answer:Consider a point (O) which is the origin. Four radius-vectors ( overrightarrow{OA}_1, overrightarrow{OA}_2, overrightarrow{OA}_3, overrightarrow{OA}_4 ) are to be drawn from (O) such that all the angles between them are congruent. 1. **Setup the coordinates**: - Let (O) be the origin ((0, 0, 0)). - These vectors are in 3D space intersecting at (O). - The vectors should be equal in length and form equal angles between each pair. 2. **Equal Angles Between Vectors**: - Because there are 6 angles formed between 4 vectors, and they must be congruent, we arrange these vectors such that the solid angle ( theta ) between each pair of vectors is the same. This implies symmetrically arranging the vectors in 3-dimensional space. 3. **Using Geometry and Symmetry**: - Envision that these vectors from point (O) reach out to 4 points (A_1, A_2, A_3, A_4) forming a regular tetrahedron. - The distances between any two vectors should be equal. 4. **Constructing the Tetrahedron**: - The distance between each vector endpoint, that is, between any two points (A_i) and (A_j), is equal. - This arrangement implies that all these vectors meet at equal angles in pairs across each edge of the tetrahedron. 5. **Checking angles**: - Consider the cosine of the angle between any two vectors. For vectors (mathbf{v}_i, mathbf{v}_j) in a regular tetrahedron, they subtend the same angle (varphi). - Using vector properties, the cosine of the angle (varphi) between two vectors in a tetrahedron is calculated. 6. **Calculation with Cosine Rule in 3D**: - The angle (varphi) can be determined using the cosine rule for tetrahedral angles, where (cos varphi = -frac{1}{3}). 7. **Obtaining (varphi) Value**: - Solving for (varphi): [ cos varphi = -frac{1}{3} ] By taking arccosine, [ varphi = arccosleft(-frac{1}{3}right) ] Which gives approximately: [ varphi approx 109^{circ} 28' 16'' ] Hence, by placing these vectors symmetrically like the vertices of a regular tetrahedron, we ensure that the 6 angles between them are congruent. # Conclusion: The problem is thereby solved by architecting the vectors symmetrically such that each pair forms an angle (varphi approx 109^{circ} 28' 16''). (blacksquare)

answer:To find the maximum number of students among whom 1008 pens and 928 pencils can be distributed equally, we need to find the greatest common divisor (GCD) of the two numbers. The GCD will give us the largest number of students that can receive the same number of pens and pencils. Let's calculate the GCD of 1008 and 928. Using the Euclidean algorithm: 1. Divide 1008 by 928 and find the remainder: 1008 ÷ 928 = 1 with a remainder of 80. 2. Now divide the previous divisor (928) by the remainder (80): 928 ÷ 80 = 11 with a remainder of 48. 3. Continue this process: 80 ÷ 48 = 1 with a remainder of 32. 48 ÷ 32 = 1 with a remainder of 16. 32 ÷ 16 = 2 with a remainder of 0. When we reach a remainder of 0, the last non-zero remainder is the GCD. In this case, the GCD is 16. Therefore, the maximum number of students among whom 1008 pens and 928 pencils can be distributed equally is boxed{16} . Each student would receive 1008 ÷ 16 = 63 pens and 928 ÷ 16 = 58 pencils.

answer:To determine how many 9-digit numbers divisible by 2 can be formed by permuting the digits of the number 131152152, we follow these steps: 1. **Identify the digits**: - The given number is 131152152. It contains the digits: 1 (three times), 3 (once), 5 (twice), and 2 (three times). 2. **Check if the numbers can be formed**: - We need to make sure the resulting numbers are 9-digits long and divisible by 2. For a number to be divisible by 2, its last digit must be even. 3. **Determine the valid arrangements**: - From the given digits (1, 1, 1, 3, 5, 5, 2, 2, 2), in order to have a number divisible by 2, the last digit must be one of the even digits (2). 4. **Count the permutations**: - Since the last digit must be 2, we fix one of the three 2's in the units place. Now we need to arrange the remaining 8 digits (1, 1, 1, 3, 5, 5, 2, 2) in the first 8 positions. 5. **Calculate the total permutations**: - The number of unique permutations of these 8 digits, where some digits are repeated, is given by: [ frac{8!}{3!2!1!} ] - Here, (8!) is the total permutations of 8 digits. - (3!) accounts for the three 1's, - (2!) accounts for the two 5's, - (1!) accounts for the digit 3 (which appears once in the remaining 8 digits). 6. **Simplify the factorial expression**: - Compute the factorials: [ 8! = 40320 ] [ 3! = 6 ] [ 2! = 2 ] [ 1! = 1 ] - Hence, [ frac{8!}{3!2!1!} = frac{40320}{6 times 2 times 1} = frac{40320}{12} = 3360 ] Conclusion: There are boxed{3360} 9-digit numbers divisible by 2 that can be formed by permuting the digits of the number 131152152.

answer:Let's denote the total number of female officers on the police force as F. According to the information given, 65% of the female officers were on duty that night. This means that 0.65 * F female officers were on duty. We also know that there were 475 officers on duty in total, and out of these, 315 were male officers. Therefore, the number of female officers on duty that night would be 475 - 315 = 160. So we have the equation: 0.65 * F = 160 To find the total number of female officers (F), we divide both sides of the equation by 0.65: F = 160 / 0.65 F = 246.15 Since the number of officers must be a whole number, we round to the nearest whole number. However, in this context, it doesn't make sense to have a fraction of an officer, so we can assume that the slight discrepancy is due to rounding during the calculation. Therefore, the total number of female officers on the police force is approximately boxed{246} .