answer:To find the correct value of ( x ) given the conditions, we analyze the problem step-by-step: 1. The conditions given are: - ( 3x ) - ( x ) is not less than 10. - ( x ) is greater than 5. - Three statements are true and two are false. - Find the value of ( x ). 2. Simplify the statements: - The first statement ( 3x ) can be interpreted as ( 3x > 30 ), but we assume it's implicitly given. - Second statement ( x ge 10 ). - Third statement ( x > 5 ). 3. Analyze the validity of each statement: - The first statement ( 3x > 30 ) simplifies to ( x > 10 ). - The second statement states ( x ge 10 ). - The third statement ( x > 5 ). 4. If ( x > 35 ) is true, then ( x ) must also satisfy ( x ge 10 ) and ( x > 5 ), making all three valid. - This provides 4 correct statements if we count all given and simplified versions, which cannot fit our requirement of exactly 3 valid statements. Hence, ( x > 35 ) cannot be the case. 5. Determine the necessary truthful statements: - Thus, ( x > 35 ) must be false, implying: - First assertion (equivalent to ( x > 35 )) is false. - The second assertion (( x ge 10 )) is true to reach the 3 true statement mark. - Third assertion (( x > 5 )) is always true if second assertion is valid and thus counted true. 6. Evaluate ( x ): - Fourth statement validity must be determined: - If it were true, the count of truthful counts would again be 4. - Hence, we'll conclude the fourth must be false. 7. Summary: - Fourth must be false, making ( x = 9 ) from integral present number bounds. 8. Conclusion: - Collecting all aspects and verifying statement matches, - Full fit is given ( x = 9 ). Hence the final solution is: [ boxed{9} ]

answer:Nous devons démontrer l'inégalité suivante pour tout entier ( n ): [ n ln 2 leq ln left( binom{2n}{n} right) leq n ln 4 ] # Étape 1 : Majoration de ( ln left( binom{2n}{n} right) ) 1. Remarquons d'abord que ( binom{2n}{n} leq 2^{2n} = 4^n ). 2. Prenons le logarithme naturel de chaque côté de l'inégalité : [ ln left( binom{2n}{n} right) leq ln (4^n) ] 3. Simplifions en utilisant les propriétés du logarithme : [ ln (4^n) = n ln 4 ] # Conclusion pour la majoration : [ ln left( binom{2n}{n} right) leq n ln 4 ] # Étape 2 : Minoration de ( ln left( binom{2n}{n} right) ) 1. Considérons maintenant la décomposition du coefficient binomial ( binom{2n}{n} ) : [ binom{2n}{n} = frac{(2n)!}{n! , n!} ] 2. Réécrivons cette expression sous forme de produit : [ binom{2n}{n} = frac{(2n)(2n-1) cdots (n+1)}{n(n-1) cdots 1} = frac{prod_{k=1}^{n} (n+k)}{prod_{k=1}^{n} k} = prod_{k=1}^{n} left( frac{n+k}{k} right) ] 3. Simplifions les termes : [ frac{n+k}{k} = frac{n}{k} + 1 ] 4. Notons que pour tout ( k ) entre 1 et ( n ), ( frac{n}{k} + 1 geq 2 ). En effet, ( frac{n}{k} ) est supérieur ou égal à 1 pour tout ( k ) entre 1 et ( n ). 5. Prenons le produit de ces inégalités positives : [ prod_{k=1}^{n} left( frac{n}{k} + 1 right) geq prod_{k=1}^{n} 2 = 2^n ] 6. Prenons le logarithme naturel de chaque côté de l'inégalité : [ ln left( binom{2n}{n} right) geq ln (2^n) ] 7. Utilisons de nouveau les propriétés du logarithme : [ ln (2^n) = n ln 2 ] # Conclusion pour la minoration : [ ln left( binom{2n}{n} right) geq n ln 2 ] # Conclusion générale : En combinant les résultats des étapes de majoration et de minoration, nous obtenons l'inégalité recherchée : [ n ln 2 leq ln left( binom{2n}{n} right) leq n ln 4 ] [ boxed{n ln 2 leq ln left( binom{2n}{n} right) leq n ln 4} ]

answer:1. **Initial Setup**: Aladdin has a set of coins with weights (1, 2, ldots, 20) grams. He can ask Genie about any two coins from the set which one is heavier, but he should pay Genie some other coin from the set before. With every question, the set of coins becomes smaller. 2. **First Comparison**: Aladdin picks an arbitrary coin (C(1)) to pay for his first comparison. 3. **Finding the Heaviest Coin Among (C(2)) to (C(10))**: - Aladdin compares coins (C(2)) and (C(3)). The losing coin is paid to the Genie. The winning coin is called the "current winner". - For (k = 4) to (10): - Aladdin compares coin (C(k)) against the current winner. - The losing coin is paid to the Genie. - The winning coin becomes the new current winner. 4. **Finding the Heaviest Coin Among (C(11)) to (C(20))**: - Aladdin compares coins (C(11)) and (C(12)). The losing coin is paid to the Genie. The winning coin is called the "current champion". - For (k = 13) to (20): - Aladdin compares coin (C(k)) against the current champion. - The losing coin is paid to the Genie. - The winning coin becomes the new current champion. 5. **Final Comparison**: At the end, Aladdin is left with two coins: - The current winner (X), which is the heaviest among the nine coins (C(2), ldots, C(10)). - The current champion (Y), which is the heaviest among the ten coins (C(11), ldots, C(20)). 6. **Argument**: - If (X > Y), then there are at least 18 coins lighter than (X) and at least 9 coins lighter than (Y). Hence, (X geq 19) and (Y geq 10), and (X + Y geq 29). - If (X < Y), then there are at least 18 coins lighter than (Y) and at least 8 coins lighter than (X). Hence, (X geq 9) and (Y geq 19), and (X + Y geq 28). Therefore, in both cases, the total weight of the two coins (X) and (Y) is at least 28 grams. (blacksquare)

answer:(1) According to the problem statement, since the point (n,S_n) lies on the curve of the function y= frac{3}{2}x^2- frac{1}{2}x, we have S_n= frac{3}{2}n^2- frac{1}{2}n. For n geq 2: a_n = S_n - S_{n-1} = left( frac{3}{2}n^2 - frac{1}{2}n right) - left( frac{3}{2}(n-1)^2 - frac{1}{2}(n-1) right) = 3n - 2. When n=1, we have: a_1 = S_1 = 1, which also fits the above formula. Therefore, the general formula for the sequence is a_n = 3n - 2 for n in mathbb{N}^*. (2) By the result from (1), we have: b_n = frac{3}{a_na_{n+1}} = frac{3}{(3n - 2)(3n + 1)} = frac{1}{3n - 2} - frac{1}{3n + 1}. Thus: T_n = 1 - frac{1}{4} + frac{1}{4} - frac{1}{7} + ldots + frac{1}{3n - 2} - frac{1}{3n + 1} = 1 - frac{1}{3n + 1}. In order for T_n < frac{m}{20} to hold, m must satisfy 1 leq frac{m}{20}, hence m geq 20. Therefore, the smallest integer m that satisfies the condition is boxed{20}.