🧠:Okay, let's see. I need to figure out how many teeth each gear has. There are three gears: the largest one touches the two smaller ones. All together, they have 60 teeth. When the largest gear is short by 20 teeth to complete four full rotations, the second and third gears make 5 and 10 rotations respectively. Hmm, okay, let's break this down step by step.First, let's denote the number of teeth on the gears. Let’s call the largest gear A, and the two smaller ones B and C. So, A + B + C = 60. That's the total teeth. Got that.Now, the problem says when the largest gear lacks 20 teeth to complete four full rotations. I need to parse this. So, if gear A needs to make four full rotations, but it's short by 20 teeth. That probably means that if gear A had 20 more teeth, it would be able to make exactly four full rotations. Wait, maybe not. Let me think again.When a gear makes a full rotation, the number of teeth that pass by a point is equal to the number of teeth it has. So, if a gear has T teeth, one full rotation would mean T teeth have passed. Therefore, if gear A has T_A teeth, then four full rotations would require 4*T_A teeth to pass by. But the problem says it lacks 20 teeth to complete four full rotations. So, maybe gear A is rotated such that it's just 20 teeth short of four full rotations? So, the number of teeth that actually passed is 4*T_A - 20. Is that right?Alternatively, maybe when the largest gear is turned such that it's 20 teeth short of four rotations, meaning it has turned (4*T_A - 20) teeth. Hmm, but how does this relate to the rotations of the other gears?Since gears B and C are meshed with gear A, the number of teeth that pass on gear A must equal the number of teeth that pass on gears B and C. Wait, but gears B and C are separate, each meshed with A. So, if gear A rotates, it will drive both gears B and C. The number of teeth engaged between A and B would be the same as the number of teeth that pass on both gears at the point of contact. Similarly for A and C. So, the rotation of gear B and gear C will depend on how many teeth have passed from gear A.Wait, but in the problem statement, when gear A is short 20 teeth for four rotations, gears B and C have made 5 and 10 rotations. So, in that scenario, gear A has rotated such that it's 20 teeth short of four full rotations, so the teeth that have passed on gear A would be 4*T_A - 20. Then, gears B and C have rotated 5 and 10 times, so the teeth that have passed on B would be 5*T_B, and on C would be 10*T_C. But since they're meshed with A, the teeth that have passed on A must equal the teeth that have passed on B and C. But wait, gear A is meshed with both B and C, so the teeth on A that have passed would be equal to the teeth on B and also equal to the teeth on C? That can't be, unless B and C have the same number of teeth, which they don't. So maybe the problem is that gears B and C are separate, each meshed with A, so the rotation of A affects both B and C independently. So, when gear A turns, it turns B and C in opposite directions, but the key point is that the number of teeth that pass between A and B must be equal, and the number of teeth that pass between A and C must be equal. Therefore, the number of teeth moved on A is equal to the number of teeth moved on B, and also equal to the number of teeth moved on C. Therefore, the teeth moved on A is equal to both teeth moved on B and on C. But in the problem, when gear A is short 20 teeth for four rotations, gears B and C have made 5 and 10 rotations. So, the teeth moved on B would be 5*T_B, and on C would be 10*T_C, and on A would be 4*T_A - 20. Therefore, we can set up equations:4*T_A - 20 = 5*T_B 4*T_A - 20 = 10*T_CBecause the teeth moved on A must equal the teeth moved on B and on C. So, these two equations. Then, we also have the total teeth:T_A + T_B + T_C = 60So, we have three equations:1) 4*T_A - 20 = 5*T_B 2) 4*T_A - 20 = 10*T_C 3) T_A + T_B + T_C = 60Okay, so let's try to solve these equations.From equations 1 and 2, we can express T_B and T_C in terms of T_A.From equation 1: 5*T_B = 4*T_A - 20 => T_B = (4*T_A - 20)/5From equation 2: 10*T_C = 4*T_A - 20 => T_C = (4*T_A - 20)/10Now, substitute T_B and T_C into equation 3:T_A + [(4*T_A - 20)/5] + [(4*T_A - 20)/10] = 60Let's compute this. To simplify, let's find a common denominator. Let's use 10.Multiply each term by 10 to eliminate denominators:10*T_A + 2*(4*T_A - 20) + (4*T_A - 20) = 600Compute each term:10*T_A + 8*T_A - 40 + 4*T_A - 20 = 600 Combine like terms:10*T_A + 8*T_A + 4*T_A = 22*T_A -40 -20 = -60So:22*T_A - 60 = 600 22*T_A = 660 T_A = 660 / 22 T_A = 30So, gear A has 30 teeth.Then, substitute back into T_B:T_B = (4*30 - 20)/5 = (120 - 20)/5 = 100 /5 = 20And T_C:T_C = (4*30 -20)/10 = (120 -20)/10 = 100/10 =10So, T_A=30, T_B=20, T_C=10. Let's check if these add up to 60.30+20+10=60. Yes.Also, check the other condition. When gear A is short 20 teeth for four rotations. So, four full rotations would be 4*30=120 teeth. Lacking 20 teeth means it moved 120-20=100 teeth.Then, gear B would have moved 100 teeth, so 100 / T_B = 100 /20=5 rotations. Gear C would have moved 100 teeth, so 100 /10=10 rotations. Perfect, that matches the problem statement.So, the answer is 30,20,10 teeth respectively.Wait, but let me just make sure. If gears A, B, and C have 30,20,10 teeth. Then, the largest gear is A with 30, which is correct. When A is rotated such that it's 20 teeth short of four rotations, meaning 4*30 -20=100 teeth have passed. Then, since B and C are meshed with A, they also have 100 teeth passing. So, B, with 20 teeth, makes 100/20=5 rotations. C, with 10 teeth, makes 100/10=10 rotations. That's exactly what the problem states. So all conditions are satisfied. Therefore, the solution is correct.Final AnswerThe largest gear has boxed{30} teeth, the second gear has boxed{20} teeth, and the third gear has boxed{10} teeth.

✅:Given three interconnected gears where the largest gear touches both smaller ones, and all three gears combined have 60 teeth. When the largest gear lacks 20 teeth to complete four full rotations, the second and third gears make 5 and 10 full rotations, respectively. We need to determine the number of teeth on each gear.1. Define variables: - Let ( T_A ) be the number of teeth on the largest gear. - Let ( T_B ) be the number of teeth on the second gear. - Let ( T_C ) be the number of teeth on the third gear.2. Total teeth equation: [ T_A + T_B + T_C = 60 ]3. Relate rotations to teeth: - When the largest gear (A) is 20 teeth short of four full rotations, the teeth moved on A is ( 4T_A - 20 ). - The second gear (B) makes 5 full rotations: ( 5T_B ). - The third gear (C) makes 10 full rotations: ( 10T_C ).4. Set up equations based on teeth moved: [ 4T_A - 20 = 5T_B ] [ 4T_A - 20 = 10T_C ]5. Solve for ( T_B ) and ( T_C ) in terms of ( T_A ): [ T_B = frac{4T_A - 20}{5} ] [ T_C = frac{4T_A - 20}{10} ]6. Substitute ( T_B ) and ( T_C ) into the total teeth equation: [ T_A + frac{4T_A - 20}{5} + frac{4T_A - 20}{10} = 60 ]7. Simplify and solve for ( T_A ): - Multiply through by 10 to clear denominators: [ 10T_A + 2(4T_A - 20) + (4T_A - 20) = 600 ] - Combine like terms: [ 10T_A + 8T_A - 40 + 4T_A - 20 = 600 ] [ 22T_A - 60 = 600 ] [ 22T_A = 660 ] [ T_A = 30 ]8. Find ( T_B ) and ( T_C ): [ T_B = frac{4(30) - 20}{5} = frac{100}{5} = 20 ] [ T_C = frac{4(30) - 20}{10} = frac{100}{10} = 10 ]9. Verify the solution: - Total teeth: ( 30 + 20 + 10 = 60 ) - Teeth moved on A: ( 4 times 30 - 20 = 100 ) - Rotations for B: ( 100 / 20 = 5 ) - Rotations for C: ( 100 / 10 = 10 ) Final AnswerThe largest gear has boxed{30} teeth, the second gear has boxed{20} teeth, and the third gear has boxed{10} teeth.

🔑:Let ( x, y, z ) be the number of teeth on the three gears, with ( x > y > z ). We are given the following conditions:1. The sum of the teeth on all three gears is 60: [ x + y + z = 60 ]2. The gears are linked such that the number of teeth meeting during rotation for each gear is the same. Given the rotation details: - The largest gear ( x ) will make less than 4 complete rotations by 20 teeth: [ 4x - 20 text{ teeth} ] - The second gear ( y ) makes exactly 5 complete rotations: [ 5y text{ teeth} ] - The smallest gear ( z ) makes exactly 10 complete rotations: [ 10z text{ teeth} ]Accordingly, we can write the following equation from the above conditions: [ 4x - 20 = 5y = 10z ]We need to solve this system of equations. # Step-by-step Solution:1. Starting with the relationship between ( y ) and ( z ): [ 5y = 10z implies y = 2z ]2. Now, use the relationship between ( x ) and ( z ): [ 4x - 20 = 10z implies 4x = 10z + 20 implies x = frac{10z + 20}{4} = frac{5z + 10}{2} = frac{5z}{2} + 5 ]3. Substitute ( y = 2z ) and ( x = frac{5z}{2} + 5 ) into the sum equation ( x + y + z = 60 ): [ left(frac{5z}{2} + 5right) + 2z + z = 60 ] [ frac{5z}{2} + 5 + 3z = 60 ]4. Combine the terms: [ frac{5z}{2} + 3z + 5 = 60 ]5. Convert ( 3z ) to a fraction with a common denominator: [ frac{5z}{2} + frac{6z}{2} + 5 = 60 ] [ frac{11z}{2} + 5 = 60 ]6. Isolate ( frac{11z}{2} ): [ frac{11z}{2} = 55 ]7. Solve for ( z ): [ 11z = 110 implies z = 10 ]8. Now, find ( y ) and ( x ): [ y = 2z = 2 times 10 = 20 ] [ x = frac{5z}{2} + 5 = frac{5 times 10}{2} + 5 = 25 + 5 = 30 ]So, the number of teeth on each gear is ( x = 30 ), ( y = 20 ), and ( z = 10 ).Conclusion:[boxed{30, 20, 10 text{ teeth}}]

🧠:Okay, so I need to find the probability that when I break down 2.5 into two non-negative numbers, round each to the nearest integer, their sum is 3. The answer choices are given, so I need to figure this out step by step.First, let me make sure I understand the problem correctly. We have to split 2.5 into two non-negative numbers, say x and y, such that x + y = 2.5. Then, we round each of x and y to the nearest integer. We need to find the probability that the sum of these two rounded integers is 3. The options are A through E, with various fractions.So, probability here would depend on the different ways we can split 2.5 into x and y, considering all possible splits, and then seeing in what fraction of those splits the rounded values add up to 3. Since the splitting is random, I assume that every possible split is equally likely. Therefore, the problem reduces to determining the measure (length, since it's a continuous interval) of the set of x and y where the sum of their rounded values is 3, divided by the total measure of all possible splits.Since x and y are non-negative and sum to 2.5, we can represent this as x ranging from 0 to 2.5, and y being 2.5 - x. So effectively, we can model this as a one-dimensional problem with x varying from 0 to 2.5. The key is to determine for each x in [0, 2.5], what the rounded values of x and y = 2.5 - x would be, and check if their sum is 3. Then, the probability would be the length of the intervals where this condition holds divided by the total length (2.5).But first, let's recall how rounding works. A number is rounded to the nearest integer. The midpoint between two integers is the point where we round up. For example, 0.5 rounds up to 1, 1.5 rounds up to 2, etc.So, for any real number z, the rounded integer is floor(z + 0.5). So, the rounding function can be defined as follows: if the decimal part is 0.5 or higher, we round up, otherwise down.Given x and y = 2.5 - x, we need round(x) + round(y) = 3.Let’s denote round(x) as a and round(y) as b. So, we want a + b = 3. Since x and y are non-negative, a and b are non-negative integers.Possible pairs (a, b) that sum to 3 are (0, 3), (1, 2), (2, 1), (3, 0). But we need to check which of these pairs are possible given that x + y = 2.5.But x and y are both non-negative and sum to 2.5. So, x can be at most 2.5, which would mean y = 0. Similarly, y can be at most 2.5. Therefore, the rounded integers a and b must satisfy certain constraints.For example, if x is 0, then y = 2.5, round(x) = 0, round(y) = 3 (since 2.5 rounds up to 3). So, sum is 0 + 3 = 3. Similarly, if x is 2.5, then y = 0, round(x) = 3 (since 2.5 rounds up to 3), round(y) = 0. So, sum is 3 + 0 = 3. So, the endpoints give sum 3.But what about the middle values? Let's consider possible pairs:1. (0, 3): This would require x to round to 0 and y to round to 3. Since y = 2.5 - x, y must be in [2.5, 3.5) to round to 3. But since x is non-negative, y = 2.5 - x ≤ 2.5. Therefore, y can't be in [2.5, 3.5) unless x is negative, which it's not. So the only y that rounds to 3 is y = 2.5 (since 2.5 rounds up to 3). But y = 2.5 implies x = 0. So, this case is only the single point x = 0. However, in continuous probability, individual points have measure zero, so this contributes nothing to the probability.Wait, but perhaps when x is very close to 0, such that y is just below 2.5, but still rounds to 2.5? Wait, y is 2.5 - x. So, if x is in [0, 0.5), then y = 2.5 - x is in (2.0, 2.5]. Then, y in (2.0, 2.5) would round to 2, since 2.5 rounds up. Wait, 2.5 is the midpoint between 2 and 3. So numbers less than 2.5 round down, and 2.5 rounds up. Wait, is that correct?Wait, the standard rounding rule is that 0.5 rounds up. So, for a number z, if the decimal part is 0.5 or higher, we round up. So, for example, 2.5 would round to 3, but 2.499... would round to 2. Similarly, 2.0 rounds to 2, 2.1, up to 2.5 would round to 3. Wait, no, 2.5 is the midpoint between 2 and 3. So, z in [2.0, 2.5) would round to 2, and z in [2.5, 3.0) would round to 3. So, 2.5 itself is included in the upper interval.Therefore, for y = 2.5 - x, when does y round to 3? y must be ≥ 2.5. But since y = 2.5 - x, y ≥ 2.5 implies that 2.5 - x ≥ 2.5 → -x ≥ 0 → x ≤ 0. But x is non-negative, so x must be 0. So, only when x = 0, y = 2.5, which rounds to 3. Similarly, when x = 2.5, y = 0, which rounds to 0. But again, these are single points.Therefore, the pair (0, 3) and (3, 0) only occur at the endpoints, which have measure zero. So, effectively, the only pairs that contribute to the probability are (1, 2) and (2, 1). So, we need to find the measure of x where round(x) = 1 and round(y) = 2, plus the measure where round(x) = 2 and round(y) = 1.So, let's first find when round(x) = 1. That occurs when x is in [0.5, 1.5). Similarly, round(y) = 2 when y is in [1.5, 2.5). But y = 2.5 - x, so y in [1.5, 2.5) translates to 2.5 - x ∈ [1.5, 2.5) → x ∈ (0.0, 1.0]. So, for round(x) = 1 and round(y) = 2, x must be in [0.5, 1.5) (for round(x)=1) AND x ∈ (0.0, 1.0] (for round(y)=2). The intersection of these intervals is [0.5, 1.0].Similarly, for round(x) = 2 and round(y) = 1. round(x) = 2 when x ∈ [1.5, 2.5). round(y) = 1 when y ∈ [0.5, 1.5). Since y = 2.5 - x, y ∈ [0.5, 1.5) translates to 2.5 - x ∈ [0.5, 1.5) → x ∈ (1.0, 2.0]. So, round(x)=2 requires x ∈ [1.5, 2.5), and x ∈ (1.0, 2.0]. The intersection is [1.5, 2.0].Therefore, the total measure where the sum is 3 is the length of [0.5, 1.0] plus the length of [1.5, 2.0]. Each interval is 0.5 in length, so total measure is 0.5 + 0.5 = 1.0. The total possible x values are from 0 to 2.5, which is 2.5 in length. Therefore, the probability would be 1.0 / 2.5 = 0.4, which is 2/5. So, the answer is B.But wait, let me double-check to make sure.First, let's confirm the intervals:Case 1: round(x) = 1 and round(y) = 2.round(x) = 1 ⇒ x ∈ [0.5, 1.5).round(y) = 2 ⇒ y ∈ [1.5, 2.5).But y = 2.5 - x, so:1.5 ≤ 2.5 - x < 2.5 ⇒ subtract 2.5:-1.0 ≤ -x < 0 ⇒ multiply by -1 (reverse inequalities):0 < x ≤ 1.0.Therefore, x ∈ (0, 1.0].But round(x) must be 1, so x ∈ [0.5, 1.5). The intersection of (0, 1.0] and [0.5, 1.5) is [0.5, 1.0]. So, length 0.5.Case 2: round(x) = 2 and round(y) = 1.round(x) = 2 ⇒ x ∈ [1.5, 2.5).round(y) = 1 ⇒ y ∈ [0.5, 1.5).y = 2.5 - x, so:0.5 ≤ 2.5 - x < 1.5 ⇒ subtract 2.5:-2.0 ≤ -x < -1.0 ⇒ multiply by -1 (reverse inequalities):1.0 < x ≤ 2.0.So x ∈ (1.0, 2.0].But round(x) = 2 requires x ∈ [1.5, 2.5). The intersection is [1.5, 2.0]. Length 0.5.Total favorable length: 0.5 + 0.5 = 1.0. Total possible length: 2.5. Probability: 1.0 / 2.5 = 2/5. So answer B.But let me think again. Is there any other case where the sum can be 3? The other possible pairs were (0, 3) and (3, 0), but those only occur at the endpoints x=0 and x=2.5, which are single points and don't contribute to the length. Therefore, they can be ignored in a continuous probability distribution.Therefore, the answer is B, 2/5.Wait, but let me check another approach for verification. Let's graph the function.Imagine x from 0 to 2.5 on a number line. For each x, y = 2.5 - x. The rounding of x and y can be visualized.For x in [0, 0.5), round(x) = 0. Then y = 2.5 - x is in (2.0, 2.5]. Round(y) is 2 if y is in [2.0, 2.5), which is x in (0, 0.5]. Wait, when x is in [0, 0.5), y is in (2.0, 2.5]. But 2.5 is included in the upper interval. Wait, y = 2.5 when x=0. So, round(y) when x is in [0, 0.5) is:y ∈ (2.0, 2.5). So, numbers from 2.0 to 2.5. The rounding of y would be:- For y in [2.0, 2.5), the rounded value is 2 if y < 2.5, and 3 if y ≥ 2.5. But y = 2.5 only when x = 0, which is a single point.Therefore, for x in [0, 0.5), y is in (2.0, 2.5). So, y rounds to 2. Therefore, sum of round(x) + round(y) is 0 + 2 = 2.At x = 0.5, x rounds to 1 (since 0.5 rounds up). Then y = 2.5 - 0.5 = 2.0. y = 2.0 rounds to 2. So sum is 1 + 2 = 3.Wait, so when x is exactly 0.5, the sum is 3. But x = 0.5 is a single point. Similarly, moving from x=0.5 upwards.Wait, perhaps my previous interval analysis was missing some edge cases?Wait, let's step through the intervals carefully:1. x ∈ [0, 0.5): - round(x) = 0. - y = 2.5 - x ∈ (2.0, 2.5]. - For y in (2.0, 2.5), round(y) = 2 (since 2.5 is the midpoint; but 2.5 is not included in this interval except when x=0, which is y=2.5. But x=0 is a single point. So, for x ∈ (0, 0.5), y ∈ (2.0, 2.5). So round(y) = 2. - Therefore, sum is 0 + 2 = 2.2. x = 0.5: - round(x) = 1. - y = 2.0, which rounds to 2. - Sum is 1 + 2 = 3.3. x ∈ (0.5, 1.5): - round(x) = 1. - y = 2.5 - x ∈ (1.0, 2.0). - For y ∈ (1.0, 2.0): - If y ∈ [1.0, 1.5), round(y) = 1. - If y ∈ [1.5, 2.0), round(y) = 2. - So, need to find for x ∈ (0.5, 1.5): - y = 2.5 - x ∈ (1.0, 2.0). - So, split into two intervals: a. When y ∈ [1.0, 1.5) → x ∈ (1.0, 2.5 - 1.5] → x ∈ (1.0, 1.0]? Wait, no. Let's compute: - y ≥ 1.0: 2.5 - x ≥ 1.0 → x ≤ 1.5. - y < 1.5: 2.5 - x < 1.5 → x > 1.0. Therefore, for y ∈ [1.0, 1.5), x ∈ (1.0, 1.5]. b. When y ∈ [1.5, 2.0): x ∈ [0.5, 1.0). - Wait, same process: y ≥ 1.5: 2.5 - x ≥ 1.5 → x ≤ 1.0. y < 2.0: 2.5 - x < 2.0 → x > 0.5. So, x ∈ (0.5, 1.0]. So, for x ∈ (0.5, 1.0], y ∈ [1.5, 2.0), round(y) = 2. So sum is 1 + 2 = 3. For x ∈ (1.0, 1.5], y ∈ [1.0, 1.5), round(y) = 1. So sum is 1 + 1 = 2. - Therefore, in x ∈ (0.5, 1.5), sum is 3 when x ∈ (0.5, 1.0], and sum is 2 when x ∈ (1.0, 1.5).4. x ∈ [1.5, 2.5): - round(x) = 2. - y = 2.5 - x ∈ (0.0, 1.0]. - For y ∈ (0.0, 1.0]: - If y ∈ [0.0, 0.5), round(y) = 0. - If y ∈ [0.5, 1.0), round(y) = 1. - y = 1.0 rounds to 1. - So, split into: a. y ∈ [0.0, 0.5): x ∈ (2.0, 2.5] - Wait, y = 2.5 - x < 0.5 → x > 2.0. - But x ∈ [1.5, 2.5), so x ∈ (2.0, 2.5). Then, round(y) = 0. - Sum is 2 + 0 = 2. b. y ∈ [0.5, 1.0]: x ∈ [1.5, 2.0]. - y ≥ 0.5 → 2.5 - x ≥ 0.5 → x ≤ 2.0. - y ≤ 1.0 → 2.5 - x ≤ 1.0 → x ≥ 1.5. - So x ∈ [1.5, 2.0]. - round(y) = 1. - Sum is 2 + 1 = 3. - Therefore, for x ∈ [1.5, 2.0], sum is 3, and for x ∈ (2.0, 2.5), sum is 2.Putting it all together:- x ∈ [0, 0.5): sum 2- x = 0.5: sum 3 (measure zero)- x ∈ (0.5, 1.0]: sum 3- x ∈ (1.0, 1.5): sum 2- x ∈ [1.5, 2.0]: sum 3- x ∈ (2.0, 2.5): sum 2Therefore, the intervals where sum is 3 are (0.5, 1.0] and [1.5, 2.0].The length of (0.5, 1.0] is 0.5, and the length of [1.5, 2.0] is 0.5. Total length: 1.0.Total interval length: 2.5.Probability: 1.0 / 2.5 = 0.4 = 2/5.So, the answer is B.I think this is thorough. Let me check if there's a different interpretation.Another way to think about this is that for the sum of the rounded numbers to be 3, one of them must round up and the other down in such a way that their sum is 3. Since the original numbers add to 2.5, which is a half-integer, maybe there's symmetry here.Alternatively, note that the total rounded sum can be 2, 3, or 4. Because the original numbers add to 2.5. When rounding, each number can either round down or up. The sum of the rounded numbers can be as low as 2 (if both round down: 0 + 2) or as high as 4 (if both round up: 3 + 1). Wait, but 2.5 split into two numbers, each can round to 0,1,2, or 3.Wait, for example, x=0.5 rounds to 1, y=2.0 rounds to 2. Sum 3.If x=1.5 rounds to 2, y=1.0 rounds to 1. Sum 3.But if x=0.6, rounds to 1, y=1.9, rounds to 2. Sum 3.If x=1.4, rounds to 1, y=1.1, rounds to 1. Sum 2.If x=2.4, rounds to 2, y=0.1, rounds to 0. Sum 2.x=2.0, rounds to 2, y=0.5 rounds to 1. Sum 3.Wait, so the intervals where x is between 0.5 to 1.0 and 1.5 to 2.0 contribute to the sum 3.Which is exactly what we found before. So, these intervals each contribute 0.5 length, totaling 1.0. Therefore, the probability is 1.0 / 2.5 = 2/5.Therefore, the answer is B.

✅:To determine the probability that the sum of the two rounded integers is 3 when 2.5 is split into two non-negative numbers (x) and (y) (where (y = 2.5 - x)), we analyze the intervals where rounding (x) and (y) results in a sum of 3.1. Rounding Rules: A number rounds to the nearest integer, with midpoints (e.g., 0.5, 1.5) rounding up. Thus: - ( text{round}(z) = k ) if ( z in [k - 0.5, k + 0.5) ) for integer (k).2. Possible Cases for Sum 3: - The pairs ((1, 2)) and ((2, 1)) are feasible since ((0, 3)) and ((3, 0)) only occur at measure-zero endpoints.3. Interval Analysis: - Case 1: ( text{round}(x) = 1 ) and ( text{round}(y) = 2 ) - ( x in [0.5, 1.5) ) and ( y = 2.5 - x in [1.5, 2.5) ). - Intersection: ( x in [0.5, 1.0] ). - Case 2: ( text{round}(x) = 2 ) and ( text{round}(y) = 1 ) - ( x in [1.5, 2.5) ) and ( y = 2.5 - x in [0.5, 1.5) ). - Intersection: ( x in [1.5, 2.0] ).4. Total Favorable Length: - Each interval ([0.5, 1.0]) and ([1.5, 2.0]) has length 0.5. - Combined length: (0.5 + 0.5 = 1.0).5. Probability Calculation: - Total interval length: (2.5). - Probability: ( frac{1.0}{2.5} = frac{2}{5} ).Answer: (boxed{B})

🔑:We need to determine the probability that the sum of two random non-negative numbers that add up to 2.5 becomes 3 when each number is rounded to the nearest integer.1. Decomposition of 2.5 into Two Non-Negative Numbers: [ x + y = 2.5 ] with x, y geq 0.2. Possible Combinations and Their Rounding: - Both numbers, x and y, will be rounded to their nearest integers. We consider all possible pairs (x, y) such that x + y = 2.5, and analyze their integer rounding: - If x in [0, 0.5), then y in (2, 2.5] rightarrow 0 + 2 = 2. - If x in [0.5, 1.5), then y in (1.0, 2.0) rightarrow 1 + 1 = 2. - If x in [1.5, 2.5], then y in (0, 1.0) rightarrow 2 + 0 = 2. - If x in [1.5, 2.5) and y = 2.5-x, both numbers will round as: - (x in [1.5, 2.5)) ensures either 2 + 0 or 2 + 1 or 1 + 2.3. Identifying Pairs that Round to a Sum of 3: - A pair (x, y) needs to be such that when rounded their sum is 3: - (2.143 + 0.357) rounds to 2 + 0 = 2 - (sqrt{3} + (2.5 - sqrt{3})) rounds to (2 + 1 = 3) - Thus, the combinations yielding x and y to sum to 3 are: [ { (1.5, 1.0), (1.0, 1.5) } ]4. Calculating the Probability: - Let's observe the intervals. Consider the line segment (0 leq x leq 2.5): - Event occur in: - 1.5 leq x < 2.5 rightarrow 1.0 < y leq 1.55. Segments to Valid Integers: - Out of these ranges, the valid combinations yielding 3 (two out of the five segments). Therefore: [ frac{text{Number of Valid Segments}}{text{Total Possible Segments}} = frac{2}{5} ]6. Conclusion: [ boxed{frac{2}{5}} ]

🧠:Okay, so I have this problem here where I need to prove that if positive integers a, b, k satisfy the equation (a² + b²)/(ab - 1) = k, then k must be 5. Hmm, interesting. Let me try to figure this out step by step.First, let me restate the equation to make sure I understand it. The equation is (a² + b²) divided by (ab - 1) equals k, and all a, b, k are positive integers. The goal is to show that k can only be 5. So, regardless of the values of a and b, as long as they're positive integers and satisfy that equation, k must be 5. That seems specific. Maybe there's some underlying structure here that forces k to be 5.Let me start by rearranging the equation. If (a² + b²)/(ab - 1) = k, then multiplying both sides by (ab - 1) gives:a² + b² = k(ab - 1)So, expanding that, we have:a² + b² = kab - kLet me bring all terms to one side to see if I can factor it or find some relationship:a² - kab + b² + k = 0Hmm, this looks a bit like a quadratic in terms of a or b. Maybe I can treat this as a quadratic equation in a and see what conditions are needed for the solutions to be integers. Let's try that.Treating it as a quadratic in a:a² - (kb)a + (b² + k) = 0Using the quadratic formula for a, we get:a = [kb ± √( (kb)² - 4*(1)*(b² + k) )]/2Simplify the discriminant:(kb)² - 4(b² + k) = k²b² - 4b² - 4k = b²(k² - 4) - 4kFor a to be an integer, the discriminant must be a perfect square. Let's denote the discriminant as D:D = b²(k² - 4) - 4kSo D must be a perfect square. That's a necessary condition. Let's see if we can manipulate this.Alternatively, maybe I can assume that a and b are positive integers and try to find possible solutions for k. Let's test small values of k and see if we can find a pattern or contradiction.Suppose k = 1:Then the equation is (a² + b²)/(ab - 1) = 1 => a² + b² = ab - 1. But since a and b are positive integers, ab - 1 must be positive, so ab > 1. However, a² + b² is at least 2 (since a and b are at least 1). Let's check a=1, b=2: Left side: 1 + 4 = 5; Right side: 2 -1 =1. Not equal. Similarly, other small a and b will result in left side being larger than right side. For example, a=2, b=2: Left=4+4=8, Right=4-1=3. Not equal. So k=1 is impossible.k=2:Then a² + b² = 2ab - 2. Rearranged: a² - 2ab + b² = -2. The left side is (a - b)^2, so (a - b)^2 = -2. Impossible since squares are non-negative. So k=2 is invalid.k=3:a² + b² = 3ab - 3. Let's rearrange: a² -3ab + b² = -3. Let's see if this can hold. Again, maybe test small values. Let a=1, b=1: Left=1+1=2, Right=3*1 -3=0. Not equal. a=1, b=2: Left=1+4=5, Right=6-3=3. Not equal. a=2, b=2: Left=4+4=8, Right=12-3=9. Close but not equal. a=2, b=3: Left=4+9=13, Right=18 -3=15. Hmm, difference of 2. Not equal. Maybe try to see if equation can be satisfied. Alternatively, think of discriminant approach. Let's write the equation as a² -3ab + b² +3=0. Treating as a quadratic in a:a² -3b a + (b² +3) =0. Discriminant D = 9b² -4(b² +3) =9b² -4b² -12=5b² -12. For D to be a perfect square, 5b² -12 must be a square. Let’s try small b:b=1: 5 -12= -7, not square.b=2: 20 -12=8, not a square.b=3: 45 -12=33, not square.b=4: 80 -12=68, nope.b=5: 125 -12=113, no.Doesn't seem like this will ever be a perfect square. So k=3 is impossible.k=4:a² + b² =4ab -4. Rearranged: a² -4ab + b² = -4. Again, left side is (a - 2b)^2 -3b². Wait, not sure. Let's compute discriminant for quadratic in a:Equation: a² -4b a + (b² +4)=0. Discriminant D = 16b² -4*(b² +4)=16b² -4b² -16=12b² -16. For D to be a square: 12b² -16 must be a square. Let’s check for small b:b=1:12 -16=-4, invalid.b=2:48 -16=32, not square.b=3:108 -16=92, no.b=4:192 -16=176, no.b=5:300 -16=284, nope.Not squares. So k=4 also impossible.k=5:Now, the target k=5. Let's check if there are solutions. The equation becomes a² + b² =5ab -5. Let's rearrange to a² -5ab + b² +5=0. Let's try some small a and b.a=1:1 -5b + b² +5 = b² -5b +6 =0. Solutions b=(5 ±√(25 -24))/2 = (5 ±1)/2. So b=3 or 2. Both positive integers. So if a=1, then b=2 or 3. Let's check:For a=1, b=2: (1 +4)/(2 -1)=5/1=5, which is k=5. Similarly, a=1, b=3: (1 +9)/(3 -1)=10/2=5. So k=5 works here. So there are solutions when k=5. So that's good. So k=5 is possible.But the problem states that if the equation holds, then k must be 5. So we need to show that for any a, b positive integers, if (a² + b²)/(ab -1)=k is integer, then k=5.So need to prove that k cannot be other numbers. So maybe assuming k is not 5, and reach a contradiction. Alternatively, show that all solutions must lead to k=5.Alternatively, let's try to analyze the equation more deeply. Let's start from the equation:a² + b² =k(ab -1)Let me rearrange this as:a² - kab + b² +k=0This is symmetric in a and b. So without loss of generality, assume that a ≥ b. Then maybe express in terms of a.Alternatively, consider this equation as a quadratic in a:a² - (kb)a + (b² +k)=0For a to be a positive integer, the discriminant must be a perfect square. As we did earlier:Discriminant D = (kb)^2 -4*(b² +k)=k²b² -4b² -4k =b²(k² -4) -4kFor D to be a perfect square, let's denote D = m², where m is a positive integer. So:b²(k² -4) -4k =m²Hmm, this seems complicated. Maybe consider specific values of k. Wait, but since we need to show that k=5 is the only possible, perhaps suppose that k is not 5 and find a contradiction.Alternatively, use Vieta jumping. Vieta jumping is a technique often used in Diophantine equations, especially symmetric ones. Since the equation is symmetric in a and b, Vieta might be useful here.Let me recall Vieta's theorem: if (a, b) is a solution, then another solution can be generated by fixing one variable and solving for the other. Let's try that.Suppose (a, b) is a solution. Then treating the equation as quadratic in a, the other root can be found via Vieta: a + a' = kb, and a*a' = b² +k.So, if a is a solution, then the other solution a' = (kb -a). Similarly, since a' must also be a positive integer. Similarly, we can apply this to b as well.So, starting from a solution (a, b), we can generate another solution (a', b) where a' = kb - a. Similarly, if we fix a and treat the equation as quadratic in b, we can generate another solution (a, b').But to apply Vieta jumping, we need to find minimal solutions and then descend to a base case.Suppose we have a minimal solution (a, b) with a ≥ b. Then using Vieta, the other root a' = kb -a. But since a' = (b² +k)/a, from Vieta. So if a' is positive integer, then (a', b) is another solution.If a is large, then a' might be smaller. So perhaps we can keep reducing until we reach a base case.For example, let's take the known solution when k=5: a=1, b=2. Then, using Vieta, if we have a=1, b=2, then the other root for a is 5*2 -1=10 -1=9. So a'=9. So (9,2) is another solution. Let's check:(9² +2²)/(9*2 -1)= (81 +4)/(18 -1)=85/17=5. Correct. Then, from (9,2), we can generate another solution. The other a would be 5*2 -9=10 -9=1. So back to (1,2). So it cycles between (1,2) and (9,2). Similarly, if we fix a=1 and solve for b, we get b=2 or 3. Then from (1,3), the other b would be 5*1 -3=2. So it cycles between (1,3) and (1,2). Interesting.So maybe all solutions can be generated by Vieta jumping from the minimal solution. But how does that help in proving that k must be 5?Alternatively, suppose that k is not 5. Let's see if there can be solutions. For example, if k=6:Equation: a² + b² =6ab -6. Let's test a=1: 1 + b² =6b -6 => b² -6b +7=0. Discriminant 36 -28=8, which is not a square. So no integer solutions. a=2: 4 + b² =12b -6 =>b² -12b +10=0. Discriminant 144 -40=104, not a square. a=3:9 + b²=18b -6 =>b² -18b +15=0. Discriminant 324 -60=264, not a square. Hmm, not promising.Alternatively, let's use the discriminant approach for k=6. The discriminant D = b²(k² -4) -4k. For k=6, D =b²*(36 -4) -24=32b² -24. So 32b² -24 must be a perfect square. Let's check for small b:b=1:32 -24=8, not square.b=2:128 -24=104, not square.b=3:288 -24=264, nope.b=4:512 -24=488, nope.Not squares. So k=6 impossible.Similarly, trying k=5, we saw that it works. Let's see if there are higher k possible. For example, k=5, works. What about k=6,7, etc.? As above, maybe not. Let's check k=5.Alternatively, maybe there is a way to bound k. Let's assume that a and b are positive integers with a ≥ b ≥1.From the equation a² + b² =k(ab -1). Let's try to bound k in terms of a and b.Note that ab -1 must divide a² + b². Let's see, ab -1 divides a² + b². Let's write this as:a² + b² ≡0 mod (ab -1)Let me see if I can manipulate this congruence. Let's compute a² ≡ -b² mod (ab -1). But ab ≡1 mod (ab -1), so a ≡1/b mod (ab -1). Hmm, not sure if this helps. Let's try expressing a² in terms of ab.From ab ≡1 mod (ab -1), so ab =1 + m(ab -1) for some integer m. Wait, perhaps not helpful.Alternatively, let's use the fact that ab -1 divides a² + b². Let me perform polynomial division or find some relation.Suppose we divide a² + b² by ab -1. Let's consider a as variable and b as constant. Then:Divide a² + b² by ab -1. The leading term is a² divided by ab, which gives (a/b). But since we are dealing with integers, maybe not helpful. Alternatively, express a² + b² as (ab -1)*q + r, but since it's divisible, r=0. So, we can write a² + b² = (ab -1)*q. But q is equal to k. So, q=k.Alternatively, let's use some inequalities. Since a and b are positive integers, ab -1 is positive, so ab >1. Also, since a and b are positive integers, ab ≥2. Let's find upper and lower bounds for k.From the equation, k = (a² + b²)/(ab -1). Let's find possible values for k.Using AM ≥ GM: a² + b² ≥2ab. So, k = (a² + b²)/(ab -1) ≥2ab/(ab -1) =2 + 2/(ab -1). Since ab ≥2, ab -1 ≥1, so 2/(ab -1) ≤2. Thus, k ≥2 + something ≤2, so k ≥2 + something, but this is not tight. Wait, perhaps:Wait, 2ab/(ab -1) =2 + 2/(ab -1). Since ab -1 ≥1, then 2/(ab -1) ≤2. Therefore, k ≥2 + 2/(ab -1) ≥2 + 2/(something). For example, if ab -1=1, then 2 + 2/1=4, so k ≥4? Wait, if ab -1=1, then ab=2, so a=1,b=2 or a=2,b=1. Then k=(1 +4)/(2-1)=5. So in that case, k=5. So that's the minimal case. But when ab is larger, say ab=3, then k=(a² +b²)/(3 -1)=(a² +b²)/2. For ab=3, possible (a,b)=(1,3) or (3,1). Then k=(1+9)/2=5. Also 5. Wait, if ab=3, then a and b are 1 and 3, and k=(1 +9)/2=5. So same result. If ab=4, possible (1,4),(2,2),(4,1). Then k=(1 +16)/(4-1)=17/3≈5.666, not integer. For (2,2): (4 +4)/(4 -1)=8/3≈2.666, not integer. So no solution. So the next possible ab is 5. ab=5, possible (1,5),(5,1). Then k=(1 +25)/(5-1)=26/4=6.5, not integer. So no solution.Wait, but earlier, when we had ab=2 (a=1,b=2), we had k=5. Similarly, ab=3 (a=1,b=3) gives k=5. So when ab=2 or 3, k=5. For ab=4 or higher, k is not integer. Hmm. Interesting. So maybe the only possible solutions are when ab=2 or 3, leading to k=5. But how to generalize this?Alternatively, consider that when a and b are such that ab -1 divides a² + b², then k=(a² + b²)/(ab -1)=5. Let me see if there's a way to show that this ratio is forced to be 5.Alternatively, suppose that we let t = ab. Then, the equation becomes a² + b² =k(t -1). But a² + b² ≥2ab =2t by AM ≥ GM. So, 2t ≤k(t -1). So, 2t ≤k(t -1). Rearranged: k ≥2t/(t -1). The function 2t/(t -1) decreases as t increases. For t=2: 4/1=4; t=3:6/2=3; t=4:8/3≈2.666; t=5:10/4=2.5; etc. So, k must be at least 4 when t=2, but k=5 is possible. As t increases, the lower bound for k decreases, but k must still be an integer. However, even if the lower bound is lower, the actual value of k=(a² + b²)/(t -1) could be higher. Wait, but if a and b are large, a² + b² grows quadratically, while t=ab grows quadratically as well. So perhaps the ratio (a² + b²)/ab approaches 2 as a and b become large (if a ≈b). So then (a² + b²)/(ab -1)≈2ab/(ab)=2. But since we subtract 1 in the denominator, it's slightly more than 2. However, if a and b are not balanced, say one is much larger than the other, then a² + b² could be larger.But in the cases where k=5, we have small a and b. Maybe for larger a and b, the equation doesn't hold. Wait, but there's the solution (9,2) which gives k=5. Let's check that: a=9, b=2. Then ab=18. (81 +4)/(18 -1)=85/17=5. So even with a larger a, we still get k=5. So there are solutions with larger a and b but still k=5. So maybe the only possible k is 5, regardless of the size of a and b.This suggests that the equation only holds when k=5, and the solutions can be generated via Vieta jumping from the minimal solutions. So perhaps we can use Vieta jumping to show that all solutions must have k=5.Assume that k ≠5. Suppose there exists a solution with k≠5. Then, by Vieta jumping, we can generate an infinite descent of solutions, but since we can't have infinitely descending positive integers, this would lead to a contradiction. However, the problem is to show that k must be 5, so maybe by Vieta jumping, all solutions must be part of a chain that leads to the minimal solution (1,2) or (1,3), which have k=5. Therefore, all solutions must have k=5.Alternatively, let's formalize this.Assume that (a, b) is a solution with minimal sum a + b. Without loss of generality, assume a ≥ b. Then, since a² + b² =k(ab -1), rearranged as a² -kab + b² +k=0. Treating this as a quadratic in a, the other root is a' =kb -a. Since a' is positive integer, we have a' = (b² +k)/a. Since a ≥b, and a' must be positive, then kb -a ≥0 => a ≤kb. But since a ≥b, then kb -a ≤kb -b =b(k -1). Not sure.If we consider the minimal solution, suppose that b=1. Then, equation becomes a² +1 =k(a -1). Then, k=(a² +1)/(a -1). Let's compute this:Divide a² +1 by a -1. Using polynomial division:a² +1 ÷ a -1. a² ÷a =a. Multiply (a -1) by a: a² -a. Subtract from a² +1: (a² +1) - (a² -a) =a +1. Now divide a +1 by a -1: 1*(a -1) + 2. So remainder 2. Therefore, (a² +1)/(a -1)=a +1 +2/(a -1). For this to be integer, 2 must be divisible by a -1. So a -1 divides 2. Hence, a -1=1 or 2, so a=2 or 3.If a=2: k=(4 +1)/1=5.If a=3: k=(9 +1)/2=10/2=5.So when b=1, only possible k=5. So minimal solutions with b=1 lead to k=5.Similarly, suppose b=2. Then the equation is a² +4 =k(2a -1). Let's solve for k:k=(a² +4)/(2a -1). Let's see if this can be integer.Perform polynomial division: divide a² +4 by 2a -1.Let’s write a² +4 = (2a -1)(a/2 + 1/4) + (1/4 +4). Not helpful for integer division.Alternatively, use the fact that 2a -1 divides a² +4. So, let's write:a² +4 ≡0 mod (2a -1). Let me express a in terms of the modulus. Let m =2a -1, then a=(m +1)/2. Substitute into the equation:[( (m +1)/2 )² +4] ≡0 mod mExpand:(m² +2m +1)/4 +4 ≡0 mod mMultiply both sides by 4:m² +2m +1 +16 ≡0 mod4mWait, maybe better to compute directly:[(m +1)/2]^2 +4 = (m² +2m +1)/4 +4 = (m² +2m +1 +16)/4 = (m² +2m +17)/4So, (m² +2m +17)/4 ≡0 mod mWhich means that m divides (m² +2m +17)/4. Multiply both sides by 4:4m divides m² +2m +17Which implies that m divides 17. Because m² +2m +17 =m(m +2) +17. Therefore, m divides 17.Since m=2a -1 and m>0 (since a≥1), m must be a positive divisor of 17. The divisors of 17 are 1 and 17.Case 1: m=1 =>2a -1=1 =>a=1. Then check k=(1 +4)/(2*1 -1)=5/1=5. So k=5.Case 2: m=17 =>2a -1=17 =>a=9. Then k=(81 +4)/(18 -1)=85/17=5. Again, k=5.Thus, for b=2, the only possible k=5.Similarly, for b=3. Let's see if there are solutions with b=3.Equation: a² +9 =k(3a -1). So k=(a² +9)/(3a -1). Let's check if this can be integer.Again, set m=3a -1. Then a=(m +1)/3. Substitute into the equation:[(m +1)/3]^2 +9 =k*mMultiply out:(m² +2m +1)/9 +9 =k*mMultiply both sides by 9:m² +2m +1 +81 =9k*mm² +2m +82 =9k*mRearranged: m² +2m +82 ≡0 mod9mBut this seems complex. Alternatively, find m such that m divides a² +9 where m=3a -1.So, 3a -1 divides a² +9. Let's write this as:a² +9 ≡0 mod(3a -1)Express a ≡(1)/3 mod m. Let me compute a in terms of m. Let m=3a -1, so a=(m +1)/3. Substitute into the congruence:[(m +1)/3]^2 +9 ≡0 mod mExpand:(m² +2m +1)/9 +81/9 ≡0 mod mCombine:(m² +2m +1 +81)/9 ≡0 mod m => (m² +2m +82)/9 ≡0 mod mThus, m divides (m² +2m +82)/9. Multiply both sides by 9:9m divides m² +2m +82Thus, m divides 82. Because m² +2m +82 =m(m +2) +82. Hence, m divides 82.Possible positive divisors of 82 are 1,2,41,82. But m=3a -1 must also satisfy m ≡2 mod3 (since m=3a -1, so m≡-1 mod3). Check which divisors of 82 are ≡2 mod3:82's divisors:1,2,41,82.1 mod3=1≠22 mod3=2 yes41 mod3=41-39=2 yes82 mod3=82-81=1 no.So possible m=2,41.Case m=2: m=2=3a -1 =>3a=3 =>a=1. Then check k=(1 +9)/(3*1 -1)=10/2=5.Case m=41: m=41=3a -1 =>3a=42 =>a=14. Then k=(14² +9)/(3*14 -1)= (196 +9)/41=205/41=5. So again, k=5.Thus, for b=3, only possible k=5.This pattern suggests that for any b, the minimal solutions generated by Vieta jumping lead to k=5. Therefore, regardless of a and b, k must be 5.To generalize, suppose there exists a solution with k≠5. Then by Vieta jumping, we can descend to a minimal solution where a or b is 1 or 2, which as shown earlier must have k=5. Hence, the original solution must also have k=5. Therefore, k must be 5.Alternatively, suppose there's a solution with k≠5. Then by Vieta jumping, we can generate a sequence of solutions with smaller and smaller a and b until we reach a minimal solution, which must have k=5. But this would imply that the original k was also 5, leading to a contradiction if we assumed k≠5. Hence, no such solutions exist for k≠5, proving that k must be 5.Thus, through analyzing specific cases, using Vieta jumping, and checking divisibility conditions, we conclude that k=5 is the only possible integer value satisfying the given equation with positive integers a and b.

✅:To prove that ( k = 5 ) is the only solution for positive integers ( a, b, k ) satisfying ( frac{a^2 + b^2}{ab - 1} = k ), we proceed as follows:1. Rearrange the Equation: [ a^2 + b^2 = k(ab - 1) implies a^2 - kab + b^2 + k = 0 ] Treat this as a quadratic in ( a ): [ a^2 - (kb)a + (b^2 + k) = 0 ] The discriminant must be a perfect square for ( a ) to be an integer: [ Delta = (kb)^2 - 4(b^2 + k) = b^2(k^2 - 4) - 4k ]2. Test Small Values of ( k ): - ( k = 1, 2, 3, 4 ) lead to contradictions (non-integer solutions or negative squares). - ( k = 5 ) yields valid solutions (e.g., ( (a, b) = (1, 2) ), ( (1, 3) ), ( (9, 2) ), etc.), verified by substitution.3. Vieta Jumping: - For a solution ( (a, b) ), Vieta's formulas generate another solution ( (kb - a, b) ). - Starting from minimal solutions like ( (1, 2) ), repeated jumping cycles through solutions all with ( k = 5 ).4. Divisibility Analysis: - For ( b = 1 ), solving ( a^2 + 1 = k(a - 1) ) gives ( k = 5 ) when ( a = 2 ) or ( 3 ). - For ( b = 2 ), solving ( a^2 + 4 = k(2a - 1) ) yields ( k = 5 ) for ( a = 1 ) or ( 9 ).5. General Case: - Any solution reduces via Vieta jumping to a minimal case where ( k = 5 ), proving no solutions exist for ( k neq 5 ).Conclusion: The only possible value for ( k ) is ( boxed{5} ).

🔑:Given the equation [frac{a^2 + b^2}{ab - 1} = k]we must show that ( k = 5 ).1. Case Analysis on ( a ) and ( b ): Suppose ( a = b ). Then, the equation becomes: [ k = frac{2a^2}{a^2 - 1} = 2 + frac{2}{a^2 - 1} ] This implies that ( (a^2 - 1) mid 2 ) because the fraction must be an integer, and since ( a^2 - 1 > 0 ) for positive integers ( a ), the possible values of ( a^2 - 1 ) are the divisors of 2. The divisors are 1 and 2, leading to: - ( a^2 - 1 = 1 implies a^2 = 2 ) (invalid since a is not an integer) - ( a^2 - 1 = 2 implies a^2 = 3 ) (invalid since a is not an integer) Therefore, ( a neq b ). Without loss of generality, assume ( a > b ).2. Solving for Specific ( b ) Values: Consider ( b = 1 ): [ k = frac{a^2 + 1}{a - 1} = a + 1 + frac{2}{a - 1} ] Then ( (a - 1) mid 2 ), and since ( a geq 2 ) (because ( a > b = 1 )), valid values for ( a ) are: - ( a - 1 = 1 implies a = 2 ) - ( a - 1 = 2 implies a = 3 ) For both cases: - If ( a = 2 ): [ k = frac{2^2 + 1}{2 - 1} = frac{5}{1} = 5 ] - If ( a = 3 ): [ k = frac{3^2 + 1}{3 - 1} = frac{10}{2} = 5 ] In both cases, ( k = 5 ).3. Reducing the problem to smallest sum pairs: Consider the pair ( (a, b) ) which gives the smallest value of ( a + b ): From the original equation ( frac{a^2 + b^2}{ab - 1} = k ), rewrite it as: [ a^2 - kba + b^2 + k = 0 ] This is a quadratic in ( a ), and by Vieta’s formulas, the roots must be ( a ) and ( kb - a ). Since ( kb - a = frac{b^2 + k}{a} > 0 ) and ( kb - a in mathbb{Z} ), it must be a positive integer. We expect ( kb - a ) to form an integer and ( kb - a in mathbb{N} ), satisfying: [ frac{(kb - a)^2 + b^2}{(kb - a)b - 1} = k ]4. Bounding ( a ) and ( b ) values: We must have [ a + b leq frac{b^2 + k}{a} + b implies a + b leq k ] Given this bound and the condition ( b leq 2 ): If ( b = 1 ), we already saw ( k = 5 ). If ( b = 2 ): [ (a - 1)(b - 1) < 2 implies (a - 1)(2 - 1) < 2 implies a - 1 < 2 implies a leq 3 ] For ( a > b ) and ( b = 2 ): - If ( a = 3 ): [ k = frac{3^2 + 2^2}{3 cdot 2 - 1} = frac{9 + 4}{6 - 1} = frac{13}{5} neq text{integer} ] This leads to a contradiction. Hence, ( b = 2 ) does not yield a valid integer ( k ).5. Conclusion: Therefore, based on the above analysis, the only valid integer solution for ( k ) while satisfying all conditions is:[boxed{5}]

🧠:Okay, let's try to tackle this problem. So, the question is asking for the sum of all positive integers that meet three conditions: their binary representation has exactly 2n digits, exactly n ones and n zeros, and the first digit isn't zero. Hmm, right. Let me break this down step by step.First, since we're dealing with binary numbers, the first digit has to be 1. Otherwise, it would start with a zero, which isn't allowed. So every number we consider here is a 2n-digit binary number starting with 1, with exactly n ones and n zeros. Our goal is to find the sum of all such numbers.Hmm, how do I approach summing all these numbers? Well, maybe instead of trying to list all the numbers and add them up, which would be impractical for large n, there's a smarter combinatorial way. Let me think.Each binary number can be considered as a string of bits, where the first bit is 1, and the remaining 2n - 1 bits consist of n - 1 ones and n zeros. Wait, because the total number of ones is n, and the first digit is already 1, so the remaining 2n - 1 digits must have n - 1 ones and n zeros. That's a key point.So, the problem reduces to finding all such binary numbers starting with 1, followed by a combination of n - 1 ones and n zeros in the remaining 2n - 1 positions. Then, summing all these numbers.But how do we calculate the sum? Maybe by considering the contribution of each bit position to the total sum. Since each bit (whether 0 or 1) contributes a different power of 2 depending on its position. For example, the leftmost bit (the first one) is 2^{2n - 1}, the next one is 2^{2n - 2}, and so on, down to the rightmost bit, which is 2^0.So, if we can compute how many times each bit position (from the second to the last) is 1 across all valid numbers, we can multiply that count by the corresponding power of 2 and sum all those contributions. Then add the contribution from the first bit, which is always 1, so it's just 2^{2n - 1} multiplied by the total number of such numbers.Wait, that makes sense. Let me formalize this idea.First, the total number of such numbers is equal to the number of ways to arrange n - 1 ones and n zeros in the remaining 2n - 1 positions. That would be the combination C(2n - 1, n - 1). So, the first bit contributes (2^{2n - 1}) * C(2n - 1, n - 1).Then, for each of the other bit positions (positions 2 to 2n), we need to calculate how many numbers have a 1 in that position. Each of these positions can be considered similarly. For a specific position i (where i ranges from 2 to 2n), how many numbers have a 1 in that position?Since the first bit is already 1, if we fix position i to be 1, then we have n - 2 ones left to distribute among the remaining 2n - 2 positions (excluding the first and the i-th). But wait, actually, the remaining positions after setting the first bit and the i-th bit to 1: that leaves (2n - 2) positions, and we need to place n - 2 ones and n zeros. Wait, but n - 2 ones plus n zeros would give a total of 2n - 2 digits, which is correct.Therefore, the number of numbers with a 1 in position i is C(2n - 2, n - 2). But wait, does this depend on which position i we are considering? For example, if position i is in the second bit, or somewhere else, does it affect the count?No, because regardless of where the fixed 1 is placed (other than the first bit), the remaining positions are all the same. So, for each of the positions 2 to 2n, the number of times they have a 1 is C(2n - 2, n - 2). Therefore, each of these bit positions contributes C(2n - 2, n - 2) * 2^{2n - i}, where i is the position index. Wait, but the exponent depends on the position. Since position 2 is the second most significant bit, contributing 2^{2n - 2}, position 3 contributes 2^{2n - 3}, and so on, down to the last position, which is 2^{0}.But perhaps instead of dealing with each position individually, we can compute the total contribution of all positions from 2 to 2n. Let's see.The total contribution of all the positions except the first is the sum over each position i (from 2 to 2n) of [number of times position i is 1] multiplied by [2^{2n - i}].Since each position i (from 2 to 2n) is 1 in C(2n - 2, n - 2) numbers, the total contribution for each position is C(2n - 2, n - 2) * 2^{2n - i}.Therefore, the total contribution from all these positions is C(2n - 2, n - 2) multiplied by the sum of 2^{2n - i} for i from 2 to 2n.Wait, let's compute that sum. Let's let k = 2n - i. When i = 2, k = 2n - 2. When i = 2n, k = 0. So the sum is 2^{0} + 2^{1} + ... + 2^{2n - 2}. That is the sum of a geometric series from 0 to 2n - 2, which is 2^{2n - 1} - 1.Wait, the sum from k=0 to m of 2^k is 2^{m+1} - 1. So here, m = 2n - 2, so the sum is 2^{2n - 1} - 1. Therefore, the total contribution from positions 2 to 2n is C(2n - 2, n - 2) * (2^{2n - 1} - 1).Therefore, the total sum S is the contribution from the first bit plus the contribution from the other bits:S = [C(2n - 1, n - 1) * 2^{2n - 1}] + [C(2n - 2, n - 2) * (2^{2n - 1} - 1)].Hmm, let's check if that makes sense. Let me verify with a small n. Let's take n=1. Then, n=1, so 2n=2 digits. The number must have exactly 1 one and 1 zero, starting with 1. The only possible number is 10, which is 2 in decimal. So the sum should be 2.Plugging into the formula:C(2*1 - 1, 1 - 1) * 2^{2*1 - 1} + C(2*1 - 2, 1 - 2) * (2^{2*1 - 1} - 1)First term: C(1, 0) * 2^1 = 1 * 2 = 2Second term: C(0, -1). Wait, C(0, -1) is zero, since combinations with negative k are zero. So the second term is 0 * (2^1 -1) = 0. Therefore, total S=2+0=2, which matches. Good.Let me try n=2. Then, 2n=4 digits. Numbers with 2 ones and 2 zeros, starting with 1. So the numbers are:1001 (9), 1010 (10), 1100 (12). So the sum is 9+10+12=31.Now, let's compute the formula:First term: C(3,1)*2^3 = 3*8=24Second term: C(2,0)*(2^3 -1)=1*(8 -1)=7So total S=24 +7=31. Perfect, that's correct. So the formula works for n=1 and n=2.Another test: n=3. Let's see. 2n=6 digits. Numbers starting with 1, with 3 ones and 3 zeros. The numbers would be all 6-digit binary numbers starting with 1, with exactly 2 more ones in the remaining 5 positions. The total number is C(5,2)=10 numbers. Calculating their sum manually would be tedious, but let's try using the formula.First term: C(5,2)*2^5=10*32=320Second term: C(4,1)*(2^5 -1)=4*(32 -1)=4*31=124Total sum S=320 +124=444. Let me check with actual numbers.List all 6-digit binary numbers starting with 1, with exactly 3 ones and 3 zeros:The remaining 5 bits must have 2 ones and 3 zeros. The positions where the two ones are placed can be in C(5,2)=10 ways.Each such number has the first bit as 1 (32), and the other ones contributing varying amounts. Let's pick a few to check:For example:101100: 32 + 8 + 4 = 44101010: 32 + 8 + 2 = 42But wait, this is going to take a while. Alternatively, note that the sum can also be calculated using the formula. But since the formula gave 444, maybe we can compute the average value multiplied by the number of terms. Wait, the average value times 10.Alternatively, let's note that each of the remaining 5 bits can be considered: each bit has a certain probability of being 1. Since there are 2 ones in 5 bits, each bit has a probability of 2/5 of being 1. Then, the average contribution from each of the bits 2 to 6 is (2/5)*2^{position}. But perhaps this is complicating. Alternatively, the formula from before: sum is 320 + 124=444. Let's check with actual computation.Alternatively, think of each number as 32 (from the first bit) plus a 5-bit number with exactly 2 ones. The sum of all such 5-bit numbers is the sum of numbers from 0 to 31 with exactly 2 ones. Wait, but they are shifted by the first bit. Wait, actually, the remaining 5 bits can be considered as 5-bit numbers where exactly 2 bits are 1. The sum of all such numbers is equal to the number of such numbers multiplied by the average value. Wait, but perhaps there's a formula for the sum of numbers with exactly k ones in their binary representation.Alternatively, the sum can be calculated by considering each bit position. For the 5-bit numbers with exactly 2 ones, each bit position (positions 4 to 0, corresponding to 16, 8, 4, 2, 1) will have a 1 in C(4,1) numbers. Because for each bit, if we fix that bit to 1, then we need to choose 1 more one from the remaining 4 bits. Therefore, each bit is 1 in C(4,1)=4 numbers. Therefore, the total sum of the 5-bit numbers is 4*(16 +8 +4 +2 +1)=4*31=124. Therefore, adding the 32 for each of the 10 numbers: 10*32=320. Then, total sum 320+124=444. So that's correct. Therefore, the formula works.So the formula seems to hold. Therefore, the general formula is:S = C(2n -1, n -1) * 2^{2n -1} + C(2n -2, n -2) * (2^{2n -1} -1)But maybe this can be simplified. Let's see.First, note that C(2n -1, n -1) is equal to C(2n -1, n). Similarly, C(2n -2, n -2) is equal to C(2n -2, n). Let me check:C(2n -1, n -1) = (2n -1)! / [(n -1)! (n)!]C(2n -2, n -2) = (2n -2)! / [(n -2)! (n)!]But not sure if that helps. Let's see if we can factor out terms.Let me express both terms with factorials:First term: C(2n -1, n -1) = (2n -1)! / [(n -1)! n!] * 2^{2n -1}Second term: C(2n -2, n -2) = (2n -2)! / [(n -2)! n!] * (2^{2n -1} -1)Hmm, perhaps we can factor out something. Let me compute the ratio between the two coefficients.C(2n -1, n -1) / C(2n -2, n -2) = [(2n -1)! / ((n -1)! n!)] / [(2n -2)! / ((n -2)! n!))] = (2n -1) * (n -1)! / (n -1)! ) * (n -2)! / (2n -2)! )Wait, maybe that's messy. Alternatively, let's note that:C(2n -1, n -1) = [(2n -1)/(n)] * C(2n -2, n -1)Wait, perhaps that's not helpful.Alternatively, let me consider that:C(2n -1, n -1) + C(2n -1, n) = C(2n, n). But not sure.Alternatively, notice that C(2n -1, n -1) = (2n -1 choose n -1) and C(2n -2, n -2) = (2n -2 choose n -2). Let's see:Let me write the coefficients:First term: (2n -1 choose n -1) * 2^{2n -1}Second term: (2n -2 choose n -2) * (2^{2n -1} -1)Is there a way to combine these terms?Alternatively, notice that (2n -1 choose n -1) = (2n -1)/(n) * (2n -2 choose n -1)Wait, maybe not.Alternatively, let's consider that (2n -2 choose n -2) = (2n -2)! / [(n -2)! n!]While (2n -1 choose n -1) = (2n -1)! / [(n -1)! n!] = (2n -1)/(n -1) * (2n -2)! / [(n -2)! n!] ) = (2n -1)/(n -1) * (2n -2 choose n -2)Therefore, C(2n -1, n -1) = [(2n -1)/(n -1)] * C(2n -2, n -2)Therefore, substituting into the first term:First term = [(2n -1)/(n -1)] * C(2n -2, n -2) * 2^{2n -1}Second term = C(2n -2, n -2) * (2^{2n -1} -1)Therefore, we can factor out C(2n -2, n -2):S = C(2n -2, n -2) * [ (2n -1)/(n -1) * 2^{2n -1} + (2^{2n -1} -1) ]Let me compute the expression inside the brackets:Let me factor out 2^{2n -1}:= C(2n -2, n -2) * [ 2^{2n -1} * [ (2n -1)/(n -1) + 1 ] -1 ]Wait, compute the coefficient:(2n -1)/(n -1) + 1 = [ (2n -1) + (n -1) ] / (n -1) = (3n -2)/(n -1)Therefore,S = C(2n -2, n -2) * [ 2^{2n -1} * (3n -2)/(n -1) -1 ]Hmm, not sure if this is helpful. Alternatively, let's compute specific values for small n and see if we can spot a pattern.For n=1:C(0, -1) is zero, but in our formula S=2, which was correct. For n=1, perhaps the formula is an edge case.For n=2:C(2,0)=1. Then,S=1 * [8*(4)/1 -1] = [32 -1] =31, which matches.Wait, but when n=2, 3n -2=4, n -1=1, so 4/1=4, so 2^{2*2 -1} *4=8*4=32, minus 1 is 31. So yes, 31. Then multiplied by C(2n -2, n -2)=C(2,0)=1, so S=31. Correct.For n=3:C(4,1)=4. Then,[2^{5} * (7)/2 -1] = 32*(7/2) -1=112 -1=111. Then S=4*111=444, which matches our previous result. So this seems correct.But perhaps this expression can be simplified further. Let's look at the term inside the brackets:2^{2n -1} * (3n -2)/(n -1) -1But (3n -2)/(n -1) can be written as 3 + (1)/(n -1). Let's check:(3n -2)/(n -1) = (3(n -1) +1 ) / (n -1) )= 3 + 1/(n -1). Yes, correct.Therefore,2^{2n -1}*(3 + 1/(n -1)) -1 = 3*2^{2n -1} + 2^{2n -1}/(n -1) -1But I don't know if this helps. Alternatively, maybe express the entire expression as a combination.Alternatively, note that C(2n -2, n -2) * (3n -2)/(n -1) = C(2n -2, n -2) * (3n -2)/(n -1). Let's see:C(2n -2, n -2) = (2n -2)! / [(n -2)! n!]Multiply by (3n -2)/(n -1):= (2n -2)! / [(n -2)! n!] * (3n -2)/(n -1)= (2n -2)! * (3n -2) / [ (n -2)! n! (n -1) ]But note that n! = n*(n -1)!, so:= (2n -2)! * (3n -2) / [ (n -2)! n*(n -1)! (n -1) ) ]Wait, this seems messy. Maybe there's another approach.Alternatively, recall that in the original problem, the numbers are binary numbers with 2n digits, starting with 1, followed by n-1 ones and n zeros. The total number of such numbers is C(2n -1, n -1). Each such number corresponds to a unique binary number, and their sum is what we need.Alternatively, maybe there's a generating function approach. Let's think.Each number can be represented as 2^{2n -1} + x, where x is a number formed by the remaining 2n -1 bits, which have exactly n -1 ones and n zeros. Wait, no: the remaining bits have exactly n -1 ones and n zeros, but they can start with zero. Wait, but the total bits are 2n -1, with n -1 ones. So x can be any number with 2n -1 bits where n -1 are ones. However, the problem is that these bits can start with zero. So x ranges from 0 to 2^{2n -1} -1.But the original numbers are 2^{2n -1} + x, where x is a number with exactly n -1 ones in its 2n -1 bit representation. Therefore, the sum we want is the sum over all such x of (2^{2n -1} + x) = number_of_x * 2^{2n -1} + sum_x.But number_of_x is C(2n -1, n -1), and sum_x is the sum of all numbers with 2n -1 bits (including leading zeros) having exactly n -1 ones.Wait, so sum_x is the sum of all numbers with exactly n -1 ones in their 2n -1 bits. How can we compute this sum?Similarly to how we computed before, considering each bit position. For each bit position in the 2n -1 bits, the number of times it is 1 is C(2n -2, n -2). Because fixing one bit as 1, the remaining n -2 ones are distributed among the remaining 2n -2 bits. Therefore, each bit contributes C(2n -2, n -2) * 2^{position}.But since the bits are from position 2n -2 down to 0 (since x is a 2n -1 bit number), the total sum_x is C(2n -2, n -2) * (2^{2n -1} -1). Wait, similar to before.Wait, the sum of 2^{k} for k from 0 to m is 2^{m+1} -1. If each bit position from 0 to 2n -2 (total 2n -1 positions) contributes C(2n -2, n -2) times 2^{k}, then the total sum is C(2n -2, n -2) * (2^{2n -1} -1). Which is exactly what we had before.Therefore, the total sum S is:C(2n -1, n -1) * 2^{2n -1} + C(2n -2, n -2) * (2^{2n -1} -1)Which is the same as:Number_of_numbers * 2^{2n -1} + sum_xWhich makes sense.But we need to see if this can be simplified further. Let me check the relationship between the two coefficients.C(2n -1, n -1) = C(2n -2, n -2) * (2n -1)/(n -1)As we saw earlier. Therefore, substituting:S = [C(2n -2, n -2) * (2n -1)/(n -1)] * 2^{2n -1} + C(2n -2, n -2) * (2^{2n -1} -1)Factor out C(2n -2, n -2):S = C(2n -2, n -2) * [ (2n -1)/(n -1) * 2^{2n -1} + 2^{2n -1} -1 ]Combine terms:= C(2n -2, n -2) * [ 2^{2n -1} * ( (2n -1)/(n -1) +1 ) -1 ]As before. Which simplifies to:= C(2n -2, n -2) * [ 2^{2n -1} * ( (2n -1 + n -1)/ (n -1) ) -1 ]= C(2n -2, n -2) * [ 2^{2n -1} * (3n -2)/(n -1) -1 ]But is there a way to write this in terms of binomial coefficients or known expressions?Alternatively, note that C(2n -2, n -2) * (3n -2)/(n -1) can be written as C(2n -1, n -1) * 3n -2/(n -1). Not sure.Alternatively, think of the term 3n -2. Maybe relate this to the derivative or some generating function.Alternatively, look for a pattern in the expressions for small n:For n=1: S=2=2^1 *1For n=2: S=31=2^5 -1 (but 31=32-1)For n=3: S=444= let's see, 444=256 + 128 + 32 + 32 - no, not obvious. Wait, 444=4*111=4*(3*37). Hmm, not helpful.Wait, 444 divided by 12 (which is C(4,1)* something?) Not sure.Alternatively, note that 31 is 2^{5} -1. Wait, but 2^{5} -1=32-1=31, which is n=2. For n=3, 2^{2n -1}=2^5=32. But 32* something minus 1 gives 111, which then multiplied by 4 gives 444. Not sure.Alternatively, think of the formula as S = C(2n -1, n -1) *2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1). Maybe factor out 2^{2n -1}:S=2^{2n -1}*(C(2n -1, n -1)+ C(2n -2, n -2)) ) - C(2n -2, n -2)But C(2n -1, n -1) + C(2n -2, n -2) = C(2n -1, n -1) + C(2n -2, n -2)= [ (2n -1)! / ( (n -1)!n! ) ] + [ (2n -2)! / ( (n -2)!n! ) ]Let me factor out (2n -2)! / ( (n -2)!n! ) ):= (2n -2)! / ( (n -2)!n! ) [ (2n -1)/(n -1) +1 ]= (2n -2)! / ( (n -2)!n! ) [ (2n -1 + n -1)/ (n -1) ]= (2n -2)! / ( (n -2)!n! ) [ (3n -2)/ (n -1) ]= [ (2n -2)! * (3n -2) ] / [ (n -2)!n! (n -1) ) ]= [ (2n -2)! * (3n -2) ] / [ (n -1)!n! ) ]Because (n -2)!*(n -1)= (n -1)! )Therefore,C(2n -1, n -1) + C(2n -2, n -2) = [ (2n -2)! * (3n -2) ] / [ (n -1)!n! ) ]But (2n -2)! / (n -1)!n! ) = C(2n -2, n -1). So,= C(2n -2, n -1)*(3n -2)Therefore,S=2^{2n -1}*C(2n -2, n -1)*(3n -2) - C(2n -2, n -2)But this might not be helpful.Wait, C(2n -2, n -1) is equal to C(2n -2, n -1) = (2n -2)! / [ (n -1)! (n -1)! ]But not sure.Alternatively, recall that C(2n -2, n -2) = C(2n -2, n). Because C(a, b)=C(a, a -b). So C(2n -2, n -2)=C(2n -2, (2n -2)-(n -2))=C(2n -2, n).But not sure if that helps.Alternatively, use generating functions.Let me consider generating functions for the sum.The sum we want is the sum over all binary numbers of length 2n starting with 1, with exactly n ones and n zeros. Each such number can be represented as 1 followed by a string of 2n -1 bits with n -1 ones and n zeros.The generating function for the numbers would be x^{2^{2n -1}} * (1 + x^{2^{2n -2}} + ... + x^{2^{0}})^{2n -1} with exactly n -1 ones. Wait, no, that's not correct.Wait, the generating function for the bits is complicated because each bit position has a different weight. So perhaps the generating function approach would involve multiplying terms for each bit position, where each term is (1 + 2^k x), where k is the bit position. But this seems complex.Alternatively, since each bit position contributes independently except for the fixed first bit, maybe we can model the sum as the first bit plus the expected value of the remaining bits multiplied by the number of numbers.Wait, but the expected value approach might work. Let's see.Each of the remaining 2n -1 bits has a certain probability of being 1. For each bit, the number of numbers where that bit is 1 is C(2n -2, n -2), as before. Therefore, the average value contributed by each bit is [C(2n -2, n -2)/C(2n -1, n -1)] * 2^{bit position}Then, the total average value per number would be 2^{2n -1} + sum_{bit=0}^{2n -2} [C(2n -2, n -2)/C(2n -1, n -1)} * 2^{bit}Then, multiply by the total number of numbers C(2n -1, n -1) to get the total sum.But this would give:Total sum = C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2) * sum_{bit=0}^{2n -2} 2^{bit}Which is the same as our original formula. So it's consistent.But since sum_{bit=0}^{2n -2} 2^{bit} = 2^{2n -1} -1, we again get:Total sum = C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)Therefore, the formula seems to be as simplified as possible unless there's a combinatorial identity that can further simplify it.Alternatively, note that the term C(2n -1, n -1)*2^{2n -1} can be written as (2n -1 choose n -1) * 2^{2n -1}, and the second term is (2n -2 choose n -2)*(2^{2n -1} -1). Let's see if these terms can be combined.Alternatively, think of the sum as:S = 2^{2n -1} * [C(2n -1, n -1) + C(2n -2, n -2)] - C(2n -2, n -2)But earlier, we saw that C(2n -1, n -1) + C(2n -2, n -2) = [C(2n -2, n -1) + C(2n -2, n -2)] * something. Wait, but C(2n -1, n -1) = C(2n -2, n -1) + C(2n -2, n -2), by the Pascal's identity.Wait, Pascal's identity says that C(a, b) = C(a -1, b) + C(a -1, b -1). So applying this to C(2n -1, n -1):C(2n -1, n -1) = C(2n -2, n -1) + C(2n -2, n -2)Therefore, substituting into S:S = 2^{2n -1}*(C(2n -2, n -1) + C(2n -2, n -2)) + C(2n -2, n -2)*(2^{2n -1} -1) - C(2n -2, n -2)Wait, no, original S was:S = C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)But substituting C(2n -1, n -1) = C(2n -2, n -1) + C(2n -2, n -2):S = [C(2n -2, n -1) + C(2n -2, n -2)]*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)= C(2n -2, n -1)*2^{2n -1} + C(2n -2, n -2)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)= C(2n -2, n -1)*2^{2n -1} + C(2n -2, n -2)*[2^{2n -1} +2^{2n -1} -1]= C(2n -2, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n} -1)But this seems to be going in circles. Maybe not helpful.Alternatively, consider that C(2n -2, n -1) + C(2n -2, n -2) = C(2n -1, n -1). Wait, that's exactly Pascal's identity. So this substitution doesn't help.At this point, it seems that the expression:S = C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)Is the simplest form unless there's a combinatorial identity that can express this sum in a closed form. Let me check the literature or known Olympiad solutions.Wait, this problem is from the 1991 Canadian Mathematical Olympiad. Looking up the solution, perhaps the answer is n * 2^{2n -1} / (2n -1 choose n -1)}? No, but maybe not. Wait, let's think again.Alternatively, recall that the total number of such numbers is C(2n -1, n -1). Each number starts with 1, and the rest have an average value. Maybe the average value of the remaining bits is (sum_x)/C(2n -1, n -1). We know sum_x = C(2n -2, n -2)*(2^{2n -1} -1). Therefore, average x is [C(2n -2, n -2)/C(2n -1, n -1)]*(2^{2n -1} -1)But C(2n -2, n -2)/C(2n -1, n -1) = [ (2n -2)! / ( (n -2)!n! ) ] / [ (2n -1)! / ( (n -1)!n! ) ] = [ (2n -2)! (n -1)! ) ] / [ (2n -1)! (n -2)! ) ] = (n -1)/(2n -1)Therefore, average x = (n -1)/(2n -1)*(2^{2n -1} -1)Therefore, average total number is 2^{2n -1} + (n -1)/(2n -1)*(2^{2n -1} -1)Multiply by the number of numbers, C(2n -1, n -1), to get the total sum:S = C(2n -1, n -1)*[2^{2n -1} + (n -1)/(2n -1)*(2^{2n -1} -1)]= C(2n -1, n -1)*[ (2^{2n -1}*(2n -1) + (n -1)(2^{2n -1} -1) ) / (2n -1) ]= [C(2n -1, n -1)/(2n -1)] * [2^{2n -1}(2n -1 + n -1) - (n -1)]= [C(2n -1, n -1)/(2n -1)] * [2^{2n -1}(3n -2) - (n -1)]But C(2n -1, n -1)/(2n -1) = [ (2n -1)! / ( (n -1)!n! ) ] / (2n -1) ) = (2n -2)! / ( (n -1)!n! ) ) = C(2n -2, n -1)Wait, C(2n -2, n -1) = (2n -2)! / [ (n -1)! (n -1)! ) ]But no, (2n -2)! / ( (n -1)!n! ) ) = C(2n -2, n -1) * n / (n -1). Not sure.Alternatively, the term [2^{2n -1}(3n -2) - (n -1)] multiplied by C(2n -1, n -1)/(2n -1). This seems too convoluted.Given that we have verified the formula for small n and it works, perhaps this is the simplest form. However, checking the original problem statement from the Canadian Mathematical Olympiad 1991, I recall that the answer is n * 2^{2n -1} * C(2n -1, n -1)} / (2n -1). Wait, no, that may not be.Wait, let me think differently. Suppose we consider each bit position. The first bit is always 1, contributing 2^{2n -1} for each of the C(2n -1, n -1) numbers, so that term is clear.For the other bit positions, each position i (from 2 to 2n) is 1 in C(2n -2, n -2) numbers. So each bit contributes C(2n -2, n -2) * 2^{2n - i}.But the sum over all positions i=2 to 2n of 2^{2n -i} is the same as the sum from k=0 to 2n -2 of 2^k, which is 2^{2n -1} -1. Therefore, the total contribution from all the other bits is C(2n -2, n -2)*(2^{2n -1} -1)Therefore, combining both contributions, the total sum is S= C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)This is as simplified as it gets. However, perhaps there is a closed-form expression.Wait, notice that C(2n -1, n -1) is equal to (1/(2))C(2n, n). Because C(2n, n) = 2n!/(n!n!), and C(2n -1, n -1)= (2n -1)! / [(n -1)!n!)] = (2n -1)/n * (2n -2)! / [ (n -1)! (n -1)! ) ] = (2n -1)/n * C(2n -2, n -1). Not sure.Alternatively, note that C(2n -1, n -1) = (1/2) * C(2n, n) * (2n)/ (2n -1). Wait, maybe not.Alternatively, consider that C(2n -1, n -1) = (2n -1)! / [(n -1)!n!] = (2n -1)/n * C(2n -2, n -1)But I think we have to accept that this is the simplest form.Alternatively, notice that the given expression can be written as:S = C(2n -1, n -1) *2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)But in the case of n=2, this becomes 3*8 +1*31=24 +31=55? Wait, no, no, wait for n=2, C(3,1)=3, 2^{3}=8, so 3*8=24, and C(2,0)=1, (2^{3}-1)=7, so 24+7=31, which is correct.Wait, perhaps there's a way to combine the terms:S = 2^{2n -1}(C(2n -1, n -1) + C(2n -2, n -2)) - C(2n -2, n -2)But using Pascal's identity: C(2n -1, n -1) = C(2n -2, n -1) + C(2n -2, n -2), so:S = 2^{2n -1}(C(2n -2, n -1) + 2*C(2n -2, n -2)) - C(2n -2, n -2)But this seems to complicate more.Alternatively, use generating functions for the sum.The generating function for binary numbers with exactly n ones and n zeros, starting with 1, is the generating function where the first term is x^{2^{2n -1}}, and the remaining terms are products of x^{2^k} for each bit set to 1. However, generating functions for such sums are difficult because the exponents are not linear.Alternatively, think of the sum as the coefficient of x^n in the generating function, but weighted by the values. This seems tricky.Alternatively, I recall that in the CMO problems, sometimes the answer is a binomial coefficient multiplied by a power of 2, but let's see.Wait, when n=1: S=2=1*2^1When n=2: S=31= not a multiple of a simple binomial coefficient.When n=3: S=444=4*111=4*3*37. Hmm, not obvious.Alternatively, check if the formula can be expressed as (n) * 2^{2n -1} * C(2n -2, n -1)/ (2n -1)}. For n=2:(2)* 8 * C(2,1)/3= 2*8*2/3=32/3 ≈10.666, no. Not matching.Alternatively, think of the sum S as:Each number is a 2n-digit binary number starting with 1, followed by a combination of n-1 ones and n zeros. Each such number corresponds to a unique subset of positions (excluding the first) where the ones are placed. The sum can be calculated as the sum over all such subsets of the value of the number.The value is 2^{2n -1} plus the sum over each bit position (from 0 to 2n -2) of 2^{bit} if the bit is set.Therefore, S = C(2n -1, n -1)*2^{2n -1} + sum_{k=0}^{2n -2} (number of subsets with bit k set) *2^{k}Where the number of subsets with bit k set is C(2n -2, n -2), as fixing one bit leaves n -2 ones to be placed in 2n -2 bits.Therefore, the sum is:C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)This is the same as before. So unless there's a way to express this combination as a fraction times a binomial coefficient, I think this is as simplified as it can get.However, looking back at the problem statement, it's from the CMO, which often has elegant answers. Maybe the answer is n * 4^{n-1} * C(2n -1, n -1)} or something similar. Let me check for n=2:n=2: n *4^{n-1}*C(3,1)=2*4*3=24, which is less than 31. Not correct.Alternatively, 31= (16 + 15). Not helpful.Alternatively, perhaps the answer is (2^{2n} -1) * C(2n -2, n -1). For n=2: (16 -1)*C(2,1)=15*2=30, close but not 31. No.Alternatively, recall that 31=2^5 -1. For n=2, 2n=4, 2^{2n -1}=8, 8*4-1=31. Wait, 8*4=32-1=31. But 4 is C(2n -2, n -2)=C(2,0)=1. Hmm, no.Wait, the formula for n=2 is 3*8 +1*7=24+7=31. Where 3=C(3,1) and 1=C(2,0). For n=3, it's 10*32 +4*31=320+124=444. It's possible that the answer can be written as C(2n -1, n -1) * 2^{2n -1} + C(2n -2, n -2) * (2^{2n -1} -1). But I need to check if this can be expressed more elegantly.Alternatively, consider that C(2n -1, n -1) + C(2n -2, n -2) = C(2n -1, n -1) + C(2n -2, n -2). But we can write this as:C(2n -1, n -1) + C(2n -2, n -2) = C(2n -1, n -1) + C(2n -2, n -2) = ?Using the identity C(n, k) + C(n, k-1) = C(n+1, k). But here, it's C(2n -2, n -2) + C(2n -2, n -1) = C(2n -1, n -1). Wait, yes:C(2n -2, n -2) + C(2n -2, n -1) = C(2n -1, n -1)Which is Pascal's identity. Therefore, the first term in S is [C(2n -2, n -1) + C(2n -2, n -2)] *2^{2n -1} = C(2n -1, n -1)*2^{2n -1}And the second term is C(2n -2, n -2)*(2^{2n -1} -1)So S = C(2n -1, n -1)*2^{2n -1} + C(2n -2, n -2)*(2^{2n -1} -1)But how does this help? Unless we can combine these terms:S = 2^{2n -1}(C(2n -1, n -1) + C(2n -2, n -2)) - C(2n -2, n -2)But from Pascal's identity, C(2n -1, n -1) = C(2n -2, n -1) + C(2n -2, n -2). Therefore:S = 2^{2n -1}(C(2n -2, n -1) + 2C(2n -2, n -2)) - C(2n -2, n -2)But this still doesn't lead to a familiar form.At this point, it might be best to accept that the expression is already in its simplest form, and perhaps the problem expects this form as the answer. However, to confirm, let's check if there's a combinatorial interpretation.Consider that each number contributes a value of 2^{2n -1} plus the sum of the remaining bits. The total contribution from the first bit is straightforward. The remaining bits are symmetric, so each bit position contributes equally. Therefore, the sum over all numbers is the first bit contribution plus the sum over all remaining bit contributions.As each remaining bit position is 1 in C(2n -2, n -2) numbers, and there are (2n -1) such positions, the total contribution from the remaining bits is C(2n -2, n -2) * sum_{k=0}^{2n -2} 2^{k} = C(2n -2, n -2) * (2^{2n -1} -1)Thus, the final answer is:S = C(2n -1, n -1) * 2^{2n -1} + C(2n -2, n -2) * (2^{2n -1} -1)This can be written using binomial coefficients as:S = binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)Alternatively, combining terms:S = binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)But perhaps there is a way to express this using a single binomial coefficient. For example, using the identity:binom{2n -1}{n -1} = binom{2n -2}{n -1} + binom{2n -2}{n -2}But even then, it doesn't simplify the expression further.Alternatively, factor out 2^{2n -1}:S = 2^{2n -1} left( binom{2n -1}{n -1} + binom{2n -2}{n -2} right) - binom{2n -2}{n -2}But since binom{2n -1}{n -1} = binom{2n -2}{n -1} + binom{2n -2}{n -2}, we have:S = 2^{2n -1} left( binom{2n -2}{n -1} + 2binom{2n -2}{n -2} right) - binom{2n -2}{n -2}But this doesn't seem helpful.Given that the problem is from a competition, it's possible that the answer is expressed in terms of binomial coefficients and powers of two, as above. Alternatively, there might be a generating function approach that gives a closed-form expression, but I can't see it immediately.Alternatively, notice that for each number, the average value per number is S divided by C(2n -1, n -1). As we calculated earlier, this average is:Average = 2^{2n -1} + [C(2n -2, n -2)/C(2n -1, n -1)]*(2^{2n -1} -1)And we found that [C(2n -2, n -2)/C(2n -1, n -1)] = (n -1)/(2n -1)Therefore:Average = 2^{2n -1} + (n -1)/(2n -1)*(2^{2n -1} -1)= [ (2n -1)*2^{2n -1} + (n -1)(2^{2n -1} -1) ] / (2n -1)= [2^{2n -1}(2n -1 + n -1) - (n -1) ] / (2n -1)= [2^{2n -1}(3n -2) - (n -1) ] / (2n -1)Therefore, multiplying by the number of numbers C(2n -1, n -1):S = [2^{2n -1}(3n -2) - (n -1) ] / (2n -1) * C(2n -1, n -1)But this can be rewritten as:S = frac{3n -2}{2n -1} cdot 2^{2n -1} cdot C(2n -1, n -1) - frac{n -1}{2n -1} cdot C(2n -1, n -1)But I don't see a clear simplification here.Alternatively, note that frac{C(2n -1, n -1)}{2n -1} = frac{(2n -1)!}{(n -1)!n!} cdot frac{1}{2n -1} = frac{(2n -2)!}{(n -1)!n!} = frac{1}{n} cdot frac{(2n -2)!}{(n -1)!(n -1)!} ) = frac{1}{n} C(2n -2, n -1)But again, this doesn't lead to a simplification.Given that this is a competition problem, and since we've verified the formula with specific cases, and given that it's expressed in terms of binomial coefficients and powers of two, which are standard in combinatorics, I think this is the expected answer.Therefore, the final answer is:S = binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)However, to check if this can be written more neatly, perhaps using the fact that binom{2n -2}{n -2} = binom{2n -2}{n} and binom{2n -1}{n -1} = binom{2n -1}{n}, but I don't think this helps.Alternatively, factor out 2^{2n -1}:S = 2^{2n -1} left( binom{2n -1}{n -1} + binom{2n -2}{n -2} right) - binom{2n -2}{n -2}But since binom{2n -1}{n -1} = binom{2n -2}{n -1} + binom{2n -2}{n -2}, as per Pascal's identity:S = 2^{2n -1} left( binom{2n -2}{n -1} + 2binom{2n -2}{n -2} right) - binom{2n -2}{n -2}= 2^{2n -1} left( binom{2n -2}{n -1} + binom{2n -2}{n -2} + binom{2n -2}{n -2} right) - binom{2n -2}{n -2}= 2^{2n -1} left( binom{2n -1}{n -1} + binom{2n -2}{n -2} right) - binom{2n -2}{n -2}But this still doesn't provide a simplification.Given the time I've spent on this and the verification with small cases, I think the formula is correct and as simplified as possible. Therefore, the answer is:boxed{binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)}But wait, let me check if this can be expressed as a single binomial coefficient multiplied by a term. For example, notice that:binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1) = binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot 2^{2n -1} - binom{2n -2}{n -2}= 2^{2n -1} left( binom{2n -1}{n -1} + binom{2n -2}{n -2} right) - binom{2n -2}{n -2}But using Pascal's identity again:binom{2n -1}{n -1} + binom{2n -2}{n -2} = binom{2n -1}{n -1} + binom{2n -2}{n -2} = binom{2n -1}{n -1} + binom{2n -2}{n -2} = binom{2n -1}{n -1} + binom{2n -2}{n -2}This doesn't help. Alternatively, note that binom{2n -2}{n -2} = frac{n -1}{2n -2} binom{2n -2}{n -1}But this may not lead anywhere.Alternatively, consider that binom{2n -1}{n -1} = frac{2n -1}{n} binom{2n -2}{n -1}Then:S = frac{2n -1}{n} binom{2n -2}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)But binom{2n -2}{n -2} = binom{2n -2}{n}And binom{2n -2}{n} = frac{n}{2n -2} binom{2n -2}{n -1}Wait, binom{2n -2}{n} = binom{2n -2}{n -2} = frac{(2n -2)!}{n! (n -2)!}While binom{2n -2}{n -1} = frac{(2n -2)!}{(n -1)! (n -1)!}So their ratio is:binom{2n -2}{n -2} / binom{2n -2}{n -1} = frac{(n -1)! (n -1)!}{n! (n -2)!} } = frac{(n -1)}{n}Therefore, binom{2n -2}{n -2} = frac{n -1}{n} binom{2n -2}{n -1}Substituting into S:S = frac{2n -1}{n} binom{2n -2}{n -1} cdot 2^{2n -1} + frac{n -1}{n} binom{2n -2}{n -1} cdot (2^{2n -1} -1)Factor out frac{binom{2n -2}{n -1}}{n}:S = frac{binom{2n -2}{n -1}}{n} [ (2n -1) cdot 2^{2n -1} + (n -1)(2^{2n -1} -1) ]Expand inside the brackets:= (2n -1)2^{2n -1} + (n -1)2^{2n -1} - (n -1)= [ (2n -1) + (n -1) ]2^{2n -1} - (n -1)= (3n -2)2^{2n -1} - (n -1)Therefore,S = frac{binom{2n -2}{n -1}}{n} [ (3n -2)2^{2n -1} - (n -1) ]Hmm, this seems more complex, but perhaps it can be expressed as:S = frac{(3n -2)2^{2n -1} - (n -1)}{n} cdot binom{2n -2}{n -1}But I don't know if this is helpful. However, notice that frac{binom{2n -2}{n -1}}{n} = frac{(2n -2)!}{n (n -1)! (n -1)!)} = frac{(2n -2)!}{n! (n -1)!} } = binom{2n -2}{n -1} / nBut not sure.At this point, after extensive analysis and verification, I conclude that the sum is best expressed as:S = binom{2n -1}{n -1} cdot 2^{2n -1} + binom{2n -2}{n -2} cdot (2^{2n -1} -1)And this is the final answer.

✅:To solve the problem of finding the sum of all positive integers with exactly (2n) binary digits, (n) ones, (n) zeros, and the first digit not being zero, we can break down the solution as follows:1. First Digit Contribution: The first digit is always 1, contributing (2^{2n - 1}) for each valid number. The number of such numbers is given by the binomial coefficient (binom{2n - 1}{n - 1}), as we need to choose (n - 1) positions for the remaining ones out of (2n - 1) positions. Therefore, the total contribution from the first digit is: [ binom{2n - 1}{n - 1} cdot 2^{2n - 1} ]2. Contribution from Remaining Digits: For each of the remaining (2n - 1) positions, the number of times a 1 appears in each position is (binom{2n - 2}{n - 2}). This is because fixing one position as 1 leaves (n - 2) ones to be distributed among the remaining (2n - 2) positions. The sum of contributions from all these positions is the sum of (2^{2n - i}) for each position (i) from 2 to (2n), which is a geometric series summing to (2^{2n - 1} - 1). Therefore, the total contribution from the remaining digits is: [ binom{2n - 2}{n - 2} cdot (2^{2n - 1} - 1) ]3. Total Sum: Combining the contributions from the first digit and the remaining digits, the total sum (S) is: [ S = binom{2n - 1}{n - 1} cdot 2^{2n - 1} + binom{2n - 2}{n - 2} cdot (2^{2n - 1} - 1) ]Thus, the final answer is:[boxed{binom{2n - 1}{n - 1} cdot 2^{2n - 1} + binom{2n - 2}{n - 2} cdot (2^{2n - 1} - 1)}]

🔑:1. Given ( n ) is a fixed positive integer. We need to find the sum of all integers that satisfy the given properties: - In binary representation, the integer is formed by ( 2n ) digits. - There are ( n ) ones ((1)) and ( n ) zeros ((0)). - The first digit is not zero.2. Case 1: When ( n = 1 ). - The only number that satisfies the condition is ( (10)_2 ), which is ( 2 ) in decimal. - There are no other possible numbers in this case. - Therefore, the sum is ( 2 ).3. Case 2: When ( n geq 2 ). - The first digit must be ( 1 ), leaving ( 2n-1 ) digits. - Out of these ( 2n-1 ) positions, there must be ( n-1 ) ones and ( n ) zeros. - The number of ways to choose ( n-1 ) positions out of ( 2n-1 ) is given by the binomial coefficient ( C_{2n-1}^{n-1} ).4. Let's denote the configurations of these remaining ( 2n-1 ) digits: - The total number of valid configurations is ( C_{2n-1}^{n-1} ).5. For each of these ( C_{2n-1}^{n-1} ) configurations, we analyze the sum of the numbers represented by these binary digits: - Contribution Analysis: - Consider each bit position from ( 2^{2n-2} ) down to ( 2^0 ). - If the digit is 1, it contributes (sum_{k=0}^{2n-2} 2^k). This sum is a geometric series which sums to ( 2^{2n-1} - 1 ). - There are ( C_{2n-2}^{n-2} ) numbers with the first digit fixed and specific ( n-1 ) remaining (1)’s, contributing ( (2^{2n-1} - 1) times C_{2n-2}^{n-1} ). - Additionally, configurations where the leading one can be any location in the first digit lead to: [ C_{2n-1}^{n-1} times 2^{2n-1} ]6. Thus, we can sum all contributions to obtain the total sum: [ begin{aligned} text{Total Sum} &= C_{2n-2}^{n-2}(2^{2n-1} - 1) &+ C_{2n-1}^{n-1} cdot 2^{2n-1} &= C_{2n-2}^{n}(2^{2n-1} - 1) + C_{2n-1}^{n-1} cdot 2^{2n-1} &Rightarrow sum = C_{2n-2}^{n} cdot (2^{2n-1} - 1) + C_{2n-1}^{n-1} cdot 2^{2n-1} end{aligned} ]7. Conclusion: Since specified valid binary numbers are formed under specific constraints resulting into the provided combinatorial setup, the final result can be computed based on the combination and sum analysis.[ boxed{ sum = C_{2n-2}^{n} cdot (2^{2n-1} - 1) + C_{2n-1}^{n-1} cdot 2^{2n-1} } ]

🧠:Okay, so I need to find the remainder when the polynomial ( x^{2008} + 2008x + 2008 ) is divided by ( x + 1 ). Hmm, let me think about how to approach this. I remember something about the Remainder Theorem, which might be useful here. Let me recall... The theorem states that if a polynomial ( f(x) ) is divided by ( x - a ), the remainder is ( f(a) ). Wait, but here we have ( x + 1 ), which is the same as ( x - (-1) ). So according to the theorem, the remainder should be ( f(-1) ), right?Let me verify that. If we divide a polynomial ( f(x) ) by ( x + 1 ), which is ( x - (-1) ), then the remainder is indeed the value of the polynomial evaluated at ( x = -1 ). That seems correct. So instead of doing polynomial long division, which would take a long time especially with such a high exponent, I can just plug in ( x = -1 ) into the polynomial and calculate the result. That should give me the remainder. Alright, let me try that. The polynomial is ( x^{2008} + 2008x + 2008 ). So substituting ( x = -1 ):First term: ( (-1)^{2008} ). Since 2008 is an even number, this should be ( 1 ).Second term: ( 2008 times (-1) = -2008 ).Third term: 2008, which is just 2008.So adding them up: ( 1 + (-2008) + 2008 ).Let me compute that step by step. ( 1 - 2008 + 2008 ). The -2008 and +2008 cancel each other out, leaving just 1. So the remainder should be 1. Wait, is that possible? Let me check my calculations again to be sure.First term: ( (-1)^{2008} ). Since the exponent is even, yes, that's 1. Correct.Second term: 2008 multiplied by -1 is indeed -2008. Third term is 2008. So adding 1 + (-2008) + 2008. Yes, the -2008 and +2008 sum to zero, so 1 remains. Therefore, the remainder is 1. But wait, maybe I missed something? Let me think again. Is there a possibility that the Remainder Theorem doesn't apply here? No, the theorem is straightforward: when dividing by ( x - a ), the remainder is ( f(a) ). Here, ( a = -1 ), so substituting ( x = -1 ) into the polynomial gives the remainder. Therefore, unless there's a mistake in substitution or arithmetic, the remainder should be 1. Alternatively, I can try using polynomial long division as a check, but with such a high exponent, that seems impractical. Alternatively, maybe synthetic division? Let me recall how synthetic division works. For dividing by ( x + 1 ), synthetic division would use -1 as the root. But even synthetic division for a degree 2008 polynomial would be tedious. So, probably, the Remainder Theorem is the best approach here. Alternatively, another way to think about it: when you divide any polynomial ( f(x) ) by ( x + 1 ), the remainder is a constant term (since the divisor is linear) which is equal to ( f(-1) ). Therefore, regardless of the degree of the polynomial, the remainder is just the value at ( x = -1 ). So, substituting ( x = -1 ) is the correct method. Therefore, my conclusion is that the remainder is 1. But let me just go through the substitution once again to make sure. First term: ( (-1)^{2008} ). Since 2008 divided by 2 is 1004, which is an integer, so exponent is even, result is 1. Correct. Second term: 2008 * (-1) = -2008. Correct.Third term: 2008, so adding those: 1 - 2008 + 2008 = 1. Yep, that's definitely right. So I think that's the answer. The remainder is 1. Maybe I can also recall that even exponents on negative numbers become positive, odd exponents stay negative. Since 2008 is even, so first term positive. The linear term is negative 2008, and the constant term is positive 2008. So they cancel each other out, leaving 1. Therefore, I'm confident the remainder is 1. Final AnswerThe remainder is boxed{1}.

✅:To find the remainder when ( x^{2008} + 2008x + 2008 ) is divided by ( x + 1 ), we can use the Remainder Theorem. The theorem states that the remainder of a polynomial ( f(x) ) divided by ( x - a ) is ( f(a) ). Here, we are dividing by ( x + 1 ), which is equivalent to ( x - (-1) ). Therefore, we need to evaluate the polynomial at ( x = -1 ).1. Substitute ( x = -1 ) into the polynomial: - The first term is ( (-1)^{2008} ). Since 2008 is even, this term is ( 1 ). - The second term is ( 2008 times (-1) = -2008 ). - The third term is ( 2008 ).2. Summing these values: [ 1 + (-2008) + 2008 = 1 - 2008 + 2008 = 1 ]Thus, the remainder when ( x^{2008} + 2008x + 2008 ) is divided by ( x + 1 ) is (boxed{1}).

🔑:1. To find the remainder when a polynomial P(x) is divided by a binomial of the form x - a, we can use the Remainder Theorem.2. The Remainder Theorem states that the remainder of the division of P(x) by x - a is P(a).3. In this problem, the polynomial is ( P(x) = x^{2008} + 2008x + 2008 ) and we are dividing by ( x + 1 ), which can be rewritten as ( x - (-1) ).4. According to the Remainder Theorem, the remainder when ( P(x) ) is divided by ( x + 1 ) is given by evaluating ( P(-1) ).[P(-1) = (-1)^{2008} + 2008(-1) + 2008]5. Calculate ( (-1)^{2008} ): - Since the exponent 2008 is even, ( (-1)^{2008} = 1 ).6. Substitute this value back into the polynomial:[P(-1) = 1 + 2008(-1) + 2008]7. Simplify the expression:[P(-1) = 1 - 2008 + 2008]8. Combining the terms gives:[P(-1) = 1 + (-2008 + 2008)][P(-1) = 1 + 0][P(-1) = 1] Conclusion:The remainder when ( x^{2008} + 2008x + 2008 ) is divided by ( x + 1 ) is:[boxed{1}]

🧠:Okay, let me try to figure out this problem. So, we have a group of seven boys, and each one has at least three brothers among the others. We need to prove that all seven are brothers. Hmm, okay. Let me start by understanding what the question is asking. First, let's parse the problem statement again. There are seven boys in total. Each boy has at least three brothers in the group. We have to show that this implies that all seven are brothers. So, brothers here probably mean siblings, right? So, if all seven are brothers, that means they all share the same parents. But the problem is to show that given each has at least three brothers, they must all be siblings. Let me think. So, if each boy has at least three brothers, that means for any boy, when we look at the other six boys, at least three of them are his brothers. Wait, but how can we model this? Maybe using graph theory? Or maybe combinatorics? Let me try to approach it step by step.First, let's consider each boy as a node in a graph. Then, the brother relationship can be represented as edges between the nodes. If two boys are brothers, there's an edge connecting them. Now, the problem states that each boy has at least three brothers. So, in graph terms, each node has a degree of at least three. But wait, in graph theory terms, if we have a graph with seven nodes where each node has degree at least three, does that imply that the graph is complete? A complete graph would mean every node is connected to every other node, which in this context would mean all seven boys are brothers. But is this necessarily true?Wait, no. A complete graph on seven nodes would have each node with degree six. But the problem states each node has degree at least three. So, the graph is not necessarily complete. So, maybe the answer isn't straightforward. Hmm. Alternatively, maybe we can model this using sets. Let's think of each boy's set of brothers. Each boy has at least three brothers, so the size of each set is at least three. However, since brotherhood is mutual, if boy A is a brother of boy B, then boy B is a brother of boy A. Therefore, the brother relationships have to form a symmetric relation. So, maybe the problem reduces to showing that in a symmetric relation over seven elements, where each element is related to at least three others, the relation must be the complete relation. But that doesn't seem right. For example, you could partition the seven boys into two groups where each group has mutual brotherhood, but then each boy would have brothers only within their group. However, the problem states each has at least three brothers. So, if we partitioned into two groups, say, four and three. Then, each boy in the four-group would have three brothers, and each boy in the three-group would have two brothers. But that doesn't satisfy the condition because the three-group would have only two brothers each. So, such a partition wouldn't work. Alternatively, if we tried a partition of five and two. Then, the five-group would have four brothers each, and the two-group would have one brother each. Still, the two-group doesn't meet the three brothers requirement. Similarly, if we tried a three-three-one partition, the singleton would have zero brothers, which is even worse. So, maybe any partition into smaller groups would result in some group having insufficient brothers? Wait, maybe the key is that if you have multiple sibling groups, the boys in each group can only count their siblings as brothers. So, if there is more than one sibling group, then each boy's brothers are limited to his own sibling group. Therefore, the size of each sibling group must be at least four, because each boy needs at least three brothers. Because if a sibling group has size k, then each boy in that group has k-1 brothers. Therefore, to have k-1 >= 3, we need k >= 4. So, each sibling group must have at least four boys. But there are seven boys in total. If there is more than one sibling group, say two sibling groups. Then, the sizes of the groups would have to be at least four and at least four. But 4 + 4 = 8, which is more than seven. Therefore, it's impossible to have two sibling groups each of size at least four. Hence, there can only be one sibling group. Therefore, all seven must be brothers. Wait, that seems like a possible line of reasoning. Let me check again. If we suppose that there are two or more sibling groups, then each sibling group must have at least four members. But since we have seven boys, two groups of four would require eight boys, which is impossible. Therefore, there can only be one sibling group. Hence, all seven are brothers. Is that correct? Let me verify step by step.Each boy must have at least three brothers. Brothers are defined as being in the same sibling group. So, if two boys are in different sibling groups, they are not brothers. Therefore, for any boy, his brothers are exactly the other boys in his sibling group. Therefore, the number of brothers a boy has is (size of his sibling group) - 1. Thus, to have at least three brothers, each boy must be in a sibling group of size at least 4 (since 4 - 1 = 3). Therefore, each sibling group must have size >= 4. If there are two sibling groups, then each must have at least four boys, but 4 + 4 = 8 > 7. Therefore, two sibling groups are impossible. If there are three or more sibling groups, each would need to have at least four boys, which is also impossible since 3 * 4 = 12 > 7. Therefore, the only possibility is one sibling group containing all seven boys. Hence, all seven are brothers. Hmm, that seems to check out. Let me see if there's another angle. Suppose we try to model it with graph theory again. If each node has degree at least three, but in this case, edges represent brotherhood. But in reality, brotherhood is transitive within a sibling group. Wait, no. Brotherhood is not transitive in reality, but in the problem's context, if we assume that all members of a sibling group are mutual brothers, then the graph would consist of one or more cliques (complete subgraphs). But in reality, siblings are mutual, so if A is a brother of B, then B is a brother of A. So the graph is undirected. But sibling groups would form cliques. So, in this problem, the graph is a union of disjoint cliques. Each clique represents a sibling group. Given that each node has degree at least three, each clique must have size at least four, since in a clique of size k, each node has degree k - 1. So, k - 1 >= 3 => k >= 4. Therefore, the entire graph is a union of cliques each of size at least four. Since total nodes are seven, and 4 + 4 = 8 > 7, only one clique is possible. Hence, the entire graph is a single clique of seven nodes. Therefore, all seven are brothers. Therefore, the answer is that all seven must be brothers. Wait, but let me test this with an example. Suppose there are two sibling groups, one of size four and another of size three. Then, the boys in the size four group have three brothers each (which meets the requirement), but the boys in the size three group have two brothers each, which does not meet the requirement. Therefore, this configuration is invalid. Similarly, any split where one group is less than four would result in those boys not having enough brothers. Therefore, the only possible way is a single group of seven. Alternatively, if there are three sibling groups, say four, two, and one. Then, the four-group is okay, but the two and one groups are not. So, that's invalid. Therefore, the conclusion holds. Another way to think about it: Suppose that there is a sibling group of size exactly four. Then, the remaining three boys must be in another sibling group(s). But any sibling group with three boys would only give two brothers per boy, which is insufficient. So, the remaining three cannot form a valid sibling group. Therefore, the only way is to have all seven in one group. Therefore, the proof is complete. I think that's a solid reasoning. Let me summarize it step by step to ensure there are no gaps.1. Assume that the group of seven boys is divided into one or more sibling groups, where each sibling group consists of mutual brothers.2. For each boy, the number of brothers he has is equal to the size of his sibling group minus one.3. Each boy must have at least three brothers, so each sibling group must have size at least four (since 4 - 1 = 3).4. The total number of boys is seven. If there are two or more sibling groups, each of size at least four, the minimum total number of boys required would be 4 + 4 = 8, which exceeds seven. Hence, this is impossible.5. Therefore, there can only be one sibling group, which contains all seven boys.6. Thus, all seven boys are brothers.Yes, that seems to cover it. Therefore, the conclusion is that all seven must be brothers.

✅:To prove that all seven boys are brothers given each has at least three brothers among the others, follow this reasoning:1. Model the Problem: Consider each boy as part of a sibling group where mutual brotherhood defines membership. Brotherhood is symmetric: if boy ( A ) is a brother of ( B ), then ( B ) is a brother of ( A ).2. Key Constraint: Each boy has at least three brothers. Thus, for any sibling group of size ( k ), each member has ( k - 1 ) brothers. To satisfy ( k - 1 geq 3 ), every sibling group must have ( k geq 4 ).3. Total Boys Constraint: There are seven boys total. If there were multiple sibling groups, each would need at least four members. However: - Two groups of four require ( 4 + 4 = 8 ) boys (impossible, as there are only seven). - Any partition into smaller groups (e.g., 4+3, 5+2) results in at least one group with fewer than four boys, violating the minimum brothers requirement.4. Conclusion: The only feasible configuration is a single sibling group containing all seven boys. Hence, all seven are brothers.Final Answer: All seven boys must be brothers, as any division into smaller sibling groups fails to satisfy the condition that each boy has at least three brothers.

🔑:1. Assumption and Initial Setup: Consider a company of seven boys. Each boy is said to have at least three other boys among these seven who are considered as brothers. 2. Proof by Contradiction: Assume, for the sake of contradiction, that not all boys in this company are brothers. 3. Selection of Two Boys: Take any two boys from the company. Let's denote them as ( B_1 ) and ( B_2 ). According to our assumption, ( B_1 ) and ( B_2 ) are not brothers.4. Possibilities for Such a Pair: - Both ( B_1 ) and ( B_2 ) have at least three brothers from the remaining five boys. Hence, we list the remaining boys as ( B_3, B_4, B_5, B_6, ) and ( B_7 ). - ( B_1 ) has at least three brothers among ( B_3, B_4, B_5, B_6, B_7 ). - Similarly, ( B_2 ) also has at least three brothers among ( B_3, B_4, B_5, B_6, B_7 ).5. Intersection of Brotherhood: - Since ( B_1 ) has at least three brothers and ( B_2 ) has at least three brothers among the same five boys, their sets of brothers must have at least one common member. Without loss of generality, assume that ( B_3 ) is a common brother to both ( B_1 ) and ( B_2 ).6. Contradiction: By the nature of the problem, ( B_3 ) considers both ( B_1 ) and ( B_2 ) as brothers. However, our initial assumption is that ( B_1 ) and ( B_2 ) are not brothers, which leads to a direct contradiction.7. Conclusion: - This contradiction implies that our initial assumption that ( B_1 ) and ( B_2 ) are not brothers is false. - Hence, it must be that any two boys in the company are indeed brothers.Thus, we conclude all seven boys in the company are brothers. [boxed{text{All seven boys are brothers}}]