answer：Alright, so I'm trying to prove these two statements about the set ( A = {2, 2^2, 2^3, dots, 2^n, dots} ). Let me start by understanding what each part is saying.**Part (1):** For every ( a in A ) and ( b in mathbb{N}^* ), if ( b < 2a - 1 ), then ( b(b+1) ) is not a multiple of ( 2a ).Okay, so ( A ) consists of powers of 2. So ( a ) is like 2, 4, 8, 16, etc. For each such ( a ), and any positive integer ( b ) less than ( 2a - 1 ), the product ( b(b+1) ) shouldn't be divisible by ( 2a ).Hmm, let's take an example. Let ( a = 2 ). Then ( 2a = 4 ), and ( 2a - 1 = 3 ). So ( b ) can be 1 or 2. Let's check ( b(b+1) ):- For ( b = 1 ): ( 1 times 2 = 2 ), which is not a multiple of 4.- For ( b = 2 ): ( 2 times 3 = 6 ), which is not a multiple of 4.Okay, that works. Let's try ( a = 4 ). Then ( 2a = 8 ), and ( 2a - 1 = 7 ). So ( b ) can be from 1 to 6.- ( b = 1 ): ( 1 times 2 = 2 ) (not multiple of 8)- ( b = 2 ): ( 2 times 3 = 6 ) (not multiple of 8)- ( b = 3 ): ( 3 times 4 = 12 ) (not multiple of 8)- ( b = 4 ): ( 4 times 5 = 20 ) (not multiple of 8)- ( b = 5 ): ( 5 times 6 = 30 ) (not multiple of 8)- ( b = 6 ): ( 6 times 7 = 42 ) (not multiple of 8)Seems to hold. Maybe this is a pattern. Since ( a ) is a power of 2, ( 2a ) is the next power of 2. So ( b ) is less than ( 2a - 1 ), which is just one less than the next power of 2.I think the key here is that ( b ) and ( b+1 ) are consecutive integers, so one is even and one is odd. Since ( 2a ) is a power of 2, the even number among ( b ) and ( b+1 ) must have enough factors of 2 to make ( b(b+1) ) divisible by ( 2a ). But since ( b < 2a - 1 ), the even number can't have enough factors of 2 to reach ( 2a ).Wait, let me think about that more carefully. If ( a = 2^k ), then ( 2a = 2^{k+1} ). So the even number between ( b ) and ( b+1 ) can have at most ( k ) factors of 2 because ( b < 2^{k+1} - 1 ). Therefore, it can't have ( k+1 ) factors of 2, which is required for ( 2a ). So ( b(b+1) ) can't be a multiple of ( 2a ).That makes sense. So for part (1), the idea is that ( b ) is too small to have enough factors of 2 in either ( b ) or ( b+1 ) to make the product divisible by ( 2a ).**Part (2):** For every ( a in bar{A} ) (the complement of ( A ) in ( mathbb{N}^* )) and ( a neq 1 ), there exists ( b in mathbb{N}^* ) such that ( b < 2a - 1 ) and ( b(b+1) ) is a multiple of ( 2a ).Alright, so ( bar{A} ) consists of all positive integers that are not powers of 2. So ( a ) could be 3, 5, 6, 7, 9, etc. For each such ( a ), we need to find a ( b ) less than ( 2a - 1 ) such that ( b(b+1) ) is divisible by ( 2a ).Let me take an example. Let ( a = 3 ). Then ( 2a = 6 ), and ( 2a - 1 = 5 ). So ( b ) can be from 1 to 4.- ( b = 1 ): ( 1 times 2 = 2 ) (not multiple of 6)- ( b = 2 ): ( 2 times 3 = 6 ) (which is a multiple of 6) Okay, so ( b = 2 ) works.Another example: ( a = 5 ). Then ( 2a = 10 ), ( 2a - 1 = 9 ). So ( b ) can be from 1 to 8.- ( b = 4 ): ( 4 times 5 = 20 ) (multiple of 10) So ( b = 4 ) works.Wait, is there a pattern here? For ( a = 3 ), ( b = 2 ); for ( a = 5 ), ( b = 4 ). It seems like ( b = a - 1 ) works in these cases.Let me test ( a = 6 ). ( 2a = 12 ), ( 2a - 1 = 11 ). ( b ) can be up to 10.- ( b = 3 ): ( 3 times 4 = 12 ) (multiple of 12) So ( b = 3 ) works. But ( a - 1 = 5 ), and ( 5 times 6 = 30 ), which is also a multiple of 12. So both ( b = 3 ) and ( b = 5 ) work.Hmm, interesting. So maybe there are multiple ( b ) that satisfy the condition, but we just need to find at least one.Another example: ( a = 7 ). ( 2a = 14 ), ( 2a - 1 = 13 ). So ( b ) can be up to 12.- ( b = 6 ): ( 6 times 7 = 42 ) (multiple of 14) So ( b = 6 ) works.Wait, ( a = 9 ). ( 2a = 18 ), ( 2a - 1 = 17 ). ( b ) can be up to 16.- ( b = 8 ): ( 8 times 9 = 72 ) (multiple of 18) So ( b = 8 ) works.It seems like ( b = a - 1 ) often works, but not always. For example, when ( a = 6 ), ( b = 3 ) also worked. So maybe ( b ) can be chosen as ( a - 1 ) or some other value depending on ( a ).But how do we prove that such a ( b ) exists for any ( a ) not a power of 2?I think the key is that since ( a ) is not a power of 2, it has some odd prime factors. So we can use the Chinese Remainder Theorem or some number theory to find such a ( b ).Let me consider that ( a ) can be written as ( a = 2^k times m ), where ( m ) is odd and greater than 1 (since ( a ) is not a power of 2). Then ( 2a = 2^{k+1} times m ).We need ( b(b+1) ) to be divisible by ( 2^{k+1} times m ). Since ( b ) and ( b+1 ) are consecutive integers, one is even and one is odd. The even one needs to be divisible by ( 2^{k+1} ), and the odd one needs to be divisible by ( m ).So, we can set up the following system:1. ( b equiv 0 mod 2^{k+1} )2. ( b + 1 equiv 0 mod m )Or alternatively,1. ( b equiv -1 mod m )2. ( b equiv 0 mod 2^{k+1} )By the Chinese Remainder Theorem, since ( m ) and ( 2^{k+1} ) are coprime (because ( m ) is odd), there exists a solution ( b ) modulo ( 2^{k+1} times m ). So the smallest positive solution ( b ) will be less than ( 2^{k+1} times m ).But ( ( 2a = 2^{k+1} times m ), so ( b < 2a ). But we need ( b < 2a - 1 ). Hmm, the smallest solution might be exactly ( 2a - 1 ), which is not allowed. So maybe we need to adjust.Wait, actually, the smallest positive solution ( b ) will be less than ( 2a ), but we need ( b < 2a - 1 ). So if the smallest solution is less than ( 2a - 1 ), we're good. If it's equal to ( 2a - 1 ), then we need to find the next solution, which would be ( b = 2a - 1 - 2a = -1 ), which is not positive. Hmm, maybe I'm missing something.Alternatively, maybe we can find a ( b ) such that ( b ) is divisible by ( m ) and ( b + 1 ) is divisible by ( 2^{k+1} ). Let's try that.Set up:1. ( b equiv 0 mod m )2. ( b + 1 equiv 0 mod 2^{k+1} )Again, since ( m ) and ( 2^{k+1} ) are coprime, there exists a solution ( b ) modulo ( 2^{k+1} times m ). The smallest positive solution ( b ) will be less than ( 2^{k+1} times m = 2a ). So ( b < 2a ). But we need ( b < 2a - 1 ). So if the smallest solution is less than ( 2a - 1 ), we're good. If it's equal to ( 2a - 1 ), then we need to find the next solution, which would be ( b = 2a - 1 - 2a = -1 ), which is not positive. Hmm, this seems tricky.Wait, maybe the smallest solution is actually less than ( 2a - 1 ). Let's see. For example, when ( a = 3 ), ( 2a = 6 ), ( 2a - 1 = 5 ). The solution ( b = 2 ) is less than 5. For ( a = 5 ), ( 2a = 10 ), ( 2a - 1 = 9 ). The solution ( b = 4 ) is less than 9. For ( a = 6 ), ( 2a = 12 ), ( 2a - 1 = 11 ). The solution ( b = 3 ) is less than 11. For ( a = 7 ), ( 2a = 14 ), ( 2a - 1 = 13 ). The solution ( b = 6 ) is less than 13.So in these examples, the smallest solution is less than ( 2a - 1 ). Maybe in general, the smallest solution is less than ( 2a - 1 ). Let me think about why.Since ( b ) is a solution to the system ( b equiv 0 mod m ) and ( b + 1 equiv 0 mod 2^{k+1} ), the smallest positive solution ( b ) is given by the Chinese Remainder Theorem as ( b = m times t ), where ( t ) is chosen such that ( m times t equiv -1 mod 2^{k+1} ).Since ( m ) is odd, it has an inverse modulo ( 2^{k+1} ). Let ( m^{-1} ) be the inverse of ( m ) modulo ( 2^{k+1} ). Then ( t equiv -m^{-1} mod 2^{k+1} ). So the smallest positive ( t ) is ( t = 2^{k+1} - m^{-1} ).Wait, this might be getting too technical. Maybe there's a simpler way to see that ( b < 2a - 1 ).Alternatively, since ( b ) is a solution to ( b equiv 0 mod m ) and ( b equiv -1 mod 2^{k+1} ), the smallest positive ( b ) is less than ( 2^{k+1} times m ), which is ( 2a ). But we need ( b < 2a - 1 ). So if the smallest solution is less than ( 2a - 1 ), we're good. If it's equal to ( 2a - 1 ), then we need to find another solution.But in our examples, the smallest solution was always less than ( 2a - 1 ). Maybe in general, the smallest solution is less than ( 2a - 1 ). Let me try to see why.Suppose ( b ) is the smallest positive solution to ( b equiv 0 mod m ) and ( b equiv -1 mod 2^{k+1} ). Then ( b = m times t ), and ( m times t equiv -1 mod 2^{k+1} ). Since ( m ) is odd, ( t ) must be odd as well because ( m times t ) is odd (since ( m ) is odd and ( t ) is odd). Wait, no, ( m times t ) is congruent to -1 modulo ( 2^{k+1} ), which is even. So ( m times t ) must be odd because ( -1 mod 2^{k+1} ) is odd. Since ( m ) is odd, ( t ) must be odd as well.But I'm not sure if this helps. Maybe I need to think differently.Another approach: Since ( a ) is not a power of 2, it has an odd prime factor. Let's say ( p ) is an odd prime dividing ( a ). Then ( p ) divides ( 2a ). So for ( b(b+1) ) to be divisible by ( 2a ), either ( p ) divides ( b ) or ( p ) divides ( b+1 ).If ( p ) divides ( b ), then ( b equiv 0 mod p ). If ( p ) divides ( b+1 ), then ( b equiv -1 mod p ).Similarly, since ( 2a ) has factors of 2, one of ( b ) or ( b+1 ) must be even, and the other must be odd. The even one needs to have enough factors of 2 to cover ( 2a ).Wait, but ( a ) is not a power of 2, so ( 2a ) has at least one odd prime factor. So we can use the Chinese Remainder Theorem to find a ( b ) that satisfies the necessary congruences.Let me try to formalize this.Let ( a = 2^k times m ), where ( m ) is odd and greater than 1. Then ( 2a = 2^{k+1} times m ).We need ( b(b+1) ) divisible by ( 2^{k+1} times m ). Since ( b ) and ( b+1 ) are consecutive, one is even and one is odd.Case 1: ( b ) is even. Then ( b ) must be divisible by ( 2^{k+1} ), and ( b+1 ) must be divisible by ( m ).Case 2: ( b+1 ) is even. Then ( b+1 ) must be divisible by ( 2^{k+1} ), and ( b ) must be divisible by ( m ).Let's consider Case 1 first.We need ( b equiv 0 mod 2^{k+1} ) and ( b equiv -1 mod m ).By the Chinese Remainder Theorem, since ( 2^{k+1} ) and ( m ) are coprime (because ( m ) is odd), there exists a unique solution modulo ( 2^{k+1} times m ).Let ( b = 2^{k+1} times t ). Then ( 2^{k+1} times t equiv -1 mod m ). Since ( 2^{k+1} ) and ( m ) are coprime, there exists an inverse of ( 2^{k+1} ) modulo ( m ). Let ( t_0 ) be such that ( 2^{k+1} times t_0 equiv -1 mod m ). Then ( t = t_0 + s times m ) for some integer ( s ).Thus, ( b = 2^{k+1} times (t_0 + s times m) ). The smallest positive ( b ) is when ( s = 0 ), so ( b = 2^{k+1} times t_0 ).Now, ( b < 2a - 1 = 2^{k+1} times m - 1 ). Since ( t_0 < m ) (because ( t_0 ) is the smallest positive solution modulo ( m )), ( b = 2^{k+1} times t_0 < 2^{k+1} times m = 2a ). But we need ( b < 2a - 1 ). If ( t_0 = m - 1 ), then ( b = 2^{k+1} times (m - 1) = 2^{k+1} times m - 2^{k+1} = 2a - 2^{k+1} ). Since ( 2^{k+1} geq 2 ), ( b leq 2a - 2 ), which is less than ( 2a - 1 ).Therefore, the smallest positive ( b ) in this case is less than ( 2a - 1 ).Similarly, for Case 2, where ( b+1 ) is divisible by ( 2^{k+1} ) and ( b ) is divisible by ( m ), we can set up the congruences:( b equiv 0 mod m ) and ( b equiv -1 mod 2^{k+1} ).Again, by the Chinese Remainder Theorem, there exists a solution ( b ) modulo ( 2^{k+1} times m ). The smallest positive solution ( b ) will be less than ( 2^{k+1} times m = 2a ). To ensure ( b < 2a - 1 ), we can use a similar argument as above.Therefore, in both cases, there exists a ( b ) such that ( b < 2a - 1 ) and ( b(b+1) ) is a multiple of ( 2a ).So, putting it all together, for any ( a ) not a power of 2, we can find such a ( b ) by solving the appropriate congruences using the Chinese Remainder Theorem, ensuring that ( b ) is less than ( 2a - 1 ).I think this covers both parts of the proof. For part (1), the key idea is that powers of 2 don't allow ( b(b+1) ) to have enough factors of 2 to be divisible by ( 2a ) when ( b < 2a - 1 ). For part (2), since ( a ) has odd prime factors, we can find a ( b ) using the Chinese Remainder Theorem that satisfies the necessary conditions.

answer：Okay, so I have this problem where a pedestrian and a cyclist start at the same point on a circular track and move in the same direction. The cyclist is faster than the pedestrian by 55%. I need to figure out how many different points on the track the cyclist will overtake the pedestrian. Hmm, let me think about how to approach this.First, I should probably define some variables to make this more concrete. Let's say the length of the circular track is L. I don't know the exact length, but maybe I can assume it's 1 unit for simplicity, or perhaps another number that might make the calculations easier. Wait, since the cyclist is 55% faster, maybe choosing a length that relates to 55 would be helpful. Let me think... If I set the length of the track to 55 units, that might make the math work out nicely because 55 is a multiple of 55%. Yeah, that seems like a good idea.So, let's assume the track is 55 units long. Now, let's denote the speed of the pedestrian as v. Then, the speed of the cyclist would be v plus 55% of v, which is v + 0.55v = 1.55v. So, the cyclist's speed is 1.55 times the pedestrian's speed.Now, since they're moving in the same direction, the relative speed between the cyclist and the pedestrian is the difference in their speeds. That would be 1.55v - v = 0.55v. This relative speed is how much faster the cyclist is compared to the pedestrian.The next thing I need to figure out is how often the cyclist overtakes the pedestrian. Since the track is circular, every time the cyclist gains a full lap on the pedestrian, they'll overtake them. So, the time it takes for the cyclist to overtake the pedestrian once is the time it takes to cover the length of the track at the relative speed.So, the time between overtakes, let's call it T, would be the length of the track divided by the relative speed. That is, T = L / (relative speed) = 55 units / 0.55v. Simplifying that, 55 divided by 0.55 is 100, so T = 100 / v. So, every 100/v units of time, the cyclist overtakes the pedestrian.Now, in this time T, how far does the pedestrian travel? The pedestrian's speed is v, so distance = speed × time, which would be v × (100 / v) = 100 units. Wait a minute, the track is only 55 units long. So, 100 units is more than one full lap. Specifically, 100 divided by 55 is approximately 1.818 laps. Hmm, that seems a bit confusing. Let me think again.Wait, no, actually, the pedestrian travels 100 units in the time it takes for the cyclist to overtake them once. But since the track is 55 units, 100 units is 1 full lap (55 units) plus an additional 45 units. So, the pedestrian has gone around the track once and then an extra 45 units. That means the overtaking point is 45 units ahead of the starting point on the track.But wait, the cyclist is faster, so shouldn't the overtaking point be somewhere behind the pedestrian? Hmm, maybe I'm mixing up the reference frames. Let me clarify.From the perspective of the pedestrian, the cyclist is approaching at a relative speed of 0.55v. So, the time it takes for the cyclist to catch up is the time to cover the entire track length relative to the pedestrian. But since the pedestrian is also moving, the distance the cyclist needs to cover to overtake is effectively the length of the track, but in the frame of reference of the pedestrian.Wait, maybe it's better to think in terms of how many times the cyclist laps the pedestrian. Since the cyclist is 55% faster, the ratio of their speeds is 1.55:1. So, the number of times the cyclist laps the pedestrian in a certain period is related to this ratio.Alternatively, maybe I can think about the number of distinct overtaking points by considering the least common multiple of their lap times or something like that. Hmm, let's see.If the pedestrian's speed is v, their time to complete one lap is L / v = 55 / v. The cyclist's time to complete one lap is L / (1.55v) = 55 / (1.55v). Let me compute that: 55 divided by 1.55 is approximately 35.5477 units of time.So, the pedestrian takes 55 / v units of time per lap, and the cyclist takes approximately 35.5477 / v units of time per lap.Now, the time between overtakes is the time it takes for the cyclist to gain a full lap on the pedestrian, which we previously calculated as 100 / v units of time. In this time, the pedestrian completes 100 / v divided by (55 / v) laps, which is 100 / 55 ≈ 1.818 laps. So, the pedestrian is 0.818 laps ahead of the starting point when the cyclist overtakes them.Wait, 0.818 laps is approximately 0.818 × 55 ≈ 45 units. So, the overtaking point is 45 units from the starting point.But then, the next overtaking would occur after another 100 / v units of time, right? So, in that time, the pedestrian would have traveled another 100 units, which is again 1 lap plus 45 units. So, the next overtaking point would be 45 units further along the track.But since the track is circular, 45 units from the previous overtaking point would be 45 + 45 = 90 units from the start. But 90 units is more than the track length of 55 units, so 90 - 55 = 35 units from the start.Wait, so the overtaking points are at 45, 35, and so on? Hmm, that seems like they're moving backward around the track each time.Wait, maybe I need to think about it differently. Since the pedestrian is moving forward, and the cyclist is lapping them, the overtaking points are actually moving backward relative to the starting point.Let me try to visualize this. Imagine the track as a circle with 55 units. The first overtaking happens at 45 units from the start. Then, the next overtaking would be 45 units further, but since the track is circular, that would be 45 + 45 = 90, which is equivalent to 90 - 55 = 35 units from the start. The next one would be 35 + 45 = 80, which is 80 - 55 = 25 units from the start. Continuing this way, the overtaking points would be at 45, 35, 25, 15, 5, 45, and so on.Wait, that seems like after 5 overtakes, we're back to 45 units, which is the starting point of the overtaking cycle. So, does that mean there are 5 distinct overtaking points?But wait, let me check. If I keep adding 45 units each time and subtracting 55 when it exceeds, the sequence would be:1. 452. 45 + 45 = 90 → 90 - 55 = 353. 35 + 45 = 80 → 80 - 55 = 254. 25 + 45 = 70 → 70 - 55 = 155. 15 + 45 = 60 → 60 - 55 = 56. 5 + 45 = 507. 50 + 45 = 95 → 95 - 55 = 408. 40 + 45 = 85 → 85 - 55 = 309. 30 + 45 = 75 → 75 - 55 = 2010. 20 + 45 = 65 → 65 - 55 = 1011. 10 + 45 = 55 → 55 - 55 = 0, which is the starting point.Wait, so after 11 overtakes, we return to the starting point. So, does that mean there are 11 distinct overtaking points?But earlier, I thought it might be 5. Hmm, maybe I made a mistake in my initial reasoning.Let me try a different approach. The number of distinct overtaking points is related to the ratio of their speeds. Since the cyclist is 55% faster, the ratio of their speeds is 1.55:1. To find the number of distinct points, we can look at the fractional part of this ratio.1.55 can be written as 1 + 0.55. The 0.55 represents how much faster the cyclist is compared to the pedestrian. To find the number of distinct overtaking points, we can consider the reciprocal of this fractional part, but I'm not sure if that's the right approach.Alternatively, the number of distinct overtaking points can be found by looking at how many times the faster runner (cyclist) laps the slower one (pedestrian) before they both align again at the starting point.The formula for the number of distinct meeting points on a circular track is given by the least common multiple (LCM) of their lap times divided by the difference in their lap times. But I'm not entirely sure about that.Wait, maybe it's better to think in terms of the number of times the cyclist laps the pedestrian before they both return to the starting point simultaneously.The time it takes for both to return to the starting point is the least common multiple of their lap times. The pedestrian's lap time is 55 / v, and the cyclist's lap time is 55 / (1.55v). Let's compute these:Pedestrian's lap time: 55 / vCyclist's lap time: 55 / (1.55v) ≈ 35.5477 / vThe LCM of 55 / v and 35.5477 / v is the LCM of 55 and 35.5477, divided by v. But 35.5477 is approximately 55 / 1.55, which is exact. So, 55 and 55 / 1.55.To find the LCM of 55 and 55 / 1.55, we can note that 55 / 1.55 = 55 / (31/20) = 55 * (20/31) ≈ 35.5477.But LCM of two numbers is usually easier when they are integers. Since 55 is 5 × 11, and 35.5477 is not an integer, this approach might not be straightforward.Maybe another way is to consider the ratio of their speeds. The ratio is 1.55:1, which can be written as 31:20 when multiplied by 20 to eliminate the decimal. So, 1.55 = 31/20.Therefore, the ratio of their speeds is 31:20. This means that for every 31 units the cyclist moves, the pedestrian moves 20 units.Now, the number of distinct overtaking points is given by the difference in the number of laps they complete before they both return to the starting point. Since their speed ratio is 31:20, the LCM of 31 and 20 is 620. So, after 620 units of time, the cyclist would have completed 620 / 31 ≈ 20 laps, and the pedestrian would have completed 620 / 20 = 31 laps.Wait, no, that doesn't seem right. Let me clarify.Actually, the number of distinct overtaking points is given by the numerator of the reduced fraction of their speed ratio minus the denominator. Since the speed ratio is 31:20, the number of overtaking points is 31 - 20 = 11.Wait, that seems to make sense. So, if the speed ratio is a reduced fraction a/b, then the number of distinct overtaking points is a - b.In this case, 31/20 is already in its simplest form, so the number of overtaking points is 31 - 20 = 11.Therefore, the cyclist will overtake the pedestrian at 11 distinct points on the track before they both return to the starting point simultaneously.Wait, let me verify this with another approach to make sure I'm not making a mistake.Another way to think about it is to calculate how many times the cyclist overtakes the pedestrian in the time it takes for both to return to the starting point.The time for both to return to the starting point is the LCM of their lap times. The pedestrian's lap time is 55 / v, and the cyclist's lap time is 55 / (1.55v) ≈ 35.5477 / v.The LCM of 55 / v and 35.5477 / v is the LCM of 55 and 35.5477 divided by v. But since 35.5477 is 55 / 1.55, which is 55 / (31/20) = 55 * (20/31) ≈ 35.5477.So, LCM of 55 and 55 * (20/31). Hmm, this seems complicated because 55 * (20/31) is not an integer. Maybe instead, I should express both lap times in terms of a common unit.Let me denote the pedestrian's lap time as T_p = 55 / v, and the cyclist's lap time as T_c = 55 / (1.55v) = (55 / v) / 1.55 = T_p / 1.55.So, T_c = T_p / 1.55.The LCM of T_p and T_c is LCM(T_p, T_p / 1.55). Since T_p is a multiple of T_p / 1.55, the LCM is T_p.Wait, that can't be right because T_p is larger than T_c, so the LCM should be a multiple of both. Hmm, maybe I'm not thinking about this correctly.Alternatively, the number of overtakes in the time it takes for both to return to the starting point is given by the difference in the number of laps each completes in that time.If the LCM of their lap times is T, then the number of laps the pedestrian completes is T / T_p, and the number of laps the cyclist completes is T / T_c.The number of overtakes is then the difference between these two, which is T / T_c - T / T_p.But since T is the LCM of T_p and T_c, T / T_p and T / T_c are integers.Given that T_c = T_p / 1.55, then T / T_c = 1.55 * (T / T_p).Since T / T_p must be an integer, let's say n, then T / T_c = 1.55n.But 1.55n must also be an integer because T / T_c is the number of laps the cyclist completes, which has to be an integer.Therefore, 1.55n must be an integer. Since 1.55 = 31/20, we have (31/20)n must be an integer. Therefore, n must be a multiple of 20 to make (31/20)n an integer.Let n = 20k, where k is an integer. Then, T / T_p = 20k, and T / T_c = (31/20)*20k = 31k.Therefore, the number of overtakes is T / T_c - T / T_p = 31k - 20k = 11k.Since we're looking for the number of distinct overtaking points before they both return to the starting point, we take k=1, which gives 11 overtakes.Thus, there are 11 distinct points on the track where the cyclist overtakes the pedestrian.Wait, that seems consistent with my earlier conclusion. So, I think the answer is 11 distinct overtaking points.Just to make sure, let me think about it in terms of modular arithmetic.If the overtaking points are determined by the position where the cyclist catches up to the pedestrian, and since the relative speed is 0.55v, the distance between overtakes is 55 units, which is the length of the track.But actually, the overtaking occurs when the cyclist has covered an extra 55 units relative to the pedestrian. So, the distance between overtakes is 55 units, but since the track is circular, this corresponds to different points each time.Given that the pedestrian is moving, the overtaking points are spaced at intervals of 55 units relative to the pedestrian's position.But since the pedestrian is moving forward, the overtaking points are actually spaced at intervals of 55 units minus the distance the pedestrian has moved in that time.Wait, no, that might complicate things. Alternatively, since the relative speed is 0.55v, the time between overtakes is 55 / 0.55v = 100 / v.In that time, the pedestrian has moved 100 units, which is 100 / 55 ≈ 1.818 laps.So, the overtaking points are spaced 0.818 laps apart, which is 0.818 × 55 ≈ 45 units.Therefore, each overtaking point is 45 units from the previous one in the direction of travel.But since the track is circular, after a certain number of overtakes, the points will repeat.The number of distinct points is the number of unique positions before the pattern repeats.Given that 45 and 55 are both multiples of 5, their greatest common divisor (GCD) is 5. Therefore, the number of distinct points is 55 / GCD(45,55) = 55 / 5 = 11.Yes, that makes sense. So, the number of distinct overtaking points is 11.I think I've confirmed it through multiple approaches: considering the speed ratio, calculating the LCM of lap times, and using modular arithmetic with the GCD. All methods point to 11 distinct overtaking points.Therefore, the final answer is 11.

answer：Alright, so I've got this question here asking which of the options A to D is an irrational number. Let me try to figure this out step by step. First, I need to remember what an irrational number is. From what I recall, an irrational number is a real number that cannot be expressed as a simple fraction. That means it can't be written as a ratio of two integers. Okay, so it's not like 1/2 or 3/4 or something like that.Now, looking at the options:A: 1/3B: sqrt(16)C: 3.1415926D: piLet's go through each one.Starting with option A: 1/3. Hmm, that's a fraction, right? So, it's a ratio of two integers, 1 and 3. That means it's a rational number. So, A is not irrational.Moving on to option B: sqrt(16). Okay, the square root of 16. I know that sqrt(16) is 4 because 4 times 4 is 16. So, sqrt(16) equals 4, which is an integer. Integers are definitely rational numbers because they can be expressed as a fraction over 1, like 4/1. So, B is also rational.Next is option C: 3.1415926. That looks familiar. Isn't that an approximation of pi? Pi is approximately 3.1415926535... and it goes on forever without repeating. But wait, this option is just 3.1415926, which is a finite decimal. Finite decimals can be expressed as fractions. For example, 0.5 is 1/2, and 0.25 is 1/4. So, 3.1415926 can be written as a fraction, making it a rational number. Therefore, C is rational.Finally, option D: pi. Pi is a well-known irrational number. It's approximately 3.1415926535..., but it continues infinitely without repeating. Because it can't be expressed as a simple fraction, it's irrational. So, D is irrational.Wait, let me double-check to make sure I didn't make a mistake. For option C, 3.1415926 is a finite decimal, right? So, it's rational. And pi is irrational because it's non-terminating and non-repeating. Yeah, that seems correct.So, out of all the options, only D is an irrational number.

answer：Alright, so I have this problem here: "Given that the terminal side of angle α passes through point p₀ (-3, -4), the value of cos(π/2 + α) is __________." Hmm, okay. Let me try to figure this out step by step.First, I remember that when dealing with angles in standard position, the terminal side passing through a specific point means that we can use that point to find the trigonometric functions related to the angle α. The point given is (-3, -4), which is in the third quadrant since both x and y are negative.Now, I need to find cos(π/2 + α). Hmm, I think there's a trigonometric identity that can help simplify this expression. Let me recall... Ah, yes! The cosine of a sum formula: cos(A + B) = cos A cos B - sin A sin B. So, applying this to cos(π/2 + α), it should be cos(π/2)cos α - sin(π/2)sin α.I know that cos(π/2) is 0 and sin(π/2) is 1. So plugging those in, we get 0 * cos α - 1 * sin α, which simplifies to -sin α. So, cos(π/2 + α) is equal to -sin α. That simplifies things a bit.Now, I need to find sin α. Since the terminal side passes through (-3, -4), I can consider this point as (x, y) on the terminal side of angle α. To find sin α, I remember that sin α = y / r, where r is the distance from the origin to the point (x, y). So, I need to find r first.Calculating r using the Pythagorean theorem: r = sqrt(x² + y²). Plugging in the values, x is -3 and y is -4, so r = sqrt((-3)² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5. Okay, so r is 5.Now, sin α = y / r = (-4) / 5 = -4/5. So, sin α is -4/5. But remember earlier, we found that cos(π/2 + α) = -sin α. So, substituting sin α, we get cos(π/2 + α) = -(-4/5) = 4/5.Wait, let me double-check that. So, cos(π/2 + α) = -sin α. Since sin α is negative (-4/5), multiplying by -1 gives a positive 4/5. That makes sense because the angle π/2 + α would be in a different quadrant, but since we're dealing with cosine, which can be positive or negative depending on the quadrant.Just to make sure I didn't make any mistakes, let me go through the steps again. The point (-3, -4) is in the third quadrant. The radius r is 5, correct. Sin α is y/r, which is -4/5. Then, using the identity, cos(π/2 + α) is -sin α, which is -(-4/5) = 4/5. Yeah, that seems right.I think I got it. So, the value of cos(π/2 + α) is 4/5.

answer：Okay, so I have this problem about Ponchik eating pies. Let me try to understand it step by step. Ponchik starts eating pies on Monday, and every day he eats the same number of pies. One day, he notices that only a quarter of his daily portion is left from the initial stock of 340 pies. I need to figure out on which day of the week this happened.First, let's break down the information:- Initial number of pies: 340- He eats the same number of pies each day.- On a certain day, only a quarter of his daily portion remains.So, let me denote some variables to make this clearer. Let's say:- Let ( n ) be the number of days he has been eating pies.- Let ( s ) be the number of pies he eats each day.According to the problem, after ( n ) days, only a quarter of his daily portion ( s ) is left. That means the total number of pies eaten in ( n ) days plus the remaining quarter of ( s ) equals the initial stock of 340 pies.So, mathematically, this can be written as:[s times n + frac{s}{4} = 340]Let me simplify this equation. First, factor out ( s ):[s left( n + frac{1}{4} right) = 340]To make it easier, I can multiply both sides by 4 to eliminate the fraction:[4s left( n + frac{1}{4} right) = 4 times 340][4s times n + s = 1360][s(4n + 1) = 1360]So, now I have the equation ( s(4n + 1) = 1360 ). This means that ( s ) multiplied by ( (4n + 1) ) equals 1360. Since both ( s ) and ( n ) are integers (you can't eat a fraction of a pie or have a fraction of a day), I need to find integer values of ( s ) and ( n ) that satisfy this equation.Let me factorize 1360 to find possible values for ( s ) and ( 4n + 1 ):[1360 = 2^4 times 5 times 17]So, the factors of 1360 are combinations of these prime factors. Let's list the factors:1, 2, 4, 5, 8, 10, 16, 17, 20, 34, 40, 68, 85, 136, 170, 272, 340, 680, 1360Now, ( s ) and ( 4n + 1 ) must be factors of 1360. Also, since ( 4n + 1 ) must be one more than a multiple of 4, it should be an odd number because 4n is even, and adding 1 makes it odd. So, among the factors of 1360, the odd ones are 1, 5, 17, 85.Let me check each of these:1. If ( 4n + 1 = 1 ): [ 4n + 1 = 1 implies 4n = 0 implies n = 0 ] But ( n = 0 ) doesn't make sense because he started eating on Monday, so he must have eaten pies for at least one day. So, this is not possible.2. If ( 4n + 1 = 5 ): [ 4n + 1 = 5 implies 4n = 4 implies n = 1 ] Then, ( s = frac{1360}{5} = 272 ) So, he eats 272 pies a day, and after 1 day, he has a quarter left, which is 68 pies. But 272 pies in one day seems a lot, but mathematically, it's possible. Let's keep this as a possibility.3. If ( 4n + 1 = 17 ): [ 4n + 1 = 17 implies 4n = 16 implies n = 4 ] Then, ( s = frac{1360}{17} = 80 ) So, he eats 80 pies a day, and after 4 days, he has 20 pies left (which is a quarter of 80). This seems more reasonable.4. If ( 4n + 1 = 85 ): [ 4n + 1 = 85 implies 4n = 84 implies n = 21 ] Then, ( s = frac{1360}{85} = 16 ) So, he eats 16 pies a day, and after 21 days, he has 4 pies left (which is a quarter of 16). This is also possible, but 21 days is quite a long time.Now, we have three possible scenarios:- ( n = 1 ) day, ( s = 272 ) pies/day- ( n = 4 ) days, ( s = 80 ) pies/day- ( n = 21 ) days, ( s = 16 ) pies/dayBut the problem says "one day he discovered that only a quarter of his daily portion remained." It doesn't specify how many days have passed, so all these scenarios are possible mathematically. However, considering the context, it's more likely that it's not 21 days because that would mean he started eating 16 pies a day and took 21 days to get to a quarter left, which seems a bit too long for the problem's context. Similarly, eating 272 pies in one day seems excessive.Therefore, the most reasonable scenario is ( n = 4 ) days, where he eats 80 pies a day and after 4 days, he has 20 pies left.Now, since he started on Monday, let's count the days:- Day 1: Monday- Day 2: Tuesday- Day 3: Wednesday- Day 4: ThursdaySo, on the 4th day, which is Thursday, he discovers that only a quarter of his daily portion remains.Wait, but let me double-check. If he started on Monday and ate 80 pies each day, then:- Monday: 80 pies eaten, remaining: 340 - 80 = 260- Tuesday: 80 pies eaten, remaining: 260 - 80 = 180- Wednesday: 80 pies eaten, remaining: 180 - 80 = 100- Thursday: 80 pies eaten, remaining: 100 - 80 = 20Yes, on Thursday, he has 20 pies left, which is a quarter of his daily portion (80 pies). So, that makes sense.Alternatively, if we consider the 21-day scenario:Starting on Monday, 21 days later would be:- Monday + 21 days = Monday + 3 weeks = Monday again.But that would mean he discovered the remaining pies on a Monday, which is the same day he started. That seems a bit odd because he just started on Monday, and 21 days later, it's another Monday. But mathematically, it's possible.However, given the context, it's more likely that the discovery happened within a week, so Thursday seems more plausible.Therefore, the day of the week when he discovered that only a quarter of his daily portion remained is Thursday.

answer：Okay, so I have this problem here: Let ( r ) be a positive constant. If the line ( x + y = r ) is tangent to the circle ( x^2 + y^2 = 4r ), find the value of ( r ). The options are A) 2, B) 4, C) 8, D) 16, E) 32.Hmm, okay. I need to find the value of ( r ) such that the line ( x + y = r ) is tangent to the circle ( x^2 + y^2 = 4r ). First, let me recall what it means for a line to be tangent to a circle. If a line is tangent to a circle, it means that the line touches the circle at exactly one point. Another way to think about it is that the distance from the center of the circle to the line is equal to the radius of the circle. So, in this case, the circle is given by ( x^2 + y^2 = 4r ). I know that the general equation of a circle is ( x^2 + y^2 = a^2 ), where ( a ) is the radius. So, comparing, the radius of this circle is ( sqrt{4r} ), which simplifies to ( 2sqrt{r} ). The center of the circle is at the origin, (0,0), because there are no shifts in the x or y terms.Now, the line given is ( x + y = r ). I can rewrite this in the standard form for a line, which is ( Ax + By + C = 0 ). So, subtracting ( r ) from both sides, I get ( x + y - r = 0 ).To find the distance from the center of the circle (which is (0,0)) to this line, I can use the formula for the distance from a point to a line. The formula is:[ d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}} ]Where ( (x_0, y_0) ) is the point, and ( Ax + By + C = 0 ) is the line. In this case, the line is ( x + y - r = 0 ), so ( A = 1 ), ( B = 1 ), and ( C = -r ). The point is (0,0), so plugging into the formula:[ d = frac{|1*0 + 1*0 - r|}{sqrt{1^2 + 1^2}} = frac{| - r |}{sqrt{2}} = frac{r}{sqrt{2}} ]Since distance can't be negative, we take the absolute value, which is just ( r ) over ( sqrt{2} ).Now, since the line is tangent to the circle, this distance ( d ) must be equal to the radius of the circle. The radius we found earlier is ( 2sqrt{r} ). So, setting them equal:[ frac{r}{sqrt{2}} = 2sqrt{r} ]Hmm, okay, now I need to solve for ( r ). Let me square both sides to eliminate the square roots. First, let me write it again:[ frac{r}{sqrt{2}} = 2sqrt{r} ]Squaring both sides:[ left( frac{r}{sqrt{2}} right)^2 = (2sqrt{r})^2 ]Calculating both sides:Left side: ( frac{r^2}{2} )Right side: ( 4r )So now, the equation is:[ frac{r^2}{2} = 4r ]Multiply both sides by 2 to eliminate the denominator:[ r^2 = 8r ]Now, subtract 8r from both sides to set the equation to zero:[ r^2 - 8r = 0 ]Factor out an r:[ r(r - 8) = 0 ]So, the solutions are ( r = 0 ) or ( r = 8 ). But since the problem states that ( r ) is a positive constant, ( r = 0 ) doesn't make sense because the radius would be zero, which isn't a circle. So, the only valid solution is ( r = 8 ).Let me double-check my steps to make sure I didn't make a mistake. 1. I found the radius of the circle correctly as ( 2sqrt{r} ).2. I converted the line equation to standard form correctly.3. I used the distance formula from a point to a line correctly, resulting in ( frac{r}{sqrt{2}} ).4. I set the distance equal to the radius, which is correct for tangency.5. I squared both sides correctly, leading to ( frac{r^2}{2} = 4r ).6. I solved the quadratic equation correctly, resulting in ( r = 0 ) or ( r = 8 ), and chose ( r = 8 ) since it's positive.Everything seems to check out. So, the value of ( r ) is 8.**Final Answer**boxed{C}