answer:To find the smallest integer ( n ) such that we can select any four distinct integers ( a, b, c, d ) from the set of ( n ) integers, and ( a + b - c - d ) will be divisible by 20. 1. **Understanding the Divisibility Condition:** For ( a + b - c - d ) to be divisible by 20, we must have: [ a + b - c - d equiv 0 pmod{20} ] 2. **Using the Possible Remainders Modulo 20:** Any integer modulo 20 can only take a remainder in the set ( {0, 1, 2, ldots, 19} ). 3. **Forming Pairs of Distinct Integers:** Take any set of ( k ) integers modulo 20. The total number of distinct pairs ( (a, b) ) that can be formed is given by: [ binom{k}{2} = frac{k(k-1)}{2} ] 4. **Ensuring at Least Two Pairs Have the Same Sum modulo 20:** - For any integer set of size ( k ), if the number of pairs exceeds 20, there must be at least two pairs ( (a, b) ) and ( (c, d) ) such that: [ a+b equiv c+d pmod{20} ] - This translates to: [ a+b - c-d equiv 0 pmod{20} ] 5. **Bounding the Number of Distinct Remainders:** - To guarantee that ( k binom{k}{2} ) is greater than 20, we must solve: [ frac{k(k-1)}{2} > 20 ] - Solving the inequality: [ k(k-1) > 40 ] - By trial: - ( k = 7 ) gives: [ 7(6) = 42 quad (text{which is greater than 40}) ] 6. **Evaluating Specific Cases for ( n ):** - By ( k geq 7 ): - We need ( n geq 9 ). - Consider a set with 9 integers. Ensure the sums of all distinct pairs of integers modulo 20 cover at least 20 different sums. - Thus, within 9 distinct integers, sufficient sums will collide, making ( a+b-c-d equiv 0 pmod{20} ). Finally, we conclude: [ boxed{9} ] This minimum ( n = 9 ) guarantees that from any 9 distinct integers, we can pick any four integers ( a, b, c, d ) such that ( a+b-c-d ) is divisible by 20.

answer:A regular tetrahedron with side length s = 3 has four equilateral triangle faces, each with area frac{s^2 sqrt{3}}{4}. The area of one such triangle is: frac{3^2 sqrt{3}}{4} = frac{9 sqrt{3}}{4}. Thus, the total surface area S of the tetrahedron is: 4 times frac{9 sqrt{3}}{4} = 9 sqrt{3}. Given that this surface area is also the surface area of the sphere, and using the formula 4pi r^2 for the surface area of a sphere, 4pi r^2 = 9 sqrt{3}. Solving for r, we find: r^2 = frac{9sqrt{3}}{4pi}, quad r = frac{3sqrt{3}}{2sqrt{pi}}. The volume V of the sphere with radius r is: V = frac{4}{3}pi r^3 = frac{4}{3}pi left(frac{3sqrt{3}}{2sqrt{pi}}right)^3 = frac{4}{3} pi frac{27sqrt{27}}{8pisqrt{pi}} = frac{9sqrt{27}}{2sqrt{pi}}. Rewriting sqrt{27} as 3sqrt{3}, we get: V = frac{27sqrt{3}}{2sqrt{pi}}, Expressing this in the form frac{Ksqrt{2}}{sqrt{pi}}, where sqrt{3} can be written as frac{sqrt{6}}{sqrt{2}}, let's attempt to rewrite: V = frac{27 cdot sqrt{6}}{2 cdot sqrt{2} cdot sqrt{pi}} = frac{27sqrt{6}}{2sqrt{2}sqrt{pi}}, Thus, boxed{K = 27}.

answer:Let's denote the total salary of the 8 workers as W and the salary of the old supervisor as S_old. From the information given, we know that the average monthly salary of the 8 workers and the old supervisor was 430. Therefore, the total monthly salary for all 9 people was: W + S_old = 9 * 430 Now, when the old supervisor retired and the new supervisor was appointed, the average monthly salary for the 9 people became 390. The new supervisor's salary is 510, so the total monthly salary for all 9 people is now: W + 510 = 9 * 390 We can set up the equations as follows: 1) W + S_old = 9 * 430 2) W + 510 = 9 * 390 Now we can solve for S_old by subtracting equation 2 from equation 1: (W + S_old) - (W + 510) = (9 * 430) - (9 * 390) S_old - 510 = 9 * (430 - 390) S_old - 510 = 9 * 40 S_old = 9 * 40 + 510 S_old = 360 + 510 S_old = 870 Therefore, the salary of the old supervisor was boxed{870} per month.

answer:1. **Define Variables:** Let s be the side length of the larger square and t be the side length of the smaller square. 2. **Relate Perimeters:** The perimeter of a square is 4 times its side length. For the larger square having 5 times the perimeter of the smaller square, we have: [ 4s = 5 cdot 4t ] Simplifying, we find: [ s = 5t ] 3. **Given Total Side Lengths:** The sum of the side lengths of both squares is given as 60 units: [ s + t = 60 ] Substituting s = 5t into the sum we get: [ 5t + t = 60 implies 6t = 60 ] Solving for t gives: [ t = 10 ] And using s = 5t: [ s = 5 times 10 = 50 ] 4. **Conclusion:** The side length of the smaller square is 10 units, and the side length of the larger square is 50 units. Hence, the answers are 10 text{units and 50 text{units}}. The final answer is boxed{C. 10 units and 50 units}