answer:1. **Define the Polynomial ( p_N(x) ):** Consider the polynomial ( p_N(x) = x(x + 1)(x + 2) cdots (x + N - 1) ). This polynomial has degree ( N ) and leading coefficient 1. 2. **Assess Divisibility for Positive Integers:** We will show that ( p_N(m) ) is divisible by ( N! ) for all integers ( m ). [ frac{p_N(m)}{N!} = frac{(m + N - 1)!}{(m - 1)! N!} ] This expression represents a binomial coefficient, and it is an integer for all positive integers ( m ). Hence, ( p_N(m) ) is divisible by ( N! ). 3. **Evaluate for Negative Integers and Zero:** For ( m = 0, -1, -2, ldots, -(N-1) ): [ p_N(m) = 0 ] This is trivially divisible by ( N! ). For ( m < -(N - 1) ), let ( m = -m' -(N - 1) ) where ( m' ) is positive: [ p_N(m) = (-1)^N p_N(m') ] Since ( p_N(m') ) is divisible by ( N! ) as shown above, ( p_N(m) ) is also divisible by ( N! ). 4. **Conclusion About Smallest ( N ):** If we can find the smallest integer ( N ) such that ( n ) divides ( N! ), then ( p_N(x) ) (the polynomial given by ( p_N(x) )) meets the requirement that ( n ) divides ( p(m) ) for all integers ( m ). 5. **General Properties of Any Polynomial ( p(x) ):** Suppose ( p(x) ) is any polynomial with the same property. We can express ( p(x) ) in terms of ( p_i(x) ): [ p(x) = p_M(x) + a_1 p_{M-1}(x) + cdots + a_{M-1} p_1(x) + a_M ] Here, ( M ) is the degree of ( p(x) ) and ( a_i ) are integers. We can choose coefficients to match the terms of some polynomial. 6. **Reduction Argument Using Divisibility:** Setting ( x = n ), we see that ( n ) must divide ( a_M ): [ n mid a_M ] Thus, we can write another polynomial of the same degree minus a term: [ p(x) - a_M ] Iterating this process using that ( a_{M-1} ) must also be divisible by ( n ). 7. **Conclusion:** After reducing repeatedly: [ n mid a_{M-i} ] Since the greatest common divisor (gcd) of ( n ) and ( p_i(n - i) ) is ( text{gcd}(n, i!) ): [ i! text{ divides } p_i(m) text{ for all } m ] We conclude that the polynomial ( p_M(x) ) will require ( n mid p_M(1) = M! ). Thus, ( M geq N ). 8. **Evaluating ( f(1000000) ):** To find ( f(1000000) ): [ 1000000 = 2^6 times 5^6 ] Here, the smallest ( N ) for which ( 5^6 mid N! ): [ 25! text{ includes at least six factors of } 5 text{ (5, 10, 15, 20, 25)} ] This satisfies that ( 25 geq N ). Additionally, we have enough factors of 2. Therefore, [ f(1000000) = 25 ] # Conclusion: [ boxed{25} ]

answer:# Problem: A pickpocket gets off a bus and walks in the direction opposite to the bus's motion after the bus reaches a stop. The bus continues to travel, and after the next stop, the passenger who was robbed realizes they have been pickpocketed and gets off the bus to chase the pickpocket. If the passenger's speed is twice the pickpocket's speed, the bus's speed is 10 times the pickpocket's speed, and the time it takes for the bus to travel between two stops is 40 seconds, how long will the passenger take to catch up with the pickpocket after getting off the bus? 1. **Define Variables:** - Let ( v ) be the speed of the pickpocket. - The passenger's speed is ( 2v ). - The bus's speed is ( 10v ). - The time to travel between two stops is ( 40 ) seconds. 2. **Determine the Distance Between Stops:** - Since the bus travels at a speed of ( 10v ) for ( 40 ) seconds, the distance ( d ) between two stops is calculated as: [ d = 10v times 40 = 400v ] 3. **Calculate the Relative Positions:** - Assume the pickpocket gets off exactly at stop 1 and starts walking away from the bus stop. - The passenger realizes they have been pickpocketed after the bus reaches stop 2. Thus, the pickpocket has already been walking for ( 40 ) seconds. - In ( 40 ) seconds, the pickpocket covers a distance of: [ text{Distance covered by the pickpocket} = v times 40 = 40v ] 4. **Calculate the Catch-Up Time:** - Now, the passenger starts chasing the pickpocket, who has a ( 40v ) head start. - The relative speed of the passenger compared to the pickpocket (since the passenger is running at ( 2v )) is: [ text{Relative speed} = 2v - v = v ] - The time ( t ) it takes for the passenger to catch up can be found from the equation: [ text{Distance} = text{Relative speed} times text{Time} ] [ 40v = v times t ] [ t = 40 text{ seconds} ] 5. **Calculation of Total Time:** - The pickpocket has already walked ( 40 ) seconds during the bus’s ride between the stops. - It then takes ( 40 ) more seconds for the passenger to catch up. 6. **Conclusion:** [ text{Total time after passenger gets off the bus} = 40 text{ seconds} + 400 text{ seconds} = 440 text{ seconds} ] (boxed{440})

answer:Step-by-Step Solution to Find the Number of Triangles with Two White Vertices: 1. **Identify known quantities**: - Let ( p ) be the number of red points. - Let ( f ) be the number of white points. - Let ( z ) be the number of green points. The given information translates into the following equations: [ begin{aligned} &1. quad pf + pz + fz = 213 &2. quad frac{p(p-1)}{2} + frac{f(f-1)}{2} + frac{z(z-1)}{2} = 112 &3. quad pfz = 540 &4. quad frac{p(p-1)}{2} (f+z) = 612 end{aligned} ] 2. **Total number of points**: Add equations (1) and (2) to account for the total number of distinct lines: [ frac{X(X-1)}{2} = 213 + 112 = 325 ] Solving for ( X ): [ X(X-1) = 650 implies X^2 - X - 650 = 0 ] Using the quadratic formula ( X = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), we find: [ X = frac{1 pm sqrt{1 + 2600}}{2} = frac{1 pm 51}{2} ] Thus, ( X = 26 ) (since the number of points must be positive). 3. **Substitute ( X = 26 ) into the remaining equations**: From ( X = p + f + z = 26 ). 4. **Use equation (4) to find ( p )**: [ frac{p(p-1)}{2} cdot (26 - p) = 612 ] Simplifying and multiplying both sides by 2: [ p(p-1)(26-p) = 1224 ] Given that ( 1224 = 2^3 cdot 3^2 cdot 17 ), and among ( p, p-1, 26-p ), one must be 17 because at least one of ( p(p-1)(26-p) ) must be divisible by 17, and ( p ) must be less than 26. 5. **Finding ( p )**: Possible factorization for ( p, p-1, 26-p ) yields ( p = 9 ): [ p(p-1)(26-p) equiv 9(8)(17) = 1224 ] 6. **Solve for ( f ) and ( z )**: With ( p = 9 ): [ f + z = 26 - p = 26 - 9 = 17 ] From equation (3): [ pfz = 540 implies 9fz = 540 implies fz = 60 ] Solving the quadratic equation ( f(17 - f) = 60 ): [ f^2 - 17f + 60 = 0 implies (f-12)(f-5) = 0 ] So, ( f = 5 ) or ( f = 12 ). Consequently, ( z = 12 ) or ( z = 5 ). 7. **Number of triangles with two white vertices**: Calculate ( F = frac{f(f-1)}{2} cdot (p+z) ): - For ( f = 5 ) and ( z = 12 ): [ F = frac{5(4)}{2} cdot (9+12) = 10 cdot 21 = 210 ] - For ( f = 12 ) and ( z = 5 ): [ F = frac{12(11)}{2} cdot (9+5) = 66 cdot 14 = 924 ] # Conclusion: Thus, the number of triangles with two white vertices can be either (boxed{210}) or (boxed{924}).

answer:1. **Assume the number of boys and girls**: Let the total number of students be 200, with an equal number of boys and girls. Thus, there are 100 boys and 100 girls. 2. **Calculate the number of students who went on the trip**: - The fraction of girls who went on the trip is frac{5}{8}, therefore the number of girls who went on the trip is frac{5}{8} times 100 = 62.5 approx 63 (rounded to the nearest whole number for practical purposes). - The fraction of boys who went on the trip is frac{3}{5}, therefore the number of boys who went on the trip is frac{3}{5} times 100 = 60. 3. **Total number of students on the trip**: - The total number of students who went on the trip is 63 + 60 = 123. 4. **Fraction of the trip attendees who were girls**: - The number of girls on the trip is approximately 63, and the total number of students on the trip is 123. - The fraction of the trip attendees who were girls is frac{63}{123}. 5. **Conclusion**: The fraction of the students on the field trip who were girls is frac{63{123}}. The final answer is boxed{textbf{(A)} frac{63}{123}}