🧠:Okay, let's try to figure out this card arrangement problem. So, there are three playing cards in a row, and we need to determine their order from left to right based on the given clues. Let me start by listing out all the information provided and then break it down step by step.First, the clues:1. The club is to the right of the heart and the diamond.2. The 5 is to the left of the heart.3. The 8 is to the right of the 4.Possible options are given from A to E. Let's analyze each clue one by one and see how they can help us eliminate options.Starting with the first clue: The club is to the right of the heart and the diamond. Hmm, so the club must be the rightmost card because it's to the right of both heart and diamond. Since there are three cards, the club can't be in the first or second position; it has to be in the third position. That means the third card is clubs. Let me check the options:A: 4♥, 5♦, 8♣ – third is clubs, okay.B: 5♦, 4♥, 8♣ – third is clubs, good.C: 8♣, 4♥, 5♦ – third is diamonds, so club is first, which violates the first clue. So C is out.D: 4♦, 5♣, 8♥ – third is hearts, so club is second, which is to the right of diamond but not heart (since heart is third). Wait, the club is to the right of both heart and diamond. In D, the club is in the middle. The heart is third, so club (second) is to the left of heart. That violates the first clue. So D is invalid.E: 5♥, 4♦, 8♣ – third is clubs, okay.So from the first clue, options C and D are eliminated. Now remaining options are A, B, E.Second clue: The 5 is to the left of the heart. So wherever the heart is, the 5 must be somewhere to the left of it. Let's check each remaining option.Option A: 4♥, 5♦, 8♣. Here, the heart is the first card (4 of hearts). The 5 is in the second position (5♦). But the 5 is to the right of the heart here, which violates the clue that 5 must be to the left of the heart. So A is invalid.Option B: 5♦, 4♥, 8♣. The heart is in the second position (4♥). The 5 is in the first position (5♦), which is to the left of the heart. That's okay.Option E: 5♥, 4♦, 8♣. Here, the heart is in the first position (5♥). The 5 is part of the heart card. The clue says the 5 must be to the left of the heart. Wait, but in this case, the 5 and the heart are the same card. So does that count as the 5 being to the left of the heart? Or is the 5 and the heart together, so the 5 isn't to the left but on the same card. Hmm, the clue says "the 5 is to the left of the heart". So if the heart is the 5 of hearts, then the 5 is part of the heart card. Therefore, the 5 isn't to the left of the heart; it's the same card. So that would not satisfy the clue. So E is invalid because the 5 is not to the left but on the heart itself.Therefore, after the second clue, options A and E are eliminated, leaving only option B.But let's confirm with the third clue to be sure.Third clue: The 8 is to the right of the 4. So wherever the 4 is, the 8 must be somewhere to the right.Looking at option B: 5♦, 4♥, 8♣. The 4 is in the second position (4♥), and the 8 is in the third position (8♣). So 8 is to the right of 4. That satisfies the third clue.Option B satisfies all three clues. Let's check once more.1. Club is third (8♣), which is to the right of heart (second position, 4♥) and diamond (first position, 5♦). So yes, club is rightmost, so that's good.2. 5♦ is first, heart is second. So 5 is left of heart. Good.3. 4♥ is second, 8♣ is third. So 8 is right of 4. Good.Therefore, the correct answer should be B.Wait, but let me double-check the options again. The possible answer B is 5 of diamonds, 4 of hearts, 8 of clubs. Yes, that's correct. The club is in the correct position, 5 is left of heart, 8 is right of 4. All clues satisfied.

✅:The correct arrangement of the cards from left to right is determined by analyzing the clues step-by-step:1. Clue 1: The club is to the right of the heart and diamond. - This means the club must be in the third position. - Eliminates C (club first) and D (club second, heart third). - Remaining options: A, B, E.2. Clue 2: The 5 is to the left of the heart. - The 5 cannot share the same card as the heart. - A invalid: 5 is to the right of the heart (4♥). - E invalid: 5 and heart are on the same card (5♥). - Only B remains: 5♦ (left) of 4♥ (heart). 3. Clue 3: The 8 is to the right of the 4. - B satisfies this: 4♥ (second) and 8♣ (third). Final Answer: (B) 5 of diamonds, 4 of hearts, 8 of clubs Step-by-Step Verification: - Clubs (♣) are third, satisfying Clue 1. - 5♦ is left of 4♥, satisfying Clue 2. - 8♣ is right of 4♥, satisfying Clue 3. All constraints are met in option B.

🔑:To determine the correct order of the cards, we should first list the clues provided in the problem and discuss each one step-by-step.1. Clue 1: "The club is to the right of the heart and the diamond."2. Clue 2: "The 5 is to the left of the heart."3. Clue 3: "The 8 is to the right of the 4 ." Step-by-Step Solution# Step 1: Analyze Clue 1First, Clue 1 states:- The club is to the right of both the heart and the diamond.- Possible orderings for this condition are: - Heart, Diamond, Club. - Diamond, Heart, Club.So, our remaining valid card position combinations are:- (Heart, Diamond, Club)- (Diamond, Heart, Club)# Step 2: Analyze Clue 2Next, Clue 2 states:- The 5 is to the left of the heart.- This tells us that the 5 card cannot be the heart.Since the heart can't be a five, we must pair the diamond with the five. This narrows our order to:- 5 of diamonds, Heart, Club.# Step 3: Analyze Clue 3Finally, Clue 3 states:- The 8 is to the right of the 4.- This tells us the heart must be 4 (since it is still in the middle position and there is an 8 card).Thus, the remaining cards must be:- 5 of diamonds, 4 of hearts, 8 of clubs. ConclusionMatching these details to the given choices, we get:- (B) 5 of diamonds, 4 of hearts, 8 of clubs.Thus, the correct order is from left to right:[ ( text{5 of diamonds}, text{4 of hearts}, text{8 of clubs} ) ]Therefore, the correct option is (boxed{B})

🧠:Okay, let's tackle this problem step by step. First, I need to understand what's being asked. We have two people, Sasha and Dania, walking towards a common intersection point C along two different roads, A and B. After passing C, they continue walking. We need to find the minimum possible distance between them at moments when both have taken an integer number of steps. Also, determine how many steps each has taken by that time.Alright, let's start by gathering all the given data:Sasha:- Walks along road A towards C.- Speed: 45 steps per minute.- Step length: 60 cm.- Initial distance from C: 290 meters.Dania:- Walks along road B towards C.- Speed: 55 steps per minute.- Step length: 65 cm.- Initial distance from C: 310 meters.Both continue walking past C without stopping. We need to record times when both have taken whole numbers of steps and find the minimum distance between them at those times. The distance is along the roads, which I assume means the sum of their distances from C along their respective roads. Wait, but roads A and B are straight lines intersecting at C. So the actual minimum distance between them in space would be a straight line, but the problem mentions "distance along the roads". Hmm, maybe that's the sum of their distances from C along their respective roads? Or is it the Euclidean distance? Wait, the problem says "minimum possible distance between them (along the roads)". So if they are on different roads, the distance along the roads would be the sum of their distances from C? Because to go from one to the other along the roads, you have to go through C. So if Sasha is x meters past C on road A, and Dania is y meters past C on road B, then the distance along the roads would be x + y. But if they haven't reached C yet, then it's the sum of their remaining distances? Wait, but they can be on either side of C. Let me think.Wait, the roads A and B intersect at C. So if Sasha is approaching C along road A and Dania is approaching C along road B, then before reaching C, their positions are on different roads approaching C. Once they pass C, Sasha continues along road A away from C, and Dania continues along road B away from C. So, depending on the time, they can be on the same side of C (both approaching) or opposite sides (one approaching, one moving away). But the distance along the roads would be the sum of their distances from C, regardless of direction. Wait, but if they are on opposite sides of C, the path along the roads would require going from one person to C and then to the other person, so yes, the total distance would be the sum of their absolute distances from C. So whether they are approaching or moving away, the distance along the roads between them is the sum of their individual distances from C.Therefore, we need to model their positions over time as functions of time, compute the sum of their distances from C at each time when both have taken an integer number of steps, and find the minimum such sum.But first, let's convert all units to be consistent. The step lengths are given in cm, and distances from C in meters. Let's convert everything to meters.Sasha's step length: 60 cm = 0.6 meters.Dania's step length: 65 cm = 0.65 meters.Their speeds are given in steps per minute. Let's convert their speeds to meters per minute.Sasha's speed: 45 steps/min * 0.6 m/step = 27 m/min.Dania's speed: 55 steps/min * 0.65 m/step = 55 * 0.65 = let's compute that: 50*0.65=32.5, 5*0.65=3.25, total 32.5 + 3.25 = 35.75 m/min.Now, initial distances from C:Sasha starts 290 meters from C, moving towards C at 27 m/min.Dania starts 310 meters from C, moving towards C at 35.75 m/min.We need to model their positions as functions of time. Let's define time t in minutes.For Sasha:Position at time t: 290 - 27t meters from C. When t is such that 290 - 27t >= 0, she is approaching C. Once 27t > 290, she has passed C and her position becomes 27t - 290 meters past C.Similarly for Dania:Position at time t: 310 - 35.75t meters from C. When 310 - 35.75t >=0, she is approaching C. After that, she is 35.75t - 310 meters past C.But since we need to consider times when both have taken an integer number of steps, we need to model the times when Sasha has taken n steps and Dania has taken m steps, where n and m are integers. The steps happen at discrete intervals.Wait, the problem says: "We record the moments in time when both Dania and Sasha have taken whole numbers of steps." So these are discrete moments when both have completed an integer number of steps. So for Sasha, each step takes a certain amount of time, and similarly for Dania. The moments when both have taken whole numbers of steps would be the least common multiples of their step intervals? Wait, no. Each of them takes steps at their own rate, so the times when both have taken integer steps are the times that are integer multiples of their individual step durations.Wait, let's think about step timing. Sasha takes 45 steps per minute, so each step takes 1/45 minutes. Similarly, Dania takes 55 steps per minute, each step takes 1/55 minutes. So Sasha's steps occur at t = k*(1/45) minutes, where k is a non-negative integer. Similarly, Dania's steps occur at t = m*(1/55) minutes, where m is a non-negative integer. The times when both have taken whole numbers of steps are the common multiples of 1/45 and 1/55 minutes. So the times t must satisfy t = k/45 = m/55 for some integers k and m. Therefore, t must be a multiple of the least common multiple (LCM) of 1/45 and 1/55. To find LCM of 1/45 and 1/55, we can think of LCM of fractions as LCM(numerators)/GCD(denominators). But here, it's LCM of periods. The period for Sasha is 1/45 min per step, for Dania 1/55 min per step. The LCM of these periods would be the smallest t such that t is a multiple of both 1/45 and 1/55. That's LCM(1/45, 1/55) = LCM(1,1)/GCD(45,55) = 1/5. Because GCD(45,55)=5. So LCM(1/45,1/55)=1/5. Therefore, every 1/5 minutes, both have taken an integer number of steps. Wait, let's check that.If LCM(1/45, 1/55) = 1/5, then 1/5 divided by 1/45 is 9, which is integer, and 1/5 divided by 1/55 is 11, which is integer. So yes, every 1/5 minutes, both have taken 9 steps (for Sasha: 1/5 min / (1/45 min/step) = 9 steps) and 11 steps for Dania (1/5 min / (1/55 min/step) = 11 steps). Therefore, the common times when both have taken integer steps are t = k*(1/5) minutes, where k is a positive integer. However, this is only true if their stepping is synchronized such that their steps occur at these common times. Wait, but LCM of the step intervals is 1/5 minutes. So the first time they both have integer steps is at 1/5 minutes, then 2/5, 3/5, etc. But is that accurate?Wait, let's see. For Sasha, each step takes 1/45 minutes. So she takes steps at t=1/45, 2/45, 3/45, ..., which simplifies to t=1/45, 2/45, 1/15, 4/45, etc. For Dania, steps at t=1/55, 2/55, 3/55, ..., etc. The common times when both have stepped an integer number of times would be the intersection of these two sequences. The LCM of 1/45 and 1/55 is indeed 1/5, since LCM(1/45,1/55) = LCM(1,1)/GCD(45,55) = 1/5. So every 1/5 minutes, both have taken 9 and 11 steps respectively. Therefore, the times we need to consider are t = 1/5, 2/5, 3/5, ..., minutes. Each time t = k/5 minutes, where k is a positive integer.Therefore, to find the minimum distance between them at such times, we can model their positions at each t = k/5 minutes, compute the sum of their distances from C (since distance along the roads is the sum), and find the minimum sum.But we need to check when they pass C. Let's compute the time when Sasha reaches C and when Dania reaches C.For Sasha:She starts at 290 m from C, moving towards C at 27 m/min. Time to reach C: 290 / 27 ≈ 10.7407 minutes.For Dania:She starts at 310 m from C, moving towards C at 35.75 m/min. Time to reach C: 310 / 35.75 ≈ 8.6713 minutes.So Sasha reaches C after approximately 10.74 minutes, Dania reaches C after approximately 8.67 minutes.Therefore, in the time interval from t=0 to t≈8.67, both are approaching C. From t≈8.67 to t≈10.74, Dania has passed C and is moving away, while Sasha is still approaching. After t≈10.74, both have passed C and are moving away.Our common times t = k/5 minutes. Let's compute k such that k/5 covers up to when both have passed. Let's see, up to, say, 15 minutes? Probably not necessary, but we need to check when the minimal distance occurs. Likely before both pass C, but we need to check.But let's formalize their positions.At time t minutes:- Sasha's distance from C: If t ≤ 290/27 ≈10.7407, then 290 - 27t meters. If t >10.7407, then 27t - 290 meters.- Dania's distance from C: If t ≤310/35.75 ≈8.6713, then 310 -35.75t meters. If t >8.6713, then 35.75t -310 meters.But we need to evaluate these positions at times t = k/5, where k is integer starting from 0.Therefore, for each k, compute t = k/5, compute Sasha's and Dania's positions, take the absolute distance from C (since after passing, it's positive in the other direction), sum them, and find the minimum sum.So steps:1. For k from 0 upwards, compute t = k/5.2. For each t, compute Sasha's position: If t ≤290/27, distance_from_C_Sasha = 290 -27t Else, distance_from_C_Sasha = 27t -290Similarly for Dania: If t ≤310/35.75, distance_from_C_Dania =310 -35.75t Else, distance_from_C_Dania=35.75t -310Total distance = |distance_from_C_Sasha| + |distance_from_C_Dania|. Wait, but since we are considering distance along the roads, and they are moving along straight lines intersecting at C, the distance between them along the roads is the sum of their distances from C if they are on opposite sides, or the absolute difference if they are on the same side. Wait, no. Wait, if both are approaching C, then the distance between them along the roads would be (distance_Sasha + distance_Dania). If one has passed C and the other hasn't, then it's (distance_Sasha_past + distance_Dania_remaining). If both have passed C, then depending on the roads' configuration. Wait, but roads A and B intersect at C. If they continue past C, they are moving away from C on their respective roads. So the distance along the roads between them is the sum of their distances from C, regardless of direction. Because to go from Sasha to Dania along the roads, you have to go from Sasha to C and then to Dania. So whether they are on the same side or opposite sides of C, the total distance is the sum of their distances from C. For example, if Sasha is 50m past C on road A, and Dania is 30m before C on road B, then the distance along the roads is 50m + 30m =80m. If both are past C, say Sasha is 50m past and Dania is 30m past, then it's 50m +30m=80m (along roads A and B through C). If both are approaching, say Sasha is 100m away and Dania is 200m away, then distance is 300m. So yes, in all cases, the distance along the roads is the sum of their absolute distances from C.Therefore, total distance at time t is |290 -27t| + |310 -35.75t| for t ≤8.6713 (both approaching). After Dania passes C but before Sasha does (t between 8.6713 and 10.7407), it's |290 -27t| + |35.75t -310|. After both pass C (t >10.7407), it's |27t -290| + |35.75t -310|.But since we are evaluating at discrete times t =k/5, we can compute the total distance as the sum of their distances from C, considering their positions (approaching or passed).Our goal is to compute this total distance at each t =k/5 and find the minimum.Given that k is an integer starting from 0, let's compute t in increments of 0.2 minutes (since 1/5 min =0.2 min). Let's convert the critical times to fractions over 1/5.First, find when Dania passes C: t=310/35.75 ≈8.6713 min. Let's see what k/5 is just after 8.6713. 8.6713 /0.2=43.3565. So k=44 would be t=44/5=8.8 min. Similarly, Sasha passes C at 290/27≈10.7407 min. 10.7407 /0.2=53.7035, so k=54 would be t=54/5=10.8 min.Therefore, the key intervals:1. For k from 0 to 43 (t=0 to t=8.6), both are approaching C.2. For k=44 to 53 (t=8.8 to10.6), Dania has passed C, Sasha hasn't.3. For k=54 onwards (t=10.8), both have passed C.We need to compute total distance for k in these intervals.But since we need the minimal distance, it's likely that the minimal occurs near when one or both pass C. Let's check.First, let's compute t in each interval:First interval: k=0 to43, t=0, 0.2, 0.4,...,8.6.In this interval, total distance is (290 -27t)+(310 -35.75t)=600 -62.75t.This is a linear function decreasing with t. So the minimal in this interval would be at the end, t=8.6. Let's verify:At k=43, t=43/5=8.6 min.Distance: 600 -62.75*8.6.Compute 62.75*8.6:62.75*8=502, 62.75*0.6=37.65, total=502+37.65=539.65.600 -539.65=60.35 meters.Wait, but when t=8.6, Dania's position is 310 -35.75*8.6. Let's check if Dania has passed C yet.35.75*8.6=35.75*8 +35.75*0.6=286 +21.45=307.45 meters.310 -307.45=2.55 meters. So Dania is still approaching C, 2.55 meters away. Therefore, total distance is (290 -27*8.6)+(310 -35.75*8.6).Compute Sasha's distance: 290 -27*8.6=290 -27*8 -27*0.6=290 -216 -16.2=290-232.2=57.8 meters.Dania's distance:310 -35.75*8.6=310 -307.45=2.55 meters.Total distance:57.8 +2.55=60.35 meters.Next, k=44, t=8.8 min.At t=8.8, Dania's position:35.75*8.8=35.75*8 +35.75*0.8=286 +28.6=314.6 meters. So Dania has passed C by 314.6 -310=4.6 meters.Sasha's position:290 -27*8.8=290 -237.6=52.4 meters from C.Total distance:52.4 +4.6=57 meters.Similarly, k=45, t=9.0 min.Dania:35.75*9.0=321.75 meters. Passed C by 11.75 meters.Sasha:290 -27*9.0=290 -243=47 meters.Total distance:47 +11.75=58.75 meters.Wait, so the distance decreased from 60.35 to57 at k=44, then increased to58.75 at k=45. Interesting.Next, k=46, t=9.2 min.Dania:35.75*9.2=35.75*(9 +0.2)=321.75 +7.15=328.9 meters. Passed C by 328.9 -310=18.9 meters.Sasha:290 -27*9.2=290 -248.4=41.6 meters.Total distance:41.6 +18.9=60.5 meters.So even higher. So between k=44 and k=45, the distance increases. Wait, but k=44 gives t=8.8 min with distance 57m, k=45 gives 58.75m, which is higher. Then k=46 gives 60.5m. So the minimal in this interval is at k=44 (t=8.8 min) with distance 57m.But wait, we need to check if there's a lower distance after Sasha passes C. Let's move to the next interval.Third interval: k=54, t=10.8 min.Both have passed C.Sasha's distance:27*10.8 -290=291.6 -290=1.6 meters past C.Dania's distance:35.75*10.8 -310=386.1 -310=76.1 meters past C.Total distance:1.6 +76.1=77.7 meters.Which is higher than previous minima.Similarly, k=55, t=11.0 min.Sasha:27*11 -290=297 -290=7m.Dania:35.75*11 -310=393.25 -310=83.25m.Total:7+83.25=90.25m. Increasing.So after both pass C, the distance keeps increasing. Hence, the minimal distance is likely in the second interval (k=44 to53), but we saw that at k=44, distance is57m, then increases. But wait, maybe there's a point where Sasha is approaching C and Dania is moving away, and their combined distance is lower. Let's check more points in the second interval.For example, k=44, t=8.8 min: distance 57m.k=45, t=9.0:58.75m.k=46, t=9.2:60.5m.k=47, t=9.4 min.Dania:35.75*9.4=35.75*9 +35.75*0.4=321.75 +14.3=336.05m. Passed by 26.05m.Sasha:290 -27*9.4=290 -253.8=36.2m.Total:36.2 +26.05=62.25m.k=48, t=9.6 min.Dania:35.75*9.6=343.2m. Passed by 33.2m.Sasha:290 -27*9.6=290 -259.2=30.8m.Total:30.8 +33.2=64m.k=49, t=9.8 min.Dania:35.75*9.8=35.75*10 -35.75*0.2=357.5 -7.15=350.35m. Passed by40.35m.Sasha:290 -27*9.8=290 -264.6=25.4m.Total:25.4 +40.35=65.75m.k=50, t=10.0 min.Dania:35.75*10=357.5m. Passed by47.5m.Sasha:290 -270=20m.Total:20 +47.5=67.5m.k=51, t=10.2 min.Dania:35.75*10.2=35.75*10 +35.75*0.2=357.5 +7.15=364.65m. Passed by54.65m.Sasha:290 -27*10.2=290 -275.4=14.6m.Total:14.6 +54.65=69.25m.k=52, t=10.4 min.Dania:35.75*10.4=35.75*10 +35.75*0.4=357.5 +14.3=371.8m. Passed by61.8m.Sasha:290 -27*10.4=290 -280.8=9.2m.Total:9.2 +61.8=70.0m.k=53, t=10.6 min.Dania:35.75*10.6=35.75*10 +35.75*0.6=357.5 +21.45=378.95m. Passed by68.95m.Sasha:290 -27*10.6=290 -286.2=3.8m.Total:3.8 +68.95=72.75m.k=54, t=10.8 min: as before,77.7m.So in the second interval (k=44 to53), the minimal distance is at k=44 (t=8.8 min) with57m. After that, the distance increases.But wait, let's check between k=43 and k=44. At k=43, t=8.6 min, both are approaching. Dania is 2.55m from C, Sasha is57.8m from C. Total distance60.35m. Then at k=44, Dania has passed C by4.6m, Sasha is52.4m from C. Total distance57m. So the distance decreased by3.35m. Then at k=45, it starts increasing.So the minimal in this interval is at k=44.But wait, is there a lower distance before both pass C? In the first interval, the total distance decreases linearly as t increases. The minimal would be at the last point before Dania passes C, which is k=43 (t=8.6 min) with total distance60.35m. But when Dania passes C at t≈8.6713, which is between k=43 (8.6) and k=44 (8.8). However, we can only consider times at k/5. So the minimal possible distance in the first interval is60.35m, and in the second interval it's57m. Then we need to check if there's a lower distance in the second interval. But according to the calculations, k=44 gives57m, which is lower than the first interval.Now, is there a k in the second interval where the total distance is less than57m? From the calculations above, no. After k=44, the distance increases. So the minimal is57m at k=44.But wait, let's check at t=8.6713, which is when Dania passes C. That time is not exactly at a k/5. The closest k would be k=43 (8.6) and k=44 (8.8). At t=8.6713, Dania is exactly at C. If we could take that exact time, the total distance would be Sasha's distance plus Dania's distance (which is0). But since we can only take times when both have taken integer steps, which are at k/5 minutes, we cannot take t=8.6713. The closest is k=44, t=8.8, where Dania is4.6m past C, and Sasha is52.4m from C. Total57m.But maybe there is a smaller distance at a different k? Let's check k=44 and see if that's indeed the minimum.Alternatively, maybe when Sasha passes C. Let's see.Sasha passes C at t≈10.7407 min. The closest k is k=54 (t=10.8). At that time, Sasha is1.6m past C, Dania is76.1m past C. Total77.7m.No, that's higher.Alternatively, between k=44 and k=54, but in the second interval, the minimal is at k=44.But wait, let's check another possibility. Suppose there's a time after Dania passes C but before Sasha does, where their paths cross such that the distance is minimized. But since they are on different roads, the minimal Euclidean distance might be when they are closest in space, but the problem asks for the distance along the roads, which is the sum of their distances from C. Therefore, the minimal sum would occur when one is as close as possible to C and the other is also as close, but considering the discrete times.But according to the previous calculations, the minimal sum is57m at k=44.However, let's verify if there's a lower sum elsewhere. For example, check k=44:t=8.8 min.Sasha's steps:45 steps/min *8.8 min=396 steps. But wait, steps must be integer. At t=8.8 min, which is k=44, t=44/5=8.8 min.Sasha's number of steps:45 steps/min *8.8 min=396 steps. Yes, integer.Dania's number of steps:55 steps/min *8.8 min=484 steps. Yes, integer.So both have taken integer steps at t=8.8 min.Total distance:52.4 +4.6=57m.Is this the minimal? Let's check a few more points around k=44.For example, k=44-1=43: t=8.6 min.Sasha steps:45*8.6=387 steps. Yes, integer.Dania steps:55*8.6=473 steps. Yes, integer.Total distance60.35m.k=44+1=45: t=9.0 min.Steps:45*9=405,55*9=495. Integer.Total distance58.75m.Wait, but earlier I thought k=45 gives58.75m, which is higher than57m.But wait, no: 58.75m is higher than57m. So the minimal is at k=44.Wait, but maybe there's another point where the sum is lower. Let's check k=44 gives57m. Let's check k=44+1=45 gives58.75m. So increasing.k=44-1=43 gives60.35m. So indeed, the minimal is at k=44.But just to be thorough, let's check k=44 in more detail.At k=44:t=44/5=8.8 min.Sasha's distance:290 -27*8.8=290 -237.6=52.4m.Dania's distance:35.75*8.8 -310=314.6 -310=4.6m.Total:52.4+4.6=57m.Yes. So this seems to be the minimal.But wait, is there a time before k=44 where the total distance is less than57m? For example, in the first interval, the total distance decreases linearly as t increases. At k=43 (t=8.6), total distance60.35m. At k=44, it drops to57m. So between t=8.6 and t=8.8, the total distance decreases from60.35 to57m. Since these are the only valid times (at k/5), the minimal is at k=44.But let's think differently: perhaps there is a time when Sasha and Dania are closest in terms of Euclidean distance, but the problem asks for the distance along the roads. However, the problem specifies to consider the distance along the roads, so our previous approach is correct.Therefore, the minimum possible distance is57 meters, achieved at t=8.8 minutes, when Sasha has taken45*8.8=396 steps and Dania has taken55*8.8=484 steps.But wait, 45 steps per minute *8.8 minutes=396 steps. Yes. Similarly,55*8.8=484 steps. Both integers. Therefore, this is valid.But let me confirm the step counts:Sasha's steps:45 steps/min *8.8 min=45*(8 +0.8)=360 +36=396 steps.Dania's steps:55 steps/min *8.8 min=55*(8 +0.8)=440 +44=484 steps.Yes, both are integers.Therefore, the answer is the minimum distance is57 meters, achieved when Sasha has taken396 steps and Dania has taken484 steps.But wait, the problem says "Find the minimum possible distance between them (along the roads) at such moments in time. Determine the number of steps each of them has taken by the time this minimum distance is achieved."Therefore, the answer is:Minimum distance:57 meters.Steps for Sasha:396.Steps for Dania:484.But let's double-check calculations.First, check positions at t=8.8 min:Sasha's position:290 -27*8.8=290 -237.6=52.4 meters from C.Dania's position:35.75*8.8=314.6 meters from start, which is314.6 -310=4.6 meters past C.Total distance:52.4 +4.6=57 meters. Correct.Steps:Sasha:45*8.8=396.Dania:55*8.8=484.Yes.But wait, let's confirm that at t=8.8, Dania has indeed taken484 steps. 484 steps at55 steps/min would take484/55=8.8 minutes. Yes.Similarly, Sasha's396 steps at45 steps/min:396/45=8.8 minutes. Correct.Therefore, this is the correct answer.However, just to ensure completeness, let's check if there's any other k where the distance is less than57m.For example, k=44 is8.8 min, distance57m.Next, k=45 is9.0 min, distance58.75m.What about k=44.5? No, k must be integer, so t must be multiples of0.2. Therefore, the next possible is k=44 and45.Therefore, no other points between.Hence, the minimum is indeed57 meters at t=8.8 min, with396 and484 steps.

✅:The minimum possible distance between Sasha and Dania along the roads at moments when both have taken whole numbers of steps is 57 meters. This occurs when Sasha has taken 396 steps and Dania has taken 484 steps.Step-by-Step Explanation:1. Convert Units and Calculate Speeds: - Sasha's speed: (45 , text{steps/min} times 0.6 , text{m/step} = 27 , text{m/min}). - Dania's speed: (55 , text{steps/min} times 0.65 , text{m/step} = 35.75 , text{m/min}).2. Determine Critical Times: - Sasha reaches (C) at (290 / 27 approx 10.74) minutes. - Dania reaches (C) at (310 / 35.75 approx 8.67) minutes.3. Identify Common Step Times: - The LCM of their step intervals ((1/45) and (1/55) minutes) is (1/5) minutes. Thus, both take integer steps at (t = k/5) minutes for integer (k).4. Evaluate Positions at Each (t = k/5): - Calculate distances from (C) for Sasha and Dania, considering their positions (approaching or passed (C)). - Sum distances to find the total along-the-road distance.5. Find Minimum Distance: - At (k = 44) ((t = 8.8) minutes): - Sasha: (290 - 27 times 8.8 = 52.4 , text{m}) from (C). - Dania: (35.75 times 8.8 - 310 = 4.6 , text{m}) past (C). - Total Distance: (52.4 + 4.6 = 57 , text{meters}).6. Verify Steps Taken: - Sasha: (45 , text{steps/min} times 8.8 , text{min} = 396 , text{steps}). - Dania: (55 , text{steps/min} times 8.8 , text{min} = 484 , text{steps}).Answer:The minimum distance is (boxed{57}) meters, achieved when Sasha has taken (boxed{396}) steps and Dania has taken (boxed{484}) steps.

🔑:We need to find the moment when Sasha and Danya covered whole steps such that the distance between them is minimized. We'll start with the given information and calculate their velocities and distances.1. Calculate the speed of Sasha in cm/min: Sasha makes 45 steps per minute, and each step is 60 cm. [ v_{Sasha} = 45 times 60 = 2700 text{ cm/min} ]2. Calculate the speed of Danya in cm/min: Danya makes 55 steps per minute, and each step is 65 cm. [ v_{Danya} = 55 times 65 = 3575 text{ cm/min} ]3. Convert initial distances from meters to centimeters: Sasha starts at 290 meters, Danya at 310 meters. [ d_{initial, Sasha} = 290 times 100 = 29000 text{ cm} ] [ d_{initial, Danya} = 310 times 100 = 31000 text{ cm} ]4. Position of Sasha and Danya at any time t (in minutes): If ( t ) minutes pass, Sasha and Danya will have traveled: [ d_{Sasha}(t) = 29000 - 2700t ] [ d_{Danya}(t) = 31000 - 3575t ]5. Finding integer steps for both Sasha and Danya: At any time ( t ), the steps taken by Sasha: [ text{Steps}_{Sasha} = P_{Sasha}(t) = 45t ] And by Danya: [ text{Steps}_{Danya} = P_{Danya}(t) = 55t ] For both Sasha and Danya to complete integer steps simultaneously, ( t ) must solve the condition where both step counts are integers.6. Finding the smallest distance: To find the minimum distance, define the total walking distances: [ text{Distance}_{Sasha} = 2700t ] [ text{Distance}_{Danya} = 3575t ] Updating our positions: [ text{Distance}_{Sasha} = 29000 - 2700t ] [ text{Distance}_{Danya} = 31000 - 3575t ] 7. Distance between Sasha and Danya: The distance between them at any time ( t ) is: [ d(t) = |29000 - 2700t - (31000 - 3575t)| ] Simplify: [ d(t) = |29000 - 31000 + (3575 - 2700)t| ] [ d(t) = |-2000 + 875t| ] We check where this distance is minimized, for the specific ( t ): [ |-2000 + 875t| = min ] Solve when the argument inside the modulus becomes zero: [ -2000 + 875t = 0 ] [ t = frac{2000}{875} approx 2.2857 text{ minutes} ]8. Check step counts: Sasha: [ text{Steps}_{Sasha} = 45 times 2.2857 approx 103 text{ steps} ] Danya: [ text{Steps}_{Danya} = 55 times 2.2857 approx 126 text{ steps} ]9. Distance check: Convert t to whole steps based on our periodic motion, find (d(t)) in smallest unit integer steps similar to common ( text{LCM} ) periodic motion, our calculation reveals exact minimum over rational transformation:Conclusion:[boxed{d_{min} = 57 text{ meters}; text{Steps}_{Sasha} = 396; text{Steps}_{Danya} = 484}]

🧠:Okay, so I need to find the smallest natural number greater than 1 such that it's at least 600 times greater than each of its prime divisors. Let me try to understand the problem first. The number we're looking for, let's call it N, must satisfy the condition that for every prime divisor p of N, N >= 600*p. So, each prime factor p of N must be <= N/600. That means all the prime factors of N have to be relatively small compared to N itself. The goal is to find the smallest such N. Let me start by breaking down the problem. Since N has to be at least 600 times each of its prime factors, the primes can't be too large. For example, if N had a prime factor p, then p <= N/600. So, if we consider the smallest possible N, we need to find a number whose prime factors are all less than or equal to N/600. This seems a bit circular because N depends on its prime factors, which in turn depend on N. Maybe I can approach this by considering possible prime factors and constructing N from them. Let's start with the smallest primes and see what N would look like.First, let's consider if N is a prime power. If N is a prime number, say p, then the only prime divisor is p itself. So, we need p >= 600*p, which is impossible unless p=0, but primes are greater than 1. So N cannot be prime. Hence, N must be composite. Next, consider N being a product of two primes. Let's say N = p*q. Then both p and q must satisfy N >= 600*p and N >= 600*q. So, p*q >= 600*p implies q >= 600, and similarly p*q >= 600*q implies p >= 600. Therefore, both p and q must be at least 600. But the smallest primes above 600 are 601 (which is prime), 607, 613, etc. So the smallest N in this case would be 601*607. Let me calculate that: 601*607. Let's see, 600*600=360,000, and then 600*7 + 1*600 + 1*7 = 4200 + 600 +7=4807. So 360,000 + 4,807 = 364,807. That's quite large. But maybe there's a smaller N with more prime factors?Alternatively, if N has more prime factors, perhaps smaller primes can be used. For example, if N is a product of small primes, each raised to some power, maybe we can get N to be smaller. Let's think about that. For instance, if N is a power of 2. Let's say N=2^k. Then 2^k must be >= 600*2=1200. So 2^k >=1200. The smallest k such that 2^k >=1200. 2^10=1024, 2^11=2048. So N=2048. Then 2048/2=1024, which is greater than 600. So 2048 is a candidate. But wait, 2048 is 2^11, so its only prime divisor is 2. Since 2048 /2 =1024 >=600, it satisfies the condition. But is 2048 the smallest such number? Let's check smaller exponents. 2^10=1024. 1024/2=512, which is less than 600. So 1024 does not satisfy. So 2048 is the first power of 2 that satisfies the condition. But maybe there's a composite number with multiple small primes that is smaller than 2048. Let's check. For example, if N is 2^a *3^b. Let's see. For N to satisfy N >=600*2=1200 and N >=600*3=1800. So N must be at least 1800. Let's see if 1800 works. The prime factors of 1800 are 2,3,5. Then 1800/2=900 >=600, 1800/3=600 >=600, 1800/5=360 <600. So 360 is less than 600, so 5 is a prime divisor where N/5=360 <600. Therefore, 1800 doesn't work. So we need N such that N/5 >=600 => N >=3000. Hmm. So if 5 is a prime factor, N must be at least 3000. But if we can avoid using 5, maybe we can get a smaller N. Alternatively, if we use 2 and 3 as prime factors. Let's see. Let's take N=2^a *3^b. We need N >=600*2=1200 and N >=600*3=1800. So N must be at least 1800. Let's check 1800. As before, but 1800 has prime factor 5, which causes a problem. Wait, 1800=2^3*3^2*5^2. So to have N with only 2 and 3, let's take N=2^a*3^b where a and b are such that 2^a*3^b >=1800. Let's compute the minimal such N. Let's find the smallest a and b where 2^a*3^b >=1800.Let me try a=10, b=2: 2^10=1024, 3^2=9, 1024*9=9216, which is way larger. Maybe a=8, b=4: 256*81=20736, still too big. Maybe smaller exponents. Let's start from 2^a*3^b >=1800.Compute 2^a *3^b:Start with a=10, 2^10=1024. 1024*3=3072, which is over 1800. So 2^10*3=3072. Alternatively, 2^9=512, 512*4=2048, but 4 is 2^2. Wait, but if we use 3's. 2^9*3=512*3=1536, which is less than 1800. 2^9*3^2=512*9=4608. That's way over. Maybe 2^8*3^3=256*27=6912. Hmm, still high. Wait, maybe there's a combination in between. Let's compute step by step.Let's see:Find the minimal a and b where 2^a*3^b >=1800.Let's fix a and find minimal b:For a=10 (1024): 1024*3^b >=1800. 3^b >=1800/1024 ≈1.7578. So b=1: 3^1=3, 1024*3=3072.For a=9 (512): 512*3^b >=1800. 3^b >=1800/512≈3.5156. So b=2: 3^2=9, 512*9=4608.a=8 (256): 256*3^b >=1800. 3^b >=7.03125. b=2: 3^2=9. 256*9=2304.a=7 (128): 128*3^b >=1800. 3^b >=14.0625. b=3: 3^3=27. 128*27=3456.a=6 (64): 64*3^b >=1800. 3^b >=28.125. b=4: 81. 64*81=5184.a=5 (32): 32*3^b >=1800. 3^b >=56.25. b=4: 81. 32*81=2592.a=4 (16): 16*3^b >=1800. 3^b >=112.5. b=5:243. 16*243=3888.So the minimal N for 2^a*3^b is 2304 when a=8, b=2. But 2304 is 2^8*3^2=256*9=2304. Let's check if this satisfies the condition. The prime factors are 2 and 3. N/2=2304/2=1152 >=600. N/3=2304/3=768 >=600. So yes, both are >=600. So 2304 is a candidate. Compare this to 2048. 2048 is smaller. So between 2048 and 2304, 2048 is smaller. So 2048 is better. But let's see if there are numbers with more prime factors that are smaller than 2048.Wait, maybe N can have three prime factors. For example, 2,3, and another prime. Let's say 2,3, and 5. But as we saw earlier, 5 would require N >=600*5=3000. So any number with 5 as a prime factor needs to be at least 3000, which is higher than 2048. So maybe using primes larger than 3 would require N to be larger. But perhaps using primes larger than 2 but avoiding 5? Let's see.Suppose N is 2^a *3^b *7^c. Then N must satisfy N >=600*2=1200, N >=600*3=1800, and N >=600*7=4200. Therefore, N must be at least 4200. So even if we include 7, the required N is 4200, which is larger than 2048, so not helpful. Similarly, if we include 11, N must be at least 600*11=6600. So primes larger than 3 will force N to be larger. Hence, including more primes is not helpful. Therefore, the minimal N is either 2048 (2^11) or 2304 (2^8*3^2). But 2048 is smaller. However, wait, 2048 is 2^11. Let's confirm N=2048. Its prime factors are only 2, and N/2=1024, which is greater than 600. So yes, it satisfies the condition. Therefore, 2048 is a candidate.But wait, maybe there's a composite number with multiple small primes where even though some prime factors are small, the overall N is still smaller. Let me check. For example, N=2^a *3^b where 2^a*3^b is less than 2048 but still meets the condition. Wait, earlier when we tried N=2^8*3^2=2304, which is bigger than 2048. So 2048 is smaller. What if we try to combine 2 and 3 with higher exponents but lower product? Wait, maybe not. Let's check.Wait, another thought: if N is a multiple of 2 and 3, but also other primes, but as we saw, including any other prime would require N to be larger. For example, if N is 2*3*5=30. But N/5=6 <600. Not good. So even if we have multiple primes, the largest prime factor's requirement (N >=600*p) could dominate. So, if we have a composite number with small primes, but one of them is too large, it won't work. Therefore, perhaps the minimal N is indeed 2048. But let's check numbers between 1200 and 2048. Wait, the condition is that N is at least 600 times each of its prime divisors. So for a number N, for every prime p dividing N, N >=600*p. So even if N has multiple primes, each p must satisfy p <= N/600. So, for example, if N is 1200, then any prime divisor p must be <= 1200/600=2. So the only allowed prime is 2. But 1200 factors into 2^4*3*5^2. So primes 2,3,5. But 3 and 5 are greater than 2, which is N/600=2. Hence, 1200 doesn't satisfy. Similarly, 1800: N/600=3. So primes must be <=3. But 1800 includes 5, which is greater than 3. So invalid. Let me check N=2048. Its prime factors are only 2. 2048/2=1024 >=600. So that's good. Is there a smaller N that is a prime power of 2? The next lower power is 1024. 1024/2=512 <600. So 1024 is invalid. So 2048 is the first power of 2 that works. But wait, perhaps a composite number with prime factors 2 and 3, but N is such that N/3 >=600. So N >=1800. Let's check N=1800. But N=1800 includes 5 as a prime factor, which we saw earlier. So N=1800 is invalid. What about N=2^a*3^b where a and b are such that N >=1800 and no other primes. Let's see if such a number exists below 2048. Let's check 2^a*3^b combinations.For example, 2^8*3^2=256*9=2304, which is above 2048. 2^7*3^3=128*27=3456. Still higher. 2^9*3=512*3=1536. 1536 is less than 1800. So 1536 is too small. 2^10*3=1024*3=3072. That's over. So perhaps there's a number between 1800 and 2048. Let's see. Let's try 2^a*3^b=1800. 1800=2^3*3^2*5^2. So can't have just 2 and 3. 2^5*3^3=32*27=864. Too small. 2^6*3^3=64*27=1728. 1728. Let's check N=1728. Prime factors 2 and 3. 1728/2=864 >=600, 1728/3=576. 576 is less than 600. So 576 <600, which violates the condition. Therefore, 1728 is invalid. Next, 2^7*3^3=128*27=3456. That's way over. 3456/3=1152 >=600. 3456/2=1728 >=600. But 3456 is larger than 2048, so not helpful. Wait, perhaps another combination. 2^10*3=1024*3=3072. As before. Too big. Alternatively, 2^a*3^b where a=9, b=2: 512*9=4608. Also too big. So it seems that the minimal N composed of 2 and 3 is 2304, which is higher than 2048. Therefore, 2048 is a better candidate. But let's check if there's another composite number with different prime factors that might be smaller. For example, N=2*7. Then N=14. 14/2=7 <600, 14/7=2 <600. Doesn't work. Similarly, N=2*3*5=30. As before. Wait, maybe higher exponents of small primes. For example, N=2^3*3^3=8*27=216. 216/2=108 <600. 216/3=72 <600. Doesn't work. How about N=2^5*3^4=32*81=2592. Then 2592/2=1296 >=600, 2592/3=864 >=600. So 2592 satisfies, but it's larger than 2048. Wait, another angle. Maybe N is not a prime power or product of two primes, but a product of more primes where each prime is very small. But the problem is that including a prime p requires N >=600*p. So if we include a prime like 3, then N must be at least 1800. If we include 5, N must be at least 3000. So including 3 might be possible if N is at least 1800, but in such a case, even if N is 1800, but 1800 has a prime factor 5 which requires N >=3000. Hence, including 3 is possible only if we don't have any primes larger than 3. Wait, but if N is composed of only 2 and 3, then N must be at least 1800 (since 3 is a prime factor, N >=600*3=1800). So any N composed of 2 and 3 needs to be at least 1800. But as we saw earlier, the smallest such N is 2304. However, if we can find a number composed of 2 and 3 that's between 1800 and 2304, which doesn't have any other prime factors, then maybe it works. Wait, but how? Let's try. For example, 2^a*3^b=1800. But 1800 factors into 2^3*3^2*5^2. So unless we can have a number that's 2^a*3^b exactly equal to 1800, which isn't possible because 1800 includes 5s. So the next number after 1800 that's a product of 2 and 3 only. Let's see. After 1728 (2^6*3^3), the next number is 1944=2^3*3^5=8*243=1944. Check if this works. Prime factors 2 and 3. N=1944. N/2=972 >=600, N/3=648 >=600. So 1944 satisfies the condition. Wait, is that correct? 1944 divided by 3 is 648, which is more than 600. So yes. Then 1944 is smaller than 2048. So 1944 might be a better candidate. Wait, how did I get 1944? Let me verify. 3^5=243, 243*8=1944. Yes. So 2^3*3^5=1944. So that's a valid number. Let's check if there are even smaller numbers composed of 2 and 3. Let's check numbers between 1800 and 1944. For example, 2^4*3^4=16*81=1296. Too small. 2^5*3^4=32*81=2592. Too big. Wait, maybe another combination. 2^a*3^b where a and b are such that the product is between 1800 and 1944. For example, 2^7*3^4=128*81=10368. No, too big. Wait, perhaps 2^a*3^b where a=5, b=5: 32*243=7776. No. Hmm. Let's try to list possible combinations:Start with a=3, then 3^b=1800/8=225. So 3^b=225. b is not an integer. 3^5=243. 243*8=1944. a=4, then 3^b=1800/16=112.5. Not an integer.a=5, 3^b=1800/32≈56.25. 3^4=81, which gives 32*81=2592.a=6, 3^b=1800/64≈28.125. 3^3=27, 64*27=1728. Then next is 3^4=81, 64*81=5184.So the next number after 1728 composed of 2 and 3 is 1944. Then 2592, etc. So 1944 is the first number after 1728. So 1944 is composed of 2 and 3, and satisfies the condition. N=1944. N/2=972 >=600, N/3=648 >=600. So yes. So 1944 is a valid candidate. And 1944 is less than 2048 (1944 vs 2048). Therefore, 1944 is a better candidate. But wait, 1944 is 2^3*3^5. Let's check if there's a smaller N between 1800 and 1944 that's composed of only 2 and 3. Let's see. The numbers between 1800 and 1944 that are products of 2 and 3. Let's see, 1800=2^3*3^2*5^2. Not allowed. 1818=2*3^2*101. No. 1824=2^5*3*19. No. 1836=2^2*3^3*17. No. 1848=2^3*3*7*11. No. 1872=2^4*3^2*13. No. 1890=2*3^3*5*7. No. 1920=2^7*3*5. No. 1944=2^3*3^5. So the first one after 1728 is 1944. So no numbers between 1800 and 1944 composed solely of 2 and 3. Hence, 1944 is the next one. Therefore, 1944 is smaller than 2048 and satisfies the condition. So 1944 is a better candidate. But wait, let's check N=1944. Is there a smaller number that uses another combination of primes? For example, maybe using 2, 3, and 7, but keeping N under 1944. Let me see. If we include 7, then N must be >=600*7=4200. So any number with 7 as a prime factor must be at least 4200, which is way over 1944. Similarly, including 5 requires N>=3000. So including any prime higher than 3 requires N to be larger than 1944. Therefore, the only possible primes are 2 and 3. So 1944 is a candidate. But is there a smaller N with primes 2 and 3? Let's check the previous numbers. After 1728 (which didn't work because 1728/3=576 <600), the next is 1944. So 1944 is the next possible. So 1944 is valid. So now, comparing 1944 and 2048. 1944 is smaller. So 1944 is the better answer. But let's confirm once again. N=1944. Prime factors 2 and 3. 1944/2=972 >=600, 1944/3=648 >=600. Correct. Is there a smaller N than 1944 that's composed of 2 and 3 and satisfies the condition? Let's see. The number before 1944 is 1728. 1728/3=576 <600. So invalid. Next is 1944. So yes, 1944 is the first number after 1728 that's a product of 2 and 3 and meets the condition. Therefore, 1944 is a valid candidate. But wait, let's check another possibility. What if N is a multiple of 2 and another prime, but that prime is small enough such that N is still less than 1944. For example, N=2^a * p where p is a prime. So if p=3, then N=2^a *3. Then N must be >=600*3=1800, so N >=1800. The smallest such N would be 2^a *3 >=1800. Let's find minimal a. 2^a >=600. 2^9=512 <600, 2^10=1024 >=600. So N=2^10*3=1024*3=3072. Which is larger than 1944. If p=5, then N must be >=600*5=3000, which is larger. If p=7, N >=4200. So no help. So any N that's a product of two primes where one is 3 requires N to be at least 3072, which is larger than 1944. Similarly, products with higher primes require even larger N. Therefore, the minimal N is 1944. But wait, let's check if there are numbers with multiple prime factors but still under 1944. For example, if we take 2^a *3^b *5^c, but 5^c must have N >=3000. So that's too big. Similarly, primes above 5 are worse. Alternatively, perhaps numbers with exponents of 2 and 3 but arranged differently. Let's see if there's a combination of exponents a and b such that 2^a *3^b is less than 1944 but still satisfies N/3 >=600. Wait, N/3 >=600 => N >=1800. So N must be at least 1800. So between 1800 and 1944, we need a number that is a product of 2 and 3 only. The first such number is 1944. Because 1800 is not a pure product of 2 and 3. So for example, 2^a*3^b=1944. Let's see if there's a smaller N. Let's try a=4, b=5: 16*243=3888. Too big. a=5, b=4:32*81=2592. Too big. a=3, b=5:8*243=1944. So that's the one. Alternatively, a=2, b=6:4*729=2916. Too big. So no, 1944 is the minimal in that case. Therefore, 1944 is the minimal number composed of 2 and 3 that is at least 600 times each prime factor. But wait, hold on. Let's check N=1920. Wait, 1920 is 2^7*3*5. So includes 5, which makes N/5=384 <600. So invalid. Similarly, 1800 includes 5. 1680=2^4*3*5*7. Also invalid. 1536=2^9*3. 1536/3=512 <600. 1728=2^6*3^3. 1728/3=576 <600. So the next candidate after 1728 is 1944. So, so far, the candidates are 1944 and 2048. Since 1944 is smaller, it's better. But let's check another angle. What if we take a higher power of 3 with a lower power of 2? Let's see. For example, 3^b*2^a where b is higher. Let's see. 3^6=729. 729*2^a. To get N >=1800. 729*2=1458 <1800. 729*4=2916 >=1800. 2916 is larger than 1944. So not helpful. Alternatively, 3^5=243. 243*8=1944. As before. So no, 1944 seems to be the minimal. Wait, another thought. What if N is 2^2*3^3=4*27=108. 108/2=54 <600. No. 2^4*3^4=16*81=1296. 1296/3=432 <600. Not good. So, stepping back, the minimal N is either 1944 or 2048. Comparing them, 1944 is smaller. So 1944 is the answer. But wait, wait. Let's check if there's a number with prime factors 2, 3, and 7, but where N is still under 1944. But as mentioned earlier, if 7 is a prime factor, N must be at least 600*7=4200, which is way above 1944. Similarly, any other prime factor would require N to be larger. Therefore, no such number exists below 1944. Hence, the minimal N is 1944. But to be thorough, let's check if there's a composite number with more than two prime factors, all of which are 2 and 3, but in different exponents. Wait, but we already considered that. For example, 2^3*3^5=1944. Alternatively, 2^4*3^5=16*243=3888. Larger. So no. Therefore, after all this analysis, the smallest natural number N greater than 1 that is at least 600 times greater than each of its prime divisors is 1944. But wait a second! Wait, I just realized that 1944 divided by 3 is 648, which is exactly 600*1.08. So it's 648, which is 48 more than 600. But is there a number between 1800 and 1944 that's a product of 2 and 3 and meets the condition? Let me check 1920, but 1920 includes 5. 1890 includes 5 and 7. 1800 includes 5. 1872=2^4*3^2*13. 1848=2^3*3*7*11. All of these include primes larger than 3, which would make N/p too small. Alternatively, 1728=2^6*3^3. 1728/3=576 <600. 1944 is next. Therefore, 1944 is indeed the minimal. Wait, but earlier I thought of 2048, which is 2^11=2048. 2048 is 2048/2=1024 >=600. So it satisfies. But 1944 is smaller than 2048, so 1944 is better. Therefore, the answer should be 1944. But let me cross-verify this with another approach. Suppose we model the problem as follows: For each prime p dividing N, N >=600p. Therefore, for all p|N, p <= N/600. Hence, all prime factors of N are <= N/600. To find the minimal such N, we can consider that the primes must be among the smallest primes, and the number N must be a multiple of these primes such that N is at least 600 times the largest prime factor. So, let's attempt to construct N step by step. Case 1: N is a power of 2. Then N=2^k. We need 2^k >=600*2=1200. The minimal k is 11, since 2^10=1024 <1200, 2^11=2048 >=1200. So N=2048. Case 2: N is a product of 2 and 3. Then the largest prime factor is 3, so N >=600*3=1800. The minimal N is 2^a*3^b >=1800 with a,b >=1. The minimal such N is 1944=2^3*3^5. Case 3: N includes higher primes. As discussed, this requires N to be larger (at least 3000 if 5 is included), which is worse. Comparing Case 1 and Case 2, 1944 <2048, so 1944 is better. Hence, the answer is 1944. Therefore, after thorough consideration, the smallest natural number N is 1944.

✅:The smallest natural number ( N ) greater than 1 that is at least 600 times greater than each of its prime divisors is found through systematic analysis:1. Key Condition: For every prime divisor ( p ) of ( N ), ( N geq 600p ). This means all prime factors of ( N ) must satisfy ( p leq frac{N}{600} ).2. Prime Power Consideration: - If ( N ) is a power of 2, the smallest ( N ) satisfying ( 2^k geq 600 times 2 = 1200 ) is ( 2^{11} = 2048 ).3. Product of Small Primes: - For ( N ) to include 3, the smallest ( N ) must be ( geq 600 times 3 = 1800 ). The minimal product of 2 and 3 meeting this is ( 2^3 times 3^5 = 1944 ), since: - ( 1944 / 2 = 972 geq 600 ) - ( 1944 / 3 = 648 geq 600 )4. Comparison: - ( 1944 ) (product of 2 and 3) is smaller than ( 2048 ) (power of 2), making it the better candidate. - Including primes larger than 3 (e.g., 5, 7) forces ( N ) to be significantly larger (e.g., ( geq 3000 )), which is not optimal.Final Answer: The smallest natural number is (boxed{1944}).

🔑:Let's carefully analyze the problem to find the smallest natural number ( N ) greater than 1 that is at least 600 times greater than each of its prime divisors.# Step 1: Define the problem conditionWe need ( N ) to satisfy:[N geq 600 times p quad text{for each prime divisor } p text{ of } N]# Step 2: Express ( N ) in terms of its prime factorizationAssume ( N ) has the prime factorization[N = p_1^{e_1} cdot p_2^{e_2} cdot ldots cdot p_k^{e_k}]where ( p_1, p_2, ldots, p_k ) are the prime divisors of ( N ).# Step 3: Apply the problem condition to each prime divisorFor each prime ( p_i ):[N geq 600 times p_i]This implies that ( N ) must be at least 600 times each of its prime factors.# Step 4: Identify the smallest candidate prime factorWe need to consider small prime numbers to minimize ( N ). Start with the smallest prime numbers and test the lowest ( i ) where this condition holds.# Step 5: Calculate candidate ( N )Let's try primes 2 and 3 as the smallest candidates that minimize ( N ).For ( p_1 = 2 ) and ( p_2 = 3 ):Potentially, consider combining the powers of small primes; start with:[N = 2^3 cdot 3^5 = 8 cdot 243 = 1944]# Step 6: Verify if the calculated ( N ) satisfies the conditionCheck the individual values:- For ( 2 ), ( 1944 geq 600 times 2 = 1200 )- For ( 3 ), ( 1944 geq 600 times 3 = 1800 )Given that both conditions are satisfied:[1944 geq 1200 quad text{and} quad 1944 geq 1800]Thus, the number 1944 satisfies the condition that it is at minimum 600 times each of its prime factors (2 and 3). Conclusion:The smallest such number ( N ) is:[boxed{1944}]

🧠:Okay, so I need to find the range of the real number ( a ) such that the inequality ( a f(x) - f(3x) leq 2 g(2x) ) holds for all ( x ) in the interval ( left[frac{1}{2}, 2right] ). Given that ( f(x) ) is an odd function and ( g(x) ) is an even function, and they satisfy ( f(x) + g(x) = 2^x ). Hmm, let's break this down step by step.First, since ( f ) is odd and ( g ) is even, I can use their properties to express ( f(x) ) and ( g(x) ) in terms of ( 2^x ). Remember, an odd function satisfies ( f(-x) = -f(x) ), and an even function satisfies ( g(-x) = g(x) ). Given ( f(x) + g(x) = 2^x ), if I substitute ( x ) with ( -x ), then the equation becomes ( f(-x) + g(-x) = 2^{-x} ). Using the properties of odd and even functions, this simplifies to ( -f(x) + g(x) = 2^{-x} ). Now, I have two equations:1. ( f(x) + g(x) = 2^x )2. ( -f(x) + g(x) = 2^{-x} )I can solve these two equations simultaneously to find expressions for ( f(x) ) and ( g(x) ). Let's add and subtract the equations. Adding equation 1 and equation 2:( (f(x) + g(x)) + (-f(x) + g(x)) = 2^x + 2^{-x} )Simplifying the left side: ( 2g(x) = 2^x + 2^{-x} ), so ( g(x) = frac{2^x + 2^{-x}}{2} ).Subtracting equation 2 from equation 1:( (f(x) + g(x)) - (-f(x) + g(x)) = 2^x - 2^{-x} )Simplifying the left side: ( 2f(x) = 2^x - 2^{-x} ), so ( f(x) = frac{2^x - 2^{-x}}{2} ).Alright, so now we have expressions for both ( f(x) ) and ( g(x) ):- ( f(x) = frac{2^x - 2^{-x}}{2} )- ( g(x) = frac{2^x + 2^{-x}}{2} )Good, that's a start. Now, let's note that these expressions are valid for all real numbers ( x ), since they were derived from the given functional equations and the properties of odd and even functions.Next, the inequality we need to consider is ( a f(x) - f(3x) leq 2 g(2x) ) for ( x in left[frac{1}{2}, 2right] ). Let's substitute the expressions for ( f(x) ), ( f(3x) ), and ( g(2x) ) into this inequality.First, compute each term:1. ( a f(x) = a cdot frac{2^x - 2^{-x}}{2} )2. ( f(3x) = frac{2^{3x} - 2^{-3x}}{2} )3. ( 2 g(2x) = 2 cdot frac{2^{2x} + 2^{-2x}}{2} = 2^{2x} + 2^{-2x} )Substituting these into the inequality:( a cdot frac{2^x - 2^{-x}}{2} - frac{2^{3x} - 2^{-3x}}{2} leq 2^{2x} + 2^{-2x} )Let me multiply both sides by 2 to eliminate the denominators:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) leq 2 cdot (2^{2x} + 2^{-2x}) )Simplify the right side:( 2 cdot 2^{2x} + 2 cdot 2^{-2x} = 2^{2x + 1} + 2^{-2x + 1} )Wait, actually, no. Wait, ( 2 cdot (2^{2x} + 2^{-2x}) = 2^{2x + 1} + 2^{-2x + 1} ). Hmm, actually, that's correct. Alternatively, we can leave it as ( 2 cdot 2^{2x} + 2 cdot 2^{-2x} ), but perhaps it's better to write as ( 2^{2x + 1} + 2^{-2x + 1} ).But maybe it's clearer to write each term separately. Let me check:Left side after multiplying by 2:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) )Right side:( 2 cdot 2^{2x} + 2 cdot 2^{-2x} = 2^{2x + 1} + 2^{-2x + 1} )Alternatively, ( 2^{2x} times 2 + 2^{-2x} times 2 ). Either way, okay.So let's rewrite the inequality:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) leq 2^{2x + 1} + 2^{-2x + 1} )Hmm, but maybe instead of expanding further, we can rearrange the terms.Bring all terms to the left side:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) - 2^{2x + 1} - 2^{-2x + 1} leq 0 )Alternatively, perhaps factor terms or express in terms of exponentials with the same base.Alternatively, maybe substitute ( t = 2^x ), which could simplify the expressions. Let's try that.Let ( t = 2^x ). Then, since ( x in left[frac{1}{2}, 2right] ), ( t in left[2^{1/2}, 2^2right] = [sqrt{2}, 4] ).Express all terms in terms of ( t ):First, ( 2^x = t ), so ( 2^{-x} = 1/t ).Similarly:- ( 2^{3x} = (2^x)^3 = t^3 )- ( 2^{-3x} = 1/t^3 )- ( 2^{2x} = t^2 )- ( 2^{-2x} = 1/t^2 )- ( 2^{2x + 1} = 2 cdot t^2 )- ( 2^{-2x + 1} = 2 cdot (1/t^2) )So substituting all into the inequality:Left side: ( a(t - 1/t) - (t^3 - 1/t^3) leq 2t^2 + 2/t^2 )So the inequality becomes:( a(t - 1/t) - (t^3 - 1/t^3) - 2t^2 - 2/t^2 leq 0 )Let me rearrange terms:( a(t - 1/t) leq (t^3 - 1/t^3) + 2t^2 + 2/t^2 )But perhaps it's better to collect all terms on the left:( a(t - 1/t) - t^3 + 1/t^3 - 2t^2 - 2/t^2 leq 0 )Alternatively, let's compute each term step by step.First, let's compute ( (t^3 - 1/t^3) ). Note that ( t^3 - 1/t^3 = (t - 1/t)(t^2 + 1 + 1/t^2) ). But not sure if that helps.Alternatively, maybe factor each term. Let's see:Looking at the left side expression:( a(t - 1/t) - t^3 + 1/t^3 - 2t^2 - 2/t^2 leq 0 )This is a bit complex, but maybe group terms:Let's write all terms with positive coefficients first:- ( a(t - 1/t) )- ( -t^3 )- ( +1/t^3 )- ( -2t^2 )- ( -2/t^2 )Alternatively, perhaps factor terms with similar powers:But perhaps not. Let's instead compute each term as is.But maybe it's better to write the entire expression in terms of ( t ) and ( 1/t ), then see if we can factor or find a common term.Alternatively, since ( t > 0 ), perhaps multiply both sides by ( t^3 ) to eliminate denominators. Let's try that.Multiply both sides by ( t^3 ):Left side:( a(t - 1/t) cdot t^3 - (t^3 - 1/t^3) cdot t^3 - 2t^2 cdot t^3 - 2/t^2 cdot t^3 leq 0 times t^3 )Simplify each term:1. ( a(t - 1/t) cdot t^3 = a(t cdot t^3 - (1/t) cdot t^3) = a(t^4 - t^2) )2. ( -(t^3 - 1/t^3) cdot t^3 = - (t^6 - 1) )3. ( -2t^2 cdot t^3 = -2t^5 )4. ( -2/t^2 cdot t^3 = -2t )So combining all terms:( a(t^4 - t^2) - (t^6 - 1) - 2t^5 - 2t leq 0 )Simplify the left side:( a t^4 - a t^2 - t^6 + 1 - 2t^5 - 2t leq 0 )Rearranging terms in descending powers of ( t ):( -t^6 - 2t^5 + a t^4 - a t^2 - 2t + 1 leq 0 )Multiply both sides by -1 (which reverses the inequality):( t^6 + 2t^5 - a t^4 + a t^2 + 2t - 1 geq 0 )Hmm, this seems more complicated. Maybe this approach isn't the best. Let me think again.Alternatively, perhaps instead of substituting ( t = 2^x ), we can analyze the original inequality in terms of ( x ). Let me check the original inequality again:( a f(x) - f(3x) leq 2 g(2x) )We have expressions for ( f(x) ), ( f(3x) ), and ( g(2x) ). Let's substitute them back into the inequality:( a cdot frac{2^x - 2^{-x}}{2} - frac{2^{3x} - 2^{-3x}}{2} leq 2 cdot frac{2^{2x} + 2^{-2x}}{2} )Simplify each term:Left side:( frac{a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x})}{2} )Right side:( 2 cdot frac{2^{2x} + 2^{-2x}}{2} = 2^{2x} + 2^{-2x} )Multiply both sides by 2 to eliminate the denominator:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) leq 2 cdot (2^{2x} + 2^{-2x}) )Now, let's move all terms to the left side:( a(2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) - 2 cdot 2^{2x} - 2 cdot 2^{-2x} leq 0 )Simplify each term:- ( 2 cdot 2^{2x} = 2^{2x + 1} )- ( 2 cdot 2^{-2x} = 2^{-2x + 1} )So the inequality becomes:( a(2^x - 2^{-x}) - 2^{3x} + 2^{-3x} - 2^{2x + 1} - 2^{-2x + 1} leq 0 )Alternatively, maybe express all terms as powers of 2:Let me list each term:1. ( a cdot 2^x )2. ( -a cdot 2^{-x} )3. ( -2^{3x} )4. ( +2^{-3x} )5. ( -2^{2x + 1} )6. ( -2^{-2x + 1} )This seems messy. Maybe the substitution ( t = 2^x ) is still a good idea here.So let's try that substitution again. Let ( t = 2^x ), so ( t in [sqrt{2}, 4] ).Then:- ( 2^x = t )- ( 2^{-x} = 1/t )- ( 2^{3x} = t^3 )- ( 2^{-3x} = 1/t^3 )- ( 2^{2x + 1} = 2 cdot t^2 )- ( 2^{-2x + 1} = 2 cdot (1/t^2) )Substituting into the inequality:Left side:( a(t - 1/t) - t^3 + 1/t^3 - 2t^2 - 2/t^2 leq 0 )Now, we can write this as:( a(t - 1/t) leq t^3 - 1/t^3 + 2t^2 + 2/t^2 )But perhaps instead of handling this directly, let's solve for ( a ). Since we need the inequality to hold for all ( t in [sqrt{2}, 4] ), we can express ( a ) in terms of ( t ) and find the maximum or minimum values required.So, starting from:( a(t - 1/t) - t^3 + 1/t^3 - 2t^2 - 2/t^2 leq 0 )Let's solve for ( a ):( a(t - 1/t) leq t^3 - 1/t^3 + 2t^2 + 2/t^2 )Therefore,( a leq frac{t^3 - 1/t^3 + 2t^2 + 2/t^2}{t - 1/t} ), provided that ( t - 1/t > 0 ).But wait, we need to check when ( t - 1/t ) is positive or negative because the direction of the inequality depends on the sign of the denominator when solving for ( a ).First, let's check the sign of ( t - 1/t ) over ( t in [sqrt{2}, 4] ).Since ( t geq sqrt{2} approx 1.414 ), so ( t > 1 ), hence ( t - 1/t > 0 ).Because if ( t > 1 ), then ( t - 1/t = (t^2 - 1)/t > 0 ).Therefore, since ( t - 1/t > 0 ) for all ( t in [sqrt{2}, 4] ), dividing both sides by ( t - 1/t ) does not reverse the inequality sign.Therefore, ( a leq frac{t^3 - 1/t^3 + 2t^2 + 2/t^2}{t - 1/t} ) for all ( t in [sqrt{2}, 4] ).Thus, the maximum value of the right-hand side (RHS) over ( t in [sqrt{2}, 4] ) will give the upper bound for ( a ). However, since the inequality must hold for all ( t ) in the interval, ( a ) must be less than or equal to the minimum of the RHS over ( t in [sqrt{2}, 4] ). Wait, no. Wait, if ( a leq RHS(t) ) for all ( t in [sqrt{2}, 4] ), then ( a ) must be less than or equal to the minimum of RHS(t) over that interval. Because if ( a ) is less than the smallest RHS(t), then it will certainly be less than all RHS(t). Conversely, the maximum possible ( a ) that satisfies the inequality for all ( t ) is the infimum (greatest lower bound) of RHS(t) over the interval.Wait, let me think again. Suppose we have ( a leq RHS(t) ) for all ( t in [sqrt{2}, 4] ). Then the maximum possible ( a ) that satisfies this inequality is the minimum of RHS(t) over the interval, because if ( a ) is equal to the minimum of RHS(t), then ( a leq RHS(t) ) for all t. If ( a ) were larger than that minimum, then at the point where RHS(t) is minimal, the inequality would fail.Therefore, the range of ( a ) is ( (-infty, min_{t in [sqrt{2},4]} RHS(t)] ). Wait, but the question says "find the range of the real number ( a )" such that the inequality holds. So we need to compute the minimum of the RHS expression over ( t in [sqrt{2}, 4] ), and then ( a ) must be less than or equal to that minimum value.Therefore, the next step is to compute ( RHS(t) = frac{t^3 - 1/t^3 + 2t^2 + 2/t^2}{t - 1/t} ) and find its minimum over ( t in [sqrt{2},4] ).Let me first simplify the expression for RHS(t). Let's compute numerator and denominator separately.Numerator: ( t^3 - frac{1}{t^3} + 2t^2 + frac{2}{t^2} )Denominator: ( t - frac{1}{t} )Let me try to factor numerator or manipulate it to see if it can be expressed in terms of the denominator.First, note that ( t^3 - 1/t^3 = (t - 1/t)(t^2 + 1 + 1/t^2) ). Let's verify that:( (t - 1/t)(t^2 + 1 + 1/t^2) = t(t^2 + 1 + 1/t^2) - (1/t)(t^2 + 1 + 1/t^2) )= ( t^3 + t + 1/t - (t + 1/t + 1/t^3) )= ( t^3 + t + 1/t - t - 1/t - 1/t^3 )= ( t^3 - 1/t^3 )Yes, that's correct.Similarly, the denominator is ( t - 1/t ).So, the numerator can be written as:( (t - 1/t)(t^2 + 1 + 1/t^2) + 2t^2 + 2/t^2 )Therefore, RHS(t) becomes:( frac{(t - 1/t)(t^2 + 1 + 1/t^2) + 2t^2 + 2/t^2}{t - 1/t} )Split the fraction:( frac{(t - 1/t)(t^2 + 1 + 1/t^2)}{t - 1/t} + frac{2t^2 + 2/t^2}{t - 1/t} )Simplify the first term:( t^2 + 1 + 1/t^2 + frac{2t^2 + 2/t^2}{t - 1/t} )So, RHS(t) = ( t^2 + 1 + frac{1}{t^2} + frac{2t^2 + 2/t^2}{t - 1/t} )Hmm, not sure if this helps. Let me check:Alternatively, factor 2 in the numerator:( frac{2(t^2 + 1/t^2)}{t - 1/t} )But perhaps we can write ( t^2 + 1/t^2 = (t - 1/t)^2 + 2 ). Let's check:( (t - 1/t)^2 = t^2 - 2 + 1/t^2 ), so ( t^2 + 1/t^2 = (t - 1/t)^2 + 2 ). Yes, that's correct.So substituting into the numerator:( 2[(t - 1/t)^2 + 2] )Therefore, the term becomes ( frac{2[(t - 1/t)^2 + 2]}{t - 1/t} )Which can be written as ( 2 cdot frac{(t - 1/t)^2 + 2}{t - 1/t} = 2 left( t - 1/t + frac{2}{t - 1/t} right) )Therefore, RHS(t) = ( t^2 + 1 + 1/t^2 + 2 left( t - 1/t + frac{2}{t - 1/t} right) )But this seems more complicated. Alternatively, maybe set ( u = t - 1/t ), then see if we can express RHS(t) in terms of ( u ).Let ( u = t - 1/t ). Then, since ( t in [sqrt{2}, 4] ), let's compute the range of ( u ).Compute ( u = t - 1/t ). When ( t = sqrt{2} approx 1.414 ), ( u = sqrt{2} - 1/sqrt{2} = sqrt{2} - sqrt{2}/2 = sqrt{2}/2 approx 0.707 ).When ( t = 4 ), ( u = 4 - 1/4 = 3.75 ).So ( u in [sqrt{2}/2, 3.75] ).Also, note that ( u = t - 1/t ), so ( u ) increases as ( t ) increases because the derivative of ( u ) with respect to ( t ) is ( 1 + 1/t^2 > 0 ). Hence, ( u ) is a strictly increasing function of ( t ) for ( t > 0 ).Therefore, as ( t ) goes from ( sqrt{2} ) to 4, ( u ) increases from ( sqrt{2}/2 ) to ( 15/4 ).Now, let's express RHS(t) in terms of ( u ):First, note that:( t^2 + 1/t^2 = (t - 1/t)^2 + 2 = u^2 + 2 )Similarly, ( t^3 - 1/t^3 = u(u^2 + 3) ). Wait, let me verify:( (t - 1/t)^3 = t^3 - 3t + 3/t - 1/t^3 )Therefore,( t^3 - 1/t^3 = (t - 1/t)^3 + 3(t - 1/t) = u^3 + 3u )But in our previous expression, we had ( t^3 - 1/t^3 = (t - 1/t)(t^2 + 1 + 1/t^2) = u(u^2 + 2 + 1) = u(u^2 + 3) ). Wait, but here, it's written as ( u^3 + 3u ). So both expressions are the same:( u(u^2 + 3) = u^3 + 3u )So that's correct.But let's go back to RHS(t). Originally, RHS(t) is:( frac{t^3 - 1/t^3 + 2t^2 + 2/t^2}{t - 1/t} = frac{u(u^2 + 3) + 2(u^2 + 2)}{u} )Wait, let's check:Numerator:( t^3 - 1/t^3 + 2t^2 + 2/t^2 = u(u^2 + 3) + 2(u^2 + 2) )Because:- ( t^3 - 1/t^3 = u(u^2 + 3) )- ( 2t^2 + 2/t^2 = 2(u^2 + 2) )Therefore, numerator = ( u(u^2 + 3) + 2(u^2 + 2) = u^3 + 3u + 2u^2 + 4 )Thus, RHS(t) = ( frac{u^3 + 2u^2 + 3u + 4}{u} )Divide each term by ( u ):( RHS(t) = u^2 + 2u + 3 + frac{4}{u} )So, RHS(t) simplifies to ( u^2 + 2u + 3 + frac{4}{u} )Therefore, in terms of ( u ), where ( u in [sqrt{2}/2, 15/4] ), the expression to minimize is ( u^2 + 2u + 3 + frac{4}{u} ).This seems more manageable. So now, the problem reduces to finding the minimum of the function ( h(u) = u^2 + 2u + 3 + frac{4}{u} ) over ( u in [sqrt{2}/2, 15/4] ).To find the minimum of ( h(u) ), we can take the derivative of ( h(u) ) with respect to ( u ), set it to zero, and solve for ( u ). Then check the critical points and endpoints.Compute ( h'(u) ):( h'(u) = 2u + 2 - frac{4}{u^2} )Set ( h'(u) = 0 ):( 2u + 2 - frac{4}{u^2} = 0 )Multiply both sides by ( u^2 ) to eliminate denominator:( 2u^3 + 2u^2 - 4 = 0 )Simplify:Divide by 2: ( u^3 + u^2 - 2 = 0 )We need to solve the cubic equation ( u^3 + u^2 - 2 = 0 ).Let's try rational roots. The possible rational roots are ( pm1, pm2 ).Testing ( u = 1 ):( 1 + 1 - 2 = 0 ). So ( u = 1 ) is a root.Therefore, we can factor the cubic as ( (u - 1)(u^2 + 2u + 2) = 0 ).Wait, let's perform polynomial division or factorization:Divide ( u^3 + u^2 - 2 ) by ( (u - 1) ).Using synthetic division:Coefficients: 1 (u^3), 1 (u^2), 0 (u), -2 (constant term)Using root u = 1:Bring down 1.Multiply 1 by 1 = 1. Add to next coefficient: 1 + 1 = 2.Multiply 2 by 1 = 2. Add to next coefficient: 0 + 2 = 2.Multiply 2 by 1 = 2. Add to last coefficient: -2 + 2 = 0. Perfect.Therefore, the cubic factors as ( (u - 1)(u^2 + 2u + 2) ).Set to zero: ( (u - 1)(u^2 + 2u + 2) = 0 )Solutions are ( u = 1 ), and ( u^2 + 2u + 2 = 0 ). The quadratic equation has discriminant ( 4 - 8 = -4 ), so complex roots. Therefore, the only real root is ( u = 1 ).Thus, the critical point occurs at ( u = 1 ).Now, we need to check whether ( u = 1 ) is in the interval ( [sqrt{2}/2, 15/4] approx [0.707, 3.75] ). Yes, ( 1 ) is within this interval.Therefore, we have critical points at ( u = 1 ), and the endpoints ( u = sqrt{2}/2 ) and ( u = 15/4 ). We need to evaluate ( h(u) ) at these points to find the minimum.Compute ( h(u) ) at each critical point and endpoints.First, at ( u = 1 ):( h(1) = (1)^2 + 2(1) + 3 + frac{4}{1} = 1 + 2 + 3 + 4 = 10 )Next, at ( u = sqrt{2}/2 approx 0.707 ):Compute ( h(sqrt{2}/2) = (sqrt{2}/2)^2 + 2(sqrt{2}/2) + 3 + 4/(sqrt{2}/2) )Calculate each term:1. ( (sqrt{2}/2)^2 = (2)/4 = 0.5 )2. ( 2(sqrt{2}/2) = sqrt{2} approx 1.414 )3. ( 3 )4. ( 4/(sqrt{2}/2) = 4 * 2/sqrt{2} = 8/sqrt{2} = 4sqrt{2} approx 5.656 )Adding them up: 0.5 + 1.414 + 3 + 5.656 ≈ 0.5 + 1.414 = 1.914; 1.914 + 3 = 4.914; 4.914 + 5.656 ≈ 10.57So approximately 10.57.At ( u = 15/4 = 3.75 ):Compute ( h(15/4) = (15/4)^2 + 2*(15/4) + 3 + 4/(15/4) )Calculate each term:1. ( (15/4)^2 = 225/16 ≈ 14.0625 )2. ( 2*(15/4) = 30/4 = 7.5 )3. ( 3 )4. ( 4/(15/4) = 16/15 ≈ 1.0667 )Adding them up: 14.0625 + 7.5 = 21.5625; 21.5625 + 3 = 24.5625; 24.5625 + 1.0667 ≈ 25.6292Therefore, h(15/4) ≈ 25.63Comparing the three values:- ( h(1) = 10 )- ( h(sqrt{2}/2) ≈ 10.57 )- ( h(15/4) ≈ 25.63 )Therefore, the minimum occurs at ( u = 1 ), with ( h(u) = 10 ).Therefore, the minimum value of RHS(t) over the interval is 10. Therefore, the maximum value that ( a ) can take is 10, and since the inequality requires ( a leq RHS(t) ) for all ( t ), the range of ( a ) is ( (-infty, 10] ).But wait, before finalizing, let's verify if there are any other critical points. We found that the only critical point is at ( u = 1 ), and the function ( h(u) ) has its minimum there. The endpoints give higher values. Therefore, yes, the minimum is 10.Therefore, the range of ( a ) is all real numbers less than or equal to 10. So ( a leq 10 ).But let me cross-verify this result by checking with specific values of ( x ).For example, when ( u = 1 ), which corresponds to ( t - 1/t = 1 ). Let's find ( t ):( t - 1/t = 1 implies t^2 - t - 1 = 0 implies t = [1 pm sqrt{5}]/2 ). Since ( t > 0 ), ( t = [1 + sqrt{5}]/2 ≈ 1.618 ), which is within our interval ( [sqrt{2} ≈ 1.414, 4] ).So at ( t ≈ 1.618 ), which corresponds to ( x = log_2 t ≈ log_2(1.618) ≈ 0.694 ), which is within ( [0.5, 2] ). Therefore, this critical point is indeed within the interval.Another check: let's pick ( x = 1 ), which is in [0.5, 2]. Then, compute ( a f(1) - f(3) leq 2 g(2) ).First, compute ( f(1) = (2^1 - 2^{-1}) / 2 = (2 - 0.5)/2 = 1.5 / 2 = 0.75 )( f(3) = (2^3 - 2^{-3}) / 2 = (8 - 0.125)/2 = 7.875 / 2 = 3.9375 )( g(2) = (2^2 + 2^{-2}) / 2 = (4 + 0.25)/2 = 4.25 / 2 = 2.125 )So the inequality at ( x = 1 ):( a*0.75 - 3.9375 leq 2*2.125 implies 0.75a - 3.9375 leq 4.25 implies 0.75a leq 8.1875 implies a leq 8.1875 / 0.75 ≈ 10.9167 )Wait, but according to our previous result, a should be ≤10. But this suggests that at x=1, a can be up to ≈10.9167. This contradicts our earlier conclusion. Therefore, there must be an error in our reasoning.Wait, this is a problem. It means that our previous conclusion that the minimum of RHS(t) is 10 is incorrect, because at x=1 (t=2), the allowed a is up to approximately 10.9167, which is greater than 10. Therefore, our previous calculation is wrong.This indicates a mistake in our analysis. Let's track back.We had:RHS(t) = u² + 2u + 3 + 4/u, where u = t - 1/t.We found critical point at u=1, h(u)=10. However, when x=1, t=2, u=2 - 1/2=1.5. Then h(u)= (1.5)^2 + 2*(1.5) +3 +4/(1.5)=2.25 +3 +3 +2.6667≈ 10.9167, which matches the manual calculation. But according to our previous calculation, the minimum h(u)=10 at u=1. However, in reality, when u=1.5 (t=2), h(u)≈10.9167, which is higher than 10, but when u=1, h(u)=10, but the value at u=1 is lower. However, in the interval [√2/2≈0.707, 3.75], the minimum at u=1 is 10. However, when x=1, the corresponding u=1.5, which gives a higher value. Therefore, the inequality requires that a is less than or equal to the minimal RHS(t) over the interval, which is 10. Therefore, even though at x=1 the allowed a is up to ~10.9167, to satisfy the inequality for all x in [0.5,2], a must be ≤10, because at x corresponding to u=1, the required a is 10, and if a is larger than 10, the inequality would fail there.But wait, when we calculated at x=1, we found that a can be up to ~10.9167. But if the minimal RHS(t) is 10, then setting a=10 would satisfy the inequality for all t, because 10 ≤ RHS(t) for all t in the interval, since the minimum RHS(t) is 10. Therefore, even though at some points you can have higher a, but to satisfy all points, a must be ≤10.Therefore, the mistake in the previous check was thinking that at x=1, a can be up to ~10.9167, but actually, if a=10, then at x=1, the inequality holds as equality when a=10.9167, but if a=10, then at x=1, the left side would be less than the right side. Wait, no, let's recast.Wait, when we set a=10, then the inequality is:10 f(x) - f(3x) ≤ 2g(2x)At x=1:Left side: 10*0.75 - 3.9375 = 7.5 -3.9375=3.5625Right side: 2*2.125=4.25So 3.5625 ≤4.25, which holds.But if we set a=10.9167, then Left side:10.9167*0.75 -3.9375≈8.1875 -3.9375=4.25, which equals the right side. Therefore, at x=1, a=10.9167 makes the inequality hold as equality. However, at u=1 (x such that t -1/t=1, t≈1.618), if a=10.9167, then:At u=1, h(u)=10, so RHS(t)=10. So the inequality requires a ≤10 at that point, but if a=10.9167, then 10.9167 >10, which violates the inequality at that point. Therefore, to satisfy the inequality for all x, a must be ≤10.Therefore, the check at x=1 was misleading because it's not the point where the minimal RHS(t) occurs. The minimal RHS(t) occurs at x corresponding to u=1, where RHS(t)=10. Therefore, the correct upper bound for a is 10.Therefore, the range of a is ( (-infty, 10] ).But let's verify with another point. Take x such that t= sqrt(2), so x=0.5, t=sqrt(2)≈1.414, u=sqrt(2)-1/sqrt(2)=sqrt(2)/2≈0.707.Compute RHS(t)= h(u)≈10.57 as before. Therefore, at x=0.5, the maximum allowable a is ~10.57. But if we set a=10, then:Left side: 10*f(0.5) - f(1.5) ≤2*g(1)Calculate f(0.5)= (2^0.5 -2^{-0.5})/2=(sqrt(2)-1/sqrt(2))/2=(sqrt(2)/2 -1/(2 sqrt(2)))= (sqrt(2)/2 - sqrt(2)/2/2)= sqrt(2)/2 - sqrt(2)/4= sqrt(2)/4≈0.3535f(1.5)= (2^{1.5}-2^{-1.5})/2=(2*sqrt(2)-1/(2*sqrt(2)))/2≈(2.8284 -0.3536)/2≈2.4748/2≈1.23742*g(1)=2*(2^1 +2^{-1})/2=2*(2 +0.5)/2=2.5So left side:10*0.3535 -1.2374≈3.535 -1.2374≈2.2976≤2.5, which holds.If we set a=10, then at x=0.5, the inequality holds (2.2976 ≤2.5). If we set a=10.57, then left side≈10.57*0.3535 -1.2374≈3.738 -1.2374≈2.5006≈2.5, which is approximately equal. So at x=0.5, a=10.57 makes it equal. But since the minimal RHS(t)=10 occurs at another x, setting a=10.57 would cause the inequality to fail at x where RHS(t)=10 (i.e., u=1, t≈1.618).Therefore, the correct upper limit for a is indeed 10.Therefore, the answer is ( a leq 10 ), so the range of ( a ) is ( (-infty, 10] ).Final AnswerThe range of the real number ( a ) is boxed{(-infty, 10]}.

✅:Given ( f(x) ) as an odd function and ( g(x) ) as an even function such that ( f(x) + g(x) = 2^x ), we derived the expressions for ( f(x) ) and ( g(x) ):[f(x) = frac{2^x - 2^{-x}}{2}, quad g(x) = frac{2^x + 2^{-x}}{2}]We need to find the range of ( a ) such that the inequality ( a f(x) - f(3x) leq 2 g(2x) ) holds for ( x in left[frac{1}{2}, 2right] ). Substituting the expressions for ( f(x) ), ( f(3x) ), and ( g(2x) ) into the inequality, we get:[a cdot frac{2^x - 2^{-x}}{2} - frac{2^{3x} - 2^{-3x}}{2} leq 2 cdot frac{2^{2x} + 2^{-2x}}{2}]Multiplying through by 2 and simplifying, we substitute ( t = 2^x ) (where ( t in [sqrt{2}, 4] )) to transform the inequality into:[a(t - 1/t) - (t^3 - 1/t^3) leq 2(t^2 + 1/t^2)]This simplifies to:[a leq frac{t^3 - 1/t^3 + 2t^2 + 2/t^2}{t - 1/t}]We introduce ( u = t - 1/t ) and rewrite the right-hand side (RHS) in terms of ( u ):[RHS(t) = u^2 + 2u + 3 + frac{4}{u}]To find the minimum value of ( RHS(t) ) over ( u in [sqrt{2}/2, 15/4] ), we solve the cubic equation ( u^3 + u^2 - 2 = 0 ), finding the critical point ( u = 1 ). Evaluating ( RHS(t) ) at critical points and endpoints, we find the minimum value is 10 at ( u = 1 ).Thus, the range of ( a ) is:[boxed{(-infty, 10]}]

🔑:1. Given the functions ( f(x) ) and ( g(x) ), defined in ( R ) as an odd function and an even function respectively, and the equation: [ f(x) + g(x) = 2^x ]2. Additionally, the conditions also give another equation since ( f(x) ) is odd and ( g(x) ) is even: [ f(-x) = -f(x) quad text{and} quad g(-x) = g(x) ]3. By using these properties, we can derive another equation by evaluating the initial equation at (-x): [ f(-x) + g(-x) = 2^{-x} implies -f(x) + g(x) = 2^{-x} ]4. Now, solve this system of equations: [ left{ begin{array}{l} f(x) + g(x) = 2^x, -f(x) + g(x) = 2^{-x} end{array} right. Rightarrow left{ begin{array}{l} f(x) = frac{2^x - 2^{-x}}{2}, g(x) = frac{2^x + 2^{-x}}{2} end{array} right. ]5. Substitute these into the inequality ( a f(x) - f(3x) leq 2 g(2x) ): [ a left( frac{2^x - 2^{-x}}{2} right) - left( frac{2^{3x} - 2^{-3x}}{2} right) leq 2 left( frac{2^{2x} + 2^{-2x}}{2} right) ]6. Simplifying, we multiply everything by 2 to eliminate the denominators: [ a (2^x - 2^{-x}) - (2^{3x} - 2^{-3x}) leq 2 (2^{2x} + 2^{-2x}) ]7. Recognize the pattern substitution ( t = 2^x - 2^{-x} ), noting that for ( x in left[ frac{1}{2}, 2 right] ), ( t in left[ frac{sqrt{2}}{2}, frac{15}{4} right] ). Then the equation converts to: [ a t - (t^3 + 3 t) leq 2 (t^2 + 2) ]8. Simplify the inequality: [ a t leq t^3 + 2 t^2 + 3 t + 4 ] [ a leq t^2 + 2 t + frac{4}{t} + 3 ]9. Define the function ( h(t) = t^2 + 2t + frac{4}{t} + 3 ) and find its derivative to check the monotony: [ h'(t) = 2t + 2 - frac{4}{t^2} = frac{2(t^3 + t^2 - 2)}{t^2} = frac{2(t - 1)(t^2 + 2t + 2)}{t^2} ]10. Analyze the derivative sign: - ( h(t) ) is decreasing on ( left( frac{sqrt{2}}{2}, 1 right) ). - ( h(t) ) is increasing on ( left( 1, frac{15}{4} right) ).11. The minimum value of ( h(t) ) is at ( t = 1 ): [ h(1) = 1^2 + 2(1) + frac{4}{1} + 3 = 1 + 2 + 4 + 3 = 10 ]12. Therefore, the range for the real number ( a ) is: [ a in (-infty, 10] ]Conclusion. (boxed{(-infty, 10]})

🧠:Okay, so I have this problem to simplify the expression: arctan[(1 + |x| - sqrt(1 - x²))/(1 + |x| + sqrt(1 - x²))] + (1/2) arccos|x|, where x is between -1 and 1. Hmm, let's try to figure this out step by step.First, since there's an absolute value of x in both terms, maybe I can consider the case when x is non-negative and when x is negative separately. But since |x| is the same for x and -x, maybe the expression is even? Let me check. If I replace x with -x, |x| remains the same, and arccos|x| is also the same. So the whole expression should be the same for x and -x. That means I can just consider x in [0,1], and whatever result I get there will apply to the whole interval [-1,1]. That might simplify things.So let's assume x is in [0,1]. Then |x| = x. So the expression becomes arctan[(1 + x - sqrt(1 - x²))/(1 + x + sqrt(1 - x²))] + (1/2) arccos x. Now, I need to simplify this.Looking at the arctan term: the argument is a fraction. Let me denote numerator as N = 1 + x - sqrt(1 - x²) and denominator as D = 1 + x + sqrt(1 - x²). So the arctan term is arctan(N/D). Maybe there's a trigonometric identity or substitution that can help here.Alternatively, perhaps a substitution like setting x = sinθ or x = cosθ. Since x is between 0 and 1, θ would be between 0 and π/2 if we use x = sinθ or x = cosθ. Let's try that. Let me set x = sinθ, where θ ∈ [0, π/2]. Then sqrt(1 - x²) = sqrt(1 - sin²θ) = cosθ. Let's see what the expression becomes.Substituting x = sinθ, numerator N = 1 + sinθ - cosθ, denominator D = 1 + sinθ + cosθ. So arctan[(1 + sinθ - cosθ)/(1 + sinθ + cosθ)]. Hmm, this looks like it could be related to a tangent of some angle. Maybe we can use some angle addition formulas or express this ratio in terms of tan(θ/2) or something like that.Alternatively, let's note that (1 + sinθ - cosθ)/(1 + sinθ + cosθ) might simplify if we multiply numerator and denominator by something. Let's try to rationalize or manipulate the fraction.First, let's compute N/D:N = 1 + sinθ - cosθD = 1 + sinθ + cosθLet me divide numerator and denominator by cosθ. Then:N/cosθ = (1/cosθ) + tanθ - 1D/cosθ = (1/cosθ) + tanθ + 1Hmm, not sure if that helps. Alternatively, maybe write 1 + sinθ as 2 sin²(θ/2) + 2 sin(θ/2)cos(θ/2) + cos²(θ/2) - sin²(θ/2). Wait, maybe another approach.Wait, 1 + sinθ can be written as [sin(θ/2) + cos(θ/2)]², right? Because sinθ = 2 sin(θ/2)cos(θ/2), so 1 + sinθ = 1 + 2 sin(θ/2)cos(θ/2) = [sin(θ/2) + cos(θ/2)]². Similarly, 1 + sinθ + cosθ = [sin(θ/2) + cos(θ/2)]² + cosθ. Wait, maybe not. Let me check:1 + sinθ = sin²(θ/2) + cos²(θ/2) + 2 sin(θ/2)cos(θ/2) = [sin(θ/2) + cos(θ/2)]². Yes, that's correct. Similarly, cosθ = cos²(θ/2) - sin²(θ/2). So:N = [sin(θ/2) + cos(θ/2)]² - [cos²(θ/2) - sin²(θ/2)] - cosθ. Wait, maybe I need to write N and D in terms of θ/2.Wait, N = 1 + sinθ - cosθ = [sin(θ/2) + cos(θ/2)]² - [cos²(θ/2) - sin²(θ/2)].Let me compute that:[sin²(θ/2) + 2 sin(θ/2)cos(θ/2) + cos²(θ/2)] - [cos²(θ/2) - sin²(θ/2)] =[1 + 2 sin(θ/2)cos(θ/2)] - [cos²(θ/2) - sin²(θ/2)] =1 + sinθ - cosθ. Hmm, not sure that's helpful.Alternatively, maybe express N and D in terms of tan(θ/2). Let t = tan(θ/2). Then sinθ = 2t/(1 + t²), cosθ = (1 - t²)/(1 + t²). Let's substitute these into N and D.N = 1 + (2t)/(1 + t²) - (1 - t²)/(1 + t²)= [ (1 + t²) + 2t - (1 - t²) ] / (1 + t²)= [1 + t² + 2t -1 + t²]/(1 + t²)= [2t² + 2t]/(1 + t²) = 2t(t + 1)/(1 + t²)Similarly, D = 1 + (2t)/(1 + t²) + (1 - t²)/(1 + t²)= [ (1 + t²) + 2t + (1 - t²) ] / (1 + t²)= [1 + t² + 2t + 1 - t²]/(1 + t²)= [2 + 2t]/(1 + t²) = 2(t + 1)/(1 + t²)So N/D = [2t(t + 1)/(1 + t²)] / [2(t + 1)/(1 + t²)] = t.Therefore, N/D = tan(θ/2). So arctan(N/D) = arctan(tan(θ/2)) = θ/2, provided that θ/2 is within the principal value of arctan, which is (-π/2, π/2). Since θ is in [0, π/2], θ/2 is in [0, π/4], which is within the principal value. So arctan(N/D) = θ/2.But θ was defined such that x = sinθ, so θ = arcsin x. Therefore, arctan(N/D) = (arcsin x)/2.So the arctan term simplifies to (1/2) arcsin x. Then the entire expression becomes:(1/2) arcsin x + (1/2) arccos x.Hmm, combining these terms. Let's recall that arcsin x + arccos x = π/2. So (1/2)(arcsin x + arccos x) = (1/2)(π/2) = π/4.Wait, but here we have (1/2) arcsin x + (1/2) arccos x = (1/2)(arcsin x + arccos x) = (1/2)(π/2) = π/4.So the entire expression simplifies to π/4 regardless of x? But let me verify this with a specific value.Take x = 0. Then arctan[(1 + 0 - 1)/(1 + 0 + 1)] + (1/2) arccos 0. Let's compute:Numerator: 1 + 0 - 1 = 0, denominator: 1 + 0 + 1 = 2. So arctan(0/2) = arctan(0) = 0. Then (1/2) arccos 0 = (1/2)(π/2) = π/4. So total expression is 0 + π/4 = π/4. Correct.Another test: x = 1. Then arctan[(1 + 1 - 0)/(1 + 1 + 0)] + (1/2) arccos 1.Numerator: 2, denominator: 2. So arctan(1) = π/4. Then (1/2) arccos 1 = (1/2)(0) = 0. So total is π/4 + 0 = π/4. Correct again.Another test: x = 1/√2. Then sqrt(1 - x²) = sqrt(1 - 1/2) = sqrt(1/2) = 1/√2.Numerator: 1 + 1/√2 - 1/√2 = 1. Denominator: 1 + 1/√2 + 1/√2 = 1 + 2/√2 = 1 + √2. So arctan(1/(1 + √2)).But 1/(1 + √2) = √2 - 1, since (1 + √2)(√2 -1 ) = 2 -1 =1. So arctan(√2 -1 ). What's arctan(√2 -1 )? Since tan(π/8) = √2 -1. So arctan(√2 -1 ) = π/8.Then (1/2) arccos(1/√2). arccos(1/√2) = π/4, so (1/2)(π/4) = π/8. Total expression: π/8 + π/8 = π/4. Correct again.So it seems that regardless of x, the expression simplifies to π/4. So the answer is π/4. Therefore, the simplified form is π/4.But let me just make sure there are no edge cases or mistakes in the substitution. When I set x = sinθ, θ is in [0, π/2], so arcsin x is in [0, π/2], and then θ/2 is in [0, π/4]. Then arctan(tan(θ/2)) is θ/2, correct. Then the arctan term is (1/2) arcsin x. Then adding (1/2) arccos x. Since arcsin x + arccos x = π/2, then their average is π/4. So yes, regardless of x, the sum is π/4.Hence, the simplified expression is π/4.Final Answerboxed{dfrac{pi}{4}}

✅:To simplify the expression (arctan frac{1+|x|-sqrt{1-x^{2}}}{1+|x|+sqrt{1-x^{2}}} + frac{1}{2} arccos |x|) for (-1 leqslant x leqslant 1), we start by considering (x) in the interval ([0, 1]) due to the symmetry of the absolute value function. Let (x = sin theta) where (theta in [0, frac{pi}{2}]). Then, (sqrt{1 - x^2} = cos theta). Substituting (x = sin theta) into the expression, we get:[arctan left( frac{1 + sin theta - cos theta}{1 + sin theta + cos theta} right) + frac{1}{2} arccos (sin theta)]We simplify the argument of the arctangent function. Let (t = tan(theta/2)). Using trigonometric identities, we find:[frac{1 + sin theta - cos theta}{1 + sin theta + cos theta} = tan left( frac{theta}{2} right)]Thus, the arctangent term simplifies to (frac{theta}{2}). Since (theta = arcsin x), this term becomes (frac{1}{2} arcsin x).The second term, (frac{1}{2} arccos (sin theta)), can be rewritten using the identity (arccos (sin theta) = frac{pi}{2} - theta). Therefore, this term becomes (frac{1}{2} left( frac{pi}{2} - arcsin x right)).Combining both terms, we have:[frac{1}{2} arcsin x + frac{1}{2} left( frac{pi}{2} - arcsin x right) = frac{1}{2} arcsin x + frac{pi}{4} - frac{1}{2} arcsin x = frac{pi}{4}]Thus, the simplified expression is:[boxed{dfrac{pi}{4}}]

🔑:1. Understand the Given Expression: The goal is to simplify the expression: [ arctan frac{1+|x|-sqrt{1-x^{2}}}{1+|x|+sqrt{1-x^{2}}} + frac{1}{2} arccos |x| quad text{for} quad -1 leqslant x leqslant 1. ]2. Introduce a Variable: Define ( theta ) as follows: [ |x| in [0,1], quad frac{1+|x|-sqrt{1-x^{2}}}{1+|x|+sqrt{1-x^{2}}} = theta. ] Then, we need to simplify the expression to ( 2theta + varphi ), where (varphi = arccos |x|).3. Complex Number Interpretation: Let: [ z = (1 + |x| + sqrt{1 - x^2}) + (1 + |x| - sqrt{1 - x^2}) mathrm{i}, ] and: [ omega = |x| + sqrt{1 - x^2} mathrm{i}. ] Calculate the argument of ( z^2 cdot omega ).4. Calculate ( z^2 cdot omega ): [ z = (1 + |x| + sqrt{1 - x^2}) + (1 + |x| - sqrt{1 - x^2}) mathrm{i}. ] Simplifying: [ z = (1 + |x| + sqrt{1 - x^2}) + (1 + |x| - sqrt{1 - x^2}) mathrm{i} ] Now, calculate ( z^2 ): [ z^2 = left[ (1 + |x| + sqrt{1 - x^2}) + (1 + |x| - sqrt{1 - x^2}) mathrm{i} right]^2 ]5. Calculate ( arg left(z^2 cdot omegaright) ): From the calculation above: [ z^2 cdot omega = 4(1 + |x|) mathrm{i}. ] Therefore, [ arg left(z^2 cdot omegaright) = frac{pi}{2}. ]6. Conclusion: Now we have: [ 2theta + varphi = frac{pi}{2}. ] Simplify the original expression: [ theta + frac{varphi}{2} = frac{2theta + varphi}{2} = frac{pi}{4}. ] Therefore, the simplified expression is: [ boxed{frac{pi}{4}}. ]

🧠:Okay, so I have this problem here about dividing balls between two boys, and I need to figure out how many ways they can do it under certain conditions. Let me start by understanding the problem step by step.First, the balls: there are 6 white, 4 black, and 8 red balls. They all differ only in color, so within each color, the balls are indistinct. That probably means that when we divide them, what matters is how many of each color each boy gets, not which specific ball. So it's a problem of distributing identical items into distinct groups with some constraints.The task is to divide all these balls between two boys, not necessarily equally, but each boy must get at least two balls of each color. Hmm. So for each color, each boy has to have at least two. Let's break that down.Let's take each color separately because the constraints are per color. For each color, we need to distribute the balls such that each boy gets at least two. But wait, the problem says "each boy gets at least two balls of each color." So for each color, each boy must have at least two. But let's check the counts:For white balls: total 6. Each boy needs at least two. So if each boy gets at least two, that's 2*2=4 white balls. But there are 6, so there's 6-4=2 extra white balls to distribute. Similarly, for black balls: total 4. Each boy must have at least two. Wait, 2*2=4, which is exactly the total. So there are zero extra black balls. And for red balls: total 8. Each boy needs at least two, so 2*2=4, leaving 8-4=4 extra red balls.Wait a second. For black balls, if each boy must have at least two, but there are only 4 in total, then each boy must get exactly two black balls. There's no flexibility there. So the distribution for black balls is fixed: each boy gets exactly two. So that's a key point. For black balls, there's only one way to distribute them: 2 and 2.For white balls, since there are 6 total, and each boy must have at least two, the possible distributions are (2,4), (3,3), (4,2). So three possibilities for white balls.For red balls, total 8, each boy must have at least two. So minimum 2 each, so the remaining 4 can be distributed. The number of ways to distribute 4 indistinct red balls between two boys is 4 + 1 = 5 ways (since each boy can get 0 to 4, but since they are indistinct, the number of non-negative integer solutions to x + y = 4, which is 5). But wait, actually, the formula is (n + k -1 choose k -1), where n is the number of items and k is the number of groups. So here, n=4, k=2, so (4 + 2 -1 choose 2 -1) = (5 choose 1) = 5. So 5 ways for red balls.Therefore, since the distributions for each color are independent, the total number of ways is the product of the number of ways for each color. So for white: 3 ways, black: 1 way, red: 5 ways. Total ways: 3*1*5 = 15. But wait, hold on. Is that correct?Wait, no. Because in combinatorics, when distributing identical objects to distinct recipients, the number of ways for each color would be as follows:For white balls: each boy must get at least two. The number of ways to distribute 6 white balls to two boys with each getting at least two. Let's denote the number of white balls each boy gets as w1 and w2, where w1 + w2 = 6, w1 >=2, w2 >=2. Let's subtract 2 from each, so w1' = w1 - 2, w2' = w2 - 2. Then w1' + w2' = 6 - 4 = 2. The number of non-negative integer solutions is (2 + 2 -1 choose 2 -1) = (3 choose 1) = 3. So that's correct, 3 ways.For black balls: each must get at least two, total 4. So w1 + w2 =4, w1 >=2, w2 >=2. Subtract 2 from each: w1' + w2' = 0. Only one solution: w1'=0, w2'=0. So original distribution is (2,2). So only 1 way.For red balls: each must get at least two, total 8. So r1 + r2 =8, r1 >=2, r2 >=2. Subtract 2 from each: r1' + r2' = 4. Number of non-negative solutions is (4 + 2 -1 choose 2 -1) = (5 choose 1) =5. So that's 5 ways.Therefore, the total number of ways is 3*1*5=15. So the answer is 15. But wait, the problem says "in how many ways can two boys divide these balls (not necessarily equally) between themselves in such a way that each boy gets at least two balls of each color?"But wait, the answer I get is 15. But let me think again. Because when you have distributions for each color, you need to consider that each color's distribution is independent. So you multiply the possibilities for each color. So for each color, the number of ways is as computed, so 3*1*5=15. That seems right.But let me check again for each color:White: 6 balls, each boy gets at least two. The possible distributions:- Boy1:2, Boy2:4- Boy1:3, Boy2:3- Boy1:4, Boy2:2So three possibilities.Black: 4 balls, each boy must get exactly two. Only one way.Red: 8 balls, each boy gets at least two. The distributions are:- Boy1:2, Boy2:6- Boy1:3, Boy2:5- Boy1:4, Boy2:4- Boy1:5, Boy2:3- Boy1:6, Boy2:2So five possibilities.Therefore, multiplying 3*1*5=15. So 15 total ways.But wait, the problem says "the balls differ only in color". So each color is a separate type. So the distributions for each color are independent. Therefore, the total number of ways is the product of the number of ways for each color. So yes, 3*1*5=15. Therefore, the answer is 15. But wait, the problem is asking for "the number of ways can two boys divide these balls...". But the balls are different only in color. So each color is a separate group. So in this case, for each color, we distribute the balls of that color between the two boys with the constraints. Since the colors are separate, the distributions are independent. Therefore, multiply the possibilities. So yes, 15.But wait, let me check again. Maybe there's a mistake here.Wait, actually, the problem might not be considering the distributions per color as independent. Because when you distribute the balls, you have to distribute all the balls. But in this case, since the balls of different colors are distinct from each other, but within a color, they are identical. So when distributing the white balls, you have three ways, black one way, red five ways. The total number of distributions is the product. So each possible combination of distributions for each color gives a unique overall distribution. So yes, 3*1*5=15. So the answer should be 15.But let me think of another approach to confirm.Alternatively, suppose we model this as a generating function. For each color, the generating function for the number of ways to distribute the balls such that each boy gets at least two. Then multiply the generating functions for each color and find the coefficient for the term where the total number of balls is... wait, but since we have to distribute all balls, it's actually the product of the individual generating functions for each color. Let me try.For white balls: each boy gets at least two. The generating function for one boy is x^2 + x^3 + ... + x^4 (since total white is 6, so maximum one boy can have is 4 if the other has 2). Wait, but generating functions for two boys, each with at least two, and total 6. So the generating function is (x^2 + x^3 + x^4)^2. But we need the coefficient of x^6. Let's compute that.(x^2 + x^3 + x^4)^2 = x^4(1 + x + x^2)^2. Expanding (1 + x + x^2)^2 = 1 + 2x + 3x^2 + 2x^3 + x^4. Multiply by x^4: x^4 + 2x^5 + 3x^6 + 2x^7 + x^8. The coefficient of x^6 is 3. So that's correct, 3 ways for white.For black balls: generating function is (x^2)^2 = x^4. So only one term, coefficient 1. So 1 way.For red balls: generating function (x^2 + x^3 + x^4 + x^5 + x^6)^2. But total red balls are 8. Each boy must get at least 2, so minimum 2, maximum 6 (since 8 - 2 =6). So generating function (x^2 + x^3 + x^4 + x^5 + x^6)^2. Factor out x^2: x^4(1 + x + x^2 + x^3 + x^4)^2. Then we need the coefficient of x^8 in this, which is the same as the coefficient of x^(8-4)=x^4 in (1 + x + x^2 + x^3 + x^4)^2.Expanding (1 + x + x^2 + x^3 + x^4)^2: Let's compute this.Multiply term by term:1*1 =11*x =x1*x^2 =x^21*x^3 =x^31*x^4 =x^4x*1 =xx*x =x^2x*x^2 =x^3x*x^3 =x^4x*x^4 =x^5x^2*1 =x^2x^2*x =x^3x^2*x^2 =x^4x^2*x^3 =x^5x^2*x^4 =x^6Similarly, proceeding up to x^4*x^4 =x^8.But since we are only interested up to x^4 in the product (since we need coefficient of x^4 in the expansion), let's compute up to that.So terms contributing to x^0: 1*1 =1x^1: 1*x + x*1 =2xx^2: 1*x^2 + x*x + x^2*1 =1 +1 +1=3x^2Wait, no. Wait, to compute the coefficient of x^4 in (1 + x + x^2 + x^3 + x^4)^2, we need to sum the products of terms whose exponents add up to 4.So the coefficient of x^4 is:1* x^4 (from 1 and x^4)x* x^3 (from x and x^3)x^2* x^2 (from x^2 and x^2)x^3* x (from x^3 and x)x^4*1 (from x^4 and 1)So each of these pairs contributes once. So how many?The pairs are:(0,4), (1,3), (2,2), (3,1), (4,0). Since the polynomial is symmetric, each pair except (2,2) has two combinations, but in the square, each term is multiplied in order. Wait, actually, in the expansion, for each i and j, the coefficient of x^{i+j} is the sum over all i and j of a_i a_j where a_i is the coefficient of x^i in the first polynomial and a_j in the second. Since the polynomial is (1 + x + x^2 + x^3 + x^4) multiplied by itself, the coefficient of x^4 is sum_{k=0}^4 a_k a_{4-k}.Since the coefficients a_0=1, a_1=1, a_2=1, a_3=1, a_4=1.Therefore, the coefficient of x^4 is:a_0*a_4 + a_1*a_3 + a_2*a_2 + a_3*a_1 + a_4*a_0=1*1 +1*1 +1*1 +1*1 +1*1=5.Wait, but hold on. That can't be right. Wait, if each a_i is 1 for i=0 to 4, then the coefficient of x^4 is indeed 5. But that seems conflicting with earlier thought.Wait, but (1 + x + x^2 + x^3 + x^4)^2 is:1 + 2x + 3x^2 + 4x^3 + 5x^4 + 4x^5 + 3x^6 + 2x^7 + x^8.Wait, no. Wait, let me actually compute the expansion step by step:Multiply 1*(1 + x + x^2 + x^3 + x^4) =1 +x +x^2 +x^3 +x^4Multiply x*(1 + x + x^2 + x^3 + x^4) =x +x^2 +x^3 +x^4 +x^5Multiply x^2*(1 + x + x^2 + x^3 + x^4)=x^2 +x^3 +x^4 +x^5 +x^6Multiply x^3*(1 + x + x^2 + x^3 + x^4)=x^3 +x^4 +x^5 +x^6 +x^7Multiply x^4*(1 + x + x^2 + x^3 + x^4)=x^4 +x^5 +x^6 +x^7 +x^8Now sum all these terms:x^0:1x^1:1x +1x=2xx^2:1x^2 +1x^2 +1x^2=3x^2x^3:1x^3 +1x^3 +1x^3 +1x^3=4x^3x^4:1x^4 +1x^4 +1x^4 +1x^4 +1x^4=5x^4x^5:1x^5 +1x^5 +1x^5 +1x^5=4x^5x^6:1x^6 +1x^6 +1x^6=3x^6x^7:1x^7 +1x^7=2x^7x^8:1x^8So the expansion is correct. Therefore, the coefficient of x^4 is 5. Therefore, in the generating function for red balls, after factoring out x^4, we need the coefficient of x^4 in (1 + x + x^2 + x^3 + x^4)^2, which is 5. Therefore, the number of ways for red balls is 5. So that's confirmed.Therefore, the total number of distributions is indeed 3*1*5=15.But wait, but the problem says "the balls differ only in color". So the balls are identical except for color, meaning that each color is a separate type. Therefore, the distributions for each color are independent. Therefore, the total number of ways is the product of the ways for each color, which is 15. Therefore, the answer is 15.But let me check with another approach. Suppose we model the problem as follows:Each color must be distributed such that each boy gets at least two. For white, black, and red separately.Since the distributions are independent, the total number is the product. So 3*1*5=15. That's straightforward.Alternatively, suppose we think in terms of choices. For white balls, each boy must get at least two. So the number of ways to distribute 6 white balls with each boy getting at least two is the same as the number of integer solutions to w1 + w2 =6 with w1 >=2 and w2 >=2. As done earlier, which is 3.For black balls, exactly 2 each, so 1 way.For red balls, similar to white, solutions to r1 + r2 =8 with r1 >=2, r2 >=2, which is 5.So total ways 3*1*5=15.Therefore, I think 15 is the correct answer. But let me check again if I didn't miss any constraints.Wait, the problem states "in such a way that each boy gets at least two balls of each color". So for each boy, not just per color. Wait, no. Wait, each boy gets at least two balls of each color. That means that for each color, each boy must have at least two balls. So for example, a boy cannot have only one white ball and three black balls, etc. No, per color, each boy must have at least two. So for each color, each boy must have at least two balls of that color. Therefore, the distribution per color must satisfy that each boy gets at least two of that color. Therefore, the per-color analysis is correct. Therefore, the total number of ways is 15.Alternatively, maybe the problem is interpreted as each boy must have at least two balls in total of each color. Wait, but the problem says "at least two balls of each color". So each color must have at least two balls for each boy. So yes, per color, each boy gets at least two. So the analysis is correct.Therefore, I think the answer is 15.But let me check with a different perspective. Suppose we list all possible distributions.For white: possible distributions (2,4), (3,3), (4,2)For black: only (2,2)For red: (2,6), (3,5), (4,4), (5,3), (6,2)So total combinations: 3 * 1 * 5 =15. Each combination corresponds to a unique way of distributing all the balls. Since the distributions are independent across colors, the total is indeed 15. Therefore, I think the answer is 15.But wait, let's think about whether the balls are distinguishable. The problem says "the balls that differ only in color". So all white balls are identical, all black identical, all red identical. Therefore, when distributing, the count is based on the number of each color given to each boy. So yes, the number of ways is the product of the number of distributions per color. Therefore, 15.I think this is correct. Unless there's a mistake in assuming independence between colors. But since each color is separate and the constraints are per color, the distributions are independent. So multiplying them is valid. Therefore, 15 is the answer.But let me check with a small example. Suppose there are 2 white, 2 black, and 2 red balls, and each boy must get at least one of each color. Then, for each color, there's only one way: each boy gets 1. So total ways 1*1*1=1. Which makes sense: each boy gets 1 of each color. But if there are, say, 3 white balls, 2 black, 2 red. Each boy must get at least one of each color. Then for white: 3-2=1, so two ways: (1,2). For black and red: each exactly 1. So total 2*1*1=2 ways. Which makes sense: the only difference is how the white balls are distributed. So this approach works.Therefore, applying the same logic, the answer should be 15.

✅:To determine the number of ways to distribute 6 white, 4 black, and 8 red balls between two boys such that each boy receives at least two balls of each color, we analyze each color separately and then multiply the results:1. White Balls (6 total): Each boy must receive at least 2. The remaining 2 balls can be distributed in: ( (2,4), (3,3), (4,2) ) → 3 ways.2. Black Balls (4 total): Each boy must receive exactly 2. There is only one distribution: ( (2,2) ) → 1 way.3. Red Balls (8 total): Each boy must receive at least 2. The remaining 4 balls can be distributed in: ( (2,6), (3,5), (4,4), (5,3), (6,2) ) → 5 ways.Total Number of Ways: Multiply the possibilities for each color: ( 3 times 1 times 5 = 15 ).Final Answer: (boxed{15})

🔑:: Given: - 6 white balls- 4 black balls- 8 red ballsEach boy must get at least 2 balls of each color.1. Step 1: Distribute the minimum required balls to both boys. - Each boy receives 2 white balls, 2 black balls, and 2 red balls. - Total balls given to each boy already: (2 + 2 + 2 = 6) balls. - Balls left to be distributed: - White balls left: (6 - 2 times 2 = 2) - Black balls left: (4 - 2 times 2 = 0) - Red balls left: (8 - 2 times 2 = 4)2. Step 2: Distribute the remaining white and red balls. - Balls left to distribute: - 2 white balls - 4 red balls - Each remaining ball of a given color can be given to either boy. 3. Step 3: Calculate the number of ways to distribute the remaining white balls. - Each of the 2 white balls can go to either of the 2 boys: [ text{Number of ways to distribute white balls} = 2 + 1 = 3 ]4. Step 4: Calculate the number of ways to distribute the remaining red balls. - Each of the 4 red balls can go to either of the 2 boys: [ text{Number of ways to distribute red balls} = 4 + 1 = 5 ] 5. Step 5: Calculate the total number of ways to distribute the balls. [ text{Total number of ways} = (text{ways to distribute white balls}) times (text{ways to distribute red balls}) ] [ = 3 times 5 = 15 ] Conclusion:Thus, there are ( boxed{15} ) ways to distribute the balls between the two boys.