interior point. The distance between the two intersection points can be found using the distance formula. Alternatively, the length of the chord can be calculated using the formula:Length = √[(x1 -x2)^2 + (y1 -y2)^2]But since the line has slope m, (y1 - y2)=m(x1 -x2). Therefore, the length can be written as √[(1 + m²)(x1 -x2)^2] = √(1 +m²)*|x1 -x2|Now, the difference x1 -x2 can be found using the quadratic formula. The roots of Ax² + Bx + C =0 are x1 and x2, so:x1 +x2 = -B/Ax1x2 = C/AThen (x1 -x2)^2 = (x1 +x2)^2 -4x1x2 = (B²/A² -4C/A) = (B² -4AC)/A²Therefore, |x1 -x2| = √(B² -4AC)/ATherefore, the length is √(1 +m²) * √(B² -4AC)/ASimplify:Length = √[(1 +m²)(B² -4AC)] / ALet me compute B² -4AC.First, compute B²:B = -6m² +2m -6B² = (-6m² +2m -6)^2Let me compute this term:First, factor out a common factor if possible. Let me see:-6m² +2m -6 = -6m² +2m -6. Not obvious. Let me compute the square:= [(-6m²) + (2m -6)]^2= (-6m²)^2 + (2m -6)^2 + 2*(-6m²)(2m -6)= 36m^4 + (4m² -24m +36) + 2*(-12m³ +36m²)= 36m^4 +4m² -24m +36 -24m³ +72m²Combine like terms:36m^4 -24m³ + (4m² +72m²) + (-24m) +36=36m^4 -24m³ +76m² -24m +36Now compute 4AC:A =1 +m²C=9m² -6m -154AC =4*(1 +m²)*(9m² -6m -15)Let me compute (1 +m²)*(9m² -6m -15):=9m²*(1 +m²) -6m*(1 +m²) -15*(1 +m²)=9m² +9m^4 -6m -6m^3 -15 -15m²Combine like terms:9m^4 -6m^3 + (9m² -15m²) -6m -15=9m^4 -6m^3 -6m² -6m -15Multiply by 4:4AC=36m^4 -24m^3 -24m² -24m -60Now compute B² -4AC:[36m^4 -24m³ +76m² -24m +36] - [36m^4 -24m^3 -24m² -24m -60]Subtract term by term:36m^4 -36m^4 =0-24m³ -(-24m³)=076m² - (-24m²) =76m² +24m²=100m²-24m -(-24m)=036 -(-60)=36 +60=96So B² -4AC=100m² +96Therefore, the length becomes:√[(1 +m²)(100m² +96)] / (1 +m²)Simplify numerator and denominator:√[(1 +m²)(100m² +96)] / (1 +m²) = √[(100m² +96)/(1 +m²)]So the length squared is (100m² +96)/(1 +m²). To minimize the length, we can minimize the square of the length, which is (100m² +96)/(1 +m²). Let's denote this expression as L² = (100m² +96)/(1 +m²). To find the minimum, take derivative with respect to m and set to zero.Alternatively, treat L² as a function of m, say f(m) = (100m² +96)/(1 +m²). To find the minimum, we can set the derivative f’(m)=0.But this might be complicated. Alternatively, we can write f(m) = 100 - (4)/(1 +m²). Wait, let me see:100m² +96 =100(m² +1) -4. Therefore:f(m) = [100(m² +1) -4]/(m² +1) =100 -4/(m² +1)So f(m)=100 -4/(1 +m²). Therefore, to minimize f(m), we need to maximize 4/(1 +m²), which occurs when m² is minimized. Since m² >=0, the maximum of 4/(1 +m²) is 4, achieved when m=0. Therefore, the minimum of f(m) is 100 -4=96. Therefore, the minimal L² is 96, so minimal L is sqrt(96)=4*sqrt(6). So the minimal length is 4√6, which matches the previous result. Therefore, the shortest chord length is indeed 4√6.Alternatively, there's a calculus-free way. Since f(m) = (100m² +96)/(1 +m²) = 100 -4/(1 +m²), as shown. Therefore, the minimal value occurs when 4/(1 +m²) is maximized, which is when m² is minimal (i.e., m=0). Therefore, the minimal length is sqrt(96)=4√6. Therefore, when m=0, the chord is horizontal, which is the case we considered earlier. So that's consistent. Therefore, the answer is 4√6.Wait, but let me confirm again. The problem says the equation of the circle is x² +y² -6x -8y=0, which we converted to (x-3)^2 + (y-4)^2=25, center (3,4), radius 5. The point (3,5) is one unit above the center. The chord through (3,5) perpendicular to OP (which is vertical) is horizontal, y=5. The length of this chord is 4√6. If we take another chord through (3,5) with a different slope, say a very steep slope, the chord would be longer. Therefore, the minimal length chord is indeed the horizontal one. Therefore, the answer should be 4√6.But wait, let me check the calculation once more. When substituting y=5 into the circle equation:(x -3)^2 + (5 -4)^2 =25(x -3)^2 +1 =25(x -3)^2=24Therefore, x=3 ±√24=3±2√6. Therefore, the two points are (3+2√6,5) and (3-2√6,5). The distance between them is sqrt[( (3+2√6 - (3-2√6))^2 + (5-5)^2 )] = sqrt[(4√6)^2 +0]=4√6. Yep, that's correct. So the length is indeed 4√6.Therefore, the answer is 4√6. I think that's solid. So the length of the shortest chord passing through (3,5) is 4√6.Final AnswerThe length of the shortest chord passing through the point ((3,5)) is boxed{4sqrt{6}}.</think>Given the equation of the circle (x^2 + y^2 - 6x - 8y = 0), we first convert it to the standard form by completing the square:[x^2 - 6x + y^2 - 8y = 0][(x - 3)^2 + (y - 4)^2 - 9 - 16 = 0][(x - 3)^2 + (y - 4)^2 = 25]The center of the circle is ((3, 4)) and the radius is 5. The point ((3, 5)) is inside the circle since the distance from the center ((3, 4)) to ((3, 5)) is 1, which is less than the radius 5.The shortest chord passing through a point inside a circle is perpendicular to the line joining the center to that point. The line joining the center ((3, 4)) to the point ((3, 5)) is vertical, so the shortest chord must be horizontal. The equation of the horizontal line passing through ((3, 5)) is (y = 5).Substituting (y = 5) into the circle's equation:[(x - 3)^2 + (5 - 4)^2 = 25][(x - 3)^2 + 1 = 25][(x - 3)^2 = 24][x - 3 = pm 2sqrt{6}]The points of intersection are ((3 + 2sqrt{6}, 5)) and ((3 - 2sqrt{6}, 5)). The length of the chord is the distance between these two points:[sqrt{[(3 + 2sqrt{6}) - (3 - 2sqrt{6})]^2 + (5 - 5)^2} = sqrt{(4sqrt{6})^2} = 4sqrt{6}]Thus, the length of the shortest chord passing through the point ((3, 5)) is (boxed{4sqrt{6}}).,<think>Okay, so I need to find the length of the shortest chord passing through the point (3,5) on the circle given by the equation x² + y² - 6x - 8y = 0. Hmm, let's start by recalling what a chord is. A chord is a straight line segment whose endpoints both lie on the circle. The shortest chord passing through a specific point on the circle... wait, but is the point (3,5) actually on the circle? Maybe I should check that first.Let me plug in x=3 and y=5 into the equation:Left side: 3² + 5² - 6*3 - 8*5 = 9 + 25 - 18 - 40 = 34 - 58 = -24. Hmm, that's not equal to 0. So the point (3,5) isn't on the circle. Then the chord passes through (3,5) but the endpoints are on the circle. So the chord is passing through an interior point or maybe an exterior point? Wait, if the point is inside the circle, the chord passes through it, but if it's outside, there might not be any chord passing through it. So first, I need to determine if (3,5) is inside or outside the circle.To do that, let's find the center and radius of the circle. The given equation is x² + y² - 6x - 8y = 0. Let me rewrite it in standard form by completing the squares.For the x terms: x² -6x. Take half of -6, which is -3, square it to get 9. Similarly for the y terms: y² -8y. Half of -8 is -4, square it to get 16. So adding and subtracting these:x² -6x +9 -9 + y² -8y +16 -16 =0Which simplifies to (x - 3)² + (y - 4)² -25 =0, so (x - 3)² + (y - 4)² =25. Therefore, the center is at (3,4) and the radius is 5.So the center is (3,4), radius 5. The point in question is (3,5). Let's see how far this point is from the center. The distance between (3,4) and (3,5) is √[(3-3)² + (5-4)²] = √[0 +1] =1. Since the radius is 5, and 1 <5, the point (3,5) is inside the circle. Therefore, there are infinitely many chords passing through (3,5). The shortest chord would be the one that is perpendicular to the line joining the center to the point (3,5). Wait, is that correct?Yes, I remember that the shortest chord through a point inside the circle is the one perpendicular to the radial line (the line from the center to the point). So the chord perpendicular to the radius at that point is the shortest. Let me verify.If you have a point inside a circle, then the chords passing through that point have different lengths. The chord perpendicular to the radius through that point is the shortest. The longest chord would be the diameter passing through that point. So yes, the shortest chord is the perpendicular one.So first, find the slope of the radial line from center (3,4) to the point (3,5). The coordinates are (3,4) to (3,5). The change in x is 0, change in y is 1. So the slope is (5-4)/(3-3) which is 1/0, undefined. So the radial line is vertical. Therefore, the shortest chord should be horizontal, since it needs to be perpendicular to a vertical line.Therefore, the chord is horizontal. So its equation is y = k. Since it passes through (3,5), then k =5. Therefore, the equation of the chord is y=5. Now, find the points where this line intersects the circle. Then compute the distance between these two intersection points, which will give the length of the chord.So let's solve the system of equations:1. (x - 3)² + (y - 4)² =252. y=5Substituting y=5 into the circle equation:(x - 3)² + (5 -4)² =25Simplify:(x -3)² +1 =25(x -3)² =24Take square roots:x -3 = ±√24 = ±2√6Therefore, x=3±2√6So the two intersection points are (3 +2√6,5) and (3 -2√6,5). The distance between these two points is the difference in x-coordinates, since y-coordinates are the same. So distance is | (3 +2√6) - (3 -2√6) | = |4√6| =4√6. Therefore, the length is 4√6. Wait, but let me check that again.Wait, if the x-coordinates are 3 +2√6 and 3 -2√6, then the distance between them is [ (3 +2√6) - (3 -2√6) ] =4√6. So yes, the length is 4√6. Is that the shortest chord?Alternatively, maybe I made a mistake here. Let me verify the steps again.1. The center is (3,4), radius 5. Point (3,5) is inside the circle. The radial line is vertical, so the chord is horizontal. Equation y=5. Plugging into the circle equation: (x-3)² + (5-4)²=25 => (x-3)²=24 => x=3±2√6. So distance between these points is 4√6.But let me see another way. Maybe using the formula for the length of a chord given the distance from the center.The formula is: length=2√(r² -d²), where d is the distance from the center to the chord.Wait, in this case, the chord is horizontal at y=5. The distance from the center (3,4) to the line y=5 is |5 -4|=1. Therefore, the length of the chord is 2√(r² -d²)=2√(25 -1)=2√24=2*2√6=4√6. Which matches the previous result. So that's correct.Therefore, the length of the shortest chord passing through (3,5) is 4√6.Wait, but let me think again. Is there another way to approach this? For example, using coordinates and parametric equations? Let me try.Suppose the chord passes through (3,5). Let the chord have a slope m. Then its equation is y -5 = m(x -3). Then we can find the points where this line intersects the circle. Then compute the distance between those two points as a function of m, then find the minimum distance. But that seems more complicated, but perhaps it can confirm the result.Let's try this method.Given the circle equation: (x -3)^2 + (y -4)^2 =25.The line equation: y = m(x -3) +5.Substitute into the circle equation:(x -3)^2 + [m(x -3) +5 -4]^2 =25Simplify:(x -3)^2 + [m(x -3) +1]^2 =25Let me let t = x -3. Then the equation becomes t² + [m t +1]^2 =25.Expand the second term: m² t² + 2m t +1.So total equation: t² + m² t² + 2m t +1 =25Combine terms:(1 + m²) t² + 2m t + (1 -25)=0Which is (1 + m²) t² + 2m t -24 =0This is a quadratic in t. Let's denote coefficients:A =1 + m²B=2mC=-24The solutions are t = [-B ±√(B² -4AC)]/(2A)But the distance between the two intersection points can be found using the distance formula. If t1 and t2 are the roots, then the corresponding x-coordinates are t1 +3 and t2 +3, and the y-coordinates are m(t1) +5 and m(t2)+5.So the distance squared between the two points is:[(t2 +3 - (t1 +3))]^2 + [m(t2) +5 - (m(t1) +5)]^2Simplify:(t2 - t1)^2 + [m(t2 - t1)]^2= (t2 -t1)^2 (1 + m²)Therefore, the distance is |t2 -t1| * √(1 +m²)But |t2 - t1| can be found from the quadratic equation. The difference of roots is √(B² -4AC)/A. Wait, for a quadratic equation Ax² + Bx + C=0, the difference of roots is √(B² -4AC)/A. Wait, actually, the roots are (-B +√(B² -4AC))/(2A) and (-B -√(B² -4AC))/(2A). So the difference is [2√(B² -4AC)]/(2A) = √(B² -4AC)/A.Therefore, |t2 - t1|=√(B² -4AC)/A.So the distance is [√(B² -4AC)/A] * √(1 +m²)Plug in A=1 +m², B=2m, C=-24:√( (2m)^2 -4*(1 +m²)*(-24) ) / (1 +m²) * √(1 +m²)Simplify numerator inside sqrt:4m² -4*(1 +m²)*(-24) =4m² +96*(1 +m²)=4m² +96 +96m²=100m² +96So numerator sqrt is √(100m² +96). Then divide by (1 +m²) and multiply by √(1 +m²):Thus, distance= [√(100m² +96) / (1 +m²)] * √(1 +m²) = √(100m² +96)/√(1 +m²)So distance squared is (100m² +96)/(1 +m²). To minimize the distance, which is equivalent to minimizing the square of the distance, let's set f(m)=(100m² +96)/(1 +m²). Find derivative and set to zero.Alternatively, maybe simplify f(m):f(m)= (100m² +96)/(m² +1) =100 - (4)/(m² +1). Wait, let me check:Wait, 100(m² +1) -4=100m² +100 -4=100m² +96. So f(m)=100 -4/(m² +1). Wait, that's not correct. Let's perform polynomial division:Divide 100m² +96 by m² +1:100m² +96 =100*(m² +1) -100 +96=100*(m² +1) -4Therefore, f(m)=100 -4/(m² +1). Therefore, to minimize f(m), we need to minimize 100 -4/(m² +1). Since the term -4/(m² +1) is maximized when m² is minimized. The minimum value of m² is 0, when m=0. Therefore, the maximum of -4/(m² +1) is -4/(0 +1)= -4. Hence, the minimum of f(m) is 100 -4=96. Therefore, the minimal distance squared is 96, so minimal distance is √96=4√6. Which matches the previous result. Therefore, the minimal chord length is indeed 4√6.Therefore, regardless of the method, the answer is 4√6. Therefore, the length of the shortest chord passing through (3,5) is 4√6.Final AnswerThe length of the shortest chord passing through the point ((3,5)) is (boxed{4sqrt{6}}).</think>Given the equation of the circle (x^2 + y^2 - 6x - 8y = 0), we need to find the length of the shortest chord passing through the point ((3, 5)).First, we rewrite the given equation in standard form by completing the square:[x^2 - 6x + y^2 - 8y = 0][(x - 3)^2 + (y - 4)^2 - 9 - 16 = 0][(x - 3)^2 + (y - 4)^2 = 25]This gives the center ((3, 4)) and radius (5).Next, we check if the point ((3, 5)) is inside the circle. The distance from the center ((3, 4)) to ((3, 5)) is:[sqrt{(3 - 3)^2 + (5 - 4)^2} = 1]Since (1 < 5), the point ((3, 5)) is inside the circle.The shortest chord through a point inside a circle is perpendicular to the radius at that point. The radius from the center ((3, 4)) to ((3, 5)) is vertical, so the shortest chord must be horizontal. The equation of the horizontal line passing through ((3, 5)) is (y = 5).Substituting (y = 5) into the circle's equation:[(x - 3)^2 + (5 - 4)^2 = 25][(x - 3)^2 + 1 = 25][(x - 3)^2 = 24][x - 3 = pm 2sqrt{6}]The points of intersection are ((3 + 2sqrt{6}, 5)) and ((3 - 2sqrt{6}, 5)). The distance between these points is:[|(3 + 2sqrt{6}) - (3 - 2sqrt{6})| = 4sqrt{6}]Thus, the length of the shortest chord passing through the point ((3, 5)) is (boxed{4sqrt{6}}).]

🔑:AnalysisFirst, find the coordinates of the center of the circle and its radius. Then, calculate the distance between the point ((3,5)) and the center of the circle ((3,4)). Finally, apply the Pythagorean theorem to obtain the result.SolutionGiven (x^{2}+y^{2}-6x-8y=0), we can rewrite it as ((x-3)^{2}+(y-4)^{2}=25),Thus, the coordinates of the center of the circle are ((3,4)), and the radius is (r=5),Since the slope of the line passing through points ((3,5)) and ((3,4)) is undefined,The slope of the shortest chord passing through ((3,5)) is (k=0),Also, the distance (d) between points ((3,5)) and ((3,4)) is 1,Therefore, the length of the shortest chord passing through ((3,5)) is[2sqrt{r^{2}-d^{2}}=2sqrt{25-1}=4sqrt{6}].Hence, the answer is (boxed{4sqrt{6}}).