answer:1. **Base Case:** - Consider the smallest case where there are 3 girls and 1 boy. In this scenario, it is easy to see that at least one girl will have a strong position. For instance, if the arrangement is ( G_1, B, G_2, G_3 ), then ( G_1 ) and ( G_3 ) both have strong positions because in either direction, the number of girls is always greater than the number of boys. 2. **Inductive Hypothesis:** - Assume that for any arrangement of ( 2k+1 ) girls and ( k ) boys, there is always at least one girl in a strong position. This is our inductive hypothesis. 3. **Inductive Step:** - We need to show that if the hypothesis holds for ( k ), it also holds for ( k+1 ). That is, we need to show that for ( 2(k+1)+1 = 2k+3 ) girls and ( (k+1) ) boys, there is always at least one girl in a strong position. 4. **Constructing the Inductive Step:** - Consider an arrangement of ( 2k+3 ) girls and ( k+1 ) boys. Select any girl ( G ) who has a boy as a neighbor. Since there are ( k+1 ) boys and ( 2k+3 ) girls, such a girl ( G ) must exist. - Starting from this girl ( G ), move around the table until you find another girl ( G' ) such that the sequence between them (inclusive) is ( G, B, ldots, B, G' ), where ( B ) represents boys. This sequence must exist because there are more girls than boys. - Remove the girl ( G ), one boy ( B ) from the sequence, and the girl ( G' ). This leaves us with ( 2k+1 ) girls and ( k ) boys. 5. **Applying the Inductive Hypothesis:** - By the inductive hypothesis, in the remaining ( 2k+1 ) girls and ( k ) boys, there is at least one girl in a strong position. - When we add back the removed girl ( G ), the boy ( B ), and the girl ( G' ), the girl who was in a strong position among the ( 2k+1 ) girls and ( k ) boys will still be in a strong position. This is because the addition of ( G, B, G' ) does not change the relative count of girls and boys in any direction from this girl. 6. **Conclusion:** - By induction, we have shown that for any number ( n ) of boys (where ( n leq 668 )) and ( 2n+1 ) girls, there is always at least one girl in a strong position. (blacksquare)

answer:**Solution**: A: A quadrilateral with perpendicular diagonals is not necessarily a rhombus; it can also be a kite. A parallelogram with perpendicular diagonals is a rhombus. Therefore, this option is incorrect; B: A quadrilateral with all four sides equal is a rhombus. Therefore, this option is correct; C: A rhombus is a parallelogram with equal adjacent sides, so a rhombus with equal diagonals is a square. Therefore, this option is correct; D: The sum of the interior angles of a quadrilateral is 360 degrees, so a quadrilateral with all four interior angles equal is a rectangle, meaning all four interior angles are right angles. Therefore, this option is correct; Therefore, the answer is: boxed{A}.

answer:1. **Total Miles Driven by All Tires**: The car now has six tires, though still only five tires are put to use at any given time. Thus, to find the total number of tire-miles, multiply the car's mileage by the number of tires used simultaneously: [ 42,000 text{ miles} times 5 = 210,000 text{ tire-miles} ] 2. **Equal Usage Among Tires**: Since each of the six tires was used for an equal amount of time, divide the total tire-miles by the number of tires: [ frac{210,000 text{ tire-miles}}{6 text{ tires}} = 35,000 text{ miles per tire} ] 3. **Conclusion**: Each tire was used for 35,000 miles within the first 42,000 miles driven by the car. Therefore, each tire was used for 35,000 miles. The final answer is C) boxed{35,000}

answer:Let's break down the solution into detailed steps: # Part 1: Minimum Value of f(m) 1. **Finding the Minimum Value of p**: - Consider a unity number m = 110, where the hundreds digit is the sum of the tens and units digit. - Moving the hundreds digit to the units place, we get n = 101. - Calculating p = m - n = 110 - 101 = 9. - Therefore, the minimum value of p is 9. 2. **Calculating the Minimum Value of f(m)**: - Given f(m) = frac{p}{9}, substituting the minimum value of p we found, we get f(m) = frac{9}{9} = 1. 3. **Conclusion for Part 1**: - The minimum value of f(m) is boxed{1}, and the corresponding unity number m is boxed{110}. # Part 2: Finding All "Zodiac Numbers" 1. **Setting Up the Equation**: - Since f(m) is a multiple of 12, let f(m) = 12x. - This implies frac{p}{9} = 12x, and hence p = 108x. 2. **Expressing m and n**: - Let m = 100a + 10b + c and n = 100b + 10c + a. 3. **Calculating p**: - p = m - n = 99a - 90b - 9c. - Since a = b + c, substituting a we get p = 99(b + c) - 90b - 9c = 9b + 90c. 4. **Simplifying the Equation**: - 9b + 90c = 108x simplifies to b + 10c = 12x. 5. **Constraints and Calculations**: - Given 1 leq b leq 9, 0 leq c < 9, and b + c leq 9. - b + 10c = 12x leq 81 implies x leq 6. 6. **Finding Solutions for x**: - For x = 1: b = 2, c = 1, then f(m) = 12. - For x = 2: b = 4, c = 2, then f(m) = 24. - For x = 3: b = 6, c = 3, then f(m) = 36. - For x = 4: b = 8, c = 4, this does not satisfy b + c leq 9. - For x = 5: No valid values of b and c. - For x = 6: b = 2, c = 7, then f(m) = 72. 7. **Conclusion for Part 2**: - The "zodiac numbers" are boxed{12, 24, 36, 72}.