answer: Problem 1:1. Label the bags from 1 to 10.2. Take 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third bag, and so on, until you take 10 coins from the tenth bag.3. Weigh all these coins together.Let ( W ) be the total weight of the coins. If all coins were honest, the total weight would be:[W_{text{honest}} = 10 times (1 + 2 + 3 + cdots + 10) = 10 times frac{10 times 11}{2} = 550 text{ grams}]Since one of the bags contains false coins weighing 11 grams each, the actual weight ( W ) will be:[W = 550 + (n - 1)]where ( n ) is the number of the bag with false coins.By comparing the actual weight ( W ) with the expected weight ( W_{text{honest}} ), we can determine the bag with false coins:[n = W - 549] Problem 2:1. Let ( a, b, c ) be the sides of the triangle, and ( h_a, h_b, h_c ) be the altitudes corresponding to these sides.2. The area ( A ) of the triangle can be expressed using any altitude:[A = frac{1}{2} a h_a = frac{1}{2} b h_b = frac{1}{2} c h_c]3. Given ( h_a, h_b, h_c < 1 ) cm, we have:[A < frac{1}{2} a times 1 = frac{a}{2}][A < frac{1}{2} b times 1 = frac{b}{2}][A < frac{1}{2} c times 1 = frac{c}{2}]4. Since ( a, b, c ) are the sides of a triangle, by the triangle inequality, the maximum possible area is less than:[A < frac{1}{2} times text{(sum of sides)} times 1 < 100 text{ cm}^2] Problem 3:1. Let ( x ) be the number. We need:[frac{x}{2} = a^2, quad frac{x}{3} = b^3, quad frac{x}{5} = c^5]2. Therefore, ( x ) must be divisible by ( 2, 3, ) and ( 5 ) raised to the appropriate powers:[x = 2^3 times 3^4 times 5^6]3. Calculating the value:[x = 8 times 81 times 15625 = 10125000] Problem 4:1. Let ( N ) be the total sum of all 6-digit numbers on lucky tickets.2. A ticket is lucky if the sum of the first three digits equals the sum of the last three digits.3. The total sum of all 6-digit numbers is:[sum_{i=000001}^{999999} i]4. The sum of the digits of each lucky ticket is divisible by 13, as the sum of the first three digits equals the sum of the last three digits.5. Therefore, the total sum ( N ) is divisible by 13.(blacksquare)The final answer is ( boxed{ x = 10125000 } )

answer:【Analysis】Find the pattern. We find: The number in the upper left corner of the square can be: 1,2,3 ; 5 sim 10 ; 12 sim 17 ; 19 sim 23 for a total of 20.

answer:For A and B, since acute angles are angles in the first quadrant, but angles in the first quadrant are not necessarily acute angles, so A is incorrect, and B is correct. For C, there are many angles with the same terminal side, and their relationship is that their difference is an integer multiple of a full angle, so C is incorrect. For D, there are both positive and negative angles in the second quadrant, and similarly, there are both positive and negative angles in each quadrant. Therefore, it is impossible to compare the sizes of angles in the first quadrant with those in the second quadrant, making D incorrect. Thus, the correct choice is boxed{text{B}}. This question examines the relationship between quadrant angles and special angles. Acute angles must be angles in each quadrant, but angles in the first quadrant are not necessarily acute angles. Angles with the same terminal side differ by an integer multiple of a full angle. Angles in the second quadrant can be negative, and there are also positive angles in the first quadrant, so it is impossible to compare the sizes of angles in the second quadrant with those in the first quadrant. The key point of this question is the relationship between quadrant angles and axial angles, examining the relationship between quadrant angles and acute angles, the size relationship between angles with the same terminal side, and the size relationship between angles in two different quadrants. It effectively checks the basic knowledge of quadrant angles and axial angles.

answer:1. Restate the problem: We need to find all odd primes ( p ) such that ( 1 + k(p-1) ) is prime for all integers ( k ) where ( 1 le k le frac{p-1}{2} ).2. Consider the case ( p ge 13 ): - Let ( q ) be an odd prime such that ( q le frac{p-1}{2} ). - We need ( 1 + k(p-1) ) to be prime for ( k = 1, 2, ldots, q ). - By the Pigeonhole Principle, there exist ( a ) and ( b ) such that ( 1 + a(p-1) equiv 1 + b(p-1) pmod{q} ). - This implies ( q mid (a-b)(p-1) ) and since ( |a-b| < q ), we have ( q mid p-1 ). - Therefore, ( q mid frac{p-1}{2} ).3. Apply Bertrand's Postulate: - Bertrand's Postulate states that for any integer ( n > 1 ), there is always at least one prime ( r ) such that ( n < r < 2n ). - Applying this to ( frac{p-1}{4} ), we get a prime ( r ) such that ( frac{p-1}{4} < r < frac{p-1}{2} ). - Since ( r ) is a prime and ( r neq 3 ), we have ( r mid frac{p-1}{2} ) and ( 3 mid frac{p-1}{2} ).4. Derive a contradiction: - If ( 3r mid frac{p-1}{2} ), then ( frac{3(p-1)}{2} < 3r le frac{p-1}{2} ), which is a contradiction because ( frac{3(p-1)}{2} ) is greater than ( frac{p-1}{2} ).5. Consider the case ( p le 12 ): - For ( p = 11 ): ( 1 + 2(11-1) = 21 ) is not prime. - For ( p = 7 ): ( 1 + (7-1) = 7 ), ( 1 + 2(7-1) = 13 ), ( 1 + 3(7-1) = 19 ) are all primes. - For ( p = 5 ): ( 1 + 2(5-1) = 9 ) is not prime. - For ( p = 3 ): ( 1 + (3-1) = 3 ) is prime.6. Conclusion: - The only odd primes ( p ) that satisfy the condition are ( p = 3 ) and ( p = 7 ).The final answer is (boxed{3, 7})

answer:[Solution] As shown in the figure, draw a line from E to the midpoint G of DC.Since EG is the line connecting the midpoints of sides AC and DC in triangle ACD, EG parallel AD, thenS_{triangle EBG}=frac{2}{3} S_{triangle EBC},and S_{triangle BDF}=frac{1}{4} S_{triangle EBG}=frac{1}{6} S_{triangle EBC}.Therefore, S_{FDC E}=frac{5}{6} S_{triangle EBC}.Thus, frac{S_{triangle BDF}}{S_{FDC E}}=frac{1}{5}.Hence, the answer is (A).

answer:A1. If at the beginning the factory produced x products, then after the equipment was updated, they produced frac{125}{100} cdot x products, and after the layoffs, they produced frac{80}{100} cdot frac{125}{100} cdot x=x products. The number of final products has therefore not changed. The correct answer is D.