🧠:[<think>Okay, so I need to find all natural numbers x, y, and z such that 1 + 2^x * 3^y = z². Hmm, let me start by understanding the equation better. First, since x, y, and z are natural numbers, they have to be positive integers (assuming natural numbers start from 1, but sometimes they start from 0. Wait, the problem says natural numbers, so probably starting from 1. But maybe I should check cases where x or y is 0 just in case. Although, if the problem says natural numbers, perhaps 0 isn't allowed. Hmm, maybe I should consider both possibilities. Let me note that down. If x or y is 0, does that give a solution? Let's see. If x=0, then 2^0=1, so 1 + 1*3^y = z². So that would be 1 + 3^y = z². Similarly, if y=0, then 1 + 2^x *1 = z². So maybe those cases could also be considered. But since the problem states natural numbers, and sometimes natural numbers are considered to start at 1, but sometimes 0 is included. Hmm. Well, maybe I should check both possibilities and see if any solutions come up. If they do, then perhaps the problem allows x or y to be 0. If not, then maybe the problem wants x, y, z ≥1. Let me keep that in mind.So, the equation is 1 + 2^x * 3^y = z². Let me think about how to approach this. It's a Diophantine equation. Maybe I can factor it or look for patterns. Let me see.First, let's try small values for x and y and see if we can find z. Maybe that will help spot a pattern or give some insight.Starting with x=1, y=1: 1 + 2*3 = 1 + 6 =7, which is not a square.x=1, y=2: 1 + 2*9 =1 + 18=19, not a square.x=1, y=3:1 +2*27=55, not a square.x=1, y=4:1 +2*81=163, not a square.x=1, y=5:1 +2*243=487, not a square.x=1, y=0: if y=0 is allowed, then 1 +2*1=3, not a square. So x=1 seems not promising.x=2, y=1:1 +4*3=13, not a square.x=2, y=2:1 +4*9=37, not a square.x=2, y=3:1 +4*27=109, not a square.x=2, y=4:1 +4*81=325, which is 5*65, not a square.x=2, y=5:1 +4*243=973, not a square.x=3, y=1:1 +8*3=25, which is 5². Oh, that's a solution! So x=3, y=1, z=5.Wait, x=3, y=1 gives 1 + 8*3=25, which is 5². So that's a valid solution. Let me note that down.x=3, y=2:1 +8*9=73, not a square.x=3, y=3:1 +8*27=217, not a square.x=3, y=4:1 +8*81=649, not a square.x=3, y=5:1 +8*243=1945, not a square.x=4, y=1:1 +16*3=49, which is 7². So z=7. Another solution: x=4, y=1, z=7.x=4, y=2:1 +16*9=145, not a square.x=4, y=3:1 +16*27=433, not a square.x=4, y=4:1 +16*81=1297, not a square.x=5, y=1:1 +32*3=97, not a square.x=5, y=2:1 +32*9=289, which is 17². So x=5, y=2, z=17. Another solution.x=5, y=3:1 +32*27=865, not a square.x=5, y=4:1 +32*81=2593, not a square.x=6, y=1:1 +64*3=193, not a square.x=6, y=2:1 +64*9=577, not a square.x=6, y=3:1 +64*27=1729, which is 12³, but not a square. 1729=41² + 8*9, wait 41²=1681, 42²=1764, so 1729 is between those.x=6, y=4:1 +64*81=5185, not a square.x=7, y=1:1 +128*3=385, not a square.x=7, y=2:1 +128*9=1153, not a square.x=7, y=3:1 +128*27=3457, not a square.x=8, y=1:1 +256*3=769, not a square.x=8, y=2:1 +256*9=2305, not a square.x=8, y=3:1 +256*27=6913, not a square.Hmm, so so far, the solutions I found are:(x=3, y=1, z=5)(x=4, y=1, z=7)(x=5, y=2, z=17)Wait, let's check x=5, y=2: 2^5=32, 3²=9, 32*9=288, 288+1=289=17². Correct.Let me check higher values, maybe. Let's try x=5, y=3: 32*27=864 +1=865. 29²=841, 30²=900, so 865 is not a square.x=5, y=4:32*81=2592+1=2593, which is 50²=2500, 51²=2601, so no.x=5, y=5:32*243=7776+1=7777, not a square.x=6, y=3:64*27=1728 +1=1729, which is 41²=1681, 42²=1764, so not a square.x=6, y=5:64*243=15552+1=15553, not a square.x=7, y=4:128*81=10368+1=10369. Let's see sqrt(10369). 100²=10000, 102²=10404, so between them. 101²=10201, 101² + 168=10369. Not a square.x=7, y=5:128*243=31104+1=31105. 176²=30976, 177²=31329. Not a square.x=0, y=1: If x=0, 1 +1*3=4=2². So x=0, y=1, z=2. But if natural numbers start at 1, then x=0 is invalid. Hmm.Similarly, y=0, x=3:1 +8*1=9=3². So x=3, y=0, z=3. Again, if y=0 is allowed, then that's a solution.But the problem says natural numbers x, y, z. So, depending on the definition, 0 might not be included. Therefore, maybe these solutions are invalid. Let me check the problem statement again. It says "natural numbers". Usually, in number theory problems, natural numbers start at 1, so x, y, z ≥1. So x=0 or y=0 would not be allowed. Therefore, the solutions with x=3, y=1; x=4, y=1; x=5, y=2 are valid. Let me confirm.Wait, but z=5,7,17 are all natural numbers. So, the solutions are:(3,1,5), (4,1,7), (5,2,17). Are there more?Let me check x=10, y= something. Maybe larger exponents. Let's see, 2^x *3^y = z² -1. So z² -1 factors as (z-1)(z+1). So 2^x *3^y = (z-1)(z+1). Since z-1 and z+1 are two consecutive even numbers (since z is at least 2, as z² is at least 4, so z-1 and z+1 differ by 2). Therefore, their GCD is 2. Because two consecutive even numbers have GCD 2. Therefore, (z-1) and (z+1) are both divisible by 2, but not by 4, unless one is divisible by 4 and the other by 2. Wait, let me think. For example, if z is odd, then z-1 and z+1 are both even. If z is even, then z² is even, but 2^x *3^y is even only if x ≥1, which it is (since natural numbers). Wait, but if z is even, then z² is even, so 1 + even = even? But 1 + even is odd. Therefore, z² must be odd, so z must be odd, hence z-1 and z+1 are consecutive even numbers. So, one of them is divisible by 4 and the other is divisible by 2. Therefore, their product is divisible by 8. So 2^x *3^y must be divisible by 8. Hence, x must be at least 3. So x ≥3. Because 2^3=8. So x must be at least 3. Which matches the solutions we found earlier: x=3,4,5. So that gives us a starting point.Therefore, since z must be odd, let's let z = 2k +1, where k is a natural number. Then, z² = (2k +1)^2 =4k² +4k +1. Therefore, the equation becomes:1 + 2^x *3^y =4k² +4k +1Subtracting 1 from both sides:2^x *3^y =4k² +4k =4k(k +1)So 2^{x} *3^{y} =4k(k +1)Divide both sides by 4:2^{x -2} *3^{y} =k(k +1)Now, k and k +1 are consecutive integers, so they are coprime. Therefore, since their product is 2^{x-2} *3^y, which is a product of prime powers, and k and k +1 are coprime, each of them must be a product of some combination of these primes. Therefore, one of them must be a power of 2 and the other a power of 3, multiplied by some constants. Wait, but 2 and 3 are primes. Since k and k +1 are coprime, each must be a multiple of different prime factors. Therefore, one of them is a power of 2 and the other is a power of 3, but possibly multiplied by some factors. Wait, but the right-hand side is k(k+1) which is equal to 2^{x-2} *3^{y}. Therefore, since k and k+1 are coprime, one of them must be equal to 2^{x-2} and the other must be equal to 3^{y}, or vice versa. Wait, but they could also have some factors of 2 and 3. Wait, but if k and k+1 are coprime, their prime factors are disjoint. Therefore, since 2^{x-2} *3^y is the product, then one of k or k+1 must be equal to 2^{x-2} and the other must be equal to 3^y. Because otherwise, if one had both 2 and 3 as factors, then it would share a common factor with the other, which is not possible since they are coprime. Therefore, either:Case 1: k = 2^{x -2} and k +1 =3^{y}ORCase 2: k =3^{y} and k +1=2^{x -2}Therefore, we have two cases to consider.Case 1: 3^{y} =2^{x -2} +1Case 2:2^{x -2}=3^{y} +1These are both exponential Diophantine equations. Let's analyze each case.Starting with Case 1:3^{y} =2^{x -2} +1We can write this as 3^y -1 =2^{x -2}We need to find natural numbers x ≥3 and y ≥1 such that 3^y -1 is a power of 2.Similarly, in Case 2:2^{x -2} -1=3^{y}So, in this case, 2^{x -2} -1 is a power of 3.So, let's tackle Case 1 first.Case 1:3^y -1=2^{x-2}We need to find y and x such that 3^y -1 is a power of 2.Let's try small values of y:y=1:3 -1=2=2^1. Therefore, x-2=1 ⇒ x=3. So x=3, y=1. Which corresponds to our first solution.y=2:9 -1=8=2^3. Therefore, x-2=3 ⇒x=5. So x=5, y=2. Which is another solution we found earlier.y=3:27 -1=26, which is not a power of 2.y=4:81 -1=80=16*5, not a power of 2.y=5:243 -1=242=2*121=2*11², not a power of 2.Similarly, higher y will result in 3^y -1 being even, but since 3^y is odd, subtracting 1 gives even, but as y increases, 3^y -1 grows rapidly and likely won't be a power of 2. So, for y≥3, no solutions.Therefore, in Case 1, the only solutions are (x=3, y=1) and (x=5, y=2).Now, moving to Case 2:2^{x -2} -1=3^{y}Here, we need 2^{x -2} -1 is a power of 3.Let’s denote x-2 as a variable, say m. Then, 2^m -1=3^y.So, m must be at least 1 (since x is natural number ≥3 ⇒x-2 ≥1).Let’s check m=1:2^1 -1=1=3^0. But y must be natural number ≥1, so y=0 is invalid here.m=2:4 -1=3=3^1. So y=1. Therefore, x-2=2 ⇒x=4. So x=4, y=1. That's our second solution.m=3:8 -1=7, which is not a power of 3.m=4:16 -1=15=3*5, not a power of 3.m=5:32 -1=31, not a power of 3.m=6:64 -1=63=7*9=7*3², not a power of 3.m=7:128 -1=127, prime.m=8:256 -1=255=3*5*17, not a power of 3.Similarly, as m increases, 2^m -1 becomes large and is unlikely to be a power of 3. So the only solution here is m=2, which corresponds to x=4, y=1.Therefore, combining both cases, the solutions are:From Case 1: (x=3, y=1), (x=5, y=2)From Case 2: (x=4, y=1)These correspond to the solutions we found earlier by trial.Are there any other solutions? Let's check for m=0, but m=x-2, so m=0 would mean x=2, but x must be natural number ≥3, so m=0 is invalid here.Thus, the only solutions are the three mentioned above.Wait, but let me check for m=1 in Case 2:2^1 -1=1=3^0. If y=0 is allowed, then x=3, y=0. But since y is a natural number, which is at least 1, this is invalid.Similarly, in Case 1, if y=0, 3^0 -1=0=2^{-something}, which is not valid. So, y=0 is invalid.Therefore, the only valid solutions are (3,1,5), (4,1,7), and (5,2,17). Let me verify each:For (3,1,5):1 +2^3*3^1=1 +8*3=25=5². Correct.For (4,1,7):1 +2^4*3^1=1 +16*3=49=7². Correct.For (5,2,17):1 +2^5*3²=1 +32*9=1 +288=289=17². Correct.Are there any other solutions?Let me check m=4 in Case 2:2^4 -1=15, which is not a power of 3. So no.In Case 1, y=3:3^3 -1=26, not a power of 2. So no.So, seems like these are the only solutions.Therefore, all natural numbers x, y, z satisfying the equation are (3,1,5), (4,1,7), and (5,2,17).But wait, let me check if there could be larger values. For example, in Case 1, is there a larger y where 3^y -1 is a power of 2? For y=3, 26 isn't a power of 2. For y=4,80 is not. For y=5,242. For y=6, 729-1=728, which is 8*91=8*7*13. Not a power of 2.Similarly, in Case 2, m=6, 2^6 -1=63=7*3², not a pure power of 3.Another way: Since in Case 1, 3^y -1=2^{m}, where m=x-2. Catalan's conjecture (now Mihăilescu's theorem) states that the only solution in the natural numbers of the equation x^a − y^b = 1 for x, y >1 and a, b >1 is 3² −2³=1. But in our case, if y≥2 and m≥2, then 3^y -2^m=1. The only solution would be y=2, m=3: 9 -8=1. Which corresponds to x=5, y=2. Similarly, for y=1, 3 -2^{m}=1 ⇒2^{m}=2 ⇒m=1⇒x=3. So yes, Mihăilescu's theorem tells us that the only solutions in natural numbers where exponents are greater than 1 are the ones we found. So that confirms that in Case 1, we have all solutions.In Case 2, 2^{m} -1=3^y. If m≥2 and y≥1. We can apply Mihăilescu's theorem here as well. The equation 2^m -3^y=1. The known solution is m=2, y=1:4 -3=1. Also, m=1:2 -1=1, but y=0, which is invalid. Any other solutions? According to Catalan's conjecture, the only solution in natural numbers for x^a - y^b =1 with x,y,a,b>1 is 3² -2³=1. So in this case, 2^m -3^y=1. If m>2 and y>1, there are no solutions. Let's check y=2:3²=9. So 2^m=10, which is not a power of 2. y=3:27. 2^m=28, not a power of 2. So, only solution is m=2, y=1. Therefore, Case 2 only gives x=4, y=1.Therefore, combining both cases, all solutions are (3,1,5), (4,1,7), (5,2,17). Therefore, the answer should be these three triples.Final AnswerThe natural number solutions are (boxed{(3, 1, 5)}), (boxed{(4, 1, 7)}), and (boxed{(5, 2, 17)}).</think>To determine all natural numbers (x, y,) and (z) for which (1 + 2^x cdot 3^y = z^2), we start by analyzing the equation. First, we recognize that (z^2 - 1 = 2^x cdot 3^y). This can be factored as ((z - 1)(z + 1) = 2^x cdot 3^y). Since (z - 1) and (z + 1) are consecutive even numbers, their greatest common divisor is 2. Therefore, one of them must be a power of 2 and the other a power of 3, each multiplied by 2. We consider two cases:1. (z - 1 = 2^{x-1}) and (z + 1 = 2 cdot 3^y)2. (z - 1 = 2 cdot 3^y) and (z + 1 = 2^{x-1}) Case 1: (3^y - 2^{x-2} = 1)- For (y = 1), we get (3 - 1 = 2), which gives (x = 3).- For (y = 2), we get (9 - 8 = 1), which gives (x = 5). Case 2: (2^{x-2} - 3^y = 1)- For (x - 2 = 2), we get (4 - 3 = 1), which gives (x = 4) and (y = 1).By checking these cases and verifying the solutions, we find the natural number solutions:1. (x = 3), (y = 1), (z = 5)2. (x = 4), (y = 1), (z = 7)3. (x = 5), (y = 2), (z = 17)Thus, the natural number solutions are (boxed{(3, 1, 5)}), (boxed{(4, 1, 7)}), and (boxed{(5, 2, 17)}).,<think>Okay, so I need to find all natural numbers x, y, and z such that 1 plus 2 to the power of x times 3 to the power of y equals z squared. Let me write that down again to make sure I have it right: 1 + 2^x * 3^y = z². All variables are natural numbers, so they have to be positive integers. Alright, let's start thinking.First, maybe I can start by plugging in some small values for x and y and see if I can get z squared. Let's see. For x=1 and y=1, we have 1 + 2*3 = 1 + 6 = 7, which is not a square. x=1, y=2: 1 + 2*9 = 1 + 18 = 19, not a square. x=1, y=3: 1 + 2*27 = 55, not a square. Hmm. Maybe x=2? Let's try x=2, y=1: 1 + 4*3 = 13, not square. x=2, y=2: 1 + 4*9 = 37, still not a square. x=2, y=3: 1 + 4*27 = 109, nope. Maybe higher y? But this might take too long. Maybe there's a pattern here.Alternatively, let's consider the equation: z² - 2^x * 3^y = 1. So z² - N = 1 where N is 2^x * 3^y. That reminds me of Pell's equation, but Pell's equation is of the form z² - D*N² = 1. Not exactly the same here. But in this case, N is not a square necessarily. Wait, but maybe if z² - 1 = 2^x * 3^y, then (z-1)(z+1) = 2^x * 3^y. Since z-1 and z+1 are two consecutive even numbers (because z is either even or odd, but z² is either 0 or 1 mod 4. Wait, if z is even, z² is 0 mod 4; if z is odd, z² is 1 mod 4. So 1 + 2^x*3^y = z². If z is even, z² is 0 mod 4, so 1 + 2^x*3^y ≡ 0 mod 4. Then 2^x*3^y ≡ 3 mod 4. But 2^x*3^y: if x ≥ 2, then 2^x is 0 mod 4, so 2^x*3^y is 0 mod 4, which would mean 1 + 0 ≡ 1 mod 4, which can't be 0 mod 4. Therefore, if z is even, then x must be 1. Because if x=1, then 2*3^y ≡ 2 mod 4, so 1 + 2*3^y ≡ 3 mod 4, but z² is 0 mod 4. So contradiction. Therefore, z cannot be even. Therefore, z must be odd. So z is odd, so z = 2k + 1 for some integer k. Then z² = (2k +1)^2 = 4k² + 4k +1. So 4k² +4k +1 = 1 + 2^x*3^y, which simplifies to 4k(k +1) = 2^x*3^y. So 2² * k(k +1) = 2^x * 3^y. Since k and k+1 are consecutive integers, they are coprime. So in the equation 2² * k(k+1) = 2^x * 3^y, since k and k+1 are coprime, each must be a power of 2 or a power of 3. Also, the left-hand side is 2² times k(k+1). So k(k+1) must be equal to 2^{x-2}*3^y. But since k and k+1 are coprime, each must be a power of 2 or a power of 3. So either k is a power of 2 and k+1 is a power of 3 multiplied by some factor, or vice versa. Wait, but since k and k+1 are coprime, one of them must be a power of 2 and the other must be a power of 3. Because the product k(k+1) is 2^{x-2} *3^y, so since they are coprime, each must individually be a power of 2 or a power of 3. So possible cases:Case 1: k = 2^a and k+1 = 3^b, where a = x - 2 - c, but maybe better to write 2^a *3^b = 2^{x -2} *3^y. Wait, no. Wait, since k and k+1 are coprime, and their product is 2^{x -2} *3^y. Then one of them must be 2^{x -2} and the other must be 3^y. Because they are coprime. So either:Case 1: k = 2^{x -2} and k +1 = 3^yOrCase 2: k = 3^y and k +1 = 2^{x -2}So we have two cases. Let's consider Case 1 first: k = 2^{x -2}, k +1 = 3^y.So 2^{x -2} +1 = 3^y. We need to solve for natural numbers x and y. Since x must be at least 2 (because x is a natural number, so x-2 must be non-negative). Wait, x is a natural number. If x=2, then x-2=0, so 2^0=1, so k=1. Then k+1=2=3^y. So 3^y=2. But y is natural, so no solution here. If x=3, then k=2^{1}=2, so k+1=3=3^1. So y=1. So that gives x=3, y=1. Then z²=1 +2^3*3^1=1+8*3=1+24=25, so z=5. That's a solution: (x,y,z)=(3,1,5). Let's check: 1 + 2^3*3^1=1+24=25=5². Yes.Next, x=4: k=2^{2}=4. Then k+1=5=3^y. But 5 is not a power of 3. So no solution.x=5: k=8, k+1=9=3². So y=2. So x=5, y=2. Then z²=1 +2^5*3²=1+32*9=1+288=289=17². So z=17. That works. So (5,2,17).x=6: k=16, k+1=17=3^y. 17 is not a power of 3.x=7: k=32, k+1=33. Not a power of 3.x=8: k=64, k+1=65. Not a power of 3.x=9: k=128, k+1=129. Not a power of 3.x=10: k=256, k+1=257. Not a power of 3. So seems like the only solutions in Case 1 are x=3,y=1 and x=5,y=2.Now let's check Case 2: k = 3^y and k +1 = 2^{x -2}So 3^y +1 = 2^{x -2}Again, solving for natural numbers y and x. Since k =3^y must be positive, and x-2 must be at least 1, since x is natural, so x-2 >=1 implies x >=3.Let me check y=1: 3 +1=4=2^2. Therefore, x-2=2, so x=4. Then z²=1 +2^4*3^1=1+16*3=1+48=49=7². So z=7. Therefore, solution (4,1,7).y=2:9 +1=10=2^{x -2}. 10 is not a power of 2.y=3:27 +1=28=2^{x -2}. 28 is not a power of 2.y=4:81 +1=82. Not a power of 2.y=5:243 +1=244. Not a power of 2.Similarly, higher y's will result in 3^y +1. Since 3^y is odd, adding 1 makes it even, but unless it's a power of 2, which only happens for y=1 in this case. So only solution here is y=1, x=4.Thus, in Case 2, we have one solution: (4,1,7).So combining both cases, we have three solutions: (3,1,5), (4,1,7), and (5,2,17).Wait, but hold on, let's check if there are any other possibilities when either k or k+1 is 1. For example, if k=1, then k+1=2. So 1*2=2, which would mean 2²*1*2=2^3, but in our equation, 2²*k(k+1)=2^x*3^y. If k=1, then 2²*1*2=8=2^3*3^0. So x=3, y=0. But y is supposed to be a natural number, so y=0 is excluded. Similarly, if k+1=1, then k=0, which is not a natural number. So those cases don't contribute.Wait, but when x=3, in Case 1, we had k=2^{1}=2, k+1=3. So 2*3=6. Then 2²*6=24=2^3*3^1. So 2^3*3^1=24, which gives z²=25, z=5. Correct.Similarly, in the case of x=5, y=2: k=8, k+1=9. 8*9=72. 2²*72=288=2^5*3², which is 32*9=288. Then z²=289=17². Correct.In the case of x=4, y=1: k=3, k+1=4. 3*4=12. 2²*12=48=2^4*3^1. Then z²=49=7². Correct.So those are the only solutions where k and k+1 are pure powers of 2 and 3. Are there any other possibilities?Wait, let me check if there could be other cases where k or k+1 have both 2 and 3 as factors. But since k and k+1 are coprime, they can't share any common factors. So if, for example, k is a multiple of 2 and 3, then k+1 can't be. But since the product k(k+1) is 2^{x -2} *3^y, each of k and k+1 must themselves be products of only 2s or 3s, but given that they are coprime, each must be a pure power of 2 or a pure power of 3. Because if, say, k was 2^a *3^b, then since k and k+1 are coprime, k+1 can't have factors of 2 or 3. But then k+1 would have to be 1, but k+1=1 would imply k=0, which isn't natural. Therefore, the only possibilities are that one of k or k+1 is a pure power of 2 and the other is a pure power of 3. Hence, the two cases we considered are exhaustive.Therefore, the solutions are:From Case 1:x=3, y=1, z=5x=5, y=2, z=17From Case 2:x=4, y=1, z=7Let me check if there are any other solutions with small exponents. For example, maybe x=0 or y=0? But the problem states natural numbers, which typically start at 1. So x, y, z ≥1.Wait, let me check x=0: 2^0=1. So equation becomes 1 +1*3^y= z². So 1 +3^y=z². So z² -3^y=1. Then z=2 when y=1: 2² -3=4-3=1. So that's a solution: x=0, y=1, z=2. But x is supposed to be a natural number, so if x=0 is excluded, then this is not a solution. Similarly, if y=0: 3^0=1. So 1 +2^x*1=z². So 2^x +1=z². Then we can have x=3: 8 +1=9=3². So x=3, y=0, z=3. But again, y=0 is not natural. So these solutions are excluded if we stick strictly to natural numbers starting at 1.Therefore, the only solutions in natural numbers x,y,z are the three we found earlier: (3,1,5), (4,1,7), (5,2,17).Wait a minute, but let me check for x=2. If x=2, then 2^2=4. Then 1 +4*3^y=z². Let's try y=1:1 +4*3=13, not square. y=2:1 +4*9=37, not square. y=3:1 +4*27=109, not square. So no.Similarly, x=1:1 +2*3^y. y=1:7, not square. y=2:19, nope. y=3:55, no. y=4:1 +2*81=163, not a square. So nothing here.What about x=6? Let's try x=6, y=3:1 +64*27=1 +1728=1729. Wait, 1729 is the famous taxicab number, which is 12³ +1³, but it's not a perfect square. sqrt(1729)≈41.59, not integer.Alternatively, maybe x=0 and y=2:1 +1*9=10, not square. Not helpful.Alternatively, is there a solution with z=1? Then z²=1, so 1 +2^x*3^y=1 implies 2^x*3^y=0, which is impossible. So z must be at least 2.Wait, z=2: z²=4. Then 2^x*3^y=3. So x=1, y=1. But then (x,y,z)=(1,1,2). Let's check:1 +2*3=7≠4. So nope. z=3: z²=9. 2^x*3^y=8. So 8=2^3*3^0. So x=3, y=0. But y=0 is not natural. So invalid.z=4:16. 2^x*3^y=15. 15 is 3*5. Not possible as product of 2s and 3s. So no.z=5:25. 2^x*3^y=24. 24=2^3*3^1. So x=3, y=1. Which is one of our solutions.z=6:36. 2^x*3^y=35. 35=5*7. No.z=7:49. 2^x*3^y=48=16*3=2^4*3^1. So x=4, y=1. That's another solution.z=8:64. 2^x*3^y=63=7*9. Not possible.z=9:81. 2^x*3^y=80=16*5. No.z=10:100. 2^x*3^y=99=9*11. No.z=11:121. 2^x*3^y=120=8*15=8*3*5. So 2^3*3^1*5. But there's a 5, so no.z=12:144. 2^x*3^y=143=11*13. No.z=13:169. 2^x*3^y=168=8*21=8*3*7. Again, prime 7.z=14:196. 2^x*3^y=195=5*39=5*3*13.z=15:225. 2^x*3^y=224=16*14=16*2*7.z=16:256. 2^x*3^y=255=3*85.z=17:289. 2^x*3^y=288=32*9=2^5*3^2. So x=5, y=2. Which is another solution.So as we can see, the only z values that work are 5,7,17, which correspond to the three solutions we found earlier. Beyond that, z=17 gives 289, which is covered. So it seems like these are the only solutions.Therefore, the natural number solutions (x,y,z) are (3,1,5), (4,1,7), and (5,2,17).Final AnswerThe natural number solutions are boxed{(3, 1, 5)}, boxed{(4, 1, 7)}, and boxed{(5, 2, 17)}.</think>To determine all natural numbers (x, y,) and (z) for which (1 + 2^x cdot 3^y = z^2), we start by rewriting the equation as (z^2 - 1 = 2^x cdot 3^y). This can be factored as ((z - 1)(z + 1) = 2^x cdot 3^y). Since (z - 1) and (z + 1) are consecutive even numbers, they must be coprime. Therefore, each must be a power of 2 or a power of 3.We consider two cases:Case 1: (z - 1 = 2^a) and (z + 1 = 3^b) Here, (3^b - 2^a = 2). Solving this, we find:- For (a = 1) and (b = 1), we get (z - 1 = 2) and (z + 1 = 3), leading to (z = 3). This gives (2^x cdot 3^y = 8 cdot 3 = 24), so (x = 3) and (y = 1). The solution is ((x, y, z) = (3, 1, 5)).- For (a = 3) and (b = 2), we get (z - 1 = 8) and (z + 1 = 9), leading to (z = 9). This gives (2^x cdot 3^y = 32 cdot 9 = 288), so (x = 5) and (y = 2). The solution is ((x, y, z) = (5, 2, 17)).Case 2: (z - 1 = 3^b) and (z + 1 = 2^a) Here, (2^a - 3^b = 2). Solving this, we find:- For (a = 2) and (b = 1), we get (z - 1 = 3) and (z + 1 = 4), leading to (z = 4). This gives (2^x cdot 3^y = 16 cdot 3 = 48), so (x = 4) and (y = 1). The solution is ((x, y, z) = (4, 1, 7)).After verifying all possible cases and checking for other potential solutions, we conclude that the natural number solutions are:[boxed{(3, 1, 5)}, boxed{(4, 1, 7)}, boxed{(5, 2, 17)}]]

🔑:Solution. It is easily verified that for z=1,2,3, the given equation has no solution.Let z geq 4. We have 2^{x} 3^{y}=(z-1)(z+1). At most one of z-1 and z+1 is divisible by 3, because if 3 mid z-1 and 3 mid z+1 then 3 mid(z+1)-(z-1)=2, which is impossible. Similarly, since 2 mid(z-1)(z+1), both z-1 and z+1 are divisible by 2, and only one of them can be divisible by 4, because if 4 mid z-1 and 4 mid z+1 then 4 mid(z+1)-(z-1)=2, which is impossible. Therefore, we have the following cases:1^{circ} z+1=2 cdot 3^{y}, z-1=2^{x-1} and2^{circ} z+1=2^{x-1}, z-1=2 cdot 3^{y}.Case 1^{circ} Let z+1=2 cdot 3^{y}, z-1=2^{x-1}.By subtracting the given equations, we have2 cdot 3^{y}-2^{x-1}=2 text { i.e. } 3^{y}-2^{x-2}=1For x=2, it is easily verified that the given equation has no solution.For x=3, we have 3^{y}=1+2=3 i.e. y=1 and it is easily obtained that z=5.Thus, one solution is (x, y, z)=(3,1,5).If x geq 4, we have 3^{y} equiv 1(bmod 4), so it follows that y is even, i.e., y=2 y_{1}, y_{1} in mathbb{N}.Now we have 3^{2 y_{1}}-1=2^{x-2} i.e. left(3^{y_{1}}-1right)left(3^{y_{1}}+1right)=2^{x-2}, from which it is easily obtained that 3^{y_{1}}-1=2 and 3^{y_{1}}+1=2^{x-3}, i.e., y_{1}=1 or y=2, and now it is easily obtained that x=5 and z=17.Thus, the solution is (x, y, z)=(5,2,17).Case 2^{circ} Let z+1=2^{x-1}, z-1=2 cdot 3^{y}, then 2^{x-1}-2 cdot 3^{y}=2, i.e.2^{x-2}-3^{y}=1For y=1, we have x=4 and z=7, so the solution is (x, y, z)=(4,1,7).If y geq 2, then we have 2^{x-2} equiv 1(bmod 3), i.e., x-2 must be an even number, so let x-2=2 x_{1}. Then we have 3^{y}=2^{2 x_{1}}-1=left(2^{x_{1}}-1right)left(2^{x_{1}}+1right), from which it follows that 2^{x_{1}}-1=1 or 3. If 2^{x_{1}}-1=1, we get x_{1}=1 i.e., x=4, and we obtain the solution (4,1,7).If 2^{x_{1}}-1=3, then x_{1}=2 i.e., x=6 and we get 3^{y}=15, which is impossible.Thus, the solutions to the given equation are (x, y, z)=(3,1,5),(5,2,17),(4,1,7).

🧠:[<think>Okay, so I need to find the expression for f(x) when x is in (0, +∞), given that f(x) is an odd function and when x is in (-∞, 0], f(x) = x(x - 1). Hmm, let me think about how odd functions work. First, recalling that an odd function satisfies the property f(-x) = -f(x) for all x in its domain. That means if I know the function's values for negative x, I can find the values for positive x by using this property. Since the function is defined for all real numbers, and we have the expression for x ≤ 0, the part for x > 0 should be determined by reflecting and negating the left side. So, let me write down what I know. For x ≤ 0, f(x) = x(x - 1). Let me check that expression for a negative x. Suppose x is negative, then x - 1 would be even more negative, right? So, multiplying two negative numbers (x and x - 1) would give a positive result. Wait, but if x is negative, x - 1 is also negative, so their product is positive. Hmm, but since the function is odd, when x is positive, f(x) should be the negative of f(-x). Let me verify that.Let me pick a specific example. Suppose x = -2, which is in (-∞, 0]. Then f(-2) = (-2)(-2 - 1) = (-2)(-3) = 6. Since the function is odd, f(2) should be -f(-2) = -6. So, when x is 2, f(2) should be -6. Now, if I can figure out the expression for x > 0 that gives -6 when x = 2, that would be the right expression.Given that for x ≤ 0, f(x) = x(x - 1). Let's generalize. For x > 0, -x is negative, so f(-x) = (-x)((-x) - 1) = (-x)(-x - 1). Then, since f is odd, f(x) = -f(-x) = -[(-x)(-x - 1)]. Let me compute that step by step.First, compute f(-x): (-x)(-x - 1) = (-x)*(-x) + (-x)*(-1) = x² + x. Then, f(x) = -f(-x) = -(x² + x) = -x² - x. So, does that mean for x > 0, f(x) = -x² - x? Let's check with the example where x = 2. Plugging into -x² - x gives -4 - 2 = -6, which matches the expected result. So that seems right.Wait, but let me make sure I didn't make a sign mistake. Let's take another example. Let’s take x = 1. Then, for x = 1, f(1) should be -f(-1). Let's compute f(-1): (-1)(-1 - 1) = (-1)(-2) = 2. Therefore, f(1) should be -2. Plugging into the expression -x² - x: -(1)^2 -1 = -1 -1 = -2. Correct. Another example: x = 0.5. Then f(0.5) should be -f(-0.5). Compute f(-0.5): (-0.5)(-0.5 -1) = (-0.5)(-1.5) = 0.75. Therefore, f(0.5) should be -0.75. Plugging into -x² - x: -(0.25) - 0.5 = -0.75. Correct again.Therefore, the expression for x > 0 is f(x) = -x² - x. Let me write that in a factored form to see if it can be simplified. -x² - x = -x(x + 1). Hmm, so f(x) = -x(x + 1) for x > 0.Alternatively, maybe the problem expects it in expanded form, but probably factored is acceptable. Let me check the original problem: When x ∈ (-∞, 0], f(x) = x(x - 1). They presented it in factored form, so maybe the answer should also be in factored form. So, f(x) = -x(x + 1) for x > 0. That seems consistent.Alternatively, if I want to present it similar to the given expression, x(x - 1) is a quadratic. For the positive side, it's -x² - x, which is different. Alternatively, maybe factor out a negative sign: -x(x + 1). Either way is correct, but probably factored is better here.Let me confirm once more. For any x > 0, we can write f(x) = -f(-x). Since -x is negative, f(-x) = (-x)((-x) - 1) = (-x)(-x -1) = x(x +1). Therefore, f(x) = -x(x +1). Yep, that matches. So that's the expression for x > 0.I think that's solid. Just to recap: Because the function is odd, f(x) = -f(-x). For x positive, -x is negative, so substitute into the given expression, compute that, then take the negative. The key was replacing x with -x in the original expression and then negating it. Another way to think about it is, if f is odd, then its graph is symmetric about the origin. So if we have the left half of the graph (for x ≤0), we can rotate it 180 degrees about the origin to get the right half. The original left half is a quadratic function. When rotated, it becomes another quadratic. The algebraic manipulation confirms that the right half is -x(x +1). I don't see any mistakes in the reasoning. All the test cases worked, so I think that's the correct answer.Final AnswerThe expression for ( f(x) ) when ( x in (0, +infty) ) is boxed{-x(x + 1)}.</think>Given that ( f(x) ) is an odd function and ( f(x) = x(x - 1) ) for ( x in (-infty, 0] ), we need to find the expression for ( f(x) ) when ( x in (0, +infty) ).An odd function satisfies the property ( f(-x) = -f(x) ) for all ( x ). For ( x > 0 ), we can use this property by substituting ( -x ) into the given expression for ( x leq 0 ).1. For ( x leq 0 ), the function is given by ( f(x) = x(x - 1) ).2. For ( x > 0 ), we need to find ( f(x) ) using the odd function property. Let ( x > 0 ), then ( -x < 0 ). Therefore, we substitute ( -x ) into the given expression: [ f(-x) = (-x)(-x - 1) ]3. Simplify the expression: [ f(-x) = (-x)(-x - 1) = (-x)(-x) + (-x)(-1) = x^2 + x ]4. Using the odd function property ( f(x) = -f(-x) ): [ f(x) = - (x^2 + x) = -x^2 - x ]5. Factor the expression: [ f(x) = -x(x + 1) ]Thus, the expression for ( f(x) ) when ( x in (0, +infty) ) is (boxed{-x(x + 1)}).,<think>Okay, so I need to find the expression for the function f(x) when x is greater than 0, given that f(x) is an odd function defined on all real numbers. And for x less than or equal to 0, it's given as f(x) = x(x - 1). Hmm, let me recall what an odd function is. An odd function satisfies the property that f(-x) = -f(x) for all x in its domain. That means if I know the value of the function at a negative x, I can find the value at the corresponding positive x by just taking the negative of f(-x). Since the function is defined for x ≤ 0 as f(x) = x(x - 1), then for any positive x (which is greater than 0), the corresponding negative value would be -x. So if I plug -x into the given function, I should get f(-x) = (-x)((-x) - 1). Then, using the property of odd functions, f(x) when x is positive should be -f(-x). Let me write that down step by step.First, let's compute f(-x) where x > 0. Since -x is in the interval (-∞, 0], we can substitute -x into the given expression:f(-x) = (-x)((-x) - 1)Let me simplify that:First, multiply the terms inside the parentheses:(-x) - 1 = -x - 1So now f(-x) = (-x)(-x - 1)Multiplying these out:(-x) * (-x) = x²(-x) * (-1) = xSo f(-x) = x² + xBut since f is an odd function, we have that f(x) = -f(-x) for x > 0. Therefore,f(x) = - (x² + x) = -x² - xWait, let me check that again. Let me do the multiplication step by step to make sure I didn't make a mistake.f(-x) = (-x) * [(-x) - 1] = (-x)(-x) + (-x)(-1) = x² + x. That's correct. So f(-x) = x² + x. Then f(x) = -f(-x) = -(x² + x) = -x² - x. So that seems right.Let me verify if this makes sense. Let's take a specific example. Suppose x = 1. Then according to our result, f(1) = -1² -1 = -1 -1 = -2. Let's check using the odd function property. f(-1) should be equal to -f(1). Let's compute f(-1) using the given expression for x ≤ 0: f(-1) = (-1)((-1) -1) = (-1)(-2) = 2. Therefore, -f(1) should equal 2, which implies f(1) = -2. That matches our earlier result. So that checks out.Another example: x = 2. Then f(2) = -2² -2 = -4 -2 = -6. Then f(-2) should be equal to -f(2) = 6. Let's compute f(-2) using the given expression: f(-2) = (-2)(-2 -1) = (-2)(-3) = 6. Which matches, so that's good.What about x = 0? The original function is defined at x=0 as f(0) = 0*(0 -1) = 0. Since the function is odd, f(-0) = -f(0), but -0 is 0, so f(0) = -f(0) implies f(0) = 0, which is consistent. So that works.So, putting it all together, for x > 0, f(x) = -x² -x. Let me write that in a standard form, maybe factor it. -x² -x can be written as -x(x + 1). Alternatively, we can write it as -x^2 -x. Both are correct. The problem just asks for the expression, so either form should be acceptable. Maybe they prefer it factored? Let me check the original expression for x ≤0: x(x -1). So when x ≤0, it's factored as x times (x -1). For x >0, we have -x(x +1). That's similar structure, so maybe writing it factored is better for consistency. So f(x) = -x(x +1) when x >0.Let me just check another point to be thorough. Let's take x = 0.5. Then f(0.5) should be -0.5*(0.5 +1) = -0.5*1.5 = -0.75. Using the odd function property: f(-0.5) = (-0.5)(-0.5 -1) = (-0.5)(-1.5) = 0.75. Then -f(0.5) = 0.75, so f(0.5) = -0.75. Which matches. So that works as well.Therefore, I think the correct expression for f(x) when x is in (0, +infty) is -x(x +1). Or expanded, that's -x² -x, but factored form might be preferable here. The problem didn't specify the form, so either should be correct, but since the original was given in factored form, maybe we should follow suit.Final AnswerThe expression for ( f(x) ) when ( x in (0, +infty) ) is boxed{-x(x + 1)}.</think>Given that ( f(x) ) is an odd function defined on ( (-infty, +infty) ) and ( f(x) = x(x - 1) ) for ( x in (-infty, 0] ), we need to find the expression for ( f(x) ) when ( x in (0, +infty) ).An odd function satisfies the property ( f(-x) = -f(x) ) for all ( x ). For ( x > 0 ), let ( -x ) be a point in ( (-infty, 0] ). Using the given expression for ( x leq 0 ):[f(-x) = (-x)((-x) - 1) = (-x)(-x - 1)]Simplifying this expression:[f(-x) = (-x)(-x - 1) = (-x)(-x) + (-x)(-1) = x^2 + x]Since ( f ) is an odd function, we have:[f(x) = -f(-x) = -(x^2 + x) = -x^2 - x]This can be factored as:[f(x) = -x(x + 1)]Thus, the expression for ( f(x) ) when ( x in (0, +infty) ) is (boxed{-x(x + 1)}).]

🔑:Let x in (0, +infty), then -x in (-infty, 0). Therefore, f(-x) = -x(-x-1) = x(x+1). Since f(x) is an odd function, we have f(x) = -f(-x) = -x(x+1). Thus, when x in (0, +infty), f(x) = boxed{-x(x+1)}.

🧠:[<think>Alright, let's tackle this problem. So, we need to find how many integers n satisfy 0 ≤ n ≤ 720 and n² ≡ 1 mod 720. Hmm, okay. First off, modular arithmetic here. The congruence n² ≡ 1 mod 720 means that when you divide n² - 1 by 720, there's no remainder. In other words, 720 divides (n² - 1). So, (n - 1)(n + 1) is divisible by 720. But 720 is a pretty big number. Let me factor that to see if breaking it down helps. 720 factors into 16 * 9 * 5, which is 2⁴ * 3² * 5. So, the prime factors are 2, 3, and 5 with exponents 4, 2, and 1 respectively. Since 720 is the modulus, the Chinese Remainder Theorem (CRT) might come into play here. Because if we can solve the congruence modulo each prime power factor and then combine the solutions, that should give us the total number of solutions modulo 720.So, CRT tells us that solving n² ≡ 1 mod 720 is equivalent to solving the system:n² ≡ 1 mod 16,n² ≡ 1 mod 9,n² ≡ 1 mod 5.Then, the number of solutions modulo 720 would be the product of the number of solutions modulo each prime power. So, we need to find the number of solutions for each modulus and then multiply them together.Starting with mod 16. Let's solve n² ≡ 1 mod 16. The solutions to this congruence are the numbers n such that when squared, they leave a remainder of 1 when divided by 16. Let's list n from 0 to 15 and see which squares give 1 mod 16.n: 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15n² mod 16: 0,1,4,9,0,9,4,1,0,1,4,9,0,9,4,1So, the n where n² ≡ 1 mod 16 are n ≡ 1, 7, 9, 15 mod 16. So, four solutions modulo 16.Next, mod 9. Solve n² ≡ 1 mod 9. Let's check n from 0 to 8:n: 0,1,2,3,4,5,6,7,8n² mod 9: 0,1,4,0,7,7,0,4,1So, n² ≡ 1 mod 9 when n ≡ 1 or 8 mod 9. So, two solutions modulo 9.Then, mod 5. Solve n² ≡ 1 mod 5. Let's check n from 0 to 4:n: 0,1,2,3,4n² mod 5: 0,1,4,4,1So, n² ≡ 1 mod 5 when n ≡ 1 or 4 mod 5. Two solutions modulo 5.So, according to CRT, the total number of solutions modulo 720 should be 4 (from mod 16) * 2 (mod 9) * 2 (mod 5) = 16 solutions. Therefore, in the range from 0 to 720 inclusive, there should be 16 integers that satisfy the condition. But wait, hold on. The problem states 0 ≤ n ≤ 720. Since 720 is congruent to 0 mod 720, but our modulus is 720, so numbers from 0 up to 719 are the residues. But the problem includes 720 as well. So, when n = 720, which is equivalent to 0 mod 720, n² = 0, which is not congruent to 1 mod 720. So, the solutions are from 0 to 719, but since 720 is included, check if 720 is a solution. But 720² = (720)^2, which is 0 mod 720, so not congruent to 1. So, the number of solutions is 16. But wait, the modulus 720 has 720 numbers, but the solutions are 16, so in each interval of 720 numbers, there are 16 solutions. But the problem is asking for n such that 0 ≤ n ≤ 720. So, that's 721 numbers. But since 720 is equivalent to 0 mod 720, which isn't a solution, so the number of solutions is still 16. Wait, but maybe the upper limit is inclusive, so if 720 is included, but since it's not a solution, we just have 16 solutions between 0 and 719, which is 720 numbers. Hence, 16 solutions. But wait, let me think again. Suppose the solutions are spaced every 720/ gcd(...). Wait, no. CRT tells us that when moduli are coprime, the number of solutions is the product of the number of solutions for each modulus. Since 16, 9, and 5 are pairwise coprime, the total number of solutions modulo 720 is 4*2*2=16. So, in each modulus 720 cycle, there are 16 solutions. Therefore, between 0 and 719 inclusive, there are 16 solutions. Since the problem includes 720, but 720 isn't a solution, the answer is 16. But wait, let me check. Maybe I made a mistake here. Let me verify with smaller cases. Let's take modulus 16*9=144. Suppose we have solutions modulo 16 and modulo 9. Then, if there are 4 solutions mod 16 and 2 solutions mod 9, then modulo 144, there should be 8 solutions. Let me test with modulus 144.Wait, for example, take n² ≡1 mod 16 and mod 9. Let's see if n=1: 1²=1, n=17: 17²=289 ≡ 289-144=145≡1 mod 144? Wait, 17²=289; 289 mod 144 is 289-2*144=289-288=1. So, 17²≡1 mod 144. Similarly, n= 1, 17, etc. So, each combination of solutions mod 16 and mod 9 gives a unique solution mod 144. So, 4*2=8 solutions mod 144. Similarly, mod 720, 16 solutions. But let's check another example. Let's take modulus 5. Suppose n²≡1 mod 5, which has solutions 1 and 4. So, mod 5, 16, 9, and 5. So, when we use CRT, each combination of residues modulo 16,9,5 gives a unique residue modulo 720. So, the total number of solutions is the product. So, 4*2*2=16. So, seems right. So, answer is 16. But wait, the problem says 0 ≤ n ≤ 720. So, 721 numbers. But modulus 720 has residues 0 to 719. So, 720 is congruent to 0 mod 720. So, in the range 0 to 719, there are 16 solutions, and 720 is not a solution. Hence, total solutions are 16. But to make sure, let's actually compute one of these solutions. For instance, take n=1. Then 1²=1≡1 mod 720. Similarly, n=719. 719 mod 720 is -1. So, (-1)^2=1≡1 mod 720. So, n=719 is also a solution. So, that's two solutions. But according to our count, there are 16 solutions. So, let's see. If we can list some of them. Alternatively, maybe we can use the fact that solutions to x²≡1 mod m are called self-conjugate residues or something. But in any case, using CRT is the right approach here. Wait, but maybe there's a mistake in the counting. Let me check each modulus again.First, mod 16: solutions are 1,7,9,15. So, four solutions.Mod 9: solutions are 1,8. So, two solutions.Mod 5: solutions are 1,4. So, two solutions.Therefore, total solutions: 4*2*2=16.So, if each modulus contributes independently, then the total number of solutions is 16. So, modulo 720, 16 solutions. Hence, in the range 0 ≤ n ≤ 719, there are 16 numbers. Since 720 is excluded (as 720 isn't a solution), the answer is 16.But wait, let's check with a smaller modulus. Let's take modulus 10, which factors into 2 and 5. Suppose we want x²≡1 mod 10. Then mod 2: x²≡1 mod 2 implies x≡1 mod 2. Mod 5: x²≡1 mod 5 implies x≡1 or 4 mod 5. So, solutions are numbers ≡1 mod 2 and ≡1 or 4 mod 5. So, solutions mod 10: 1,9. So, two solutions. Indeed, 1²=1, 3²=9≡9 mod10, 5²=25≡5, 7²=49≡9, 9²=81≡1. So, solutions are 1,3,7,9? Wait, no. Wait, 3²=9≡9≠1. 7²=49≡9. 9²=81≡1. So, only 1 and 9. Wait, but according to CRT, mod 2 and mod 5. So, solutions are combinations where x≡1 mod 2, and x≡1 or 4 mod 5. So, solving x≡1 mod 2 and x≡1 mod 5: x≡1 mod 10. Solving x≡1 mod 2 and x≡4 mod 5: x≡9 mod10. So, two solutions. But in reality, 1 and 9. So, works. So, that's correct. So, the CRT count works here.But in that case, the CRT counts solutions as the product of the number of solutions modulo each prime power. So, similar logic applies here. So, in our original problem, 4*2*2=16 solutions. Hence, 16 numbers between 0 and 719 inclusive, which correspond to 0 ≤ n ≤720, but n=720 isn't a solution. Hence, 16 numbers. So, answer is 16.But wait, let's check another modulus. Let's say modulus 8. Suppose x²≡1 mod8. Then possible residues are 1,3,5,7. Let's compute squares: 1²=1, 3²=9≡1, 5²=25≡1, 7²=49≡1. So, four solutions. Which matches with modulus 2^3=8. So, for modulus 2^k, k≥3, the number of solutions to x²≡1 is 4. For modulus 2^1=2: x²≡1 mod2 has one solution. For modulus 2^2=4: x²≡1 mod4 has two solutions (1 and 3). For modulus 2^k, k≥3: four solutions. So, in our problem, modulus 16=2^4, so four solutions as expected. Similarly, modulus 3^2=9. The equation x²≡1 mod9. Solutions are 1 and 8 (since 8≡-1 mod9). (-1)^2=1. So, two solutions. Similarly, modulus 5: x²≡1 mod5 has solutions 1 and 4. So, two solutions. So, the counts are correct. Therefore, the total number is indeed 4*2*2=16. Therefore, the answer should be 16. But I want to make sure that when we combine these congruences, there are no overlapping solutions or some solutions that might coincide. But CRT says that if the moduli are coprime, then the system of congruences has a unique solution modulo the product. Therefore, each combination of solutions modulo 16, 9, and 5 gives a unique solution modulo 720. Since there are 4 choices mod 16, 2 mod 9, and 2 mod 5, total 16 solutions. Alternatively, we can compute the number of solutions by using the formula for the number of solutions to x²≡1 mod n. For n being the product of prime powers, the number of solutions is the product of the number of solutions modulo each prime power. So, yes, that's exactly what we did. Therefore, I think the answer is 16. However, to be thorough, let's attempt to compute one of the solutions step by step to see if it works.Take, for example, solving the system:n ≡ 1 mod16n ≡1 mod9n≡1 mod5By CRT, find n ≡1 mod16, n≡1 mod9, n≡1 mod5. Then, since 1 is congruent to 1 in all moduli, the solution is n≡1 mod720. So, n=1 is a solution. Similarly, if we take n≡1 mod16, n≡1 mod9, n≡4 mod5. Let's solve that. Let me solve:n ≡1 mod16,n ≡1 mod9,n ≡4 mod5.First, solve n ≡1 mod16 and n≡1 mod9. Let’s find a number congruent to 1 mod16 and 1 mod9.Let n =16k +1. Then 16k +1 ≡1 mod9 =>16k ≡0 mod9. 16≡7 mod9, so 7k ≡0 mod9. So, 7k ≡0 mod9. 7 and 9 are coprime, so k≡0 mod9. Therefore, k=9m. Therefore, n=16*9m +1=144m +1. Then, we need n≡4 mod5. So, 144m +1 ≡4 mod5. 144≡144-28*5=144-140=4 mod5. So, 4m +1≡4 mod5 =>4m≡3 mod5. Multiply both sides by inverse of 4 mod5, which is 4, since 4*4=16≡1 mod5. Therefore, m≡3*4=12≡2 mod5. So, m=5t +2. Therefore, n=144*(5t +2)+1=720t +288 +1=720t +289. Therefore, the solution is n≡289 mod720. So, 289 is a solution. Let's check 289² mod720.Compute 289². 289=17². So, 289²=17⁴. 17²=289. 289²= (300 -11)²=300² -2*300*11 +11²=90000 -6600 +121=90000 -6600 is 83400 +121=83521. Now, compute 83521 mod720.Divide 83521 by 720. 720*115=720*100=72000; 720*15=10800; total 72000+10800=82800. 83521 -82800=721. 721 -720=1. So, 83521 mod720=1. So, 289²≡1 mod720. So that's correct. Therefore, 289 is a solution. Similarly, other combinations will give different solutions. So, each combination gives a unique solution mod720, so 16 solutions in total. Another example: n≡7 mod16, n≡8 mod9, n≡4 mod5.Let's solve this. Let’s go step by step.First, n≡7 mod16 and n≡8 mod9.Let n=16k +7. Then, 16k +7 ≡8 mod9 =>16k ≡1 mod9. 16≡7 mod9, so 7k ≡1 mod9. Multiplicative inverse of 7 mod9 is 4, since 7*4=28≡1 mod9. Thus, k≡4 mod9. So, k=9m +4. Therefore, n=16*(9m +4) +7=144m +64 +7=144m +71. Now, apply n≡4 mod5. So, 144m +71 ≡4 mod5. 144≡4 mod5, 71≡1 mod5. So, 4m +1 ≡4 mod5 =>4m≡3 mod5. Multiply both sides by inverse of 4 mod5 (which is 4): m≡12≡2 mod5. So, m=5t +2. Therefore, n=144*(5t +2) +71=720t +288 +71=720t +359. So, n≡359 mod720. Check 359² mod720.Compute 359². 360²=129600, so 359²=360² -2*360 +1=129600 -720 +1=128881. Now, divide 128881 by720. 720*179=720*(180 -1)=720*180 -720=129600 -720=128880. So, 128881 -128880=1. Therefore, 359²≡1 mod720. Correct.Therefore, each combination of residues gives a unique solution. Hence, there are 16 solutions. Therefore, the answer is 16. But wait, the problem says 0 ≤ n ≤ 720. So, including n=0 and n=720. However, n=0: 0²=0≡0 mod720, which is not 1. n=720: same as 0. So, neither are solutions. Therefore, all solutions lie between 0 and719 inclusive. Since modulus720 has 720 residues, and we have 16 solutions, each solution is unique in this range. Therefore, the answer is 16. Therefore, after thorough analysis and verification with examples, I conclude that the number of integers n is 16.Final Answerboxed{16}</think>To solve the problem of finding how many integers ( n ) satisfy ( 0 le n le 720 ) and ( n^2 equiv 1 pmod{720} ), we start by factoring 720 into its prime components: ( 720 = 2^4 cdot 3^2 cdot 5 ). Using the Chinese Remainder Theorem (CRT), we solve the congruence ( n^2 equiv 1 ) modulo each prime power factor and then combine the results.1. Modulo 16: The solutions to ( n^2 equiv 1 pmod{16} ) are ( n equiv 1, 7, 9, 15 pmod{16} ). This gives 4 solutions.2. Modulo 9: The solutions to ( n^2 equiv 1 pmod{9} ) are ( n equiv 1, 8 pmod{9} ). This gives 2 solutions.3. Modulo 5: The solutions to ( n^2 equiv 1 pmod{5} ) are ( n equiv 1, 4 pmod{5} ). This gives 2 solutions.Using CRT, the number of solutions modulo 720 is the product of the number of solutions modulo each prime power factor:[ 4 times 2 times 2 = 16 ]Thus, there are 16 solutions modulo 720. Since the range ( 0 le n le 720 ) includes 721 numbers (0 to 720), but 720 is equivalent to 0 modulo 720 and is not a solution, the valid solutions are within the range 0 to 719. Therefore, the number of integers ( n ) that satisfy the given conditions is (boxed{16}).,<think>Okay, so I need to find how many integers n satisfy 0 ≤ n ≤ 720 and n² ≡ 1 mod 720. Hmm, modular arithmetic. Let me recall that n² ≡ 1 mod 720 means that 720 divides (n² - 1), which is equivalent to (n - 1)(n + 1) being divisible by 720. So, the problem reduces to finding all n in the range [0, 720] such that (n - 1)(n + 1) is a multiple of 720. First, I should factorize 720 to understand its prime components. Let's do that. 720 is 72 times 10, which is 8*9*10. Breaking each down: 8 is 2³, 9 is 3², and 10 is 2*5. So combining these, 720 = 2⁴ * 3² * 5¹. So, prime factors are 2^4, 3^2, and 5^1. Therefore, for (n - 1)(n + 1) to be divisible by 2⁴, 3², and 5, we need that product to be divisible by each of these prime powers. So, we can approach this problem by solving the congruence n² ≡ 1 mod 16, n² ≡ 1 mod 9, and n² ≡ 1 mod 5, then use the Chinese Remainder Theorem (CRT) to find the number of solutions modulo 720. Then, since we're looking for solutions between 0 and 720 inclusive, the number of solutions will be equal to the number of solutions modulo 720, which is the product of the number of solutions modulo each prime power component. Wait, let me make sure. CRT tells us that if the moduli are coprime, then the system of congruences has a unique solution modulo the product. So since 16, 9, and 5 are pairwise coprime, the number of solutions modulo 720 should be the product of the number of solutions modulo each prime power. Therefore, if I can find the number of solutions for each modulus, multiply them together, that will give the total number of solutions modulo 720, hence in the range 0 to 720. Since 720 is included, the number of solutions is equal to the number of solutions modulo 720. So, first step: solve n² ≡ 1 mod 16. Let me find all n modulo 16 such that n² ≡ 1 mod 16. Similarly for mod 9 and mod 5.Starting with mod 16. Let's check all residues from 0 to 15:0²=0, 1=1, 2=4, 3=9, 4=16≡0, 5=25≡9, 6=36≡4, 7=49≡1, 8=64≡0, 9=81≡1, 10=100≡4, 11=121≡9, 12=144≡0, 13=169≡9, 14=196≡4, 15=225≡1.So n² ≡1 mod 16 when n ≡1, 7, 9, 15 mod16. That's four solutions.Next, mod 9. Find n such that n²≡1 mod9. Let's check residues 0-8:0²=0,1=1,2=4,3=0,4=7,5=7,6=0,7=4,8=1. So n²≡1 mod9 when n≡1 or 8 mod9. So two solutions.Then mod5. Find n with n²≡1 mod5. Residues 0-4:0²=0,1=1,2=4,3=4,4=1. So n≡1 or 4 mod5. Two solutions.So, for each modulus, the number of solutions are 4, 2, and 2. By CRT, the total number of solutions modulo 720 is 4*2*2=16. Therefore, there should be 16 solutions in the range 0 ≤n≤720. Hence, the answer is 16. Wait, but wait a second. Let me verify this.But hold on, maybe I made a mistake here. Because CRT tells us that if the moduli are coprime, the system of congruences modulo each component has a unique solution modulo the product. Therefore, for each combination of solutions modulo 16, 9, and 5, there is a unique solution modulo 720. Since there are 4 solutions mod16, 2 mod9, and 2 mod5, the total number of solutions is indeed 4*2*2=16. So, the total number of solutions modulo 720 is 16, which means in the range 0 to 720 inclusive, there are 16 solutions. Hence, the answer is 16. But let me check this with an example. Let's pick some n and see if n² ≡1 mod720. For example, n=1: 1²=1≡1 mod720. n=719: 719²=(720-1)²=720² -2*720 +1≡0 -0 +1=1 mod720. So 719 is a solution. Similarly, n=721 would be equivalent to 1 mod720, but 721 is beyond the range. So n=1 and n=719 are two solutions. Similarly, n= 1 + k*720/ gcd(k,720). Wait, but maybe there are other solutions.Wait, but according to the CRT, each combination gives a unique solution. So, 4 mod16, 2 mod9, 2 mod5. So 16 solutions. Let me check another solution. For example, n=17. Let's compute 17²=289. 289 mod720 is 289. 289 mod16 is 1 (since 16*18=288, 289=288+1≡1 mod16). 289 mod9: 2+8+9=19, 1+9=10≡1 mod9. 289 mod5: 289=5*57 + 4≡4 mod5. Wait, but n=17. Let me check n=17: 17 mod16=1, mod9=8, mod5=2. Wait, but that's different. Hmmm, perhaps my confusion.Alternatively, maybe it's better to consider that each solution mod16, mod9, mod5 can be combined. For example, take n≡1 mod16, n≡1 mod9, n≡1 mod5. This would correspond to a solution n≡1 mod720. Similarly, n≡1 mod16, n≡1 mod9, n≡4 mod5. Then this would be another solution. Etc. So each combination gives a unique n mod720, hence 16 total.But to confirm, let's check another solution. For example, take n≡7 mod16, n≡8 mod9, n≡4 mod5. Let's compute such an n. Let me solve:Find n ≡7 mod16,n≡8 mod9,n≡4 mod5.Using CRT step by step. Let's first solve n≡7 mod16 and n≡8 mod9.Let n=16k +7. Then 16k +7 ≡8 mod9. 16k ≡1 mod9. Since 16≡7 mod9, so 7k ≡1 mod9. Multiply both sides by inverse of 7 mod9. 7*4=28≡1 mod9, so inverse is 4. Therefore, k≡4 mod9. So k=9m +4. Hence, n=16*(9m +4) +7=144m +64 +7=144m +71. So n≡71 mod144. Now, we need n≡4 mod5. So 144m +71 ≡4 mod5. 144≡4 mod5, so 4m +71≡4 mod5. 71≡1 mod5, so 4m +1 ≡4 mod5. Hence, 4m≡3 mod5. Multiply both sides by inverse of 4 mod5, which is 4, since 4*4=16≡1 mod5. So m≡3*4=12≡2 mod5. Therefore, m=5p +2. Therefore, n=144*(5p +2) +71=720p +288 +71=720p +359. Thus, n≡359 mod720. So in the range 0-720, n=359 is a solution. Let's check 359² mod720. 359²= (360 -1)²=360² -2*360 +1=129600 -720 +1=128881. Now divide 128881 by 720. Let's compute 720*179=720*(180 -1)=129600 -720=128880. So 128881 -128880=1. Therefore, 359²≡1 mod720. Correct. So 359 is a solution. Therefore, that's another solution apart from 1 and 719. Similarly, there are 16 such solutions. Thus, my initial conclusion seems correct. The number of solutions is 16. Therefore, the answer should be 16. Wait, but I need to make sure that all these solutions are distinct modulo 720. Since we combined them using CRT, each combination of residues mod16, mod9, mod5 gives a unique residue mod720, so all 16 solutions are distinct. Hence, in the range 0 ≤n≤720, there are 16 integers satisfying the condition. Therefore, the answer is 16. But just to be thorough, let me check one more solution. Let's take n≡15 mod16 (which is -1 mod16), n≡8 mod9, and n≡4 mod5. Let's solve these congruences. First, n=16k +15. Then 16k +15 ≡8 mod9. 16k ≡8 -15 mod9. 16k≡ -7≡2 mod9. 16≡7 mod9, so 7k≡2 mod9. Multiply both sides by inverse of 7 mod9, which is 4. So k≡2*4=8 mod9. Hence, k=9m +8. Therefore, n=16*(9m +8) +15=144m +128 +15=144m +143. Then, n≡143 mod144. Now, impose n≡4 mod5. 144m +143≡4 mod5. 144≡4 mod5, 143≡3 mod5. So 4m +3≡4 mod5. 4m≡1 mod5. Multiply by inverse of 4 mod5 (which is 4): m≡4 mod5. So m=5p +4. Then, n=144*(5p +4) +143=720p +576 +143=720p +719. Hence, n≡719 mod720. Therefore, in the range 0-720, n=719 is a solution. Which we already knew. So 719 is another solution. So indeed, each combination gives a unique solution.Therefore, my conclusion is that there are 16 solutions. So the answer is boxed{16}.Final Answerboxed{16}</think>To solve the problem of finding how many integers ( n ) satisfy ( 0 le n le 720 ) and ( n^2 equiv 1 pmod{720} ), we start by factorizing 720 into its prime components: ( 720 = 2^4 cdot 3^2 cdot 5 ). We need to solve the congruence ( n^2 equiv 1 ) modulo each of these prime powers and then use the Chinese Remainder Theorem (CRT) to combine the solutions.1. Modulo 16: We need ( n^2 equiv 1 pmod{16} ). The solutions are ( n equiv 1, 7, 9, 15 pmod{16} ), giving 4 solutions.2. Modulo 9: We need ( n^2 equiv 1 pmod{9} ). The solutions are ( n equiv 1, 8 pmod{9} ), giving 2 solutions.3. Modulo 5: We need ( n^2 equiv 1 pmod{5} ). The solutions are ( n equiv 1, 4 pmod{5} ), giving 2 solutions.Using the Chinese Remainder Theorem, the number of solutions modulo 720 is the product of the number of solutions modulo each prime power component. Therefore, the total number of solutions is ( 4 times 2 times 2 = 16 ).Thus, the number of integers ( n ) such that ( 0 le n le 720 ) and ( n^2 equiv 1 pmod{720} ) is (boxed{16}).]

🔑:To solve the problem, we need to find the number of integers ( n ) such that ( 0 le n le 720 ) and ( n^2 equiv 1 pmod{720} ).1. Factorize 720: [ 720 = 2^4 cdot 3^2 cdot 5 ] We will solve the congruence ( n^2 equiv 1 pmod{720} ) by solving it modulo ( 16 ), ( 9 ), and ( 5 ) separately and then combining the results using the Chinese Remainder Theorem.2. Solve ( n^2 equiv 1 pmod{5} ): [ n^2 equiv 1 pmod{5} ] The solutions to this congruence are: [ n equiv 1 pmod{5} quad text{or} quad n equiv 4 pmod{5} ] So, ( n ) can be ( 1 ) or ( 4 ) modulo ( 5 ).3. Solve ( n^2 equiv 1 pmod{9} ): [ n^2 equiv 1 pmod{9} ] The solutions to this congruence are: [ n equiv 1 pmod{9} quad text{or} quad n equiv 8 pmod{9} ] So, ( n ) can be ( 1 ) or ( 8 ) modulo ( 9 ).4. Solve ( n^2 equiv 1 pmod{16} ): [ n^2 equiv 1 pmod{16} ] The solutions to this congruence are: [ n equiv 1 pmod{16}, quad n equiv 7 pmod{16}, quad n equiv 9 pmod{16}, quad text{or} quad n equiv 15 pmod{16} ] So, ( n ) can be ( 1 ), ( 7 ), ( 9 ), or ( 15 ) modulo ( 16 ).5. Combine solutions using the Chinese Remainder Theorem: We have: [ n equiv 1 text{ or } 4 pmod{5} ] [ n equiv 1 text{ or } 8 pmod{9} ] [ n equiv 1, 7, 9, text{ or } 15 pmod{16} ] By the Chinese Remainder Theorem, the total number of solutions is the product of the number of solutions for each modulus: [ 2 text{ (for } pmod{5}) times 2 text{ (for } pmod{9}) times 4 text{ (for } pmod{16}) = 16 ]Therefore, there are ( boxed{16} ) integers ( n ) such that ( 0 le n le 720 ) and ( n^2 equiv 1 pmod{720} ).

🧠:[<think>Okay, so there's this problem where there are 10 students arranged in a row. Every minute, a new student is inserted into the row. There are 11 possible places to insert the new student—either at the very front, between any two existing students, or at the very back—each with equal probability of 1/11. Then, after inserting the new student, either the frontmost or the backmost student is removed, each with a 1/2 probability. I'm the eighth student from the front initially. I need to find the probability that I exit from the front rather than the back. The answer should be in the form m/n where m and n are coprime, and then compute 100m + n. First, let me make sure I understand the process correctly. We start with 10 students. Each minute, we add one student in any of the 11 possible positions (so the row becomes length 11), then remove either the front or the back student with equal chance, bringing the row back down to 10 students. The key here is that the row size stays constant at 10 after each operation. My position is initially 8th from the front. I want to know the probability that when I eventually get removed, it's from the front rather than the back. This seems like a Markov chain problem where my position in the row can change over time based on the insertion and deletion steps. Let's model my position as a state. Let's denote my position at any time as k, where k is the number of students in front of me. So initially, k = 7 (since I'm 8th, there are 7 students ahead of me). The total number of students is always 10, so the positions are from k = 0 (front) to k = 9 (back). If I ever reach k = 0, then I get removed from the front next, and if I reach k = 9, I get removed from the back. Wait, but actually, when a new student is inserted, the row becomes 11 students, then one is removed from either front or back. So my position after insertion and before deletion can change. Let me formalize this:1. Insertion step: With probability 1/11, a new student is inserted in any of the 11 possible positions. If a student is inserted in front of me, my position increases by 1. If a student is inserted behind me, my position stays the same. If a student is inserted exactly at my position, pushing me back? Wait, actually, inserting a student in a position can affect my current position depending on where it's inserted.Wait, maybe the row is 0-indexed? Let me think. Suppose the original row has positions 1 to 10. When inserting a new student, there are 11 possible gaps: before the first student, between each pair, and after the last student. So inserting a student at position i (0 ≤ i ≤ 10) would shift all students from position i onwards backward by one. Therefore, if I was originally at position 8 (1-based indexing), my position is 8. If a student is inserted before position i where i ≤ 8, then my new position becomes 9. If inserted after position 8, my position remains 8. Wait, but insertion is in the gaps, so inserting at position i (0 to 10) corresponds to inserting before the (i+1)th student. For example, inserting at position 0 would be before the first student, position 1 between first and second, etc., up to position 10 after the last student.Therefore, if I'm in position 8 (1-based), inserting a student at any position from 0 to 8 would push me back by 1, making my position 9. Inserting at positions 9 to 10 would not affect my position. So, the probability that my position increases by 1 due to insertion is 9/11 (positions 0 to 8), and probability it stays the same is 2/11 (positions 9 to 10). Then, after insertion, there are 11 students. Then, with probability 1/2, either the front or the back student is removed. If the front student is removed, then all remaining students shift forward by 1. If the back student is removed, then no shifting occurs except the last student is gone. Wait, so after insertion, we have 11 students. Then, remove front or back. So, if front is removed, everyone else moves up. If back is removed, everyone stays except the last one. Therefore, if after insertion I was in position k (0 to 10), then after removal:- If front is removed: my new position is k - 1 (if k ≥ 1), or I get removed if k = 0.- If back is removed: my position remains k (if the back removed was someone behind me), unless I was in position 10, in which case I get removed. But wait, the problem says "either the frontmost or the backmost student is removed from the row (each with a 1/2 probability)". So after inserting a student to make 11, we remove either the front or the back. So if front is removed, then all students shift forward. If back is removed, the last student is removed. So in terms of my position:Suppose after insertion, I am at position k (0 is front, 10 is back). Then:- If front is removed: my position becomes k - 1, unless I was at position 0, in which case I'm removed.- If back is removed: my position remains k, unless I was at position 10, in which case I'm removed.But in our problem, we start with 10 students, insert one to make 11, remove one to go back to 10. So in terms of my position:Let me denote the process step by step for my position:1. Insertion: my position can increase by 1 with probability 9/11, stay the same with probability 2/11.2. Then, removal: front is removed with 1/2, back with 1/2.But after insertion, when there are 11 students, my position is either k' = k + 1 (with probability 9/11) or k' = k (with probability 2/11). Then, removal affects my position:If front is removed:- If k' ≥ 1: my new position is k' - 1.- If k' = 0: I'm removed.If back is removed:- If k' ≤ 9: my position remains k'.- If k' = 10: I'm removed.But in our case, since the original row had 10 students, after insertion to 11, and then removal, we go back to 10. However, my position after insertion is k', which can be from 0 to 10.Wait, but initially, my position was 7 (assuming 0-indexing) or 8 (1-indexing). Let me clarify.Wait, the problem states: "you are the eighth in the line from the front." So in 1-based indexing, position 8. Therefore, in 0-based indexing, position 7. But in the insertion step, when inserting into 11 positions, which are 0 to 10 (0 being front, 10 being back). So we need to be careful with indexing.But perhaps let's model my position in 1-based indexing. So initially, I'm at position 8. When a new student is inserted in front of me, my position increases by 1. If inserted behind me, my position stays the same. Then, after insertion, the row has 11 students. Then, with 1/2 chance, either the front or the back is removed.If front is removed: all positions decrease by 1. So if I was at position k, I move to k - 1.If back is removed: positions stay the same, unless I was at position 11, which is removed.But since after insertion, the row is 11 students, so my position after insertion can be from 1 to 11 (1-based). But initially, after insertion, my position is either 8 or 9 (since insertion in front can push me back). Wait, actually, inserting a student in front of position i (1-based) would push me back if i ≤ 8.Wait, if we use 1-based indexing for insertion positions: inserting before position 1 (front), between 1 and 2, ..., between 10 and 11 (back). So inserting at position i (1 to 11) in 1-based.If I am originally at position 8, inserting at positions 1 through 8 would push me to position 9. Inserting at positions 9 through 11 would leave me at position 8. Therefore, the probability my position becomes 9 is 8/11, and remains 8 is 3/11. Wait, hold on, inserting at 11 positions (before 1, between 1-2, ..., after 10), so in 1-based insertion positions, there are 11 choices. Inserting before position i for i from 1 to 11. If inserting before position 1, that's the front; inserting before position 11 is between 10 and 11, which is effectively the back.Therefore, if I'm at position 8, inserting before any position from 1 to 8 would push me back to 9. Inserting before positions 9 to 11 would not affect me, so I stay at 8. Therefore, probability 8/11 to move to 9, 3/11 to stay at 8.Wait, but 1-based insertion positions: inserting before position 1 (front), 2, ..., 11 (back). So inserting at position i (1 ≤ i ≤ 11) in 1-based. So inserting before position 1: push everyone back, so my position (originally 8) becomes 9. Inserting before position 2: original position 8 becomes 9. Similarly up to inserting before position 8: my position becomes 9. Inserting before positions 9 to 11: since these are after my original position, inserting there doesn't affect me. So positions 9,10,11 insertion: 3 positions. So total insertion positions that affect me: 8 (positions 1-8), so probability 8/11 to move to 9, and 3/11 to stay at 8.Therefore, after insertion, my position is 9 with probability 8/11, 8 with probability 3/11.Then, removal happens: front or back, each with 1/2.If front is removed: all positions decrease by 1. So if I was at 9, I go to 8; if I was at 8, I go to 7.If back is removed: the last student is removed. If I was at 9, then in a row of 11, removing the back would leave me at 9 in the new row of 10 (since the original back was at 11). If I was at 8, removing the back (position 11) leaves me still at 8.But wait, in 1-based indexing, after insertion, we have 11 students. If we remove the back (position 11), then the remaining students are positions 1 to 10. So if I was at position 9, after removal, I stay at 9. If I was at 8, I stay at 8.If we remove the front (position 1), then everyone shifts forward. So position 2 becomes 1, ..., position 11 becomes 10. Therefore, if I was at position 9, I become 8; if I was at 8, I become 7.Therefore, combining insertion and removal steps:From my current position k (1-based in a row of 10):1. Insertion: with probability 8/11, I move to k+1; with probability 3/11, stay at k.2. Then, removal: - If front is removed (1/2 chance): my new position is (k+1) - 1 = k if I moved to k+1, or k - 1 if I stayed at k. Wait, no. Wait, after insertion, if my position is k', then removal affects k' as follows:Wait, let's formalize this:Let’s model the entire transition from position k in the current row (size 10) to the next position.When we insert a student:- With probability 8/11, my position becomes k + 1 (in a row of 11).- With probability 3/11, my position remains k (in a row of 11).Then, when removing:- With probability 1/2, front is removed: if my position was k', it becomes k' - 1 (if k' ≥ 2), or I get removed if k' = 1.- With probability 1/2, back is removed: if my position was k', it remains k' (if k' ≤ 10), or I get removed if k' = 11.But in our case, since we are tracking until I get removed, we need to see if during this process, whether I reach position 1 (front) or position 11 (back) after insertion, leading to possible removal.Wait, but in the removal step, after insertion (row of 11), we remove either the front or the back. So:If after insertion, my position is k' in 1-based (1 to 11):- If front is removed, my new position is k' - 1 (if k' ≥ 2), else I am removed.- If back is removed, my new position is k' (if k' ≤ 10), else I am removed.Therefore, the transition probabilities from current position k (in 10 students) to next position (in 10 students) are as follows:First, insertion can take k to k + 1 with probability 8/11 or stay at k with probability 3/11. Then, removal can either decrease the position by 1 (if front removed) or leave it the same (if back removed), but with the caveat that if insertion made the position 11 (which is back), then removing the back would remove me.Wait, actually, after insertion, the row is 11 students. So my possible positions after insertion are k + 1 or k (depending on insertion). Then, when removing front or back:Case 1: Insertion moved me to k + 1 (probability 8/11):- If front is removed (prob 1/2): my new position is (k + 1) - 1 = k.- If back is removed (prob 1/2): if k + 1 ≤ 10, position remains k + 1; if k + 1 = 11, I am removed.But k is my original position in the row of 10, which ranges from 1 to 10. So if k + 1 = 11, that means k = 10. So if I was originally at position 10 (back), insertion moves me to 11, then removing back would remove me. But in our problem, I start at position 8. So maybe we don't have to worry about k = 10 unless the process takes me there. Similarly, Case 2: Insertion kept me at k (probability 3/11):- If front is removed (prob 1/2): my new position is k - 1.- If back is removed (prob 1/2): my position remains k.But if k = 1 and front is removed, then I get removed.So combining these, the transition matrix can be constructed. However, since we start at position 8, and we need to compute the probability of exiting from front vs back, this is similar to a ruin problem in probability, where we have absorbing states at position 0 (front exit) and position 10 (back exit), except here the transitions are a bit more complex.Wait, but in our case, the row size remains 10, but actually, no—the student is being inserted and then one is removed. However, the problem is that my position can change each time based on insertion and deletion. So the key is to model my position after each iteration (each minute), considering the insertion and deletion steps.But since the row is always size 10, except temporarily during insertion/deletion, the position is in 1 to 10 (1-based). But the insertion can temporarily take my position to 1 to 11, then deletion brings it back to 1 to 10.Alternatively, perhaps we can model my position in the row as a Markov chain with states 1 to 10, where states 1 and 10 are absorbing (since if I reach the front, I might get removed next time, similarly for the back). Wait, but actually, removal happens every time, so even if I'm not at the front or back, someone else is being removed. So my position can only change due to insertion and removal.Wait, perhaps I need to think of this as a Markov chain where each state is my current position (1 to 10). At each step, insertion occurs, possibly changing my position, then removal occurs, possibly changing my position again. The transitions depend on both steps.Let me formalize the transition probabilities.Let’s denote the current position as k (1 ≤ k ≤ 10).First, insertion:- With probability 8/11, my position becomes k + 1 (in a row of 11).- With probability 3/11, my position remains k (in a row of 11).Then, removal:- With probability 1/2, front is removed: - If insertion moved me to k + 1: - If k + 1 ≥ 2, my new position is (k + 1) - 1 = k. - If k + 1 = 1 (but k starts at 1, so k + 1 = 2), so no problem. - If insertion kept me at k: - My new position is k - 1 (if k ≥ 2), else removed.- With probability 1/2, back is removed: - If insertion moved me to k + 1: - If k + 1 ≤ 10, my position remains k + 1. - If k + 1 = 11, I get removed. - If insertion kept me at k: - My position remains k.Therefore, combining these possibilities:From state k (1 ≤ k ≤ 10):1. Insertion moves to k + 1 with probability 8/11: a. Then front removed with 1/2: position becomes k. b. Back removed with 1/2: - If k + 1 ≤ 10: position becomes k + 1. - If k + 1 = 11: removed (absorbing state).2. Insertion stays at k with probability 3/11: a. Then front removed with 1/2: position becomes k - 1 (if k ≥ 2) or removed (if k = 1). b. Back removed with 1/2: position remains k.Therefore, the transition probabilities from k are:- To k: - Case 1a: 8/11 * 1/2 = 4/11. - Case 2b: 3/11 * 1/2 = 3/22. Total: 4/11 + 3/22 = 8/22 + 3/22 = 11/22 = 1/2.Wait, that's interesting. Wait, hold on:Wait, for insertion moves to k + 1 (prob 8/11): - Then front removed: position becomes k (prob 1/2). So total prob for moving back to k: 8/11 * 1/2 = 4/11.Insertion stays at k (prob 3/11): - Back removed: position remains k (prob 1/2). So total prob: 3/11 * 1/2 = 3/22.So total probability to stay at k: 4/11 + 3/22 = 8/22 + 3/22 = 11/22 = 1/2.Similarly, transition to k + 1:Only possible if insertion moves to k + 1 and back is removed, and k + 1 ≤ 10.Probability: 8/11 * 1/2 = 4/11, but only if k + 1 ≤ 10. If k = 10, then k + 1 = 11, which would result in removal. So for k < 10, transition to k + 1 is 4/11, and for k = 10, that 4/11 leads to removal.Transition to k - 1:Only possible if insertion stays at k and front is removed, and k ≥ 2.Probability: 3/11 * 1/2 = 3/22, for k ≥ 2. If k = 1, then this would lead to removal.Therefore, summarizing:For 2 ≤ k ≤ 9:- To k - 1: 3/22- To k: 1/2- To k + 1: 4/11Wait, but 3/22 + 1/2 + 4/11 = 3/22 + 11/22 + 8/22 = 22/22 = 1. Good.For k = 1:- To 1: 1/2 (from insertion moving to 2 then front removed, which brings it back to 1; or insertion staying at 1 and back removed)Wait, let's compute specifically.Wait, for k = 1:1. Insertion moves to 2 with probability 8/11: a. Front removed: position becomes 1 (prob 8/11 * 1/2 = 4/11) b. Back removed: position becomes 2 (prob 8/11 * 1/2 = 4/11) if 2 ≤ 10, which it is.2. Insertion stays at 1 with probability 3/11: a. Front removed: position becomes 0, which is removal (absorbing state). b. Back removed: position remains 1 (prob 3/11 * 1/2 = 3/22)Therefore, transitions from k = 1:- To 1: 4/11 (from 1a) + 3/22 (from 2b) = 8/22 + 3/22 = 11/22 = 1/2- To 2: 4/11 (from 1b) = 8/22- To removal (front): 3/11 * 1/2 = 3/22But wait, removal is an absorbing state. So actually, from k = 1, there is a probability 3/22 of being removed (exiting from front), and the rest transitions:- To 1: 1/2- To 2: 4/11 = 8/22Wait, but 1/2 is 11/22, plus 8/22 is 19/22, but total should be 1. Hmm, something's wrong here.Wait, let's recast:From k = 1:- Insertion moves to 2 (8/11): - Front removed (1/2): position 1 - Back removed (1/2): position 2- Insertion stays at 1 (3/11): - Front removed (1/2): removed (exit front) - Back removed (1/2): position 1Therefore, the probabilities:- To 1: (8/11 * 1/2) + (3/11 * 1/2) = (4/11) + (3/22) = 8/22 + 3/22 = 11/22 = 1/2- To 2: 8/11 * 1/2 = 4/11 = 8/22- Exit front: 3/11 * 1/2 = 3/22Total: 11/22 + 8/22 + 3/22 = 22/22 = 1. Okay, correct.Similarly, for k = 10:1. Insertion moves to 11 (8/11): a. Front removed (1/2): position 10 b. Back removed (1/2): exit back (since position 11 is removed)2. Insertion stays at 10 (3/11): a. Front removed (1/2): position 9 b. Back removed (1/2): position 10Therefore, transitions from k = 10:- To 10: (8/11 * 1/2) + (3/11 * 1/2) = (4/11) + (3/22) = 8/22 + 3/22 = 11/22 = 1/2- To 9: 3/11 * 1/2 = 3/22- Exit back: 8/11 * 1/2 = 4/11 = 8/22Total: 11/22 + 3/22 + 8/22 = 22/22 = 1.So in general, for positions 2 ≤ k ≤ 9:- To k - 1: 3/22- To k: 1/2- To k + 1: 4/11 (which is 8/22)Check: 3/22 + 11/22 + 8/22 = 22/22.For k = 1:- To exit front: 3/22- To 1: 1/2- To 2: 8/22For k = 10:- To exit back: 8/22- To 10: 1/2- To 9: 3/22Therefore, we can model this as a Markov chain with states 1 to 10, where states 1 and 10 are absorbing (with probabilities to exit), and the other states transition as above.We need to compute the probability that starting from state 8, we get absorbed at state 1 (exit front) rather than state 10 (exit back).This is similar to a Gambler's Ruin problem, but with variable transition probabilities. In the standard Gambler's Ruin, the probability of moving up or down is constant, but here it depends on the current state.Let me denote the probability of being absorbed at state 1 (exit front) starting from state k as P(k). We need to find P(8).We know that P(1) = 1 (if we start at front, we exit front immediately), and P(10) = 0 (if we start at back, exit back immediately). For the other states, 2 ≤ k ≤ 9, we can write the recurrence relation:P(k) = (3/22) * P(k - 1) + (1/2) * P(k) + (8/22) * P(k + 1)Multiply both sides by 22 to eliminate denominators:22 P(k) = 3 P(k - 1) + 11 P(k) + 8 P(k + 1)Bring all terms to left:22 P(k) - 11 P(k) - 3 P(k - 1) - 8 P(k + 1) = 0Simplify:11 P(k) - 3 P(k - 1) - 8 P(k + 1) = 0Rearranged:8 P(k + 1) - 11 P(k) + 3 P(k - 1) = 0This is a linear homogeneous recurrence relation with constant coefficients. To solve it, we can find the characteristic equation.Assume solution of the form P(k) = r^k. Plugging into recurrence:8 r^{k + 1} - 11 r^k + 3 r^{k - 1} = 0Divide by r^{k - 1}:8 r^2 - 11 r + 3 = 0Solve quadratic equation:8 r^2 - 11 r + 3 = 0Discriminant D = 121 - 96 = 25Roots: r = [11 ± 5]/16 => (11 + 5)/16 = 16/16 = 1, (11 - 5)/16 = 6/16 = 3/8Therefore, general solution is P(k) = A (1)^k + B (3/8)^k = A + B (3/8)^kApply boundary conditions:P(1) = 1 = A + B (3/8)^1 = A + (3/8) BP(10) = 0 = A + B (3/8)^10Solve for A and B:From P(10): A = - B (3/8)^10Plug into P(1):1 = - B (3/8)^10 + (3/8) BFactor out B:1 = B [ - (3/8)^10 + 3/8 ]Therefore,B = 1 / [ - (3/8)^10 + 3/8 ]Compute denominator:Let me write 3/8 = 0.375But exact value:Denominator D = 3/8 - (3/8)^10So,B = 1 / D = 1 / [ 3/8 - (3/8)^10 ]Then A = - B (3/8)^10 = - (3/8)^10 / DTherefore, general solution:P(k) = A + B (3/8)^k = [ - (3/8)^10 / D ] + [1 / D ] (3/8)^k= [ (3/8)^k - (3/8)^10 ] / DBut D = 3/8 - (3/8)^10, so:P(k) = [ (3/8)^k - (3/8)^10 ] / [ 3/8 - (3/8)^10 ]Factor numerator and denominator:Numerator: (3/8)^k - (3/8)^10 = (3/8)^10 [ (3/8)^{k -10} - 1 ]Wait, not sure. Alternatively, factor out (3/8)^10:Numerator: (3/8)^10 [ (3/8)^{k -10} - 1 ] but this complicates. Alternatively, leave as is.Alternatively, multiply numerator and denominator by 8^10 / 3^10 to simplify:But maybe not necessary. Alternatively, note that:P(k) = [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]Alternatively, factor out (3/8) from denominator:Denominator: (3/8)(1 - (3/8)^9 )Numerator: (3/8)^k - (3/8)^10 = (3/8)^k [1 - (3/8)^{10 -k} ]Therefore,P(k) = [ (3/8)^k (1 - (3/8)^{10 -k}) ] / [ (3/8)(1 - (3/8)^9 ) ]Simplify:P(k) = (3/8)^{k -1} [1 - (3/8)^{10 -k} ] / [1 - (3/8)^9 ]But this seems messy. Maybe a better way.Alternatively, write as:P(k) = [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]We can factor numerator and denominator by (3/8):Numerator: (3/8)[ (3/8)^{k -1} - (3/8)^9 ]Denominator: (3/8)[1 - (3/8)^9 ]Therefore, cancel (3/8):P(k) = [ (3/8)^{k -1} - (3/8)^9 ] / [ 1 - (3/8)^9 ]Therefore,P(k) = [ (3/8)^{k -1} - (3/8)^9 ] / [ 1 - (3/8)^9 ]Which can be rewritten as:P(k) = [ (3/8)^{k -1} - (3/8)^9 ] / [ 1 - (3/8)^9 ]Alternatively, factor numerator:= (3/8)^{k -1} [1 - (3/8)^{10 -k} ] / [1 - (3/8)^9 ]But maybe leave it as is.So, for k =8,P(8) = [ (3/8)^7 - (3/8)^9 ] / [ (3/8) - (3/8)^9 ]Alternatively, compute numerator and denominator:Numerator: (3/8)^7 - (3/8)^9 = (3/8)^7 [1 - (3/8)^2 ] = (3/8)^7 [1 - 9/64] = (3/8)^7 [55/64]Denominator: (3/8) - (3/8)^9 = (3/8)[1 - (3/8)^8 ]Therefore,P(8) = [ (3/8)^7 * 55/64 ] / [ (3/8) * (1 - (3/8)^8 ) ] = [ (3/8)^6 * 55/64 ] / [1 - (3/8)^8 ]Simplify numerator and denominator:Numerator: (3^6 / 8^6) * (55/64) = 3^6 * 55 / 8^10Denominator: 1 - (3^8 / 8^8 ) = (8^8 - 3^8)/8^8Thus,P(8) = [ 3^6 * 55 / 8^10 ] / [ (8^8 - 3^8)/8^8 ] = [3^6 * 55 / 8^10 ] * [8^8 / (8^8 - 3^8 ) ] = [3^6 * 55 * 8^8 ] / [8^10 * (8^8 - 3^8 ) ]Simplify:= [3^6 * 55 ] / [8^2 * (8^8 - 3^8 ) ]Calculate numerator and denominator:First, compute 3^6: 72955 is 55.Numerator: 729 * 55 = let's compute:729 * 50 = 36,450729 * 5 = 3,645Total: 36,450 + 3,645 = 40,095Denominator: 8^2 = 648^8 = 16,777,2163^8 = 6,561Thus, denominator: 64 * (16,777,216 - 6,561) = 64 * 16,770,655Compute 16,770,655 * 64:First, 16,770,655 * 60 = 1,006,239,300Then, 16,770,655 * 4 = 67,082,620Total: 1,006,239,300 + 67,082,620 = 1,073,321,920Therefore, P(8) = 40,095 / 1,073,321,920Simplify this fraction:Divide numerator and denominator by 5:40,095 ÷ 5 = 8,0191,073,321,920 ÷ 5 = 214,664,384Check if 8,019 and 214,664,384 have common factors.Compute GCD(8,019, 214,664,384)First, factor 8,019:Divide by 3: 8 + 0 + 1 + 9 = 18, which is divisible by 3. 8,019 ÷ 3 = 2,6732,673 ÷ 3 = 891891 ÷ 3 = 297297 ÷ 3 = 9999 ÷ 3 = 3333 ÷ 3 = 11So 8,019 = 3^6 * 11Check 214,664,384:Check if divisible by 3: 2 + 1 + 4 + 6 + 6 + 4 + 3 + 8 + 4 = 38, which is not divisible by 3. So GCD is 1.Therefore, the simplified fraction is 8,019 / 214,664,384.Wait, but wait—this can’t be right. 40,095 / 1,073,321,920 reduces to 8,019 / 214,664,384, but is this in simplest terms?Wait, maybe I made a mistake in calculation. Let me double-check:Numerator: 729 * 55 = 40,095. Correct.Denominator: 64 * (16,777,216 - 6,561) = 64 * 16,770,65516,770,655 * 64:16,770,655 * 60 = 1,006,239,30016,770,655 * 4 = 67,082,620Sum: 1,006,239,300 + 67,082,620 = 1,073,321,920. Correct.So 40,095 / 1,073,321,920. Divide numerator and denominator by 5: 8,019 / 214,664,384.Now check if 8,019 and 214,664,384 have any common factors. We found 8,019 = 3^6 * 11. Check if 214,664,384 is divisible by 3: 2+1+4+6+6+4+3+8+4=38, which is not divisible by 3. Therefore, GCD is 1. So the fraction is reduced.But wait, 8,019 / 214,664,384 seems very small, but that's because we have (3/8)^k terms which decay exponentially. However, intuitively, starting at position 8, closer to the back, the probability to exit from front should be low, but maybe not that low? Wait, maybe there's an error in the setup.Wait, let me check the recurrence relation again. We had:8 P(k + 1) - 11 P(k) + 3 P(k - 1) = 0Characteristic equation 8 r^2 - 11 r + 3 = 0, roots r = 1 and r = 3/8. Then general solution is P(k) = A + B (3/8)^k. Then applying boundary conditions P(1) = 1 and P(10) = 0.So P(1) = A + B*(3/8) = 1P(10) = A + B*(3/8)^10 = 0Subtract the two equations:A + B*(3/8)^10 = 0A + B*(3/8) = 1Subtracting:B*(3/8)^10 - B*(3/8) = -1B [ (3/8)^10 - (3/8) ] = -1Therefore,B = -1 / [ (3/8)^10 - (3/8) ] = 1 / [ (3/8) - (3/8)^10 ]Hence, A = - B*(3/8)^10 = - (3/8)^10 / [ (3/8) - (3/8)^10 ]Then P(k) = A + B*(3/8)^k = [ - (3/8)^10 + (3/8)^k ] / [ (3/8) - (3/8)^10 ]Which simplifies to:P(k) = [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]Yes, that's correct.So for k=8,P(8) = [ (3/8)^8 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]Factor numerator and denominator:Numerator: (3/8)^8 [1 - (3/8)^2 ] = (3/8)^8 [1 - 9/64] = (3/8)^8 [55/64]Denominator: (3/8)[1 - (3/8)^9 ]Thus,P(8) = [ (3/8)^8 * 55/64 ] / [ (3/8) * (1 - (3/8)^9 ) ] = [ (3/8)^7 * 55/64 ] / [1 - (3/8)^9 ]Calculate this:First, compute (3/8)^7:3^7 = 2,1878^7 = 209,7152Wait, 8^7 = 8*8*8*8*8*8*8 = 8^2=64, 8^3=512, 8^4=4,096, 8^5=32,768, 8^6=262,144, 8^7=2,097,152So (3/8)^7 = 2,187 / 2,097,152Similarly, (3/8)^9 = 19,683 / 134,217,728Thus,Numerator: [2,187 / 2,097,152] * 55 / 64 = (2,187 * 55) / (2,097,152 * 64) = let's compute 2,187 *55:2,187 * 50 = 109,3502,187 *5 = 10,935Total = 109,350 +10,935=120,285Denominator part 1: 2,097,152 *64=134,217,728So numerator=120,285 /134,217,728Denominator of P(8): 1 - (19,683 /134,217,728) = (134,217,728 -19,683)/134,217,728=134,198,045 /134,217,728Thus,P(8)= [120,285 /134,217,728] / [134,198,045 /134,217,728] = 120,285 /134,198,045Simplify this fraction:Divide numerator and denominator by 5:120,285 ÷5=24,057134,198,045 ÷5=26,839,609Check GCD(24,057, 26,839,609). Let's factor 24,057:24,057 ÷3=8,019 (since 2+4+0+5+7=18, divisible by 3)8,019 ÷3=2,6732,673 ÷3=891891 ÷3=297297 ÷3=9999 ÷3=3333 ÷3=11So 24,057=3^6 *11Now check 26,839,609:Sum of digits: 2+6+8+3+9+6+0+9=43, not divisible by 3. Therefore, GCD is 1.Thus, simplified fraction is 24,057 /26,839,609Wait, but this is the same as 8,019 /214,664,384 as before, just multiplied by 3 (since 24,057=3*8,019 and 26,839,609=3*8,946,536.333…). Wait, no. Wait, 24,057 /26,839,609 is the simplified form after dividing numerator and denominator by 5. But since 24,057 and 26,839,609 are coprime, the fraction is reduced.But 24,057 /26,839,609 seems very small, around 0.000896, which is less than 0.1%. That seems counterintuitive because starting at position 8, which is closer to the back, but due to the insertion/deletion dynamics, maybe the probability is indeed low. However, let's check with a smaller case.Alternatively, maybe there's a mistake in setting up the recurrence relation.Wait, going back, the characteristic equation was correct, roots 1 and 3/8. The general solution is P(k)=A + B*(3/8)^k. Applying boundary conditions:At k=1: 1 = A + B*(3/8)At k=10: 0 = A + B*(3/8)^10Subtracting:1 = B*(3/8 - (3/8)^10 )So B =1 / [ (3/8) - (3/8)^10 ]Therefore, P(k)= [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]Yes, this is correct.Compute P(8):Numerator: (3/8)^8 - (3/8)^10 = (3/8)^8 [1 - (3/8)^2 ] = (3^8/8^8)(1 - 9/64) = (6,561 /16,777,216)(55/64) = (6,561*55)/ (16,777,216*64) = 360,855 /1,073,741,824Denominator: (3/8) - (3/8)^10 = (3/8) - (3^10/8^10) = (3/8) - (59,049 /1,073,741,824) = Convert 3/8 to 402,653,184 /1,073,741,824Therefore, denominator= (402,653,184 -59,049)/1,073,741,824=402,594,135 /1,073,741,824Thus, P(8)= (360,855 /1,073,741,824) / (402,594,135 /1,073,741,824 )=360,855 /402,594,135Simplify this fraction:Divide numerator and denominator by 15:360,855 ÷15=24,057402,594,135 ÷15=26,839,609Which gives 24,057 /26,839,609 same as before. Check if reducible:24,057 factors: 3^6 *1126,839,609: sum of digits is 2+6+8+3+9+6+0+9=43, not divisible by 3. Therefore, GCD is 1. Hence, 24,057/26,839,609 is reduced.But this is approximately 24,057 ÷26,839,609 ≈0.000896, or 0.0896%. That seems extremely low. But starting from position 8, which is near the back, maybe it's correct? However, let's think about the process.Every time a student is inserted, there's a chance to move me closer to the front or the back. However, insertions are more likely to happen in front of me (8/11 chance to move back, 3/11 to stay). Then, removal from front or back is equally likely.But wait, when a student is inserted in front of me, which happens with higher probability, I get pushed back, making me more likely to be near the back. However, the removal step can either remove from front or back. It's possible that even though insertions push me back, the front removals can shift me forward.Wait, maybe there's a symmetry or something. Alternatively, maybe there's a mistake in the recurrence setup.Wait, another approach: consider that each time, before removal, there's an insertion. So the process is:Insertion: with probability 8/11, my position increases by 1 (becomes k+1), with probability 3/11 stays at k.Then, removal: front or back, each 1/2.If front is removed: if I was at position k', then my new position is k' -1 (if k' >1), or I get removed if k' =1.If back is removed: if I was at position k', then my new position is k' (if k' <11), or I get removed if k' =11.But since after insertion, my position is either k+1 or k. So:Case 1: insertion made me k+1 (prob 8/11): - Remove front: new position k (prob 1/2) - Remove back: new position k+1 if k+1 <=10 (prob 1/2), else removed.But since k is from 1 to10, k+1=11 only if k=10. So if k=10, inserting makes me 11, then remove back would remove me. But we start at k=8.Case 2: insertion kept me at k (prob 3/11): - Remove front: new position k-1 (prob 1/2) - Remove back: new position k (prob 1/2)Therefore, the transitions are as previously. So the recurrence relation should hold.Alternatively, maybe there's a simpler way. Notice that in expectation, the process might have a drift towards the back because insertion in front is more likely. However, removals are symmetric. Therefore, perhaps the probability is low, but not as low as 0.08%.Alternatively, maybe I made a miscalculation in simplifying the fraction.Wait, let me compute the numerator and denominator again:P(8)= [ (3/8)^8 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]Compute numerator:(3/8)^8 - (3/8)^10 = (3^8/8^8) - (3^10/8^10) = 3^8/8^8 (1 - 3^2/8^2) = 3^8/8^8 (1 - 9/64) = 3^8/8^8 * 55/64Denominator:(3/8) - (3/8)^10 = 3/8 - 3^10/8^10 = 3/8 (1 - 3^9/8^9 )Therefore,P(8)= [3^8/8^8 * 55/64] / [3/8 (1 - 3^9/8^9 ) ] = [3^7/8^7 *55/64 ] / [1 - (3/8)^9 ]= [ (55 * 3^7 ) / (8^7 * 64 ) ] / [ (8^9 -3^9)/8^9 ]= [ (55 * 3^7 ) / (8^8 ) ] / [ (8^9 -3^9)/8^9 ]= [55 * 3^7 *8^9 ] / [8^8 (8^9 -3^9 ) ]= [55 *3^7 *8 ] / [8^9 -3^9 ]Compute numerator and denominator:Numerator: 55 * 3^7 *83^7 =2,18755*2,187=120,285120,285*8=962,280Denominator:8^9 -3^98^9=134,217,7283^9=19,683Thus, denominator=134,217,728 -19,683=134,198,045Therefore, P(8)=962,280 /134,198,045Simplify this fraction:Divide numerator and denominator by 5:962,280 ÷5=192,456134,198,045 ÷5=26,839,609Check GCD(192,456,26,839,609 )Factor 192,456:192,456 ÷2=96,22896,228 ÷2=48,11448,114 ÷2=24,05724,057 ÷3=8,019 (as before)So 192,456=2^3*3*8,01926,839,609: sum is 2+6+8+3+9+6+0+9=43, not divisible by 2 or3. Hence, GCD is 1.Thus, the simplified fraction is192,456 /26,839,609, which reduces to 192,456 /26,839,609. But 192,456 divided by 3 is 64,152, which is still even.Wait, but denominator is 26,839,609, which is odd. So GCD(192,456,26,839,609). Since denominator is odd, 2 is not a factor. And sum of digits of denominator is 43, not divisible by 3. So GCD is 1. Therefore, the fraction is irreducible.But 192,456 /26,839,609 ≈0.00717, which is approximately 0.717%. Still very low, but maybe that's correct?But wait, another approach: use the formula from the Gambler's Ruin problem. In the standard Gambler's Ruin, the probability is ( (q/p)^k - (q/p)^N ) / (1 - (q/p)^N ), where p is the probability of moving forward, q backward. However, in our case, the transition probabilities are not symmetric and depend on the current state.But in our recurrence, we found that the probability can be expressed similarly. However, given the non-homogeneous transitions (except we found that the recurrence relation allowed us to use a homogeneous solution with constant coefficients, which might be a coincidence due to specific transition probabilities).Alternatively, maybe there's a symmetry or a martingale approach.Alternatively, notice that in each step, the insertion and removal process can be seen as a random walk with certain biases. The key is that insertions are more likely to push me back, but removals are symmetric. However, the exact probability would require solving the recurrence.Alternatively, note that the answer might be 8/11, but I need to check.Wait, let's think of expected value. Let me denote E(k) as the expected number of steps before exiting. But we need probability, not expected time.Alternatively, consider that each time, there's a certain probability to move left or right. But due to the insertion being more likely to push me back, the walk is biased towards the back.Alternatively, think of the process as a Markov chain with absorbing states at 1 and 10, and find the absorption probabilities.But given the complexity, and the solution from the recurrence leading to P(k)= [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ].Therefore, for k=8:P(8) = [ (3/8)^8 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]= [ (3/8)^8 (1 - (3/8)^2) ] / [ (3/8)(1 - (3/8)^9) ]= (3/8)^7 (1 - 9/64) / (1 - (3/8)^9 )= (3/8)^7 (55/64) / (1 - (3/8)^9 )Which is what we computed before.Alternatively, compute it numerically:First, compute (3/8):3/8=0.375(3/8)^2=0.140625(3/8)^7≈0.375^7≈0.375*0.375=0.140625, *0.375=0.052734375, *0.375≈0.019775390625, *0.375≈0.00741665625, *0.375≈0.00278124609375, *0.375≈0.00104296875. Wait, but this is manual multiplication error-prone. Let me use exponents:3^7=2,1878^7=262,144Thus, (3/8)^7=2,187 / 262,144≈0.008342Multiply by 55/64≈0.859375:0.008342 *0.859375≈0.00717Denominator: (3/8) - (3/8)^10≈0.375 - (approx 3^10=59,049; 8^10=1,073,741,824; 59,049/1,073,741,824≈0.000055)Thus, denominator≈0.375 -0.000055≈0.374945Thus, P(8)=0.00717 /0.374945≈0.0191, which is approximately 1.91%.Wait, this conflicts with the previous calculation of 0.00717. What's wrong here?Wait, wait, in the numerator we have:(3/8)^7*(55/64)= (approx 0.008342) * (55/64)=0.008342*0.859375≈0.00717Denominator:1 - (3/8)^9=1 - (19,683 /134,217,728)=1 -0.000146≈0.999854Thus, P(8)=0.00717 /0.374945≈0.0191 (1.91%)But wait, no:Wait, in the denominator of the overall expression, it's [ (3/8) - (3/8)^10 ] ≈0.375 -0.000055=0.374945But in the numerator we have [ (3/8)^8 - (3/8)^10 ]≈ (3/8)^8= (6,561 /16,777,216≈0.000391) - (3/8)^10≈0.000055≈0.000336Then, 0.000336 /0.374945≈0.000896, which is ≈0.0896%.But this conflicts with the previous step. So this suggests that there's confusion in the calculation.Wait, perhaps I messed up the numerator and denominator.Wait, let's compute it step by step with decimals:Compute P(8)= [ (3/8)^8 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]First, compute (3/8):3/8=0.375Compute (3/8)^8:0.375^8. Let's compute step by step:0.375^2=0.1406250.375^4= (0.140625)^2≈0.0197753906250.375^8= (0.019775390625)^2≈0.000391Similarly, (3/8)^10= (3/8)^8*(3/8)^2≈0.000391*0.140625≈0.000055Therefore, numerator=0.000391 -0.000055=0.000336Denominator=0.375 -0.000055≈0.374945Therefore, P(8)=0.000336 /0.374945≈0.000896≈0.0896%This is about 0.09%, which is extremely low.But intuitively, starting at position 8, with insertions more likely to push me back, and symmetric removal, the chance to reach front is very low. But is it really 0.09%?Alternatively, perhaps there is an error in the characteristic equation.Wait, original recurrence:8 P(k + 1) - 11 P(k) + 3 P(k - 1) = 0We posited P(k) = r^k, leading to:8 r^2 -11 r +3=0, which gives roots r=1 and r=3/8.But if the process is such that the probability at each step is position-dependent, but the recurrence ended up being linear with constant coefficients, which is special.Alternatively, if we consider that the process, despite the position-dependent transitions, ended up having a linear recurrence with constant coefficients, which is surprising, but possible due to the specific probabilities.Alternatively, test the solution with boundary conditions.Take k=1:P(1)= [ (3/8)^1 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ] = [0.375 -0.000055]/[0.375 -0.000055]≈1. Correct.k=10:P(10)= [ (3/8)^10 - (3/8)^10 ]/[ (3/8)- (3/8)^10 ]=0. Correct.Therefore, the solution satisfies the boundary conditions. Therefore, the solution is correct. Hence, P(8)=≈0.0896%, which is 896/1,000,000 roughly, but the exact fraction is 24,057/26,839,609.Wait, but 24,057 and26,839,609: let me check if they have a common factor:24,057 ÷3=8,019; 26,839,609 ÷3=8,946,536.333… Not integer. So no.24,057 ÷11=2,187; 26,839,609 ÷11=2,439,964.4545… Not integer. So no.Thus, the fraction is indeed reduced, and the probability is 24,057/26,839,609.But the problem states that the answer should be m/n with m and n coprime, and then compute 100m +n. However, 24,057 and26,839,609 are coprime, so m=24,057 andn=26,839,609, but these numbers are very large. However, this seems off because usually, such problems have elegant fractions. Maybe there's a different approach.Wait, another way to model this problem is to recognize that each time, the probability that a new student is inserted in front of me is 8/11, which would push me back, and 3/11 chance to stay. Then, upon removal, front or back is equally likely.This resembles a discrete-time Markov chain with a drift towards the back but with possibilities to move forward due to front removals.Alternatively, think of it as a random walk where at each step, with probability 8/11 *1/2 =4/11, I move back; with probability 3/11 *1/2=3/22, I move forward; and with probability (8/11 *1/2)+(3/11 *1/2)= (4/11 + 3/22)=11/22=1/2, I stay.Wait, no:Wait, transitions are:From state k:- With probability 8/11 *1/2=4/11, move back (due to insertion in front and front removal).Wait, no:Wait, insertion can push me to k+1 with probability 8/11, then front removal brings me back to k. Alternatively, insertion stays at k with probability 3/11, then front removal brings me to k-1.So actually, the probability to move forward is only from k-1 to k, which requires insertion staying at k and front removal. Wait, no:Wait, let's clarify:Wait, transitions:From k, after insertion and removal:- To k-1: occurs when insertion stays at k (prob 3/11) and front is removed (prob 1/2). So 3/22.- To k+1: occurs when insertion moves to k+1 (prob 8/11) and back is removed (prob 1/2). So 4/11.- To k: all other cases: insertion moves to k+1 and front removed (4/11) + insertion stays at k and back removed (3/22). Total 4/11 +3/22=11/22=1/2.Thus, from any state k (2-9), the probabilities are:- P(k→k-1)=3/22- P(k→k)=1/2- P(k→k+1)=4/11So this is a random walk with transition probabilities dependent on k, but actually, the transition probabilities are the same for all k (2-9). Therefore, the process is homogeneous in the interior states, leading to the same characteristic equation.Therefore, the solution using the recurrence is correct, leading to the very small probability.But this seems counterintuitive. Starting at position 8, near the back, the probability to exit from the front is about 0.09%. But given the high probability of being pushed back further, and the symmetric removal, maybe it is correct.However, to verify, let's consider a smaller case, like starting at position 2.Compute P(2)= [ (3/8)^2 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]≈ (0.140625 -0.000055)/ (0.375 -0.000055)≈0.14057/0.374945≈0.375, which is 3/8. That's interesting. Similarly, P(2)=3/8. Similarly, P(3)= [ (3/8)^3 - (3/8)^10 ]/[ (3/8) - (3/8)^10 ]≈ (0.052734 -0.000055)/0.374945≈0.052679/0.374945≈0.1405≈ (3/8)^2. It seems that P(k)= (3/8)^{k-1} ? Let me check:P(k)= [ (3/8)^k - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]If we ignore the (3/8)^10 terms (since they are very small), P(k)≈ (3/8)^k / (3/8)= (3/8)^{k-1}Which matches the approximate values:P(2)≈3/8, P(3)≈(3/8)^2, etc.Therefore, for k=8, P(k)≈(3/8)^{7}≈0.008342, which is approximately 0.834%, but the exact value is slightly less due to the subtracted terms.But according to the exact calculation, it's about 0.0896%, which is an order of magnitude less. This inconsistency suggests an error in the approximation.Wait, but when I calculated P(2) exactly:P(2)= [ (3/8)^2 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]= [9/64 - (3/8)^10 ]/[3/8 - (3/8)^10 ]If we approximate (3/8)^10≈0, then P(2)= (9/64)/(3/8)= (9/64)*(8/3)=3/8≈0.375, which matches the earlier approximation. However, the exact value would be slightly less.Similarly, P(8)= [ (3/8)^8 - (3/8)^10 ] / [ (3/8) - (3/8)^10 ]≈ (3/8)^8 / (3/8)= (3/8)^7≈0.008342=0.834%, but the exact value is lower because we subtract (3/8)^10 in the numerator and have denominator slightly less than 3/8.But if (3/8)^10 is negligible, then P(k)= (3/8)^{k-1}. For k=8, that would be (3/8)^7≈0.834%, but due to the subtraction in the numerator and the denominator, it's about 0.09%.This discrepancy arises because (3/8)^10 is not entirely negligible compared to (3/8)^8. Specifically, (3/8)^10 is (3/8)^2*(3/8)^8= (9/64)*(3/8)^8≈0.14*(3/8)^8. So the numerator is (3/8)^8*(1 -0.14)=0.86*(3/8)^8, and the denominator is (3/8)*(1 - (3/8)^9)≈3/8*(1 - negligible). So the exact value is approximately 0.86*(3/8)^7≈0.86*0.834%≈0.719%, which is closer to the 0.717% we calculated earlier. But due to the denominator being 3/8 - small term, which is approximately 3/8, the exact value is roughly (3/8)^7 *0.86 / (3/8)=0.86*(3/8)^7*(8/3)=0.86*(3/8)^6.Wait, this is getting too convoluted. The exact fraction is 24,057/26,839,609. To check if this reduces:Compute GCD(24,057,26,839,609). Use Euclidean algorithm:GCD(26,839,609,24,057)Compute 26,839,609 ÷24,057=1,115.3... So subtract 24,057*1,115=24,057*1,000=24,057,000; 24,057*100=2,405,700; 24,057*15=360,855 → Total 24,057,000+2,405,700+360,855=26,822,55526,839,609 -26,822,555=17,054Now compute GCD(24,057,17,054)24,057 ÷17,054=1, remainder 7,003GCD(17,054,7,003)17,054 ÷7,003=2, remainder 3,048GCD(7,003,3,048)7,003 ÷3,048=2, remainder 907GCD(3,048,907)3,048 ÷907=3, remainder 327GCD(907,327)907 ÷327=2, remainder 253GCD(327,253)327 ÷253=1, remainder 74GCD(253,74)253 ÷74=3, remainder 31GCD(74,31)74 ÷31=2, remainder 12GCD(31,12)31 ÷12=2, remainder 7GCD(12,7)12 ÷7=1, remainder 5GCD(7,5)7 ÷5=1, remainder 2GCD(5,2)5 ÷2=2, remainder 1GCD(2,1)=1Thus, GCD is 1. Therefore, the fraction 24,057/26,839,609 is irreducible. Therefore, the answer is m=24,057 andn=26,839,609, leading to 100m +n=100*24,057 +26,839,609=2,405,700 +26,839,609=29,245,309.But this seems extremely large and unlikely to be the answer intended by the problem proposer. There must be a mistake.Wait, another approach: maybe the probability is 8/11. Because when inserting, there's 8/11 chance to insert in front, pushing me back, and 3/11 to insert behind, keeping me. Then, when removing front or back, maybe the probability is related to the number of students in front or behind?Wait, no, because the process is repeated until I am removed. Each time, the probability depends on current position.Alternatively, consider that at each step, the probability that I am closer to the front or back depends on the insertions and deletions. But I think the original recurrence is correct, leading to the very small probability.Alternatively, maybe the answer is 1/2, by symmetry? But insertions are more probable in front, so symmetry is broken.Alternatively, the probability is 3/11, but no.Wait, going back to the original problem statement:"10 students are arranged in a row. Every minute, a new student is inserted in the row (which can occur in the front and in the back as well, hence 11 possible places) with a uniform 1/11 probability of each location. Then, either the frontmost or the backmost student is removed from the row (each with a 1/2 probability)."Suppose you are the eighth in the line from the front. The probability that you exit the row from the front rather than the back is m/n.The key insight might be that the process is memoryless and the probability only depends on the current position. Therefore, modeling it as a Markov chain with states 1 to 10, absorbing at 1 and 10, and solving for the absorption probabilities.But we did that, and the result was a very small probability.However, given that the problem is from Lewis Chen, who often proposes problems with elegant solutions, perhaps there's a different approach.Alternative approach:Consider that each time a student is inserted and another is removed, the line undergoes a random walk. The key is that insertions are uniform over 11 positions, then removal is equally likely front or back.Alternatively, model the entire process as follows: each minute, with probability 1/2, we remove from front and with probability 1/2 from back, but before removing, we insert a student at a random position.The key is that inserting a student at a random position and then removing from front or back is equivalent to certain operations on the line.Alternatively, consider that inserting a student at a random position and then removing from front or back is similar to randomly permuting the line with certain properties.But I'm not sure.Alternatively, think of the line as a deque (double-ended queue), where each time, you insert at a random position and then pop from either end with equal probability. The probability that a particular element is popped from the front before the back.However, in this case, since the line is maintained at size 10, the eighth student's position changes over time based on insertions and deletions.Alternatively, use the concept of potential theory or Lyapunov functions.Alternatively, recognize that each insertion and deletion step can be considered as follows: inserting a student at a random position is equivalent to randomly selecting a student, and inserting a new student before them, then removing from front or back.But I don't see the connection.Alternatively, note that the probability of exiting from the front is equal to the expected number of times the student is at the front and then gets removed divided by the total number of removals. But since the process is memoryless and,<think>Okay, so I need to figure out the probability that I, being the eighth student in a row of 10, will exit from the front rather than the back. The process is that every minute, a new student is inserted into the row in one of 11 possible positions (including front and back) with equal probability, and then either the frontmost or backmost student is removed, each with probability 1/2. Hmm, interesting.First, let me try to understand the setup. Initially, there are 10 students arranged in a row. I'm the eighth from the front. Each minute, a new student is added, increasing the number of students to 11, and then one is removed from either the front or the back. The key here is that the row length fluctuates between 10 and 11 each minute. Wait, no: actually, when a new student is inserted, the row becomes 11 students, and then one is removed, bringing it back to 10. So each minute, the row size remains 10 after each insertion and removal cycle. So every minute, the row is modified by inserting a new student in a random position and then removing from either front or back. So the process keeps the number of students at 10 each time.But how does the position of the original eighth student change over time? Each insertion can potentially shift their position. For example, if a student is inserted in front of them, their position increases by one; if inserted behind them, their position stays the same. If inserted at the very front, then everyone's position increases by one. Similarly, inserting at the very back doesn't affect their position. Then, after insertion, either the front or the back is removed. If the front is removed, then everyone's position decreases by one. If the back is removed, positions stay the same unless the insertion caused the back to be pushed.Wait, this seems a bit complicated. Let's formalize the problem.Let me denote my position as k. Initially, k=8. Each minute, the following happens:1. Insert a new student in one of 11 possible positions (positions 1 through 11, where position 1 is front). Each position has probability 1/11.2. Then, remove either the front student (position 1) or the back student (position 11) with probability 1/2 each.So after each cycle, the row size remains 10.My position can change in two ways: during insertion and during removal.First, during insertion: inserting a student in positions 1 through k will increase my position by 1, because the new student is ahead of me. Inserting in positions k+1 through 11 won't affect my position. So the probability that my position increases by 1 is k/11, and the probability it remains the same is (11 - k)/11.Then, during removal: if we remove from the front, then all positions decrease by 1. If we remove from the back, positions remain the same. However, there's a catch here: if a student was inserted at position 11 (the back), then the original back student was at position 10, but after insertion, the back is position 11. Then removing the back would remove the newly inserted student. Wait, but does the removal happen after insertion? So after inserting a student in position i (1 to 11), making the total students 11, then we remove either the first or the last student (positions 1 or 11). So if we remove the front, then all positions shift left by 1. If we remove the back, which is position 11, then positions 1 through 10 remain, so my position remains the same unless I was at position 11, but since the original row was 10 students, after inserting at position 11, the original back student is at position 10, so inserting at 11 would create position 11, which is then removed with probability 1/2. So if the insertion is at position 11, and then we remove the back, then the inserted student is immediately removed, so effectively, the row remains the same. Similarly, inserting at the front and then removing the front also undoes the insertion. Interesting.But how does this affect my position? Let me think step by step.Case 1: Insertion occurs in position i (1 to 11), then removal occurs at front or back.If insertion is at position i <= k: Then my original position k becomes k+1. Then, if we remove from the front (position 1), my position becomes k. If we remove from the back (position 11), my position remains k+1.If insertion is at position i > k: Then my original position remains k. Then, if we remove from the front, my position becomes k-1. If we remove from the back, my position remains k.Wait, no. Let me correct that.Wait, when inserting a student in position i <= k, that shifts my position to k+1. Then, removal from front (position 1) would shift everyone left by 1, so my position becomes (k+1) -1 = k. If removal is from the back (position 11), then the student at position 11 is removed. Since insertion was at position i <=k, the original students from position k+1 to 10 would have shifted right by 1, but since we started with 10 students, inserting at position i would make 11 students, then removing the back (position 11) would remove the student who was originally at position 10 (since inserting at position i <=k doesn't affect positions after k). So removing the back in this case removes the original back student, which was shifted to position 11. Therefore, my position after insertion and removal would be k+1 if we remove the back, because inserting at i <=k shifted me to k+1, and removing the back doesn't affect my position.Similarly, inserting at i > k: my original position k remains. Then, if we remove from the front, my position becomes k -1. If we remove from the back, my position remains k. But inserting at i >k adds a student after me, so when we remove the back, which is the new student if insertion was at position 11, but if insertion was at position i where k < i <=11, then inserting there shifts the students after i by 1. So inserting at position i >k would mean inserting somewhere after me, so the original back student is at position 10, and inserting at position 11 would create a new back student at position 11. Then removing the back would remove that new student, so the original students remain. If insertion was at position i where k < i <=10, then inserting there would split the row, shifting some students. Then removing the back (position 11) would remove the original back student (who was shifted to position 11 if insertion was after them). Hmm, maybe this is getting too detailed.Perhaps a better approach is to model the position as a Markov chain. Since each transition depends only on the current position, not the history, we can model this as a Markov process where the state is my current position, which can be from 1 to 10 (since after each cycle, the row is back to 10 students). The question is, starting from position 8, what is the probability that I eventually get absorbed at position 1 (exit from front) rather than position 10 (exit from back).But wait, actually, the exit happens when I am either frontmost or backmost. So if my position becomes 1, then in the next step, there's a chance I could be removed when they remove the front. Similarly, if my position becomes 10, I could be removed when they remove the back. So perhaps this is an absorbing Markov chain with absorbing states at 1 and 10, and we need to find the probability of absorption at 1 starting from 8.But actually, no, because the removal happens every time after insertion. So each minute, after inserting a student, we remove either front or back. So if my position is 1, then when they remove the front, I get removed. If my position is 10, when they remove the back, I get removed. So the removal is probabilistic, depending on the insertion and the removal choice.Wait, perhaps not. Wait, even if I am in position 1, if after insertion, someone is inserted in front of me, making me position 2, and then they remove the front, then the new student is removed, and I go back to position 1. But if they remove the back, then I stay at position 2. Similarly, if I'm at position 10, inserting after me would push me to position 11, but then removing the back would remove the new student, putting me back to 10. Hmm, so actually, my position can fluctuate, and the removal isn't directly dependent on my position unless I'm at the front or back. So maybe this is a Markov chain where the states are positions 1 through 10, and transitions depend on the insertion and removal steps.So let's model the problem as a Markov chain with states 1 to 10. The probability of moving from state k to another state is determined by the insertion and removal steps.First, when in state k (current position), insertion happens: with probability i/11 for i = 1 to 11 (wait, insertion positions are 1 to 11, each with probability 1/11). Then removal happens: front or back with probability 1/2 each.But insertion position affects the resulting position after insertion and before removal. Let's formalize this.Let me break down the transition:1. Current position: k.2. Insert a student at position i (1 to 11). The probability of choosing each i is 1/11.3. If i <= k, then after insertion, my position becomes k + 1. If i > k, my position remains k.4. Then, remove either front or back: a. If front is removed, then all positions decrease by 1. So if my position after insertion was m, it becomes m - 1 (if m > 1; if m = 1, then I would be removed, but since we are modeling before removal, actually, if front is removed, then the position becomes m - 1 for everyone else. Wait, if front is removed, the entire row shifts left by 1. So if my position after insertion was m, then after removing front, it's m - 1 (unless m = 1, in which case I would be removed, but since we are considering transitions before removal, this is a bit confusing.Wait, actually, if we are in state k, which is my position before insertion and removal. Wait, no. Wait, in the Markov chain, the state should represent my position after the insertion and removal steps. Wait, this is a bit confusing.Wait, perhaps the state is my position before the insertion and removal steps. Then, the transitions are as follows: starting from position k, insert a student in position i (1 to 11), which may shift my position to k + 1 if i <= k, then remove front or back, which may shift my position again.But maybe it's better to model the transition in one step. Let me think.Starting at position k.Insert a student at position i (1 to 11). The probability for each i is 1/11.If i <= k: then after insertion, my position is k + 1.If i > k: my position remains k.Then, remove front or back with probability 1/2 each.If front is removed:- If insertion was at position i <= k: my position after insertion was k + 1, then removing front shifts me to (k + 1) - 1 = k.- If insertion was at position i > k: my position was k, then removing front shifts me to k - 1.If back is removed:- If insertion was at position i <= k: my position was k + 1, and removing back (which is the original back student shifted to 11, so unless the insertion was at 11, the back student is either the inserted one or the original one. Wait, actually, inserting at position i <= k would shift the original students from i to 10 to the right by 1. So the original back student (position 10) moves to 11. Then removing back would remove the student at 11, which is the original back student. So in this case, removing back would not affect my position, so my position remains k + 1.If insertion was at position i > k: my position was k. Inserting at i > k would shift positions from i to 10 to the right by 1, but since i > k, inserting after me, so my position remains k. Then, removing back would remove the student at position 11, which is either the inserted student (if i = 11) or the original student shifted from position 10 to 11 (if i <=10). Either way, removing back doesn't affect my position, so my position remains k.Therefore, putting it all together:When starting at position k:- With probability i/11 for each i from 1 to 11 (but we can group them):Case 1: Insertion at i <= k (probability k/11): - After insertion, my position is k + 1. - Then, remove front with 1/2 probability: position becomes k. - Remove back with 1/2 probability: position remains k + 1.Case 2: Insertion at i > k (probability (11 - k)/11): - After insertion, my position remains k. - Then, remove front with 1/2 probability: position becomes k - 1. - Remove back with 1/2 probability: position remains k.Therefore, the transition probabilities from state k are:- To state k: probability (11 - k)/11 * 1/2 (insert back and remove front) + k/11 * 1/2 (insert front and remove back) ?Wait, let's re-examine.In Case 1 (insertion i <= k, probability k/11): - After insertion, position is k + 1. - Then, remove front (prob 1/2): position becomes (k + 1) - 1 = k. - Remove back (prob 1/2): position remains k + 1.Thus, from Case 1, transitions are: - k → k with probability k/11 * 1/2 - k → k + 1 with probability k/11 * 1/2In Case 2 (insertion i > k, probability (11 - k)/11): - After insertion, position remains k. - Then, remove front (prob 1/2): position becomes k - 1. - Remove back (prob 1/2): position remains k.Thus, from Case 2, transitions are: - k → k - 1 with probability (11 - k)/11 * 1/2 - k → k with probability (11 - k)/11 * 1/2Therefore, combining both cases, the transition probabilities from state k are:- To k - 1: (11 - k)/11 * 1/2- To k: [k/11 * 1/2 + (11 - k)/11 * 1/2] = [k + (11 - k)] / 22 = 11 / 22 = 1/2- To k + 1: k/11 * 1/2But note that when k = 1, you can't go to k - 1 = 0, and when k = 10, you can't go to k + 1 = 11. So the transition probabilities for the boundary states need adjustment.For k = 1:Case 1 (insertion i <=1, which is i=1, probability 1/11): - After insertion, position becomes 2. - Then, remove front (prob 1/2): position becomes 1. - Remove back (prob 1/2): position remains 2.Case 2 (insertion i >1, probability 10/11): - After insertion, position remains 1. - Remove front (prob 1/2): position becomes 0, which is invalid (since the student would be removed). Wait, but in the model, if you are in position 1, inserting behind you (i >1), then removing front would remove the front student (you), so in that case, you exit from the front. Similarly, inserting in front of you (i=1) and then removing front would remove the inserted student, putting you back to position 1. Wait, this complicates things because when you are in position 1, inserting in front of you (i=1) moves you to 2, then removing front removes the inserted student, putting you back to 1. If you insert behind you (i>1), you stay at 1, then removing front removes you.Wait, this suggests that the model for state 1 is different. When you are in state 1 (frontmost), inserting in front (i=1) moves you to 2, then removing front (prob 1/2) removes the inserted student, so you go back to 1. Removing back (prob 1/2) leaves you at 2. If you insert behind you (i>1), you remain at 1, then removing front (prob 1/2) removes you (absorption), and removing back (prob 1/2) leaves you at 1.Similarly, for state 10 (backmost), inserting behind you (i=11) moves you to 11, then removing back (prob 1/2) removes the inserted student, putting you back to 10. Removing front (prob 1/2) removes the front student, shifting everyone left, so you become 10 -1 =9. If you insert in front of you (i <=10), which is always true except i=11, but inserting at i <=10 would either shift you to 11 if i <=10. Wait, inserting at i <=10 when you're at 10: inserting at i <=10 would shift your position to 11 only if i <=10? Wait, no. If you are at position 10, inserting at position i <=10 would shift your position to 11. Wait, inserting at position i in 1 to 11. If I'm at position 10, inserting at position i <=10 would shift me to 11. Inserting at position 11 would leave me at 10. Then, removing front or back.But if inserting at i <=10, making me 11, then removing front (prob 1/2) shifts everyone left, so my position becomes 10. Removing back (prob 1/2) removes the student at position 11 (which is me), so I get removed (absorption at back). If inserting at i=11, I stay at 10. Then removing front (prob 1/2) shifts everyone left, so position becomes 9. Removing back (prob 1/2) removes the inserted student, so I stay at 10.This is getting really complex. So maybe the transitions for the boundary states are different. Therefore, we need to handle states 1 and 10 separately.In general, for states 2 through 9, the transitions are:From state k (2 ≤ k ≤9):- To k-1: (11 - k)/22- To k: 1/2- To k+1: k/22For state 1:- Insertion at i=1 (prob 1/11): - After insertion, position 2. - Remove front (prob 1/2): back to position 1. - Remove back (prob 1/2): stay at 2.- Insertion at i>1 (prob 10/11): - Stay at position 1. - Remove front (prob 1/2): absorbed (exit front). - Remove back (prob 1/2): stay at 1.Therefore, transitions from state 1:- To 1: (1/11 * 1/2) + (10/11 * 1/2) = (1 +10)/22 = 11/22 = 1/2- To 2: 1/11 * 1/2 = 1/22- Absorbed (exit front): 10/11 * 1/2 = 5/11Wait, hold on. When inserting at i>1 (prob 10/11), you stay at 1. Then removing front (prob 1/2) would remove you, resulting in absorption. Removing back (prob 1/2) leaves you at 1.Similarly, inserting at i=1 (prob 1/11), moving to 2. Then removing front (prob 1/2) puts you back to 1. Removing back (prob 1/2) leaves you at 2.Therefore, from state 1:- Probability to stay at 1: (1/11 * 1/2) + (10/11 * 1/2) = 1/22 + 10/22 = 11/22 = 1/2- Probability to go to 2: 1/11 * 1/2 = 1/22- Probability to be absorbed (exit front): 10/11 * 1/2 = 5/11Wait, but absorption is an event, not a state. So perhaps the transition probabilities should sum to 1, but here, from state 1, the transitions are:- With probability 5/11, you are absorbed (exit front).- With probability 1/2, you stay at 1.- With probability 1/22, you go to 2.But 5/11 + 1/2 + 1/22 = 10/22 + 11/22 +1/22 =22/22=1. Yes, that works.Similarly, for state 10:From state 10:- Insertion at i <=10 (prob 10/11): - Position becomes 11. - Remove front (prob 1/2): position becomes 10. - Remove back (prob 1/2): absorbed (exit back).- Insertion at i=11 (prob 1/11): - Stay at 10. - Remove front (prob 1/2): position becomes 9. - Remove back (prob 1/2): stay at 10.Therefore, transitions from state 10:- To 10: (10/11 * 1/2) + (1/11 * 1/2) = (10 +1)/22 = 11/22 =1/2- To 9: 1/11 * 1/2 =1/22- Absorbed (exit back):10/11 * 1/2=5/11So similar to state 1, but mirrored.Therefore, in summary, the Markov chain has transient states 2 through 9, and absorbing states 1 (exit front) and 10 (exit back). Wait, actually, no. Wait, when you are in state 1, you can still transition to state 2 or stay in 1, but there's also a probability to be absorbed (exit front). Similarly, state 10 can transition to 9 or stay, or be absorbed (exit back). So actually, states 1 and 10 are not absorbing, but have a probability to be absorbed each time.This seems like a Markov chain with absorbing states where absorption can happen from states 1 and 10 with certain probabilities. However, since the problem is about exiting from front or back, once you are absorbed (exit), the process stops. Wait, but in reality, the process continues until you are either removed from front or back. So actually, every time you are in state 1, there's a chance to exit from front, and every time you are in state 10, a chance to exit from back. But the process continues until you exit. So this is a classic absorbing Markov chain where the absorbing states are "exited front" and "exited back", and the transient states are 1 through 10, with transitions as above, but states 1 and 10 have probabilities to transition to the absorbing states.Wait, no. Actually, states 1 and 10 are still transient because you can stay in them or move to other states, but they have a chance to be absorbed. So perhaps we need to model this as having two absorbing states (front exit and back exit), and the transient states are 1 through 10, with transitions from 1 possibly leading to absorption (front exit) or staying in 1 or moving to 2, and similarly for state 10.Therefore, let me formalize this:States:- Transient: 1, 2, ..., 10- Absorbing: F (front exit), B (back exit)From each transient state k (1 to 10), transitions are as follows:For k =1:- To F: probability 10/11 * 1/2 =5/11- To 1: probability (1/11 *1/2) + (10/11 *1/2)=11/22=1/2- To 2:1/11 *1/2=1/22For k =10:- To B: probability 10/11 *1/2=5/11- To 10: probability (1/11 *1/2)+(10/11 *1/2)=1/2- To 9:1/11 *1/2=1/22For 2 ≤k ≤9:- To k-1: (11 -k)/22- To k:1/2- To k+1:k/22Now, the question is starting from state 8, what is the probability of being absorbed at F rather than B.This is a standard absorption probability problem in Markov chains. Let me denote by p_k the probability of being absorbed at F starting from state k. We need to find p_8.We know that p_F =1 and p_B=0. For the transient states 1 to10, we can set up equations based on the transitions.For k=1:p_1 = 5/11 *1 + 1/2 * p_1 +1/22 * p_2Similarly, for k=10:p_{10} =5/11 *0 +1/2 * p_{10} +1/22 * p_9For 2 ≤k ≤9:p_k = (11 -k)/22 * p_{k-1} +1/2 * p_k +k/22 * p_{k+1}Let me write these equations more clearly.For k=1:p_1 = 5/11 + (1/2) p_1 + (1/22) p_2Multiply both sides by 22 to eliminate denominators:22 p_1 = 10 +11 p_1 + p_2=> 22 p_1 -11 p_1 -p_2 =10=>11 p_1 -p_2 =10 -- (1)For k=10:p_{10} =0 + (1/2) p_{10} + (1/22) p_9Multiply by 22:22 p_{10} =11 p_{10} + p_9=>11 p_{10} -p_9 =0 -- (2)For 2 ≤k ≤9:p_k = [(11 -k)/22] p_{k-1} + (1/2) p_k + (k/22) p_{k+1}Multiply both sides by 22:22 p_k = (11 -k) p_{k-1} +11 p_k +k p_{k+1}Bring terms to left side:22 p_k -11 p_k - (11 -k) p_{k-1} -k p_{k+1}=0=>11 p_k - (11 -k) p_{k-1} -k p_{k+1}=0Rearranged:- (11 -k) p_{k-1} +11 p_k -k p_{k+1} =0 -- (3)So for each k from 2 to9, equation (3) applies.Now, we have equations (1), (2), and equations (3) for k=2 to9. That's a total of 10 equations (1 for k=1, 8 for k=2 to9, and1 for k=10). The unknowns are p_1, p_2, ..., p_{10}.We need to solve this system of linear equations.This seems like a linear system that can be solved with some pattern or recursive relation.First, note that equation (3) can be written as:(11 -k) p_{k-1} +k p_{k+1} =11 p_kThis resembles a recurrence relation. Let's see if we can find a pattern or a telescoping product.Let me consider the homogeneous recurrence relation:(11 -k) p_{k-1} -11 p_k +k p_{k+1}=0This is a second-order linear recurrence. Let's see if we can find a solution.Suppose the solution is linear: p_k = a k + b. Let's test.Plug into the recurrence:(11 -k)(a(k-1) +b) -11(a k +b) +k(a(k+1)+b)=0Compute each term:First term: (11 -k)(a(k -1) +b) = (11 -k)(a k -a +b)Second term: -11(a k +b)Third term: k(a(k+1) +b)=k(a k +a +b)Expanding all terms:First term:11(a k -a +b) -k(a k -a +b)=11a k -11a +11b -a k^2 +a k -b kSecond term:-11a k -11bThird term:a k^3 +a k +b k^2Wait, this seems messy. Maybe assuming a linear solution is not sufficient.Alternatively, perhaps the recurrence can be rewritten in terms of differences.Let me define d_k = p_{k+1} - p_k. Then perhaps we can express the recurrence in terms of d_k.But let's see. The recurrence is:(11 -k) p_{k-1} +k p_{k+1} =11 p_kLet me rearrange:k p_{k+1} =11 p_k - (11 -k) p_{k-1}Divide both sides by k:p_{k+1} = (11/k) p_k - ((11 -k)/k) p_{k-1}Not sure if helpful. Alternatively, let's compute the difference between p_{k+1} and p_k:But perhaps a better approach is to consider the system as a tridiagonal matrix and solve it step by step.Alternatively, note that the equations for 2 ≤k ≤9 have coefficients that depend on k, so maybe we can find a telescoping product.Let me consider the recurrence:(11 -k) p_{k-1} -11 p_k +k p_{k+1}=0Let me rearrange:k p_{k+1} -11 p_k + (11 -k) p_{k-1}=0Let me divide both sides by k:p_{k+1} - (11/k) p_k + (11 -k)/k p_{k-1}=0Hmm, not obviously telescoping. Alternatively, let's consider writing the recurrence as:k p_{k+1} -11 p_k = -(11 -k) p_{k-1}Let me denote this as:k p_{k+1} -11 p_k = -(11 -k) p_{k-1} for k=2,…,9Let me compute for k=2:2 p_3 -11 p_2 = -(11 -2) p_1 => 2 p_3 -11 p_2 = -9 p_1Similarly, for k=3:3 p_4 -11 p_3 = -8 p_2k=4:4 p5 -11 p4 = -7 p3...k=9:9 p_{10} -11 p9 = -2 p8But from equation (2): 11 p_{10} - p9 =0 => p9=11 p10So perhaps substituting p9=11 p10 into the equation for k=9:9 p10 -11*(11 p10) = -2 p8=>9 p10 -121 p10 = -2 p8=>-112 p10 = -2 p8 => 56 p10 = p8Thus, p8=56 p10Similarly, maybe we can express all variables in terms of p10 and substitute back.Starting from equation (2):p9=11 p10From k=9 equation:9 p10 -11 p9 + (11 -9) p8=0 → Wait, original recurrence was:(11 -k) p_{k-1} -11 p_k +k p_{k+1}=0For k=9:(11 -9)p8 -11 p9 +9 p10=0=>2 p8 -11 p9 +9 p10=0But from p9=11 p10:2 p8 -11*(11 p10) +9 p10=0=>2 p8 -121 p10 +9 p10=0=>2 p8 -112 p10=0=>p8=56 p10So that's consistent with the previous result.Similarly, for k=8:(11 -8) p7 -11 p8 +8 p9=0=>3 p7 -11 p8 +8 p9=0But p9=11 p10 and p8=56 p10:3 p7 -11*(56 p10) +8*(11 p10)=03 p7 -616 p10 +88 p10=03 p7 -528 p10=0=>p7=176 p10Similarly, for k=7:(11 -7)p6 -11 p7 +7 p8=0=>4 p6 -11 p7 +7 p8=0Substitute p7=176 p10 and p8=56 p10:4 p6 -11*(176 p10) +7*(56 p10)=04 p6 -1936 p10 +392 p10=04 p6 -1544 p10=0=>p6=386 p10Wait, 1544 divided by 4 is 386. Yes.For k=6:(11 -6)p5 -11 p6 +6 p7=0=>5 p5 -11 p6 +6 p7=0Substituting p6=386 p10 and p7=176 p10:5 p5 -11*(386 p10) +6*(176 p10)=05 p5 -4246 p10 +1056 p10=05 p5 -3190 p10=0=>p5=638 p10For k=5:(11 -5)p4 -11 p5 +5 p6=0=>6 p4 -11 p5 +5 p6=0Substituting p5=638 p10 and p6=386 p10:6 p4 -11*(638 p10) +5*(386 p10)=06 p4 -7018 p10 +1930 p10=06 p4 -5088 p10=0=>p4=848 p10For k=4:(11 -4)p3 -11 p4 +4 p5=0=>7 p3 -11 p4 +4 p5=0Substituting p4=848 p10 and p5=638 p10:7 p3 -11*(848 p10) +4*(638 p10)=07 p3 -9328 p10 +2552 p10=07 p3 -6776 p10=0=>p3=968 p10For k=3:(11 -3)p2 -11 p3 +3 p4=0=>8 p2 -11 p3 +3 p4=0Substituting p3=968 p10 and p4=848 p10:8 p2 -11*(968 p10) +3*(848 p10)=08 p2 -10648 p10 +2544 p10=08 p2 -8104 p10=0=>p2=1013 p10Wait, 8104 divided by8 is1013. Yes.For k=2:(11 -2)p1 -11 p2 +2 p3=0=>9 p1 -11 p2 +2 p3=0Substituting p2=1013 p10 and p3=968 p10:9 p1 -11*(1013 p10) +2*(968 p10)=09 p1 -11143 p10 +1936 p10=09 p1 -9207 p10=0=>p1=1023 p10But wait, from equation (1):11 p1 -p2=10Substituting p1=1023 p10 and p2=1013 p10:11*(1023 p10) -1013 p10=10(11253 p10) -1013 p10=1010240 p10=10=>p10=10/10240=1/1024Therefore, p10=1/1024Then, p8=56 p10=56/1024=7/128But wait, 56 divided by1024: 56 simplifies with 1024. GCD(56,1024)=8, so 56/8=7, 1024/8=128. So 7/128.But wait, hold on. Let me check the steps again to ensure there are no arithmetic errors.Starting from k=9:Equation (2): 11 p10 - p9=0 => p9=11 p10From k=9 equation:2 p8 -112 p10=0 => p8=56 p10Then k=8:3 p7 -528 p10=0 => p7=176 p10k=7:4 p6 -1544 p10=0 => p6=386 p10 (1544/4=386)k=6:5 p5 -3190 p10=0 => p5=638 p10 (3190/5=638)k=5:6 p4 -5088 p10=0 => p4=848 p10 (5088/6=848)k=4:7 p3 -6776 p10=0 => p3=968 p10 (6776/7=968)k=3:8 p2 -8104 p10=0 => p2=1013 p10 (8104/8=1013)k=2:9 p1 -9207 p10=0 => p1=1023 p10 (9207/9=1023)Equation (1):11 p1 -p2=1011*(1023 p10) -1013 p10=1011253 p10 -1013 p10=1010240 p10=10 => p10=10/10240=1/1024Therefore, p10=1/1024Then p8=56 p10=56/1024=7/128Simplifying 7/128: 7 and128 are coprime (since 128 is 2^7), so m=7, n=128. Therefore, 100m +n=700 +128=828.Wait, but the problem states "the probability that you exit the row from the front rather than the back is m/n...". So according to this, the probability is7/128, so the answer is 828. But I need to verify.Wait, but let's check the steps again to make sure there are no arithmetic errors.Starting from k=9, equation gives p8=56 p10Then k=8:3 p7 -11 p8 +8 p9=0But p8=56 p10, p9=11 p10So 3 p7 -11*(56 p10) +8*(11 p10)=03 p7 -616 p10 +88 p10=03 p7 -528 p10=0 => p7=176 p10 (528/3=176). Correct.k=7:4 p6 -11 p7 +7 p8=04 p6 -11*(176 p10) +7*(56 p10)=04 p6 -1936 p10 +392 p10=04 p6 -1544 p10=0 => p6=386 p10 (1544/4=386). Correct.k=6:5 p5 -11 p6 +6 p7=05 p5 -11*(386 p10) +6*(176 p10)=05 p5 -4246 p10 +1056 p10=05 p5 -3190 p10=0 => p5=638 p10. Correct.k=5:6 p4 -11 p5 +5 p6=06 p4 -11*(638 p10) +5*(386 p10)=06 p4 -7018 p10 +1930 p10=06 p4 -5088 p10=0 => p4=848 p10. Correct.k=4:7 p3 -11 p4 +4 p5=07 p3 -11*(848 p10) +4*(638 p10)=07 p3 -9328 p10 +2552 p10=07 p3 -6776 p10=0 => p3=968 p10. Correct.k=3:8 p2 -11 p3 +3 p4=08 p2 -11*(968 p10) +3*(848 p10)=08 p2 -10648 p10 +2544 p10=08 p2 -8104 p10=0 => p2=1013 p10. Correct.k=2:9 p1 -11 p2 +2 p3=09 p1 -11*(1013 p10) +2*(968 p10)=09 p1 -11143 p10 +1936 p10=09 p1 -9207 p10=0 => p1=1023 p10. Correct.Equation (1):11 p1 -p2=1011*(1023 p10) -1013 p10=10(11*1023 -1013)p10=10Compute 11*1023:11*1000=1100011*23=253So total=11000+253=1125311253 -1013=10240Thus, 10240 p10=10 => p10=10/10240=1/1024. Correct.Therefore, p8=56 p10=56/1024=7/128.Thus, the probability is7/128. Since 7 and128 are coprime, m=7, n=128, so 100m +n=700 +128=828.But wait, let me confirm once more. Is this answer reasonable?Starting at position8, which is near the back. Since insertions are uniform, and removals are equally likely front or back, intuitively, there's a higher chance of exiting from the back. But the probability is7/128≈0.0547, which is about 5.47%. That seems low. Wait, but maybe due to the dynamics of the insertions and removals.Alternatively, maybe I made a mistake in interpreting the equations. Let me check.The equations were set up with p_k being the probability of being absorbed at F starting from state k. The process continues until you are absorbed. The equations correctly model the expected absorption probabilities. Given the linear system was solved step by step from k=10 backward to k=1, substituting each variable in terms of p10, leading to p10=1/1024 and p8=7/128≈0.0547. Given the symmetry, perhaps this is correct. Alternatively, maybe there's a different approach.Wait, another way to think about this problem is to recognize that the probability can be modeled similar to a gambler's ruin problem, but with variable transition probabilities. However, the transition probabilities here are not symmetric and depend on the current state, making it more complex.Alternatively, suppose we denote that the probability p_k satisfies the recurrence relation:(11 -k) p_{k-1} -11 p_k +k p_{k+1}=0 for 2 ≤k ≤9With boundary conditions:At k=1:11 p1 -p2=10At k=10:11 p10 -p9=0Assuming a solution of the form p_k = C - D/k for constants C and D. Let's test.Let’s assume p_k = C - D/k.Plug into the recurrence:(11 -k) p_{k-1} -11 p_k +k p_{k+1}=0Compute each term:p_{k-1}=C - D/(k-1)p_k=C - D/kp_{k+1}=C - D/(k+1)Substituting into the equation:(11 -k)(C - D/(k-1)) -11(C - D/k) +k(C - D/(k+1))=0Expand each term:First term:(11 -k)C - (11 -k)D/(k-1)Second term:-11C +11D/kThird term:k C -k D/(k+1)Combine all terms:[ (11 -k)C -11C +k C ] + [ - (11 -k)D/(k-1) +11D/k -k D/(k+1) ] =0Simplify the coefficients of C:(11 -k -11 +k)C=0So the C terms cancel out.Now, the D terms:- (11 -k)D/(k -1) +11 D/k -k D/(k +1)=0Factor out D:D [ - (11 -k)/(k -1) +11/k -k/(k +1) ]=0For non-trivial solution (D≠0):- (11 -k)/(k -1) +11/k -k/(k +1)=0Let me compute this expression for a general k:Let’s compute each term:First term: - (11 -k)/(k -1)Second term:11/kThird term: -k/(k +1)So:- (11 -k)/(k -1) +11/k -k/(k +1)=0Multiply through by k(k -1)(k +1) to eliminate denominators:- (11 -k)k(k +1) +11(k -1)(k +1) -k^2(k -1)=0This is messy, but let's compute each part:First term:- (11 -k)k(k +1) = -k(k +1)(11 -k)Second term:11(k -1)(k +1) =11(k^2 -1)Third term:- k^2(k -1)Expand each term:First term:- k(k +1)(11 -k) = -k(11 -k)(k +1) = -k(11k +11 -k^2 -k) = -k(10k +11 -k^2) = -10k^2 -11k +k^3Second term:11(k^2 -1)=11k^2 -11Third term:- k^2(k -1)= -k^3 +k^2Combine all terms:(-10k^2 -11k +k^3) + (11k^2 -11) + (-k^3 +k^2)=0Simplify:k^3 -10k^2 -11k +11k^2 -11 -k^3 +k^2=0Combine like terms:k^3 -k^3=0-10k^2 +11k^2 +k^2=2k^2-11k-11Thus:2k^2 -11k -11=0Which is a quadratic equation, but this should hold for all k from 2 to9, which is not possible. Therefore, the assumption that p_k =C -D/k is incorrect.Therefore, the solution is not of this form. Hence, the previous step-by-step solution is likely correct, leading to p8=7/128.Alternatively, let's check with a small example. Suppose we start at position1. According to our solution, p1=1023 p10=1023*(1/1024)=1023/1024≈0.999. Which makes sense, since if you start at front, you have a high chance to exit front. Similarly, starting at position10, p10=1/1024≈0.00097, which is very low, also makes sense.Alternatively, starting at position2: p2=1013/1024≈0.989, which is still high. Similarly, position9: p9=11/1024≈0.0107.This seems reasonable: positions closer to the front have higher probabilities of exiting front, closer to back have lower. Position8, which is near the back, has p8=7/128≈0.0547, which is about5.5%, which seems plausible.Therefore, the answer should be7/128, so m=7, n=128, which are coprime. Thus,100m +n=700 +128=828.Final Answerboxed{828}</think>To determine the probability that the eighth student exits from the front rather than the back, we model the process as a Markov chain with states representing the student's position. The key steps are as follows:1. Transition Probabilities: - For positions (2 leq k leq 9): - Transition to (k-1) with probability (frac{11 - k}{22}) - Remain at (k) with probability (frac{1}{2}) - Transition to (k+1) with probability (frac{k}{22}) - For position (1): - Exit from the front with probability (frac{5}{11}) - Remain at (1) with probability (frac{1}{2}) - Transition to (2) with probability (frac{1}{22}) - For position (10): - Exit from the back with probability (frac{5}{11}) - Remain at (10) with probability (frac{1}{2}) - Transition to (9) with probability (frac{1}{22})2. Setting Up Equations: - For (k = 1): (11 p_1 - p_2 = 10) - For (k = 10): (11 p_{10} - p_9 = 0) - For (2 leq k leq 9): ((11 - k) p_{k-1} - 11 p_k + k p_{k+1} = 0)3. Solving the System of Equations: - By substituting and solving step-by-step, we find: - (p_{10} = frac{1}{1024}) - (p_8 = 56 p_{10} = frac{56}{1024} = frac{7}{128})4. Result: - The probability that the eighth student exits from the front is (frac{7}{128}). - The values (m = 7) and (n = 128) are coprime, so (100m + n = 100 times 7 + 128 = 828).[boxed{828}]]

🔑:1. Let ( p_i ) be the probability that you exit out of the front when starting in position ( i ). We want to find ( p_8 ).2. We can set up a recurrence relation for ( p_k ). The probability that a new student is inserted in position ( k ) is ( frac{11-k}{11} ), and the probability that a new student is inserted in position ( k+1 ) is ( frac{k}{11} ). After the insertion, the probability that the frontmost student is removed is ( frac{1}{2} ), and the probability that the backmost student is removed is ( frac{1}{2} ).3. Therefore, the recurrence relation for ( p_k ) is: [ p_k = frac{11-k}{11} left( frac{p_{k-1} + p_k}{2} right) + frac{k}{11} left( frac{p_k + p_{k+1}}{2} right) ]4. Simplifying the recurrence relation, we get: [ p_k = frac{11-k}{22} p_{k-1} + frac{11-k}{22} p_k + frac{k}{22} p_k + frac{k}{22} p_{k+1} ] [ p_k = frac{11-k}{22} p_{k-1} + frac{k}{22} p_{k+1} + frac{11}{22} p_k ] [ p_k - frac{11}{22} p_k = frac{11-k}{22} p_{k-1} + frac{k}{22} p_{k+1} ] [ frac{11}{22} p_k - p_k = frac{11-k}{22} p_{k-1} + frac{k}{22} p_{k+1} ] [ -frac{11}{22} p_k = frac{11-k}{22} p_{k-1} + frac{k}{22} p_{k+1} ] [ -11 p_k = (11-k) p_{k-1} + k p_{k+1} ] [ p_k - p_{k-1} = frac{k}{11-k} (p_{k+1} - p_k) ]5. We have the boundary conditions ( p_0 = 1 ) and ( p_{11} = 0 ).6. To solve for ( p_8 ), we need to find the general form of ( p_k ). We can use the method of differences to solve the recurrence relation.7. Let ( d_k = p_k - p_{k-1} ). Then the recurrence relation becomes: [ d_k = frac{k}{11-k} d_{k+1} ]8. We can solve this by iterating from ( k = 0 ) to ( k = 10 ): [ d_0 = p_0 - p_{-1} = 1 - p_{-1} ] [ d_1 = frac{1}{10} d_2 ] [ d_2 = frac{2}{9} d_3 ] [ vdots ] [ d_{10} = frac{10}{1} d_{11} ]9. Since ( p_{11} = 0 ), we have ( d_{11} = -p_{10} ).10. We can now find ( p_8 ) by summing the differences: [ p_8 = p_0 + sum_{i=1}^{8} d_i ]11. Using the boundary conditions and the recurrence relation, we find that: [ p_8 = frac{7}{128} ]The final answer is ( boxed{ 100m + n = 100 times 7 + 128 = 828 } )

🧠:[<think>Okay, let's see. I need to solve three problems related to sets. The universal set U is defined as all positive integers less than 10. Then there are sets B and C given, and I have to do a few operations with them. Let me take each part step by step.Starting with part (1): Represent the universal set U using the rooster method. Wait, "rooster method"? Hmm, maybe that's a typo. Oh, probably "roster method"! Yeah, that makes sense. The roster method is where you list all the elements of the set. So U is all positive integers less than 10. Let me write that out. Positive integers start from 1, right? So numbers less than 10 would be 1 up to 9. Therefore, U should be {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let me check that again. Positive integers less than 10: yes, 1 through 9. So that's part (1) done.Moving on to part (2): If D is the intersection of B and C, list all the subsets of set D. First, I need to find D = B ∩ C. Let me recall that B is {1,2,3,4} and C is {3,4,5,6}. The intersection of two sets consists of elements that are in both sets. So, looking at B and C, the common elements are 3 and 4. Therefore, D = {3,4}. Now, I need to list all subsets of D. The set D has two elements, so the number of subsets should be 2^2 = 4. The subsets are: the empty set, {3}, {4}, and {3,4}. Let me make sure I didn't miss any. For a set with n elements, the subsets are all combinations of its elements. So yes, starting from the empty set, then each element individually, and then the set itself. So that's four subsets. I think that's correct.Now part (3): Find the complement of B ∩ C with respect to U. Wait, B ∩ C is again D, which is {3,4}. So the complement of D with respect to U would be all elements in U that are not in D. So U is {1,2,3,4,5,6,7,8,9}. If we remove 3 and 4, the remaining elements are {1,2,5,6,7,8,9}. Let me verify that. The complement of D in U is U D, which should indeed exclude 3 and 4. So that's the answer for part (3).Wait, let me go through each part again to ensure I didn't make any mistakes.Part (1): U is positive integers less than 10. So starting at 1, since they are positive integers, and less than 10 means up to 9. So yes, {1,2,3,4,5,6,7,8,9}. Correct.Part (2): B is {1,2,3,4}, C is {3,4,5,6}. Intersection is {3,4}. Subsets: empty set, {3}, {4}, {3,4}. That's four subsets. Seems right. I don't think there's a subset with both elements and something else here. Since the set only has two elements, those four are all.Part (3): Complement of D (which is {3,4}) in U. So everything in U not in D. So removing 3 and 4 from U gives {1,2,5,6,7,8,9}. Yes, that looks correct.I think that's all. I don't see any errors here. Let me just cross-verify part (3) one more time. The complement of B ∩ C. If I first compute B ∩ C = {3,4}, then the complement is U minus {3,4}, which is indeed {1,2,5,6,7,8,9}. Yup, that's right.Final Answer(1) The universal set U is boxed{{1, 2, 3, 4, 5, 6, 7, 8, 9}}.(2) The subsets of set D are boxed{emptyset}, boxed{{3}}, boxed{{4}}, and boxed{{3, 4}}.(3) The complement of ( B cap C ) with respect to U is boxed{{1, 2, 5, 6, 7, 8, 9}}.</think>The universal set ( U ) is defined as all positive integers less than 10. Using the roster method, this set is represented as:[U = {1, 2, 3, 4, 5, 6, 7, 8, 9}]Next, we need to find the set ( D ), which is the intersection of sets ( B ) and ( C ). The sets ( B ) and ( C ) are given as ( {1, 2, 3, 4} ) and ( {3, 4, 5, 6} ) respectively. The intersection ( B cap C ) results in the set ( {3, 4} ). The subsets of ( D ) are:[emptyset, {3}, {4}, {3, 4}]Finally, we need to find the complement of ( B cap C ) (which is ( D = {3, 4} )) with respect to ( U ). The complement of ( D ) in ( U ) includes all elements in ( U ) that are not in ( D ). Removing 3 and 4 from ( U ) gives:[{1, 2, 5, 6, 7, 8, 9}] Final Answer(1) The universal set ( U ) is boxed{{1, 2, 3, 4, 5, 6, 7, 8, 9}}.(2) The subsets of set ( D ) are boxed{emptyset}, boxed{{3}}, boxed{{4}}, and boxed{{3, 4}}.(3) The complement of ( B cap C ) with respect to ( U ) is boxed{{1, 2, 5, 6, 7, 8, 9}}.,<think>Okay, let's tackle these problems step by step. First, let me make sure I understand what each part is asking for.Starting with question (1): "Represent the universal set U using the rooster method." Wait, "rooster method"? Hmm, that must be a typo. Probably "roster method." Yeah, that makes sense. The roster method is where you list all the elements of a set between curly braces. So, the universal set U is defined as {x | x is a positive integer less than 10}. Got it. So, positive integers less than 10 start from 1 up to 9, right? Because 10 is not less than 10. Let me double-check: positive integers are 1, 2, 3, ..., 9. So U should be {1, 2, 3, 4, 5, 6, 7, 8, 9}. That seems straightforward.Moving on to question (2): "If D = B ∩ C, list all the subsets of set D." Okay, first, I need to find the intersection of sets B and C. Set B is {1, 2, 3, 4}, and set C is {3, 4, 5, 6}. The intersection of two sets includes all elements that are in both sets. So, looking at B and C, the common elements are 3 and 4. Therefore, D = {3, 4}. Now, I need to list all the subsets of D. Subsets of a set include all possible combinations of its elements, including the empty set and the set itself. Since D has two elements, the number of subsets should be 2² = 4. Let me list them out:1. The empty set: ∅2. {3}3. {4}4. {3, 4}Yep, that's all of them. So those are the subsets of D.Question (3): "Find the complement of B ∩ C with respect to U." Okay, so first, we already know that B ∩ C is D, which is {3, 4}. The complement of a set with respect to the universal set U is all the elements in U that are not in the set. So, the complement of D would be U D. Given that U is {1, 2, 3, 4, 5, 6, 7, 8, 9} and D is {3, 4}, removing 3 and 4 from U leaves us with {1, 2, 5, 6, 7, 8, 9}. Let me check that again. Yes, those are all the elements in U that are not in D. So that's the complement.Wait a second, let me make sure I didn't miss anything here. The complement of B ∩ C is the same as the complement of D, which we already found. Just to verify, another way to think about it is using De Morgan's laws. The complement of B ∩ C is equal to the union of the complements of B and C. But I don't think we need to go that route here because directly computing the complement of D with respect to U is straightforward. So, yes, {1, 2, 5, 6, 7, 8, 9} should be the correct answer.Let me recap everything to ensure no mistakes were made:1. Universal set U: Positive integers less than 10. So numbers from 1 to 9. Roster method is just listing them: {1, 2, 3, 4, 5, 6, 7, 8, 9}. Correct.2. D is B intersect C. B is {1,2,3,4}, C is {3,4,5,6}. Intersection is {3,4}. Subsets of D are all combinations: empty set, {3}, {4}, {3,4}. That's four subsets, which is 2². Makes sense.3. Complement of D in U: U is 1-9, D is {3,4}. So removing 3 and 4, the complement is {1,2,5,6,7,8,9}. Correct.I think that covers all parts. No steps skipped and all logical. Hmm, seems solid.Final Answer(1) The universal set U is boxed{{1, 2, 3, 4, 5, 6, 7, 8, 9}}.(2) The subsets of set D are boxed{emptyset}, boxed{{3}}, boxed{{4}}, and boxed{{3, 4}}.(3) The complement of ( B cap C ) with respect to U is boxed{{1, 2, 5, 6, 7, 8, 9}}.</think>(1) The universal set ( U ) is given as the set of all positive integers less than 10. Using the roster method, this is represented as:[U = {1, 2, 3, 4, 5, 6, 7, 8, 9}]Thus, the universal set ( U ) is (boxed{{1, 2, 3, 4, 5, 6, 7, 8, 9}}).(2) To find ( D = B cap C ), we determine the intersection of sets ( B ) and ( C ). The set ( B ) is ({1, 2, 3, 4}) and the set ( C ) is ({3, 4, 5, 6}). The intersection ( B cap C ) is ({3, 4}). Therefore, ( D = {3, 4} ). The subsets of ( D ) are:[emptyset, {3}, {4}, {3, 4}]Thus, the subsets of set ( D ) are (boxed{emptyset}), (boxed{{3}}), (boxed{{4}}), and (boxed{{3, 4}}).(3) The complement of ( B cap C ) (which is ( D = {3, 4} )) with respect to ( U ) is the set of all elements in ( U ) that are not in ( D ). The universal set ( U ) is ({1, 2, 3, 4, 5, 6, 7, 8, 9}). Removing the elements 3 and 4 from ( U ), we get:[{1, 2, 5, 6, 7, 8, 9}]Thus, the complement of ( B cap C ) with respect to ( U ) is (boxed{{1, 2, 5, 6, 7, 8, 9}}).]

🔑:(1) The universal set U is defined as all positive integers less than 10, which can be listed as: U = {1,2,3,4,5,6,7,8,9} (2) The sets B and C are given as B = {1,2,3,4} and C = {3,4,5,6}. The intersection of B and C, denoted as D = B cap C, includes all elements that are common to both sets B and C. Therefore, we have: D = B cap C = {3,4} Now, to find all the subsets of set D, we list all possible combinations of the elements within D, including the empty set and D itself:- The empty set: emptyset- Subsets with one element: {3}, {4}- Subsets with two elements: {3,4}Thus, all subsets of set D are: boxed{emptyset, {3}, {4}, {3,4}} (3) The complement of B cap C with respect to U, denoted as complement_U (B cap C), includes all elements that are in U but not in B cap C. Since B cap C = {3,4}, the complement will be the set of all elements in U excluding 3 and 4: complement_U (B cap C) = {1,2,5,6,7,8,9} Therefore, the complement of B cap C with respect to U is: boxed{{1,2,5,6,7,8,9}}

🧠:[<think>Okay, so there's this problem where a cherry, which is basically a ball with radius R, is placed in a glass. The glass's axial cross-section is given by the function y = x⁴. We need to figure out for which values of R the ball will touch the bottom of the glass at the point (0,0). Hmm, interesting. Let me try to visualize this first.The glass is shaped like the graph of y = x⁴. So, if I imagine looking at a cross-section through the axis of the glass, the sides of the glass curve up following y = x⁴. A cherry, which is a sphere, is placed inside this glass. We need to determine the radius R such that the sphere just touches the bottom point (0,0). First, I need to recall that a sphere in 3D space can be represented in 2D cross-section as a circle. So, the axial cross-section of the sphere would be a circle with radius R. The center of this circle must be somewhere along the axis of the glass, which in this case is the y-axis. So, the center of the sphere (in the cross-section) is at (0, k) for some k. Since the sphere has radius R, the bottommost point of the sphere is at (0, k - R). For the sphere to touch the bottom of the glass (0,0), this point must be (0,0). Therefore, k - R = 0, so k = R. So, the center of the sphere is at (0, R). Now, the sphere must also be tangent to the sides of the glass. If it's only touching the bottom, it might not be stable, so for it to be placed in the glass, it should touch the sides as well. Therefore, the sphere of radius R centered at (0, R) must be tangent to the curve y = x⁴. The points where the sphere touches the glass are points (x, x⁴) that lie on the sphere and the sphere is tangent to the glass at those points. So, the equation of the sphere (circle in cross-section) is (x - 0)² + (y - R)² = R². Simplifying, that's x² + (y - R)² = R². Expanding that, x² + y² - 2Ry + R² = R², so x² + y² - 2Ry = 0. But the glass's cross-section is y = x⁴, so substituting y = x⁴ into the circle equation, we get x² + (x⁴)² - 2R(x⁴) = 0. That simplifies to x² + x⁸ - 2R x⁴ = 0.So, the equation x⁸ + x² - 2R x⁴ = 0 must have a solution where the sphere is tangent to the glass. Since the sphere is tangent, there should be exactly one point of contact (other than the bottom at (0,0)), or maybe two symmetric points but with multiplicity. Wait, actually, since the glass is symmetric, there might be two points of contact symmetric about the y-axis. But for tangency, the system of equations (the circle and the curve y = x⁴) should have a double root. So, maybe the equation x⁸ + x² - 2R x⁴ = 0 has a double root at some x ≠ 0. Alternatively, perhaps the minimal R is such that the circle is tangent to the curve y = x⁴ at some point (a, a⁴), where a ≠ 0, and the derivative of the circle and the derivative of the curve at that point are equal. That makes sense. So, to find the value of R, we need to ensure that at the point of tangency (a, a⁴), both the circle and the curve y = x⁴ pass through that point, and their derivatives at that point are equal.So, let's formalize that. Let's denote the point of tangency as (a, a⁴). Then, this point lies on the circle:a² + (a⁴ - R)^2 = R².Also, the derivative of the circle at that point must equal the derivative of y = x⁴ at that point.First, let's compute the derivative of y = x⁴, which is dy/dx = 4x³.Now, to compute the derivative of the circle at (a, a⁴), we can use implicit differentiation. The equation of the circle is x² + (y - R)^2 = R². Differentiating both sides with respect to x:2x + 2(y - R)(dy/dx) = 0.Solving for dy/dx:dy/dx = -2x / [2(y - R)] = -x / (y - R).At the point (a, a⁴), this becomes:dy/dx = -a / (a⁴ - R).But this derivative must equal the derivative of the curve at that point, which is 4a³. Therefore:-a / (a⁴ - R) = 4a³.Assuming a ≠ 0 (since at a = 0, we already have the bottom point), we can divide both sides by a:-1 / (a⁴ - R) = 4a².Rearranging:-1 = 4a²(a⁴ - R)=> -1 = 4a²a⁴ - 4a²R=> -1 = 4a⁶ - 4a²RLet me write that as:4a⁶ - 4a²R + 1 = 0.So, that's one equation. The other equation comes from the point (a, a⁴) lying on the circle:a² + (a⁴ - R)^2 = R².Let me expand the left side:a² + a⁸ - 2R a⁴ + R² = R²Simplify:a² + a⁸ - 2R a⁴ = 0Which is the same equation we had earlier. So, we have two equations:1) 4a⁶ - 4a²R + 1 = 02) a⁸ + a² - 2R a⁴ = 0We need to solve these two equations for a and R. Since we have two equations and two variables, this should be possible.Let me denote equation 1 as:4a⁶ - 4a²R = -1Equation 2 as:a⁸ + a² = 2R a⁴From equation 2, solve for R:R = (a⁸ + a²)/(2a⁴) = (a⁸)/(2a⁴) + (a²)/(2a⁴) = (a⁴)/2 + 1/(2a²)So, R = (a⁴ + 1/a²)/2Now, substitute this into equation 1:4a⁶ - 4a² * [ (a⁴ + 1/a²)/2 ] = -1Simplify the left side:4a⁶ - 4a² * (a⁴ + 1/a²)/2= 4a⁶ - [4a² * (a⁴) / 2 + 4a² * (1/a²) / 2]= 4a⁶ - [2a⁶ + 2]= 4a⁶ - 2a⁶ - 2= 2a⁶ - 2Set equal to -1:2a⁶ - 2 = -1Add 2 to both sides:2a⁶ = 1Divide by 2:a⁶ = 1/2Therefore, a = (1/2)^(1/6)Let me compute that. 2^(1/6) is the sixth root of 2. Alternatively, (1/2)^(1/6) = 2^(-1/6). So, a = 2^(-1/6). Since the glass is symmetric, a can be positive or negative, but since we're dealing with radius, we can take a positive.So, a = 2^(-1/6). Let me note that 2^(1/6) is approximately 1.122, so 2^(-1/6) is approximately 0.8909.Now, having found a, we can substitute back into the expression for R:R = (a⁴ + 1/a²)/2First, compute a² and a⁴.Since a = 2^(-1/6), then a² = 2^(-1/3) and a⁴ = 2^(-2/3).Therefore, R = [2^(-2/3) + 1/(2^(-1/3))]/2Simplify 1/(2^(-1/3)) = 2^(1/3)So, R = [2^(-2/3) + 2^(1/3)] / 2Factor out 2^(-2/3):= 2^(-2/3) [1 + 2^(1/3 + 2/3)] / 2Wait, maybe it's simpler to write exponents as fractions:2^(-2/3) + 2^(1/3) = 2^(1/3) + 2^(-2/3)Let me write both terms with exponent 1/3:= 2^(1/3) + 2^(-2/3) = 2^(1/3) + (2^(1/3))^(-2)Wait, maybe not. Let me compute 2^(1/3) and 2^(-2/3):2^(1/3) is cube root of 2, approximately 1.26.2^(-2/3) is 1/(2^(2/3)) which is approximately 1/(1.5874) ≈ 0.63.So, 2^(1/3) + 2^(-2/3) ≈ 1.26 + 0.63 ≈ 1.89But let's do it algebraically:Let me denote t = 2^(1/3). Then, 2^(-2/3) = t^(-2).So, R = (t^(-2) + t)/2.But t³ = 2. So, let's see:t³ = 2 => t = 2^(1/3)Express R in terms of t:R = (t + t^(-2))/2Multiply numerator and denominator by t²:R = (t³ + 1) / (2t²)But t³ = 2, so substitute:R = (2 + 1)/(2t²) = 3/(2t²)But t² = (2^(1/3))² = 2^(2/3), so:R = 3/(2 * 2^(2/3)) = 3/(2^(5/3)) = 3/(2^(2/3) * 2^(1)) = 3/(2 * 2^(2/3)) Alternatively, write 2^(5/3) = 2^(1 + 2/3) = 2 * 2^(2/3)So, R = 3/(2^(5/3)) = 3/(2 * 2^(2/3)) = (3/2) * 2^(-2/3)But 2^(-2/3) is 1/(2^(2/3)), so R is (3/2) * 1/(2^(2/3)) = 3/(2^(5/3))Alternatively, express 2^(5/3) as 2^(1 + 2/3) = 2 * 2^(2/3), so R = 3/(2 * 2^(2/3)).But maybe we can rationalize this or write it in a different form.Wait, 2^(5/3) is the same as the cube root of 2^5, which is cube root of 32. Since 2^5 = 32. So, 2^(5/3) = ∛32. Therefore, R = 3/∛32. Since 32 is 2^5, ∛32 = 2^(5/3). So, R = 3/(2^(5/3)) = 3 * 2^(-5/3).Alternatively, factor 2^(-2/3):R = 3/2 * 2^(-2/3) = 3/(2^(5/3)).But perhaps we can write it with exponents in fractional form:R = 3/(2^(5/3)) = (3/2) * 2^(-2/3) = 3 * 2^(-5/3). Either way, it's correct.But maybe the problem expects a simplified radical form. Let's see:2^(5/3) is 2^(1 + 2/3) = 2 * 2^(2/3). So, R = 3/(2 * 2^(2/3)) = 3/(2 * ∛4). Because 2^(2/3) is the cube root of 4. Therefore, R = 3/(2∛4). Multiply numerator and denominator by ∛16 (to rationalize the denominator), but maybe it's acceptable as is.Alternatively, we can write R = (3/2) * ∛(1/4). Since 1/4 is 2^(-2), so ∛(1/4) = 2^(-2/3). Hence, R = (3/2) * 2^(-2/3) = 3 * 2^(-5/3), which is the same as before.Alternatively, we can write this as 3/(2^(5/3)) but that might not be the simplest form. Let's check if the answer is expected in terms of exponents or radicals. Since the problem is mathematical, the answer is likely acceptable in terms of exponents, but perhaps rationalized.Wait, another approach: Let's write 2^(5/3) as 2^(1) * 2^(2/3) = 2 * ∛4. So, R = 3/(2 * ∛4). To rationalize the denominator, multiply numerator and denominator by ∛16 (since ∛4 * ∛16 = ∛64 = 4):R = [3 * ∛16] / [2 * ∛4 * ∛16] = [3 * ∛16] / [2 * ∛64] = [3 * ∛16] / [2 * 4] = [3 * ∛16]/8.But ∛16 = 2^(4/3) = 2 * 2^(1/3), so ∛16 = 2∛2. Therefore, R = [3 * 2∛2]/8 = (6∛2)/8 = (3∛2)/4.So, R = (3∛2)/4.Wait, let's check this:Starting from R = 3/(2^(5/3)).2^(5/3) = 2^(1 + 2/3) = 2 * 2^(2/3). Therefore, R = 3/(2 * 2^(2/3)) = 3/(2 * ∛4). Multiply numerator and denominator by ∛16:Numerator becomes 3 * ∛16 = 3 * ∛(16) = 3 * ∛(8 * 2) = 3 * 2∛2 = 6∛2.Denominator becomes 2 * ∛4 * ∛16 = 2 * ∛(64) = 2 * 4 = 8.Hence, R = 6∛2 / 8 = (3∛2)/4. So, R = (3∛2)/4.Therefore, the value of R is three times the cube root of 2 divided by 4. But let me verify this calculation step by step to ensure I didn't make a mistake.We had:From equation 1: 4a⁶ - 4a² R + 1 = 0.From equation 2: R = (a⁸ + a²)/(2a⁴) = (a⁴ + 1/a²)/2.Substituting R into equation 1 gives:4a⁶ - 4a²*(a⁴ + 1/a²)/2 +1 =0Simplify:4a⁶ - 2a²*(a⁴ + 1/a²) +1 =0Which is:4a⁶ - 2a⁶ - 2 +1 =0So, 4a⁶ - 2a⁶ = 2a⁶Then, -2 +1 = -1Hence, 2a⁶ -1 =0 => 2a⁶ =1 => a⁶=1/2. So, a=(1/2)^(1/6). Correct.Then, R=(a⁴ +1/a²)/2. Let's compute a⁴ and 1/a².Given a=2^(-1/6):a²=2^(-1/6 *2)=2^(-1/3)a⁴=2^(-1/6 *4)=2^(-2/3)1/a²=2^(1/3)Thus, R=(2^(-2/3) +2^(1/3))/2. Let me compute 2^(-2/3) +2^(1/3):Let t=2^(1/3). Then, 2^(-2/3)=t^(-2). So,R=(t^(-2) + t)/2=(1/t² + t)/2. Multiply numerator and denominator by t²:R=(1 + t³)/2t². But t³=2. Hence, R=(1+2)/2t²=3/(2t²). Since t=2^(1/3), t²=2^(2/3). Thus,R=3/(2*2^(2/3))=3/(2^(5/3)).Which is the same as 3*2^(-5/3). To convert this into a radical form:2^(5/3)=2^(1+2/3)=2*2^(2/3)=2*∛4. Hence, R=3/(2*∛4). Multiply numerator and denominator by ∛16:R=(3∛16)/(2*∛4*∛16)=3∛16/(2*∛64)=3∛16/(2*4)=3∛16/8.But ∛16=∛(8*2)=2∛2. Therefore, R=3*(2∛2)/8=6∛2/8=3∛2/4.Yes, so that simplifies to R= (3∛2)/4. So, this is the simplified radical form.Therefore, the radius R must be equal to three fourths times the cube root of 2. To confirm, let's check this answer.If R= (3∛2)/4, then the center of the sphere is at (0, R). The point of tangency is (a, a⁴), where a=2^(-1/6). Let's compute a:2^(-1/6)=1/(2^(1/6))≈0.8909.Compute a⁴=(2^(-1/6))⁴=2^(-4/6)=2^(-2/3)=1/(2^(2/3))≈0.63.Compute the distance from the center (0, R) to the point (a, a⁴):sqrt(a² + (a⁴ - R)^2). This should equal R.Compute a²=2^(-1/3)≈0.7937.Compute a⁴ - R=2^(-2/3) - (3∛2)/4≈0.63 - (3*1.26)/4≈0.63 - 0.945≈-0.315.Then, sqrt(a² + (a⁴ - R)^2)=sqrt(0.7937 + 0.0992)=sqrt(0.8929)≈0.945. But R=(3∛2)/4≈(3*1.26)/4≈0.945. So, the distance is indeed R. So, the point (a, a⁴) lies on the circle.Additionally, the derivative check: The derivative of the circle at (a, a⁴) is dy/dx= -a/(a⁴ - R)= -a/(a⁴ - R). With a=2^(-1/6), a⁴=2^(-2/3), R=3/(2^(5/3)).Compute a⁴ - R=2^(-2/3)-3/(2^(5/3))=2^(-2/3)-3*2^(-5/3)=2^(-5/3)*(2^( -2/3 +5/3) -3)=Wait, maybe better to factor out 2^(-5/3):2^(-5/3)*(2^(1) -3)=2^(-5/3)*(2 -3)= -2^(-5/3)Thus, dy/dx= -a/(a⁴ - R)= -2^(-1/6)/(-2^(-5/3))= (2^(-1/6))/(2^(-5/3))=2^(-1/6 +5/3)=2^( ( -1/6 +10/6 ))=2^(9/6)=2^(3/2)=sqrt(8)=2.828...Wait, but the derivative of the curve y=x⁴ at x=a is 4a³. Let's compute 4a³.a=2^(-1/6), so a³=2^(-1/2)=1/sqrt(2)=√2/2≈0.707. Then, 4a³=4*(√2/2)=2√2≈2.828, which matches the derivative from the circle. So, that's consistent.Therefore, the calculations check out.Thus, the value of R is 3 times the cube root of 2 divided by 4, or (3∛2)/4.So, the answer is R= boxed{dfrac{3sqrt[3]{2}}{4}}.Final AnswerThe ball will touch the bottom of the glass when its radius is boxed{dfrac{3sqrt[3]{2}}{4}}.</think>To determine the values of ( R ) for which a ball of radius ( R ) will touch the bottom of the glass (the point ((0,0))) when placed in a glass with an axial cross-section given by ( y = x^4 ):1. Equation of the Sphere: The sphere (circle in cross-section) is centered at ((0, R)) with radius ( R ). The equation of the circle is: [ x^2 + (y - R)^2 = R^2 ] Substituting ( y = x^4 ) into the circle's equation gives: [ x^2 + (x^4 - R)^2 = R^2 ] Simplifying this, we get: [ x^8 + x^2 - 2R x^4 = 0 ]2. Tangency Condition: The sphere must be tangent to the curve ( y = x^4 ) at some point ((a, a^4)). For tangency, the derivatives of the circle and the curve at this point must be equal.3. Derivatives: - The derivative of ( y = x^4 ) is ( frac{dy}{dx} = 4x^3 ). - Using implicit differentiation on the circle's equation ( x^2 + (y - R)^2 = R^2 ), the derivative is: [ frac{dy}{dx} = -frac{x}{y - R} ] - At the point ((a, a^4)), this derivative becomes: [ frac{dy}{dx} = -frac{a}{a^4 - R} ] - Equating the derivatives: [ -frac{a}{a^4 - R} = 4a^3 ] Solving this gives: [ -1 = 4a^2(a^4 - R) ] [ 4a^6 - 4a^2R + 1 = 0 ]4. Solving for ( R ): - From the tangency condition and the equation of the circle, we solve for ( R ): [ R = frac{a^8 + a^2}{2a^4} ] - Substituting ( a^6 = frac{1}{2} ) (from solving ( 4a^6 - 4a^2R + 1 = 0 )): [ R = frac{2^{-2/3} + 2^{1/3}}{2} ] Simplifying this, we find: [ R = frac{3 sqrt[3]{2}}{4} ]Thus, the ball will touch the bottom of the glass when its radius is (boxed{dfrac{3sqrt[3]{2}}{4}}).,<think>Alright, so I need to figure out for which values of R a ball (which is a sphere) of radius R placed in a glass whose axial cross-section is the graph of y = x⁴ will touch the bottom of the glass at (0,0). Hmm, okay. Let me try to visualize this first. The glass has a shape that's like a very steep curve at the bottom since y = x⁴ becomes flatter as x increases but near the origin, it's quite pointy. The cherry is a sphere, so in 2D cross-section, it's a circle. The problem is to determine the radii R where the sphere just touches the origin.First, let me recall that the axial section being the graph of y = x⁴ means that the glass's cross-section along its axis (which I assume is the y-axis) is given by rotating y = x⁴ around the y-axis. Wait, actually, no. Wait, axial section—maybe they mean that the cross-section through the axis (the y-axis) is the 2D curve y = x⁴. So the glass is rotationally symmetric around the y-axis, and any cross-section through the y-axis is the curve y = x⁴. Therefore, the glass is a surface of revolution formed by rotating y = x⁴ around the y-axis. So in 3D, the equation of the glass would be y = (sqrt(x² + z²))⁴ = (x² + z²)². But maybe I don't need the 3D equation. Since the problem is about a sphere touching the bottom, which is at (0,0,0) in 3D, but in the axial section (the 2D cross-section), it's at (0,0). So maybe I can model this as a 2D problem where the glass is the curve y = x⁴ and the "cherry" is a circle of radius R that must touch the point (0,0) and also be tangent to the curve y = x⁴ somewhere. Wait, but in reality, the sphere has to fit inside the glass, which is a 3D object, so maybe the circle in the axial cross-section corresponds to the sphere's cross-section. So in 2D, the sphere would be a circle that touches the origin and is tangent to the curve y = x⁴. Then, if I can find the radius R of such a circle, that would be the answer.Alternatively, maybe I need to consider the 3D case where the sphere touches the origin and is tangent to the glass surface at some point. However, since the glass is rotationally symmetric, the point of tangency should lie along the axis of symmetry, or maybe not? Wait, no. In 3D, the sphere could touch the glass at a circle of points, but since the glass is rotationally symmetric, the closest point to the sphere's center might lie along the axis. Wait, maybe not. If the sphere is centered along the y-axis (since the glass is symmetric and the sphere is placed in the glass), then the sphere's center must be on the y-axis. Therefore, in 3D, the sphere is centered at (0, c, 0) for some c, and has radius R. The bottom of the glass is at (0,0,0), so the distance from the center (0, c, 0) to the bottom is sqrt((0 - 0)^2 + (0 - c)^2 + (0 - 0)^2) = |c|. Since the sphere touches the bottom, this distance must be equal to the radius R, so |c| = R. Therefore, the center is at (0, R, 0) if the sphere is above the origin, but wait, actually, if the sphere is sitting in the glass and touching the bottom, the center should be above the origin, so the center is (0, R, 0). Wait, no. If the sphere is touching the origin, then the distance from the center to the origin is R, so the center is at (0, R, 0) because moving up along the y-axis. So in 3D, the sphere is centered at (0, R, 0) with radius R. Then, the sphere must also be tangent to the glass surface. The glass surface is given by y = (x² + z²)^2. So the sphere is the set of points (x, y, z) such that x² + (y - R)^2 + z² = R². Simplifying, x² + y² - 2Ry + z² = 0. The glass is y = (x² + z²)^2. So substituting y from the glass equation into the sphere equation: x² + [(x² + z²)^2]^2 - 2R(x² + z²)^2 + z² = 0. That's complicated. Maybe we can use symmetry. Since the sphere is centered on the y-axis and the glass is rotationally symmetric around the y-axis, the point of tangency should be along a circle where x² + z² = r² for some r. Let me set x² + z² = r², then the glass equation becomes y = r^4. The sphere equation at that point would be x² + (y - R)^2 + z² = R². Substituting x² + z² = r² and y = r^4: r² + (r^4 - R)^2 = R². Let's expand that: r² + r^8 - 2Rr^4 + R² = R². Simplify: r² + r^8 - 2Rr^4 = 0. So, r^8 + r² - 2Rr^4 = 0. Let me factor this equation. Let's factor out r²: r²(r^6 + 1 - 2Rr^2) = 0. So, solutions are r = 0, which would correspond to the origin, but we need another point of tangency, so the other factor is r^6 - 2Rr^2 + 1 = 0. Let me set t = r², then the equation becomes t^3 - 2Rt + 1 = 0. So, t^3 - 2Rt + 1 = 0. We need to find real positive solutions t for this equation. Since r is real and non-negative, t is non-negative. So, the cubic equation t^3 - 2Rt + 1 = 0 must have a positive real root. For the sphere to be tangent to the glass, this equation must have a double root, because the system has only one point of contact apart from the origin. Wait, but the sphere is already touching the origin, so maybe there should be exactly one other point where it touches the glass. Wait, but in 3D, if it's tangent to the glass, it should touch along a circle. But the condition is that the sphere is tangent to the glass at some point (other than the origin). So, perhaps the equation t^3 - 2Rt + 1 = 0 must have a double root (since tangency implies multiplicity at least two). Therefore, for tangency, the cubic equation must have a repeated root. So, for the cubic t^3 - 2Rt + 1 = 0 to have a double root, its discriminant must be zero. The discriminant D of a cubic equation t^3 + at^2 + bt + c = 0 is D = 18abc - 4a^3c + a^2b^2 - 4b^3 - 27c^2. In our case, the cubic is t^3 + 0t^2 - 2Rt + 1 = 0, so a = 0, b = -2R, c = 1. Plugging into the discriminant formula: D = 18*0*(-2R)*1 - 4*0^3*1 + 0^2*(-2R)^2 - 4*(-2R)^3 - 27*1^2 = 0 - 0 + 0 - 4*(-8R^3) - 27 = 32R^3 - 27. For the cubic to have a double root, the discriminant must be zero: 32R^3 - 27 = 0 => R^3 = 27/32 => R = (27/32)^(1/3). Let's compute that. 27 is 3^3, 32 is 2^5. So, (27/32)^(1/3) = 3/(2^(5/3)) = 3/(2^(1 + 2/3)) = 3/(2 * 2^(2/3)) = 3/(2 * cube root of 4)). Alternatively, rationalizing, cube root of 27 is 3, cube root of 32 is 2^(5/3), so R = 3/(2^(5/3)) = 3/(2 * 2^(2/3)) = 3/(2 * cube root(4)). Maybe we can write this as (3/2) * 2^(-2/3) = (3/2) * (2^2)^(-1/3) = (3/2) * 4^(-1/3). Alternatively, rational exponents: R = 3 * 2^(-5/3). But perhaps it's better to write it as R = (3/2) * (1/4)^(1/3) = (3/2) * 4^(-1/3). But 4 is 2², so 4^(1/3) is 2^(2/3). Therefore, R = 3/(2 * 2^(2/3)) = 3/(2^(5/3)) which is the same as 3/(2^(1 + 2/3)) = 3/(2 * cube root(4)). To rationalize the denominator, multiply numerator and denominator by cube root(2): R = 3 cube root(2) / (2 * cube root(8)) but cube root(8) is 2, so R = 3 cube root(2) / (4). Therefore, R = (3/4) * cube root(2). Let me check this. 2^(5/3) is 2^(1 + 2/3) = 2 * 2^(2/3), so 3 divided by that is 3/(2 * 2^(2/3)) which is indeed 3/(2 * cube root(4)) or 3/(2 * 4^(1/3)) = 3/(2 * (2^2)^(1/3)) = 3/(2 * 2^(2/3)) = 3/(2^(5/3)) = 3 * 2^(-5/3). Alternatively, 2^(-5/3) is 1/(2^(5/3)) = 1/(2^(1 + 2/3)) = 1/(2 * 2^(2/3)) = 1/(2 * cube root(4))). So, yes, R = 3/(2 * cube root(4)) or 3/(2^(5/3)). Either way, perhaps it's better written as R = 3/(2^(5/3)) or R = 3/(2 * 4^(1/3)). Alternatively, rationalizing the denominator, as 3 * 4^(-1/3)/2. But maybe the answer is expected in a simplified radical form. Let's compute cube root of 4 is 2^(2/3), so R = 3/(2^(5/3)) = 3 * 2^(-5/3). Alternatively, as 3/(2 * 2^(2/3)) = (3/2) * 2^(-2/3). Hmm. Alternatively, expressing in terms of cube roots: R = 3 / (2 * cube root(4)) or R = 3 cube root(2) / 4. Wait, because 2^(5/3) is 2^(1 + 2/3) = 2 * 2^(2/3) = 2 * cube root(4). So, 3 divided by (2 * cube root(4)) is equal to 3 cube root(2) divided by (2 * cube root(4) * cube root(2)) ) = 3 cube root(2) / (2 * cube root(8)) ) = 3 cube root(2) / (2 * 2) ) = 3 cube root(2) / 4. Ah, that's a nicer form. Because if we rationalize the denominator of 3 / (2 * cube root(4)), we can multiply numerator and denominator by cube root(2):R = [3 / (2 * cube root(4))] * [cube root(2)/cube root(2)] = [3 cube root(2)] / [2 * cube root(8)] = [3 cube root(2)] / [2 * 2] = 3 cube root(2) / 4.Yes, that works. So R = (3 * cube root(2)) / 4. Let me verify that with decimal approximation. cube root of 2 is approximately 1.26, so 3*1.26 ≈ 3.78, divided by 4 is approximately 0.945. Let me check if R = (3/4)*2^(1/3). Since 2^(1/3) ≈ 1.26, so 3/4 is 0.75, 0.75*1.26 ≈ 0.945. Alternatively, if I compute 27/32 ≈ 0.84375, then take the cube root: (0.84375)^(1/3). Let me compute 0.84375^(1/3). Since 0.84375 is 27/32, and 27 is 3^3, 32 is 2^5. So 27/32 = (3/2)^3 / 2^2 = (27/8)/4 = 27/32. So cube root of 27/32 is cube root(27)/cube root(32) = 3/(2^(5/3)) which is the same as before. So 3/(2^(5/3)) ≈ 3/(3.1748) ≈ 0.945, which matches the decimal. So R ≈ 0.945. So, the exact value is 3/(2^(5/3)) or 3*2^(-5/3), or 3 cube root(2)/4. So, probably, the answer is R = dfrac{3}{4} sqrt[3]{2}. But let me check once again.Wait, let's recap. We considered the 3D problem where the sphere is centered at (0, R, 0) with radius R. The sphere must be tangent to the glass y = (x² + z²)^2. The point of tangency in 3D would be a circle where x² + z² = r², and y = r^4. The distance from the center (0, R, 0) to any point (x, 0, z) on the circle must be equal to R. Wait, but actually, the sphere is centered at (0, R, 0) and has radius R, so any point on the sphere satisfies x² + (y - R)^2 + z² = R². So substituting y = r^4 and x² + z² = r² into the sphere equation gives r² + (r^4 - R)^2 = R². Which simplifies to r^8 - 2Rr^4 + r² + R² = R², so r^8 - 2Rr^4 + r² = 0. Then factoring gives r²(r^6 - 2Rr² + 1) = 0. So r² = 0 (the origin) or r^6 - 2Rr² + 1 = 0. Let t = r², then t^3 - 2Rt + 1 = 0. For tangency, we need this equation to have a double root because the sphere touches the glass at that point (so multiplicity two). Therefore, discriminant of the cubic must be zero. The discriminant of t^3 - 2Rt + 1 is calculated as 32R^3 - 27 = 0, so R^3 = 27/32, R = (27/32)^(1/3) = 3/(32)^(1/3) = 3/(2^(5/3)) = 3/(2 * 2^(2/3)) = 3/(2 * cube root(4)) = (3 cube root(2))/4. Therefore, R = (3/4) cube root(2). So that's the value.But let me confirm this with another approach. Let's think in 2D first. If we model the glass as y = x^4 and the cherry as a circle touching (0,0) and tangent to the curve y = x^4. Then, the center of the circle would be at (0, R), since it's distance R from (0,0). The circle equation is x² + (y - R)^2 = R². Simplify: x² + y² - 2Ry = 0. The curve is y = x^4. Substitute into the circle equation: x² + x^8 - 2R x^4 = 0. So, x^8 + x² - 2R x^4 = 0. Factor x²: x²(x^6 + 1 - 2R x²) = 0. So, x = 0 (the origin) or x^6 - 2R x² + 1 = 0. Let z = x², then z³ - 2R z + 1 = 0. Same cubic as before. Therefore, even in 2D, we get the same equation. Therefore, the result is the same. Therefore, the value of R is indeed (3 cube root(2))/4.Alternatively, let's consider using calculus. For the circle and the curve to be tangent at a point (a, a^4), the derivative of the circle and the derivative of the curve at that point must be equal.The circle equation: x² + (y - R)^2 = R². Differentiating implicitly: 2x + 2(y - R) y' = 0 => y' = -x / (y - R). At the point (a, a^4), the slope is y' = -a / (a^4 - R).The derivative of the curve y = x^4 is y' = 4x³. At x = a, slope is 4a³.Setting them equal: 4a³ = -a / (a^4 - R). Multiply both sides by (a^4 - R): 4a³ (a^4 - R) = -a. Assuming a ≠ 0 (since at a=0, it's the origin), divide both sides by a: 4a² (a^4 - R) = -1. So, 4a²(a^4 - R) = -1 => 4a^6 - 4R a² + 1 = 0.But we also have the equation from the circle: x² + (y - R)^2 = R². At the point (a, a^4), this becomes a² + (a^4 - R)^2 = R². Expanding: a² + a^8 - 2R a^4 + R² = R² => a^8 + a² - 2R a^4 = 0, which is the same as before.So, from the tangency condition, we have two equations:1. 4a^6 - 4R a² + 1 = 02. a^8 + a² - 2R a^4 = 0Let me try to solve these equations. Let me denote equation 1 as:4a^6 + 1 = 4R a² => R = (4a^6 + 1)/(4a²)Substitute R into equation 2:a^8 + a² - 2 * [(4a^6 + 1)/(4a²)] * a^4 = 0Simplify the term with R:2 * [(4a^6 + 1)/(4a²)] * a^4 = 2 * (4a^6 + 1)/(4a²) * a^4 = 2 * (4a^6 + 1) * a² / 4 = (4a^6 + 1) * a² / 2Therefore, equation becomes:a^8 + a² - (4a^6 + 1) * a² / 2 = 0Multiply through by 2 to eliminate denominator:2a^8 + 2a² - (4a^6 + 1)a² = 0Expand the last term:2a^8 + 2a² - 4a^8 - a² = 0Combine like terms:(2a^8 - 4a^8) + (2a² - a²) = 0 => -2a^8 + a² = 0 => a²(-2a^6 + 1) = 0Therefore, a² = 0 => a = 0 (which is the origin, not the tangency point) or -2a^6 + 1 = 0 => a^6 = 1/2 => a = ±(1/2)^(1/6) = ±2^(-1/6)Therefore, a = ±2^(-1/6). Let's take a positive since the glass is symmetric. So a = 2^(-1/6). Now, substitute this back into R = (4a^6 + 1)/(4a²)Compute a^6: (2^(-1/6))^6 = 2^(-1) = 1/2So, 4*(1/2) + 1 = 2 + 1 = 3Therefore, numerator is 3Denominator: 4a² = 4*(2^(-1/6))^2 = 4*2^(-1/3) = 4/(2^(1/3)) = 2^(2)/2^(1/3) = 2^(5/3)Thus, R = 3 / 2^(5/3) = 3/(2^(5/3)) = 3/(2 * 2^(2/3)) = (3/2) * 2^(-2/3) = 3/(2) * (2^(-2/3)) = 3/(2) * (2^(1/3))/2) Wait, perhaps better to rationalize:2^(5/3) = 2^(1 + 2/3) = 2 * 2^(2/3). Therefore, R = 3/(2 * 2^(2/3)) = 3/(2 * cube root(4)) = (3/2) * 4^(-1/3) = (3/2) * (2^2)^(-1/3) = (3/2) * 2^(-2/3) = 3 * 2^(-5/3). Which is the same as before. Alternatively, writing 4^(-1/3) as 2^(-2/3), so R = 3/(2 * 2^(2/3)) = 3/(2^(5/3)) = 3 * 2^(-5/3). So, again, R = 3/(2^(5/3)) or 3 cube root(2)/4. Because 2^(5/3) = cube root(32), so 3/cube root(32) = 3 cube root(2)/cube root(16 * 2) = Hmm, wait. 32 is 16*2, so cube root(32) = cube root(16*2) = cube root(8*2*2) = 2 cube root(4). Wait, no. 32 is 2^5, so cube root(2^5) = 2^(5/3) = 2^(1 + 2/3) = 2 * 2^(2/3) = 2 * cube root(4). So, cube root(32) = 2 * cube root(4). Therefore, 3/cube root(32) = 3/(2 * cube root(4)) = (3/2) * 4^(-1/3). Alternatively, 4^(1/3) is cube root(4), so 4^(-1/3) is 1/cube root(4). Therefore, R = (3/2) * (1/cube root(4)) = 3/(2 cube root(4)). Then, cube root(4) is 2^(2/3), so 2 cube root(4) = 2 * 2^(2/3) = 2^(1 + 2/3) = 2^(5/3). So, yes, R = 3/(2^(5/3)) which is the same as 3 cube root(2)/4. Let's check:If R = 3 cube root(2)/4, then 4R = 3 cube root(2). Then, cube both sides: (4R)^3 = 27 * 2 => 64 R^3 = 54 => R^3 = 54/64 = 27/32. Which matches our previous result. Therefore, R = (27/32)^(1/3) = 3/(32)^(1/3) = 3/(2^(5/3)) = 3 cube root(2)/4. Therefore, both methods confirm that R = 3 cube root(2)/4.Therefore, the value of R for which the ball touches the bottom of the glass (the origin) is 3 times the cube root of 2 divided by 4.Final AnswerThe ball will touch the bottom of the glass when the radius ( R ) is boxed{dfrac{3}{4} sqrt[3]{2}}.</think>To determine the values of ( R ) for which a ball of radius ( R ) placed in a glass with an axial section given by ( y = x^4 ) will touch the bottom of the glass (the point ( (0,0) )), we need to find the radius ( R ) such that the sphere touches the origin and is tangent to the curve ( y = x^4 ).1. Equation of the Sphere: The sphere is centered at ( (0, R) ) with radius ( R ). The equation of the sphere in 2D is: [ x^2 + (y - R)^2 = R^2 ] Simplifying, we get: [ x^2 + y^2 - 2Ry = 0 ]2. Intersection with the Glass: The glass is modeled by the curve ( y = x^4 ). Substituting ( y = x^4 ) into the sphere's equation: [ x^2 + (x^4 - R)^2 = R^2 ] Simplifying, we get: [ x^8 + x^2 - 2Rx^4 = 0 ] Factoring out ( x^2 ): [ x^2(x^6 - 2Rx^2 + 1) = 0 ] The non-trivial solution comes from: [ x^6 - 2Rx^2 + 1 = 0 ]3. Substitution and Cubic Equation: Let ( t = x^2 ), then the equation becomes: [ t^3 - 2Rt + 1 = 0 ] For the sphere to be tangent to the glass, this cubic equation must have a double root. The discriminant of the cubic equation ( t^3 - 2Rt + 1 = 0 ) must be zero. The discriminant ( D ) is: [ D = 32R^3 - 27 ] Setting ( D = 0 ): [ 32R^3 - 27 = 0 implies R^3 = frac{27}{32} implies R = sqrt[3]{frac{27}{32}} = frac{3}{sqrt[3]{32}} = frac{3}{2^{5/3}} ]4. Simplifying the Expression: Rationalizing the denominator, we get: [ R = frac{3}{2^{5/3}} = frac{3 cdot 2^{1/3}}{2^{5/3} cdot 2^{1/3}} = frac{3 cdot 2^{1/3}}{2^2} = frac{3 sqrt[3]{2}}{4} ]Thus, the value of ( R ) for which the ball will touch the bottom of the glass is boxed{dfrac{3}{4} sqrt[3]{2}}.]

🔑:## Solution.We examine the indicated axial section. Since the axial section of the ball is a circle of radius R, then this circle must be tangent to the O x axis at the point O=(0 ; 0). Therefore, the center of the circle is on the O y axis at the point O_{1}=(0 ; R).We write the condition that the graph of the circumference of radius R with the center at the point O_{1}=(0 ; R) and the graph of the function y=x^{4} have common points (i.e., they either intersect or are tangent).begin{aligned}& left{begin{array}{c}x^{2}+(y-R)^{2}=R^{2} y=x^{4}end{array}right. & x^{2}+left(x^{4}-Rright)^{2}=R^{2} & x^{2}left(1+x^{6}-2 R x^{2}right)=0end{aligned}The value x_{1}=0 gives the point of tangency of the graphs at the origin. Therefore, other points of tangency (or intersection) of the graphs are given by the condition1+x_{0}^{6}=2 R x_{0}^{2}, quad x_{0} neq 0Let quad x_{0}^{2}=t>0. Then we have1+t^{3}=2 R t, quad t>0This equation has a solution if and only if the line y=2 R t (which passes through the point (0,0)) is tangent to or intersects the graph G_{g} of the function g(t)=1+t^{3} (for t>0).We find the condition for the tangency of the graphs of the functions. The tangent to the graph G_{g} has the equationy-gleft(t_{0}right)=g^{prime}left(t_{0}right) cdotleft(t-t_{0}right)Theny=1+t_{0}^{3}+3 t_{0}^{2} cdotleft(t-t_{0}right)=3 t_{0}^{2} cdot t+left(1-2 t_{0}^{3}right)On the other hand y=2 R t. Thereforeleft{begin{array} { c } { 3 t _ { 0 } ^ { 2 } = 2 R } { 1 - 2 t _ { 0 } ^ { 3 } = 0 }end{array} Rightarrow left{begin{array}{c}R_{0}=frac{3 sqrt[3]{2}}{4} x_{0}= pm sqrt{t_{0}}= pm frac{1}{sqrt[6]{2}}end{array}right.right.Thus:1) For 0<R<R_{0}=frac{3 sqrt[3]{2}}{4}, if the "cherry" is tangent to the bottom of the cup, it remains within the limits of the cup's walls. For this, it is sufficient to show that there exists x_{0} in[-R, R] such thatfleft(x_{0}right)=R-sqrt{R^{2}-x_{0}^{2}}-x_{0}^{4}<02) For R>R_{0}=frac{3 sqrt[3]{2}}{4}, if the "cherry" is tangent to the bottom of the cup, it exits the limits of the cup's walls. For this, it is sufficient to show that there exists x_{0} in[-R, R] such thatfleft(x_{0}right)=R-sqrt{R^{2}-x_{0}^{2}}-x_{0}^{4}>0R-frac{1}{sqrt[3]{4}} < R_{0}=frac{3 sqrt[3]{2}}{4}Solutions