answer：First, I need to determine the total number of women participating in the tournament. There are three teams, each consisting of two women, so there are a total of 3 × 2 = 6 women.Next, I'll calculate the total number of possible handshakes if every woman shook hands with every other woman. This can be done using the combination formula C(n, 2), where n is the number of women. So, C(6, 2) = (6 × 5) / 2 = 15 handshakes.However, each woman does not shake hands with her own partner. Since there are three teams, there are three pairs of partners who do not shake hands. Therefore, I need to subtract these three handshakes from the total.Finally, the total number of handshakes that occurred is 15 - 3 = 12.

answer：Okay, so I have this problem here: The set ( A = {x | x^2 - 4x + m = 0} ) is a singleton set, which means it has exactly one element. I need to find the value of ( m ). Hmm, let me think about how to approach this.First, I remember that a singleton set for a quadratic equation means that the equation has exactly one solution. In other words, the quadratic equation ( x^2 - 4x + m = 0 ) has only one root. I recall that for a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( D ) determines the nature of the roots. The discriminant is given by ( D = b^2 - 4ac ).If the discriminant is positive, there are two distinct real roots. If it's zero, there's exactly one real root (a repeated root), and if it's negative, there are no real roots (but two complex roots). Since the set ( A ) is a singleton, that means the equation has exactly one real root, so the discriminant must be zero.Alright, so let's compute the discriminant for the given equation. Here, ( a = 1 ), ( b = -4 ), and ( c = m ). Plugging these into the discriminant formula:( D = (-4)^2 - 4(1)(m) )Calculating that, ( (-4)^2 ) is 16, and ( 4(1)(m) ) is ( 4m ). So,( D = 16 - 4m )Since we need the discriminant to be zero for the equation to have exactly one real root, we set ( D = 0 ):( 16 - 4m = 0 )Now, solving for ( m ):( 16 = 4m )Divide both sides by 4:( m = 4 )Wait, let me double-check that. If ( m = 4 ), then the quadratic equation becomes ( x^2 - 4x + 4 = 0 ). Let me factor that:( x^2 - 4x + 4 = (x - 2)^2 = 0 )Yes, that gives a repeated root at ( x = 2 ), so the set ( A ) would indeed be ( {2} ), which is a singleton set. That makes sense.Just to make sure I didn't make any mistakes, let me go through the steps again. The discriminant must be zero for a singleton set. Calculated discriminant as ( 16 - 4m ). Set equal to zero, solved for ( m = 4 ). Plugging back in, equation becomes a perfect square, so only one solution. Yep, that all checks out.I think that's solid. So, ( m ) must be 4.**Final Answer**The value of ( m ) is boxed{4}.

answer：Okay, so I have this problem about three journalists who wrote articles about the German hockey team's performance in the World Hockey Championship. Each journalist made a different claim about the number of goals the team scored. The twist is that only two of them were correct, and I need to figure out how many goals the German team could have scored. Let me try to break this down step by step.First, let me list out what each journalist said:1. The first journalist wrote: "The German team scored more than 10 but less than 17 goals throughout the championship."2. The second journalist wrote: "The German team scored more than 11 but less than 18 goals throughout the championship."3. The third journalist wrote: "The German team scored an odd number of goals throughout the championship."So, we have three statements, each with a range or a condition on the number of goals. The key point is that only two of these statements are true, and one is false. Therefore, I need to find all possible numbers of goals that satisfy exactly two of the three statements.Let me denote the number of goals scored by the German team as G. Then, I can translate each statement into mathematical inequalities or conditions:1. First journalist: 10 < G < 172. Second journalist: 11 < G < 183. Third journalist: G is oddSo, G must satisfy exactly two of these three conditions. Let me analyze each condition and see how they overlap and where they differ.First, let's consider the ranges given by the first two journalists. The first says G is between 11 and 16 (since it's more than 10 and less than 17). The second says G is between 12 and 17 (more than 11 and less than 18). So, the overlap between these two ranges is from 12 to 16. That means if G is between 12 and 16, both the first and second journalists are correct. If G is 11, only the first journalist is correct because 11 is within the first range but not the second. If G is 17, only the second journalist is correct because 17 is within the second range but not the first.Now, the third journalist's statement is about the parity of G. If G is odd, then the third journalist is correct; otherwise, incorrect.Our task is to find all G such that exactly two of the three statements are true. So, let's consider different scenarios:**Case 1: The first and second journalists are correct, but the third is incorrect.**In this case, G must satisfy both 10 < G < 17 and 11 < G < 18, which simplifies to 11 < G < 17. Additionally, G must be even because the third journalist is incorrect.So, G must be an even number between 12 and 16 inclusive. Let's list those:12, 14, 16.So, these are possible values.**Case 2: The first and third journalists are correct, but the second is incorrect.**Here, G must satisfy 10 < G < 17 and G is odd. Additionally, G must not satisfy 11 < G < 18. Wait, but 10 < G < 17 is a subset of 11 < G < 18. So, if G is between 11 and 17, it automatically satisfies the second journalist's condition. Therefore, it's impossible for G to satisfy the first and third but not the second because the first's range is entirely within the second's. So, this case is not possible. Therefore, no solutions here.Wait, hold on. Let me think again. The first journalist's range is 10 < G < 17, which is 11 to 16. The second journalist's range is 11 < G < 18, which is 12 to 17. So, if G is 11, it satisfies the first journalist but not the second. So, if G is 11, then the first journalist is correct, the second is incorrect, and if G is 11, which is odd, then the third journalist is correct. So, in this case, G=11 would make the first and third correct, and the second incorrect. So, that is a valid case.Wait, so my initial thought was wrong. Let me correct that.So, if G is 11, it satisfies the first journalist (10 < 11 <17), it does not satisfy the second journalist (11 is not greater than 11), and it satisfies the third journalist (11 is odd). So, G=11 would make the first and third correct, and the second incorrect. So, that is a valid case.Similarly, if G is 17, it satisfies the second journalist (11 <17 <18), does not satisfy the first journalist (17 is not less than 17), and 17 is odd, so the third journalist is correct. So, G=17 would make the second and third correct, and the first incorrect.But wait, in the first case, G=11, which is in the first journalist's range but not in the second's. So, that's a valid case where first and third are correct, second is incorrect.Similarly, G=17 is in the second journalist's range but not in the first's, and it's odd, so second and third are correct, first is incorrect.So, actually, there are two more cases:**Case 2: The first and third journalists are correct, but the second is incorrect.**So, G must satisfy 10 < G <17, G is odd, and G does not satisfy 11 < G <18.But wait, 10 < G <17 is 11 to 16, and 11 < G <18 is 12 to 17. So, the only G that satisfies 10 < G <17 but not 11 < G <18 is G=11.Because G=11 is in the first range but not in the second. So, G=11 is the only number here.Similarly, for the third case:**Case 3: The second and third journalists are correct, but the first is incorrect.**G must satisfy 11 < G <18, G is odd, and G does not satisfy 10 < G <17.But 11 < G <18 is 12 to 17, and 10 < G <17 is 11 to 16. So, the overlap is 12 to 16. The numbers that satisfy 11 < G <18 but not 10 < G <17 would be G=17.Because G=17 is in the second range but not in the first. So, G=17 is the only number here.So, summarizing:- Case 1: G is even and between 12 and 16 inclusive. So, G=12,14,16.- Case 2: G=11.- Case 3: G=17.Therefore, the possible values of G are 11,12,14,16,17.Wait, let me check each of these:- G=11: First journalist correct (10 <11 <17), second incorrect (11 is not >11), third correct (11 is odd). So, two correct, one incorrect. Good.- G=12: First correct (10 <12 <17), second correct (11 <12 <18), third incorrect (12 is even). So, two correct, one incorrect. Good.- G=14: Same as G=12. First correct, second correct, third incorrect. Two correct, one incorrect. Good.- G=16: Same as above. First correct, second correct, third incorrect. Good.- G=17: First incorrect (17 is not <17), second correct (11 <17 <18), third correct (17 is odd). So, two correct, one incorrect. Good.So, all these values satisfy exactly two correct statements.Wait, but let me check if there are any other numbers that I might have missed.For example, what about G=13? Let's see:- First journalist: 10 <13 <17: correct.- Second journalist: 11 <13 <18: correct.- Third journalist: 13 is odd: correct.So, all three correct. But the problem states only two were correct, so G=13 is invalid.Similarly, G=15:- First: correct.- Second: correct.- Third: correct.Same issue.G=10: First journalist: 10 <10? No. Second: 11 <10? No. Third: 10 is even. So, only third correct. Not enough.G=18: First: 10 <18? Yes, but 18 <17? No. Second: 11 <18 <18? No. Third: 18 is even. So, only first correct? Wait, 10 <18 is true, but 18 <17 is false, so first statement is "more than 10 but less than 17", so 18 is not less than 17, so first statement is false. Second statement: more than 11 but less than 18: 18 is not less than 18, so false. Third: 18 is even, so correct. So, only third correct. Not enough.G=9: First: 10 <9? No. Second: 11 <9? No. Third: 9 is odd. So, only third correct. Not enough.G=19: First: 10 <19? Yes, but 19 <17? No. Second: 11 <19 <18? No. Third: 19 is odd. So, only third correct. Not enough.So, seems like the only possible G are 11,12,14,16,17.Wait, but let me check G=17 again. First journalist: 10 <17 <17? No, because 17 is not less than 17. So, first is incorrect. Second: 11 <17 <18? Yes. Third: 17 is odd. So, two correct, one incorrect. Correct.Similarly, G=11: first correct, second incorrect, third correct. So, two correct.G=12: first correct, second correct, third incorrect. Two correct.Same with 14 and 16.So, I think that's all. Therefore, the possible number of goals are 11,12,14,16,17.Wait, but let me make sure that there are no other numbers in the ranges that might satisfy two conditions.For example, G=13: as above, all three correct.G=15: same.G=19: only third correct.So, no, I think that's all.So, the possible goals are 11,12,14,16,17.**Final Answer**The possible number of goals the German team could have scored are boxed{11}, boxed{12}, boxed{14}, boxed{16}, and boxed{17}.

answer：Alright, so I have this problem: Among all the simple fractions with a numerator and denominator that are two-digit numbers, find the smallest fraction greater than 3/4. I need to provide its numerator in the answer. Hmm, okay, let me try to figure this out step by step.First, let me understand the problem. I need to find a fraction a/b, where both a and b are two-digit numbers (so from 10 to 99), such that a/b is just a bit larger than 3/4. And among all such fractions, I need the smallest one, meaning the one closest to 3/4 but still larger. Then, I have to report the numerator of that fraction.So, my goal is to find the minimal a/b > 3/4 where a and b are two-digit numbers. I guess the minimal such fraction would be the one where a/b is just barely larger than 3/4. So, I need to find the fraction a/b with the smallest possible value greater than 3/4, given that a and b are two-digit numbers.Let me think about how to approach this. One way is to consider fractions a/b where a and b are two-digit numbers, and a/b is just slightly larger than 3/4. Since 3/4 is 0.75, I need a fraction a/b such that 0.75 < a/b < 1, and a/b is as small as possible.Alternatively, I can think of it as a/b > 3/4, so cross-multiplying, 4a > 3b. Therefore, 4a - 3b > 0. So, I need to find the smallest a/b where 4a - 3b is positive, and a and b are two-digit numbers.But how do I find the minimal a/b? Maybe I can express this as a Diophantine equation? Or perhaps think in terms of continued fractions or Farey sequences? Hmm, Farey sequences might be helpful here because they deal with fractions between 0 and 1, ordered by size, and neighboring fractions have certain properties.Wait, Farey sequences. Let me recall. The Farey sequence of order n is the set of reduced fractions between 0 and 1 with denominators less than or equal to n, arranged in order. So, if I consider the Farey sequence of order 99, since both numerator and denominator are two-digit numbers, which can go up to 99. But 99 is a pretty large denominator, so the Farey sequence would be extensive.But perhaps I can use the mediant property. The mediant of two fractions a/b and c/d is (a+c)/(b+d). If two fractions are neighbors in a Farey sequence, their mediant is the next fraction that would be inserted between them when increasing the order. So, if I can find the fraction just above 3/4 in the Farey sequence, that might be the minimal a/b I'm looking for.But 3/4 is already a fraction with denominator 4, which is less than 99, so in the Farey sequence of order 99, 3/4 would have neighbors. The fraction immediately after 3/4 would be the minimal fraction greater than 3/4 with denominator up to 99. So, perhaps that's the one I need.Alternatively, maybe I can use continued fractions to find the minimal solution. Let me think.Another approach is to consider that I need a fraction a/b such that a = floor(3b/4) + 1, but since a must be a two-digit number, I need to find the minimal a/b where a is just above 3b/4.Wait, let's formalize this. For a given denominator b, the minimal numerator a that makes a/b > 3/4 is a = floor(3b/4) + 1. Then, for each b from 10 to 99, compute a = floor(3b/4) + 1, check if a is a two-digit number, and then compute a/b, and find the minimal such a/b.But since both a and b are two-digit numbers, a must be at least 10, so for each b, a must be at least 10, so floor(3b/4) + 1 must be at least 10. So, 3b/4 + 1 >= 10 => 3b/4 >= 9 => b >= 12. So, b must be at least 12.But wait, b is a two-digit number, so b can be from 10 to 99. But for b=10, a would be floor(30/4) + 1 = 7 + 1 = 8, which is a single-digit number, so that's invalid. Similarly, for b=11, a would be floor(33/4) + 1 = 8 + 1 = 9, still single-digit. So, starting from b=12, a becomes 10, which is two-digit.So, for b from 12 to 99, compute a = floor(3b/4) + 1, and then compute a/b, and find the minimal a/b.But computing this for all b from 12 to 99 would be tedious, but maybe we can find a pattern or a formula.Alternatively, since we need the minimal a/b greater than 3/4, it's equivalent to finding the minimal (a - 3b/4)/b, which is the same as minimal (4a - 3b)/(4b). So, to minimize a/b, we need to minimize (4a - 3b)/(4b). Since 4a - 3b must be positive, we need the minimal positive value of (4a - 3b)/(4b). Which is equivalent to minimizing (4a - 3b)/b, which is 4(a/b) - 3.Wait, maybe another approach: Let me consider that 4a - 3b = k, where k is a positive integer. Then, a = (3b + k)/4. Since a must be an integer, (3b + k) must be divisible by 4. So, 3b + k ≡ 0 mod 4. Therefore, k ≡ (-3b) mod 4. Since k is positive, the minimal k is 1, 2, 3, etc., depending on b.So, for each b, find the minimal k such that k ≡ (-3b) mod 4, and then compute a = (3b + k)/4, which must be an integer. Then, among all such a and b, find the minimal a/b.But this might be a bit involved, but perhaps manageable.Alternatively, since a and b are two-digit numbers, and a/b > 3/4, we can write 4a > 3b, so 4a - 3b > 0. Let me denote d = 4a - 3b, which must be at least 1. So, d >=1.We need to minimize a/b, which is equivalent to minimizing (3b + d)/4b = 3/4 + d/(4b). So, to minimize a/b, we need to minimize d/(4b). Since d is at least 1, the minimal d/(4b) occurs when d is minimal (d=1) and b is maximal (b=99). But wait, if d=1, then a=(3b +1)/4 must be integer, so 3b +1 must be divisible by 4. So, 3b ≡ -1 mod 4 => 3b ≡ 3 mod 4 => b ≡ 1 mod 4. So, b must be congruent to 1 modulo 4.So, the maximal b is 99, which is 99 mod 4 is 3, so 99 is not congruent to 1 mod 4. So, the next lower b congruent to 1 mod 4 is 97, which is 97 mod 4 is 1. So, b=97, then a=(3*97 +1)/4=(291 +1)/4=292/4=73. So, a=73, b=97, so 73/97≈0.752577... which is just above 3/4.But is this the minimal a/b? Because if we take a smaller b, say b=93, which is 1 mod 4, then a=(3*93 +1)/4=(279 +1)/4=280/4=70. So, 70/93≈0.752688... which is larger than 73/97≈0.752577, so 73/97 is smaller.Wait, so as b increases, a increases as well, but a/b might decrease. So, to get the minimal a/b, we need the largest possible b such that b ≡1 mod4, because that would give the minimal a/b.Wait, but is 73/97 the minimal a/b? Let's check with b=97, a=73. Is there a fraction with a smaller a/b?Wait, let's see. If I take d=1, then a=(3b +1)/4, and b must be 1 mod4. So, the largest b is 97, giving a=73. Next, b=93, a=70, which is 70/93≈0.752688, which is larger than 73/97≈0.752577.Similarly, b=89, a=(267 +1)/4=268/4=67, so 67/89≈0.752809, which is larger. So, as b decreases, a/b increases. So, 73/97 is the minimal a/b when d=1.But wait, maybe for d=2, we can get a smaller a/b? Let's check.If d=2, then 4a -3b=2, so a=(3b +2)/4. So, 3b +2 must be divisible by 4, so 3b ≡ -2 mod4 => 3b≡2 mod4 => b≡2 mod4 (since 3 inverse mod4 is 3, because 3*3=9≡1 mod4, so multiplying both sides by 3: b≡6≡2 mod4). So, b must be 2 mod4.So, the largest b is 98, which is 2 mod4. Then, a=(3*98 +2)/4=(294 +2)/4=296/4=74. So, a=74, b=98, so 74/98≈0.755102, which is larger than 73/97≈0.752577.Wait, but maybe a smaller b gives a smaller a/b? Let's check b=94, which is 2 mod4. Then, a=(3*94 +2)/4=(282 +2)/4=284/4=71. So, 71/94≈0.755319, which is larger than 73/97.Similarly, b=90, a=(270 +2)/4=272/4=68. 68/90≈0.755555, which is still larger. So, 74/98 is the smallest a/b for d=2, but it's still larger than 73/97.So, d=1 gives a smaller a/b.What about d=3? Let me check.d=3, so 4a -3b=3, so a=(3b +3)/4. So, 3b +3 must be divisible by4, so 3b ≡ -3 mod4 => 3b≡1 mod4 => b≡3 mod4 (since 3*3=9≡1 mod4). So, b must be 3 mod4.The largest b is 99, which is 3 mod4. So, a=(3*99 +3)/4=(297 +3)/4=300/4=75. So, a=75, b=99, which is 75/99≈0.757575... which is larger than 73/97.Similarly, b=95, a=(285 +3)/4=288/4=72. 72/95≈0.757895, which is larger.So, d=3 gives a larger a/b than d=1.Similarly, d=4: 4a -3b=4, so a=(3b +4)/4. So, 3b +4 must be divisible by4, so 3b≡0 mod4 => b≡0 mod4.So, b must be 0 mod4. The largest b is 96. Then, a=(3*96 +4)/4=(288 +4)/4=292/4=73. So, a=73, b=96. 73/96≈0.760417, which is larger than 73/97.Wait, but 73/96 is larger than 73/97, which is obvious because the denominator is smaller.So, d=4 gives a larger a/b.Therefore, so far, d=1 gives the minimal a/b as 73/97≈0.752577.But wait, is this the minimal? Maybe for d=1, but perhaps with smaller d, but d must be at least1.Wait, but is there a fraction a/b with a and b two-digit numbers, such that a/b is just slightly larger than 3/4, but with a smaller a/b than 73/97?Alternatively, maybe we can find a fraction a/b where 4a -3b=1, but with a and b two-digit numbers, but a/b is smaller than 73/97.Wait, 73/97 is approximately 0.752577. Let me see if there's a fraction a/b with a smaller value but still greater than 3/4.Wait, another approach: Let's consider fractions a/b where a and b are two-digit numbers, and a/b > 3/4. We can write a/b > 3/4 => 4a > 3b => 4a -3b >=1.So, 4a -3b must be at least1. So, the minimal value of 4a -3b is1, which is what we considered earlier.So, if 4a -3b=1, then a=(3b +1)/4, which must be integer, so 3b +1 must be divisible by4, so b≡1 mod4.Therefore, the possible b's are 13,17,...,97.Wait, but earlier I considered b=97, which gives a=73. But what about b=13: a=(39 +1)/4=40/4=10. So, 10/13≈0.76923, which is larger than 73/97.Similarly, b=17: a=(51 +1)/4=52/4=13. So, 13/17≈0.7647, still larger.So, as b increases, a increases, but a/b decreases.So, the minimal a/b in this case is when b is as large as possible, which is 97, giving a=73, so 73/97≈0.752577.But let's check if there's a fraction a/b with 4a -3b=1, but with a and b two-digit numbers, and a/b smaller than 73/97.Wait, but 73/97 is the minimal such fraction because as b increases, a increases, but a/b decreases. So, 73/97 is the minimal a/b for d=1.But wait, is there a fraction with d=1 but a smaller a/b? For example, if b is larger than 97, but b is two-digit, so maximum b is99. But 99 is 3 mod4, so 3b +1=297 +1=298, which is not divisible by4, since 298/4=74.5, which is not integer.Similarly, b=98: 3*98 +1=295, which is not divisible by4.b=97: 3*97 +1=292, which is divisible by4, giving a=73.So, 73/97 is the minimal a/b for d=1.But wait, maybe for d=5, we can get a smaller a/b? Let me check.Wait, d=5: 4a -3b=5 => a=(3b +5)/4. So, 3b +5 must be divisible by4 => 3b≡-5≡3 mod4 => b≡1 mod4.So, same as d=1, but with a=(3b +5)/4. So, for b=97, a=(291 +5)/4=296/4=74. So, 74/97≈0.762886, which is larger than 73/97.Similarly, for b=93, a=(279 +5)/4=284/4=71, so 71/93≈0.763441, which is larger.So, d=5 gives a larger a/b.Similarly, d=6: 4a -3b=6 => a=(3b +6)/4. So, 3b +6≡0 mod4 => 3b≡-6≡2 mod4 => b≡2 mod4.So, b=98: a=(294 +6)/4=300/4=75. 75/98≈0.765306, which is larger.So, seems like d=1 gives the minimal a/b.But wait, let me think differently. Maybe instead of fixing d=1, I can consider fractions a/b where a and b are two-digit numbers, and a/b is just above 3/4.Alternatively, perhaps the minimal a/b is 73/97, but let me check if there's a fraction with a smaller a/b.Wait, let me think of continued fractions. The continued fraction expansion of 3/4 is [0;1,3], but that might not help directly.Alternatively, perhaps I can look for fractions a/b close to 3/4, with a and b two-digit numbers.Wait, another approach: Let me consider that 3/4 is 0.75. So, I need a fraction a/b where a/b is just above 0.75, with a and b two-digit numbers.So, perhaps I can look for fractions where a = floor(0.75*b) +1, and then compute a/b, and find the minimal such a/b.But this is similar to what I did earlier.Alternatively, perhaps I can iterate over possible denominators b from 10 to99, compute the minimal a for each b such that a/b >3/4, then compute a/b, and find the minimal one.But since I can't compute all 90 denominators manually, perhaps I can find a pattern or find the minimal a/b by considering the difference.Wait, let me think about the difference between a/b and 3/4. We need the minimal difference such that a/b >3/4.So, the difference is a/b - 3/4 = (4a -3b)/4b. We need this difference to be minimal, but positive.So, to minimize (4a -3b)/4b, we need to minimize (4a -3b)/b, since 4 is a constant factor.Which is equivalent to minimizing (4a -3b)/b = 4(a/b) -3.But since a/b >3/4, 4(a/b) -3 >0.So, to minimize 4(a/b) -3, which is equivalent to minimizing a/b.Wait, that's circular. So, perhaps another approach.Alternatively, since 4a -3b must be at least1, as we saw earlier, the minimal value of 4a -3b is1, which gives the minimal difference.So, for 4a -3b=1, we have a=(3b +1)/4, which must be integer, so b≡1 mod4.So, the minimal a/b is when b is as large as possible, which is97, giving a=73, so 73/97≈0.752577.But wait, let me check if there's a fraction a/b with a smaller a/b, but with 4a -3b=1, but with a and b two-digit numbers.Wait, for example, b=97, a=73: 73/97≈0.752577.Is there a fraction with a smaller a/b? Let's see.Wait, if I take b=97, a=73: 73/97≈0.752577.If I take b=93, a=70: 70/93≈0.752688, which is larger.Similarly, b=89, a=67: 67/89≈0.752809, which is larger.So, 73/97 is indeed the minimal a/b for d=1.But wait, let me check for d=1 and b=97, a=73. Is 73/97 in lowest terms? Yes, because 73 is a prime number, and 73 doesn't divide 97.So, 73/97 is a reduced fraction.But is there a fraction with a smaller a/b? Let me think.Wait, suppose I have a fraction a/b where a and b are two-digit numbers, and a/b is just above 3/4. Maybe a fraction with a smaller numerator and denominator, but still just above 3/4.Wait, for example, 74/98 is 37/49≈0.755102, which is larger than 73/97.Similarly, 75/99=25/33≈0.757576, which is larger.Wait, so 73/97 is the minimal a/b for d=1.But let me check if there's a fraction with d=1 but a smaller a/b.Wait, if I take b=97, a=73: 73/97≈0.752577.If I take b=97, a=73, and check if there's a smaller a/b, say, 72/96=3/4, which is exactly 0.75, so it's not greater.Similarly, 71/94≈0.755319, which is larger than 73/97.Wait, so 73/97 is indeed the minimal a/b.But wait, let me check another approach. Let me consider that 3/4 is 0.75, so I can look for fractions a/b where a/b is just above 0.75, with a and b two-digit numbers.So, perhaps I can write a program or use some systematic way to check fractions near 0.75.But since I'm doing this manually, let me think of fractions just above 0.75.For example, 0.752577 is 73/97.Is there a fraction with a smaller a/b? Let me think.Wait, 0.752577 is approximately 73/97.Wait, 0.752577 is about 73/97, which is approximately 0.752577.Wait, let me compute 73*4=292, and 97*3=291, so 73/97 is just barely above 3/4, since 73*4=292 and 97*3=291, so 292>291.So, 73/97 is the minimal a/b where a and b are two-digit numbers, and a/b >3/4.But wait, let me check another fraction: 74/98=37/49≈0.755102.Wait, 74/98 is larger than 73/97.Similarly, 75/99=25/33≈0.757576.So, 73/97 is indeed smaller.Wait, but let me check another fraction: 73/97≈0.752577.Is there a fraction a/b with a smaller a/b? Let me think of a=73, b=97.Wait, 73 is a prime number, so 73/97 cannot be simplified.Wait, let me check if there's a fraction with a smaller numerator and denominator but still just above 3/4.Wait, for example, 55/73≈0.753424, which is larger than 73/97.Similarly, 56/74≈0.756757, which is larger.Wait, so 73/97 is indeed the minimal a/b.Wait, but let me check another approach. Let me consider the continued fraction expansion of 3/4, but that might not help directly.Alternatively, perhaps I can use the concept of continued fractions to find the minimal solution.Wait, 3/4 is 0.75, so I can look for fractions a/b such that a/b is just above 0.75, with a and b two-digit numbers.Wait, another way is to consider that 73/97 is the minimal a/b, but let me check if there's a fraction with a smaller a/b.Wait, for example, 73/97≈0.752577.Is there a fraction a/b with a smaller a/b, say, 74/98≈0.755102, which is larger.Wait, so 73/97 is the minimal.Wait, but let me check another fraction: 73/97≈0.752577.Is there a fraction a/b with a smaller a/b, say, 73/97 is the minimal.Wait, perhaps I can check for b=97, a=73.Is there a fraction with a smaller a/b? For example, 73/97 is approximately 0.752577.Wait, let me check 73/97 vs 74/98.Wait, 73/97≈0.752577, 74/98≈0.755102.So, 73/97 is smaller.Similarly, 75/99≈0.757576.So, 73/97 is the minimal.Wait, but let me check another fraction: 73/97 vs 74/97.Wait, 74/97≈0.762886, which is larger.Wait, so 73/97 is the minimal.Wait, but let me check if there's a fraction with a smaller a/b, say, 73/97 vs 73/96.Wait, 73/96≈0.760417, which is larger.So, 73/97 is smaller.Wait, so I think 73/97 is indeed the minimal a/b.But wait, let me check another approach. Let me consider the difference between a/b and 3/4.So, a/b - 3/4 = (4a -3b)/4b.We need this difference to be minimal, but positive.So, to minimize (4a -3b)/4b, we need to minimize (4a -3b)/b, which is 4(a/b) -3.Since a/b >3/4, 4(a/b) -3 >0.So, the minimal value occurs when 4(a/b) -3 is minimal.Which is equivalent to 4(a/b) being minimal above 3.So, 4(a/b) is minimal above 3, so a/b is minimal above 3/4.So, the minimal a/b is the minimal fraction greater than 3/4 with a and b two-digit numbers.Which, as we saw, is 73/97.Therefore, the numerator is 73.Wait, but let me check if there's a fraction with a smaller a/b.Wait, for example, 73/97≈0.752577.Is there a fraction a/b with a smaller a/b, say, 73/97 is the minimal.Wait, let me think of another fraction: 73/97 is approximately 0.752577.Is there a fraction a/b with a smaller a/b, say, 73/97 is the minimal.Wait, perhaps I can check for b=97, a=73, and see if there's a smaller a/b.Wait, 73/97 is the minimal a/b.So, I think the answer is 73.**Final Answer**The numerator of the smallest fraction greater than frac{3}{4} with two-digit numerator and denominator is boxed{73}.

answer：Okay, so I have this problem where I need to find the 100th number in a sequence of natural numbers where all squares and cubes have been crossed out. Hmm, let me think about how to approach this.First, natural numbers are just 1, 2, 3, 4, 5, and so on. So, the original sequence is straightforward. But we need to remove any number that is a perfect square or a perfect cube. That means numbers like 1 (1²), 4 (2²), 8 (2³), 9 (3²), 16 (4²), 25 (5²), 27 (3³), etc., are all out of the picture.So, the remaining numbers are those that are neither squares nor cubes. My task is to find what number is in the 100th position after removing all these squares and cubes.I think the way to go is to figure out how many numbers are removed up to a certain point and then adjust accordingly. Maybe I can find a number N such that when I subtract the count of squares and cubes up to N, the result is 100. That is, N minus (number of squares + number of cubes - number of sixth powers) equals 100. Wait, why sixth powers? Because some numbers are both squares and cubes, like 64 (which is 8² and 4³). So, to avoid double-counting, we subtract the count of sixth powers.Let me write this down as a formula:Number of remaining numbers up to N = N - (number of squares + number of cubes - number of sixth powers)So, we have:N - (floor(sqrt(N)) + floor(N^(1/3)) - floor(N^(1/6))) = 100We need to solve for N such that this equation holds. Hmm, this seems a bit tricky because N is on both sides inside these floor functions. Maybe I can approximate N first and then adjust.Let me denote:- A = floor(sqrt(N)) = number of squares up to N- B = floor(N^(1/3)) = number of cubes up to N- C = floor(N^(1/6)) = number of sixth powers up to NSo, the equation becomes:N - (A + B - C) = 100I need to find N such that this is true.Since A is about sqrt(N), B is about N^(1/3), and C is about N^(1/6), the equation approximately becomes:N - sqrt(N) - N^(1/3) + N^(1/6) ≈ 100This is a bit complicated, but maybe I can estimate N.Let me try plugging in some numbers. Let's start with N=100. Then:A = floor(sqrt(100)) = 10B = floor(100^(1/3)) ≈ 4 (since 4³=64, 5³=125)C = floor(100^(1/6)) ≈ 2 (since 2^6=64, 3^6=729)So, the number of remaining numbers would be 100 - (10 + 4 - 2) = 100 - 12 = 88. Hmm, that's less than 100, so we need a larger N.Let me try N=120.A = floor(sqrt(120)) ≈ 10 (since 10²=100, 11²=121)B = floor(120^(1/3)) ≈ 4 (since 4³=64, 5³=125)C = floor(120^(1/6)) ≈ 2 (since 2^6=64, 3^6=729)So, remaining numbers: 120 - (10 + 4 - 2) = 120 - 12 = 108. Hmm, that's more than 100. So, the 100th number is somewhere between 100 and 120.Wait, but when N=100, remaining numbers are 88, and at N=120, it's 108. So, the 100th number is somewhere in this range. Let's see how much we need to increase N beyond 100 to get 100 remaining numbers.Wait, actually, the number of remaining numbers increases as N increases, so we need to find N such that the count is exactly 100. So, perhaps we can use linear approximation or something.But maybe a better approach is to iterate N starting from 100 upwards until the count reaches 100.Alternatively, let me think about how many numbers are removed up to N. So, the count of removed numbers is A + B - C, as above.So, if I can find N such that N - (A + B - C) = 100, then N is the number we need.Alternatively, maybe I can model this as N ≈ 100 + A + B - C.But since A, B, C depend on N, it's a bit circular. Maybe I can make an initial guess for N, compute A, B, C, then adjust N accordingly.Let me try N=110.A = floor(sqrt(110)) ≈ 10B = floor(110^(1/3)) ≈ 4 (since 4³=64, 5³=125)C = floor(110^(1/6)) ≈ 2So, remaining numbers: 110 - (10 + 4 - 2) = 110 - 12 = 98. Still less than 100.Next, N=111.A = floor(sqrt(111)) ≈ 10B = floor(111^(1/3)) ≈ 4C = floor(111^(1/6)) ≈ 2Remaining: 111 - 12 = 99N=112:Same A, B, C: 112 - 12 = 100.Wait, so at N=112, the count is 100. So, does that mean 112 is the 100th number? Wait, but hold on. Because when N=112, the count is 100, but we need to make sure that 112 itself is not a square or cube.Is 112 a square? sqrt(112) ≈ 10.58, not integer. Is it a cube? 112^(1/3) ≈ 4.82, not integer. So, 112 is included in the remaining numbers. Therefore, the 100th number is 112.Wait, but let me double-check.At N=112, the number of squares is floor(sqrt(112))=10, since 10²=100, 11²=121>112.Number of cubes: floor(112^(1/3))=4, since 4³=64, 5³=125>112.Number of sixth powers: floor(112^(1/6))=2, since 2^6=64, 3^6=729>112.So, total removed numbers: 10 + 4 - 2 = 12.Thus, remaining numbers: 112 - 12 = 100.Therefore, the 100th number is 112.But wait, let me think again. Because when N=112, the count is 100, but does that mean 112 is the 100th number? Or is it the 100th number is 112?Yes, because up to 112, there are 100 numbers remaining. So, 112 is the 100th number.But let me check N=111:floor(sqrt(111))=10, floor(111^(1/3))=4, floor(111^(1/6))=2.So, removed numbers: 10 + 4 - 2 = 12.Remaining numbers: 111 - 12 = 99.So, at N=111, we have 99 numbers remaining. Therefore, the next number, 112, is the 100th number.Therefore, the answer is 112.Wait, but let me make sure that 112 is not a square or cube, which it isn't, as we saw earlier. So, yes, 112 is included.Alternatively, maybe I can test with N=112:List of numbers from 1 to 112, remove squares and cubes.Number of squares: 10 (1,4,9,...,100)Number of cubes: 4 (1,8,27,64)Number of sixth powers: 2 (1,64)So, total removed: 10 + 4 - 2 = 12.Thus, remaining numbers: 112 - 12 = 100.Therefore, 112 is indeed the 100th number.Wait, but hold on. When we remove numbers, the count of remaining numbers is 100 at N=112, which would mean that 112 is the 100th number. So, yes, 112 is the answer.But just to be thorough, let me check N=113.floor(sqrt(113))=10, floor(113^(1/3))=4, floor(113^(1/6))=2.Removed numbers: 10 + 4 - 2 = 12.Remaining: 113 - 12 = 101.So, at N=113, we have 101 numbers remaining. Therefore, 112 is the 100th number.Yes, that seems correct.Alternatively, another way to think about it is that for each number from 1 to N, we check if it's a square or cube, and count how many are left. So, if N=112, and we have 100 numbers left, then 112 is the 100th number.Therefore, I think the answer is 112.**Final Answer**The number in the 100th position is boxed{112}.

answer：Okay, so Joy has these 30 rods, each of integer lengths from 1 cm to 30 cm. She's already placed three rods on the table: 3 cm, 7 cm, and 15 cm. Now, she needs to pick a fourth rod such that these four rods can form a quadrilateral with positive area. The question is asking how many of the remaining rods she can choose for this purpose, and the options are from A to E, with the answer being one of 16, 17, 18, 19, or 20.Alright, so first, I remember that for four lengths to form a quadrilateral with positive area, the sum of any three sides must be greater than the fourth side. This is similar to the triangle inequality, but extended to four sides. So, in other words, for four lengths a, b, c, d, the following must hold:1. a + b + c > d2. a + b + d > c3. a + c + d > b4. b + c + d > aSince all four conditions must be satisfied, but in reality, if the sum of the three smaller sides is greater than the largest side, then all other conditions will automatically hold. So, maybe I can focus on that.Let me note down the lengths we have: 3 cm, 7 cm, 15 cm, and then the fourth rod, let's call its length x cm. So, the four sides are 3, 7, 15, and x.First, I need to figure out the possible values of x such that the sum of the three smaller sides is greater than the largest side. So, depending on where x falls in the order, the largest side could be either 15 or x.Case 1: x is the longest side.In this case, the sum of the other three sides must be greater than x. So, 3 + 7 + 15 > x. Let's compute that: 3 + 7 is 10, plus 15 is 25. So, 25 > x, which means x < 25.Case 2: 15 is the longest side.In this case, the sum of the other three sides must be greater than 15. So, 3 + 7 + x > 15. Let's compute that: 3 + 7 is 10, so 10 + x > 15. Therefore, x > 5.So, combining both cases, x must be greater than 5 and less than 25. So, x must satisfy 6 ≤ x ≤ 24.But wait, hold on. The original lengths are from 1 to 30, and we already have rods of 3, 7, and 15. So, x cannot be 3, 7, or 15 because those are already on the table. So, we need to exclude those values from our possible choices.But in our range, 6 ≤ x ≤ 24, excluding 3, 7, and 15. However, 3 is already below 6, so we don't have to worry about that. But 7 and 15 are within the range, so we need to exclude them.So, the possible x values are integers from 6 to 24 inclusive, excluding 7 and 15. Let's compute how many integers that is.From 6 to 24 inclusive, that's 24 - 6 + 1 = 19 numbers. Then, subtract the two numbers we need to exclude: 7 and 15. So, 19 - 2 = 17.Wait, but hold on. Let me double-check that.Wait, 6 to 24 inclusive: 6,7,8,...,24. So, the number of integers is 24 - 6 + 1 = 19. Then, excluding 7 and 15, so 19 - 2 = 17. So, 17 possible values.But hold on, let me think again. Is that all?Wait, the problem says "the remaining rods." She already placed three rods: 3,7,15. So, the remaining rods are 1,2,4,5,6,8,9,...,30, excluding 3,7,15. So, total remaining rods are 30 - 3 = 27 rods.But we are to choose x from these remaining rods, such that x is between 6 and 24 inclusive, excluding 7 and 15.Wait, but 6 to 24 inclusive is 19 numbers, but excluding 7 and 15, so 17 numbers.But hold on, is that correct? Let me list them out.From 6 to 24 inclusive: 6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24. That's 19 numbers.Excluding 7 and 15: So, 6,8,9,10,11,12,13,14,16,17,18,19,20,21,22,23,24. That's 17 numbers.So, 17 possible choices.Wait, but hold on, let me think again. Is there any other condition?Because sometimes, when dealing with quadrilaterals, it's not just the sum of three sides greater than the fourth, but also each side must be positive, which they are, since all lengths are positive.But also, is there a lower bound on x? Because if x is too small, say, x is 1, then 1 + 3 + 7 = 11, which is less than 15, so that wouldn't satisfy the quadrilateral condition. But in our earlier analysis, we considered that if 15 is the largest side, then the sum of the other three sides must be greater than 15, so x must be greater than 5. So, x must be at least 6.But wait, if x is 6, then 3 + 7 + 6 = 16, which is greater than 15, so that works. Similarly, if x is 24, 3 + 7 + 15 = 25, which is greater than 24, so that works.But what about x being 5? 3 + 7 + 5 = 15, which is equal to 15, so that would not satisfy the strict inequality, so x must be greater than 5, so x ≥ 6.Similarly, x must be less than 25, so x ≤ 24.So, x must be from 6 to 24 inclusive, excluding 7 and 15, so 17 choices.But wait, hold on, let me check the problem statement again."Joy has 30 thin rods, one each of every integer length from 1 cm through 30 cm. She places the rods with lengths 3 cm, 7 cm, and 15 cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area."So, the fourth rod must be one of the remaining rods, which are 1,2,4,5,6,8,9,...,30, excluding 3,7,15.So, the possible x is from 1 to 30, excluding 3,7,15.But x must satisfy 6 ≤ x ≤24, excluding 7 and 15. So, 6,8,9,...,24. So, 17 numbers.But hold on, let me think again: 6 to 24 inclusive is 19 numbers, minus 2 (7 and 15) is 17. So, 17 rods.But wait, the answer choices are 16,17,18,19,20. So, 17 is an option, which is B.But wait, hold on, let me think again. Maybe I made a mistake in the initial analysis.Wait, in the quadrilateral condition, it's not just that the sum of the three smaller sides is greater than the largest side, but actually, for all four sides, the sum of any three sides must be greater than the fourth.But in reality, if the sum of the three smaller sides is greater than the largest side, then all other conditions are automatically satisfied because the sum of any three sides will include the largest side plus two others, which will definitely be greater than the remaining side.Wait, let me test that.Suppose we have sides a ≤ b ≤ c ≤ d.Then, the conditions are:a + b + c > da + b + d > ca + c + d > bb + c + d > aBut since a ≤ b ≤ c ≤ d, the last three inequalities are automatically satisfied because:a + b + d > c: since d ≥ c, and a + b are positive, so yes.a + c + d > b: same logic.b + c + d > a: same.So, the only non-trivial condition is a + b + c > d.Therefore, in our case, the four sides are 3,7,15,x.We need to arrange them in order.Case 1: x is the largest side.So, 3,7,15,x with x ≥15.Then, 3 + 7 +15 >x => 25 >x. So, x <25.But since x is the largest, x must be ≥15, so 15 ≤x <25.But x cannot be 15 because that's already on the table, so x must be from 16 to24 inclusive.Case 2: 15 is the largest side.So, x ≤15.Then, the sum of the other three sides: 3 +7 +x >15.So, 10 +x >15 =>x>5.So, x must be >5.But x is ≤15, so 6 ≤x ≤15.But x cannot be 7 or 15 because those are already on the table.So, x must be from 6 to15 inclusive, excluding 7 and15.So, x can be 6,8,9,10,11,12,13,14.So, that's 8 numbers.Wait, hold on, 6 to15 inclusive is 10 numbers, excluding 7 and15, so 8 numbers.So, in total, from case1 and case2:Case1: x from16 to24 inclusive: 24 -16 +1=9 numbers.Case2: x from6 to15 inclusive, excluding7 and15: 8 numbers.So, total possible x is 9 +8=17.So, 17 possible choices.Therefore, the answer is 17, which is option B.But wait, hold on, let me just make sure.Wait, in case1, x is from16 to24 inclusive: 16,17,18,19,20,21,22,23,24. That's 9 numbers.In case2, x is from6 to15 inclusive, excluding7 and15: 6,8,9,10,11,12,13,14. That's 8 numbers.So, 9 +8=17.Yes, that's correct.Alternatively, another way to think about it is that x must satisfy 6 ≤x ≤24, excluding7 and15. So, 24 -6 +1=19, minus 2 gives 17.Yes, that's consistent.Therefore, the answer is 17, which is option B.**Final Answer**boxed{B}