answer:1. **Consider the lemma**: For any integer ( n geq 4 ) and real numbers ( r_i in [0, 1] ) for ( 1 leq i leq n ), satisfying ( sum_{i=1}^{n} r_i = r ) where ( r in mathbb{N} ) and ( r leq n-1 ), it is true that [ sum_{i=1}^{n} r_i r_{i+1} leq r - frac{3}{4}. ] 2. **Proof of the lemma**: - If ( r = 0 ), the conclusion is obvious because all ( r_i = 0 ), leading to ( sum_{i=1}^{n} r_i r_{i+1} = 0 leq 0 - frac{3}{4} ). - Assume ( r in mathbb{Z}^+ ). Define the function ( f(r_1, r_2, ldots, r_n) = sum_{i=1}^{n} r_i r_{i+1} ). This function ( f ) is continuous on the closed bounded set ([0,1]^n ), so it achieves a maximum value ( S ). - Without loss of generality, consider ( f ) achieves maximum ( S ) for ( (r_1, r_2, ldots, r_n) ), and assume that the count of values 0 or 1 among ( r_1, r_2, ldots, r_n ) is maximized. 3. **Consider any two non-adjacent ( r_i ) and ( r_j )**: - For non-adjacent ( r_i ) and ( r_j ) (i.e., ( i-j notequiv pm 1 pmod{n} )), at least one of ( r_i ) or ( r_j ) must be either 0 or 1. 4. **Details on maximum value condition**: - For any two non-adjacent ( r_i ) and ( r_j ), set ( t = r_i + r_j ), and let ( x = r_i ). Then ( sum_{i=1}^{n} r_i r_{i+1} ) is a quadratic function of ( x ). When achieving maximum value, ( x ) must be ( 0 ) or (min {1, t}). Hence, either ( r_i ) or ( r_j ) must be 0 or 1. 5. **Implications for ( r_1, r_2, ldots, r_n )**: - There can be at most two values not belonging to ({0,1}), and these two must be adjacent. Assume ( r_{1}, r_{2} in (0,1) ), then the rest are either 0 or 1. - Let ( r_1 + r_2 = 1 ) and the remaining ( n-2 ) values have exactly ( r-1 ) ones. Therefore, [ sum_{i=1}^{n} r_{i} r_{i+1} leq r_1 r_2 + r_2 + r - 2 + r_1. ] 6. **Combining terms and bounding the sum**: - Substituting ( r_1 + r_2 = 1 ), [ sum_{i=1}^{n} r_i r_{i+1} leq r_1 r_2 + (r-1) - 1 + r_1(1-r_1)r_2(1-r_2) leq r_1 r_2 + r - 3 + frac{(1+r_1 + 1+r_2)^2}{4}. ] - Which simplifies to, [ sum_{i=1}^{n} (r_i r_{i+1}) leq r - frac{3}{4}. ] 7. **Conclusion of lemma**: [ sum_{i=1}^{n} r_i r_{i+1} leq r - frac{3}{4}. ] 8. **Returning to the original problem**: - Let ( [a_i]=b_i ) and ( {a_i}=r_i ) for ( 1 leq i leq n ). Note that ( sum_{i=1}^{n} r_i = r ) where ( r leq n-1 ). - According to the lemma: [ sum_{i=1}^{n} {a_i} a_{i+1} = sum_{i=1}^{n} r_i (b_{i+1} + r_{i+1}) = sum_{i=1}^{n} r_i b_{i+1} + sum_{i=1}^{n} r_i r_{i+1}. ] - Thus, [ sum_{i=1}^{n} {a_i} a_{i+1} leq sum_{i=1}^{n} (b_{i+1} + r) - frac{3}{4} = n - frac{3}{4}. ] 9. **Considering a specific case**: - Choose ( a_1 = frac{3}{2} + (n-2)epsilon, a_2 = frac{1}{2}, a_3 = a_4 = ldots = a_n = 1 - epsilon ). As ( epsilon to 0^+ ), [ sum_{i=1}^{n} {a_i} a_{i+1} = left(frac{1}{2} + (n-2)epsilonright) cdot frac{1}{2} + frac{1}{2} cdot (1 - epsilon) + (n-3)(1-epsilon)^2 + (1-epsilon) cdot left(frac{3}{2} + (n-2)epsilonright). ] - Simplifying and taking the limit: [ sum_{i=1}^{n} {a_i} a_{i+1} to n - frac{3}{4}. ] 10. **Conclusion**: - Thus, the minimum value of ( lambda(n) ) is: [ boxed{n - frac{3}{4}}. ]

answer:1. **Define the Problem**: We are given a finite set of points, some of which are connected by line segments. We need to prove that there exist at least two points in this set such that the number of line segments emerging from these two points are equal. 2. **Number of Points**: Suppose there are ( n ) points in total. 3. **Range of Degrees**: The number of line segments (or degree of the point) emerging from any point can range from ( 0 ) to ( n-1 ). 4. **Possible Degrees**: - Each point can have a degree (number of line segments leaving the point) that must be one of the integers in the set ( {0, 1, 2, ldots, n-1} ). 5. **Applying Pigeonhole Principle**: - The Pigeonhole Principle states that if you have more pigeons than pigeonholes and you assign each pigeon to a pigeonhole, at least one pigeonhole must contain more than one pigeon. - In this context, the pigeons are the ( n ) points, and the pigeonholes are the ( n ) possible degrees (0 through ( n-1 )). 6. **Degree Assignment**: - Each of the ( n ) points is assigned one of the ( n ) degrees from 0 to ( n-1 ). 7. **Consideration of Extremes (0 and ( n-1 ))**: - If there exists a point with degree ( 0 ) (isolated point with no line segments), it means it is not connected to any other points. - If there exists a point with degree ( n-1 ), that point is connected to every other point in the set. 8. **Contradiction Argument**: - Consider if one point had degree ( n-1 ) (connected to all points), then every other point must have at least degree ( 1 ) since they are connected to this point. - In a situation where one point has 0 connections, no other point can have ( n-1 ) connections, because the isolated point cannot be part of another point’s connections. 9. **Conclusion via Pigeonhole Principle**: - Since having both a point with degree 0 and another with degree ( n-1 ) simultaneously is impossible, the degrees must be shared among the points in such a way that at least two points must have the same degree. Therefore, by the Pigeonhole Principle and the limitations on the degrees, we conclude that there exist at least two points in the set that have the same number of line segments emerging from them. (blacksquare)

answer:Given the conditions about the ages of five people, we need to determine their ages in ascending order. Let's denote the ages of the five people as (a_1, a_2, a_3, a_4, a_5 ) where (a_1 < a_2 < a_3 < a_4 < a_5). 1. **Step 1**: Analyze the first condition: The average age of the younger three (i.e., (a_1, a_2, a_3)) is 18 years. [ frac{a_1 + a_2 + a_3}{3} = 18 ] Multiplying both sides by 3: [ a_1 + a_2 + a_3 = 54 ] 2. **Step 2**: Analyze the second condition: The age difference between the older two (i.e., (a_4) and (a_5)) is 5 years. [ a_5 - a_4 = 5 ] 3. **Step 3**: Analyze the third condition: The average age of the older three (i.e., (a_3, a_4, a_5)) is 26 years. [ frac{a_3 + a_4 + a_5}{3} = 26 ] Multiplying both sides by 3: [ a_3 + a_4 + a_5 = 78 ] 4. **Step 4**: Analyze the fourth condition: The age difference between the younger two (i.e., (a_1) and (a_2)) is 7 years. [ a_2 - a_1 = 7 ] 5. **Step 5**: Analyze the fifth condition: The average age of the oldest and youngest (i.e., (a_1) and (a_5)) is 22 years. [ frac{a_1 + a_5}{2} = 22 ] Multiplying both sides by 2: [ a_1 + a_5 = 44 ] 6. **Step 6**: Solve the system of linear equations: We have: [ a_1 + a_2 + a_3 = 54 quad quad (1) ] [ a_5 - a_4 = 5 quad quad (2) ] [ a_3 + a_4 + a_5 = 78 quad quad (3) ] [ a_2 - a_1 = 7 quad quad (4) ] [ a_1 + a_5 = 44 quad quad (5) ] From equation (1): [ a_1 + a_2 + a_3 = 54 quad quad (i) ] From equation (3): [ a_3 + a_4 + a_5 = 78 quad quad (ii) ] From equation (4): [ a_2 = a_1 + 7 quad quad (iii) ] From equation (5): [ a_5 = 44 - a_1 quad quad (iv) ] Let's substitute (a_5) from equation ((iv)) into equation ((ii)): [ a_3 + a_4 + (44 - a_1) = 78 ] [ a_3 + a_4 + 44 - a_1 = 78 ] [ a_3 + a_4 - a_1 = 34 quad quad (vi) ] And from equation (3): [ a_5 = a_4 + 5 quad quad (vii) ] First, let's solve (a_5) and (a_1): Plug (a_1) into both ((i)) and ((iv)): Let's subtract equation ((i)) from the total indicating [ a_4 - a_1 = 34 - 54 ] Introducing age related to older and younger: Thus: boxed{13, 20, 21, 26, 31}

answer:Let's denote the higher selling price as ( P ). When the article is sold for Rs. 340, the gain is: ( text{Gain} = text{Selling Price} - text{Cost Price} ) ( text{Gain} = 340 - 200 ) ( text{Gain} = 140 ) Now, we are told that 5% more is gained by selling it for the higher price ( P ). So, the gain in this case would be: ( text{New Gain} = 140 + frac{5}{100} times 140 ) ( text{New Gain} = 140 + 7 ) ( text{New Gain} = 147 ) Therefore, when the article is sold for the higher price ( P ), the gain is Rs. 147. So, we can find the higher selling price by adding this gain to the cost price: ( P = text{Cost Price} + text{New Gain} ) ( P = 200 + 147 ) ( P = 347 ) Hence, the higher selling price is Rs. boxed{347} .